Operation Management-ICAI-Inter-Resource Management

4.0 Input-output Ratio
Input-output analysis reflects a general theory of production based on the idea of economic interdependence.
Many input - output models are useful in forecasting. Input - output analysis takes into consideration the
interdependence of the different sectors in the economy. This is because; the input to one sector is output
of another sector. For example, the output of coal industry will be an input to the steel plant and the output
of steel industry is an input to the construction industry and so on. There are many such cyclic relations
within the various sectors of economy. Taking the data of outputs and inputs and studying the relationship
between these two, we will be in a position to analyse the total demand for a product and the output
required from industrial units. This type of analysis is very important because it takes into account all the
intricate relationships in the economy. One of the limitations in this method is that the utility of an output
is restricted to economic analysis, not considering the other business, governmental, technological and
internal factors. It is limited but useful analysis. The analysis need not be limited to macro-level, speaking
only in terms steel sector and coal sector etc. It may be mote ‘micro’, by considering the inputs and outputs
within a general product group in the total economy. This type of analysis is very much used and is found
more beneficial and useful. Three major assumptions in developing this technique are:
1. The total output of an industry is consumed as input by all industries for a time period, under
consideration.
2. The input bought by each industry has usually been made dependent only on the industry’s level of
output.
3. The ration of an industry’s input to its output, once established is fixed. This ratio is known as inputoutput
number or production coefficient.
Input-output analysis is a mathematical study of an economy in which different production sectors such
as agriculture, industry and services etc. have interdependence. Thus, in input-output analysis, we try to
analyse quantitatively the interdependence of inputs and outputs of various industries and find the
equilibrium between the inputs and output of each industry, plant, sector or economy. The output of any
industry depends very much on its inputs and these inputs are the outputs of other industries. This analysis
signifies to foretell the total production per sector and its demand in other industries.
It can be presented as: Total output of all sectors = Total inputs of all sectors.
or, output of an industry = Total inputs of that industry.
This input-output analysis is also known as ‘analysis of inter-industry or inter-sector flows or deliveries or
analysis of inter-industry relations.’
Input-output analysis can be used in the following cases:
1. The input-output analysis can be used for the study of variation in productivity of an enterprise. This
analysis is based on index-numbers, using value added measure for output and the measures of
inputs all at constant prices. The value added in each industry can be obtained by subtracting the
value of inputs used from the value of gross output.
2. Input-output table has a historic importance showing a record of past and can be a very useful
instrument in economic analysis and framing economic policy. In developed countries, this analysis
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is used for economic forecasting and in developing countries, for economic planning or programming.
In business world, business forecasting can be done by this analysis.
3. This analysis can be very usefully used for a comparative study of growth by examining. By examining
country-wise or sector or region-wise or plant, firm or industry-wise differences in input-output
coefficients, the process of growth can be more easily interpreted.
4. This analysis can be utilised to determine the impact of some major exogamous changes in the economy
on other factors of production.
4.1 Linear Programming
The solution technique for linear optimisation models is linear programming.
Linear programming is very important tool of operation research which has recently been started in the
field of production management. It is new mathematical technique developed while the World War II was
in progress. Before the inception of this technique in production management, every problem was tackled
by trial and error method, the result of which was not so sure and best to the satisfaction. By introducing
the technique, the solution of every problem now does not depend on chance; it is now tackled with some
certainty by the use of this method.
Every business has limited resources (men materials, labour, money etc.) but wants to achieve its main
objective of getting the maximum profit. Maximisation of profit of the business depends upon a number of
variables so there may be a number of alternatives before the managing executives to maximise the profits.
The business executives are, therefore, busy in choosing the most appropriate (best possible) action among
a number of alternative courses of action. As there are a large numbers of variables to be considered for
each alternative, the situation becomes more complicated. The management is to choose the best possible
combination of variables. Even if information about variables is available, the problem is not so easy when
we go in detail. It is infrequently so intricate the detail and in the interrelationships involved as to make it
possible to use in its complexity. Previously these combinations were determined by trial and error method
and no one could ascertain whether a best solution had been obtained or not. The linear programming has
made it very simple to know about the best possible solution of any problem. Linear programming is
applied to problems which involves interaction between alternatives.
Linear programming is a recently-developed technique which is being used in production management. It
is technique of choosing a best possible alternative among the various alternatives available difficult because
each alternative consists of a large number of variables, giving different results. Linear programming
technique has considerably simplified the problem.
Linear programming consists of two words—’linear’ and ‘programming’. The word ‘linear’ establishes
certain relationships among different variables whereas the word ‘programming’ indicates a way to get
desired results by the optimum use of available resources. Thus linear programming is a mathematical
technique for allotting limited resources in an optimum manner. Hence, it is of great practical significance
for the management for achieving the prime objective of the business, i.e., maximisation of profits. It is an
undisputable fact that resources of a business are limited setting limit or restriction on the smooth functioning
of the business. The problem confronting the management is to decide the manner in which these limited
resources are to be earmarked for various’ uses so as to have the maximum profit. Linear programming
technique solves this problem.
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Linear programming is a planning technique that permits some objective functions to be minimised or
maximised within the framework of given situational restrictions.
It is a planning technique by using the mathematical equations. It can be said as a technique of selecting the
best possible (optimal) strategy among a number of alternatives. The chosen strategy is said to be the best
because it involves maximisation or minimisation of some desired action, e.g., maximisation of profits or
minimisation of costs.
Uses or Application of Linear Programming:
The linear programming technique is useful in the following cases:
(i) It helps in determining optimum combination of several variables with given constraints and thus
selecting the best possible strategy among various alternatives available.
(ii) Linear programming provides additional information for proper planning and control over various
operations in the organisation.
(iii) The management must understand the activities of the organisation for constructing suitable
mathematical model visualising the relationship between variables, if any, and making improvement
over them. The linear programming helps in better understanding the phenomenon.
(iv) Linear programming contributes to the development of executives through the techniques of model
building and their interpretations.
(v) Linear programming provides standards for the management problems by defining (a) the objectives
to be pursued, (b) various restrictions to be imposed, (c) various alternatives available and relationship
between them, (d) the contribution of each alternative to the objectives.
Application: The linear programming technique may be fruitfully applied in the following spheres:
(i) The linear programming can be used in production scheduling and inventory control so as to produce
the maximum out of the resources available to satisfy the needs of the public by minimising the cost
of production and the cost of inventory control.
(ii) The technique of linear programming can also be fruitfully used in solving the blending problems.
Where basic components are combined to produce a product that has certain set of specifications and
one may calculate the best possible combination of these compounds which maximise the profits or
minimise the costs.
(iii) Other important applications of linear programming can be in purchasing, routing, assignment and
other problems having selection problems such as(a) selecting the location of plant, (b) deciding the
transportation route within the organisation, (c) utilising the godowns and other distribution centres
to the maximising, (d) preparing low-cost production schedules, (e) determining the most profitable
product-mix , and (f) analysing the effects of changes in purchase prices and sale prices.
Limitations of Linear Programming: Linear programming, as the name suggests, presents the relations
between different variables in a line. This presumption, however, is not correct. There are many limitations
of linear programming which can be given below:
(i) The technique solves the problems of linear nature whereas, in practice, most of the business problems
are of non-linear nature. The solution of non-linear problems is not possible through this technique.
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(ii) This technique cannot provide solutions to problems which involve variables not capable of being
expressed quantitatively.
(iii) Under this technique, uncertainties are not considered whereas in business, every problem is full of
uncertainties.
(iv) One of the limitations of linear programming is that the results under this technique are not necessarily
to be in whole numbers. Sometimes results in points give a misleading picture. For example, if it
results by this technique that 1.6 machines are to be purchased, it becomes very difficult to ascertain
whether to purchase one machine or two machines because machines cannot be purchased in fraction.
Formulation of an L.P problem: In formulating the linear programming problem, the basic step is to set
up some mathematical model. For this purpose, the following considerations should be kept in mind: (i)
unknown variables, (ii) objectives and (iii) constraints. This can be done with the help of the following
terms:
1. The objective function: An objective function is some sort of mathematical relationship between
variables under consideration. Under linear programming this relationship is always linear. The
construction of objective function mainly depends upon abstraction. It is a process whereby the most
important features of a system are considered.
The objective function is always positive. The coefficients a1, a2 are certain constants and are known as
prices associated to the variables X1, X2.
2. Constraints on the variables of the objective function: In practice, the objective function is to bee
optimised under certain restraints imposed on die variables or some combination of few or all the
variables occurring in the objective function. These restrictions, in most of the cases, are never exact.
Had these been exact, the objective function would have been easily optimised by die use of differential
calculus. The restraints should be known and expressed in terms of linear algebraic expression.
3. Feasible solution: One of the essential features of linear programming problem is optimisation of
linear objective function Z. It is subject to the linear constraints on the variables of the objective function.
A set of values of X1, X2………which satisfies the constraints and the non- negativity restrictions are
called Feasible solution. A feasible solution which optimises the objective function is known as
Optimalsolution. Thus, a linear programming problem can be formulated in this way.
Simplex Method of Linear Programming
There are number of ways of finding the optimal solution for a given linear programming problems. The
following three methods are mainly used for this purpose.
(1) Graphic Method
(2) Simplex Method
(3) Transportation Method.
1. Graphic method. This method is generally used for solving the problems having two or three variables.
Due to this limitation of handling only two or three variables at a time this method has limited
application in industrial problems. In practice, two variable cases are easy to solve by this method
because three dimensional geometry becomes too complicated to find accurate results.
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2. Simplex Method. This is the most powerful and popular method for solving linear programming
problems. Any problem can be solved by this method which satisfies the conditions of linearity and
certainty irrespective of the number of variables. It is an iterative procedure which ultimately gives
the optimal solution.
3. Transportation Method. This method is used to know the minimum cost of transportation of a product
from various origins to different distribution and consumption centres.
Simplex method of linear programming. Simplex method of linear programming is a very important
technique to solve the various linear problems. Under this method, algebraic procedure is used to solve
any problem which satisfies the test of linearity and certainty. It is an iterative procedure which ultimately
gives the optimal solution. Several variables can be used under this method. However, the simplex method
is more complex and involves somewhat unsophisticated complex mathematics.
Features. There are two characteristic features of the simplex method which we can cite them at the very
outset.
First, we have the computational routine in the iterative process. Iteration means the repetition. Therefore,
in working towards the optimal solution, the computational routine if repeated again and again, following
a standard pattern, successive solutions are developed in a systematic pattern until the best solution is
arrived.
Secondly, such new solution will give a profit larger than the previous solution. This important characteristic
assures us that we are always moving closer to the optimal solution.
There are two types of linear programming problems to be solved by this method—
(1) Maximisation of objective function, and (2) Minimisation of objective function.
1. Maximisation of objective function. The following procedural steps are taken for getting the optimum
solution by simplex method, which maximise the objective function. We shall explain the procedure of
maximisation of objective function by simplex method with the help of the following illustration.
Illustration. A small manufacturing firm produces two types of gadgets, A and B, which are first processed
in the foundry, then sent to the machine for finishing. The number of man-hours of labour required in each
shop for the production of each unit of A and B and the number of man-hours the firm has available per
week are as follows:
Foundry Machine shop
Gadget A 10 5
Gadget B 6 4
Firm’s capacity per week (in hours) 1000 600
Const ruct the objective function and the corresponding constraints for calculating that how many units
should be produced per week so that the profit is maximum. The profit on the sale of A is Rs. 30 per unit as
compared to B’s Rs. 20 per unit.
Taking this illustration, we can explain the following steps:
1. The first step is to know the objective function and the linear constraints for the given problem.
2. Slack variables. The next step is to change the inequalities for two constraints into equations. The
inequalities are converted into equations only by the device of introducing slack variables. A slack
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variable represents costless process whose function is to ‘use up’ otherwise unused capacity, say
machine time or warehouse capacity. Effectually, the slack variable represents unused capacity, and
it will be zero, only if production facilities or capacities are fully utilised. In each constrained equation,
the variables used in other equations are also introduced but with zero coefficients. The same slack
variables are also introduced in the objectives function but by the time it is maximum, all their coefficient
will be zero. The equation can be developed as follows:
Z = 30X + 20Y + 0S1 +0S2;
Subject to:
10X + 6Y + 1S1+ 0S2 =1000
5X + 4Y +0S1 +1S2 = 600
Here, S1 and S2 are slack variables (unused time on machine 1 and 2 respectively). Slack variables are
always non-negative.
3. Developing initial simplex tableau.We can now set out this whole problem in a simplex tableau and
read it for a basic feasible solution. This table consists of rows and columns of figures and is also
known a simplex matrix. Each table has the following columnwise heading from left to right:
(i) (ii) (iii) (iv) (v) (vi)
Unit profit
(Cj)
Basis or
programme
Quantity or
Constant
Slack
Variables
Real Variables Ratio
The following entries are made in each column as:
(i) The first column (unit-profit column) has zero entry in each row in the initial stage.
(ii) The second column (basis or programme column) shows the first feasible solution and contains
various slack or artificial variables from top to bottom. The first feasible solution consists of
those slack or artificial variables for which the corresponding column is an identity vector.
(iii) The third column (the quantity or constant column) will have the constant right hand side value
of each constrained equation in the respective row.
(iv) The fourth column (slack variables column shall have the respective coefficients of the given
slack variables in different constrained equations for each row. These coefficients will be either 1
or 0 depending whether the slack variables of the column appears in the corresponding constraint
equation or not.
(v) The fifth column (real variables column) shall have the respective coefficients of the given real
variables in different constraint equations for each row.
(vi) The six and last column (ratio column).
Now with the help of the given illustration, we can prepare the initial simplex tableau, illustrating the
above steps:
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Initial Simplex Tableau
j 1 2 3 4 5 6
Cj 0 0 0 30 20
Constant S1 S2 X Y Ratio
Unit
profit
Programme
0 S1 1000 1 0 10 6 100
Key Row 0 S2 600 0 1 5 4 120
Base Zj 0 0 0 0 0
Row Values Zj-Cj 0 0 0 -30 -20
Key
Column
Negative largest values
Top j 1 2 3 4 5 6
Cj 0 0 0 30 20
Row No.(l) Unit
Profit
Programme Constant Slack Variables Real Variables Ratio
S1 S2 X Y
1 0 S1 1000 1 0 10 6
2 0 S2 600 0 1 5 4
It can be observed from the above table that the entries corresponding to the columns for slack variables
S1, and S2 are (1,0) and (0,1) respectively. These are unit vector. Hence the programme column of the
above table contains the slack variables S1 and S2 as its first feasible solution.
Thus Row 1 corresponds to constraint equation 10X + 6Y + S1 + 0.S2 =1000
and Row 2 corresponds to the constraint equation 5X + 4Y + 0S1 + 1S2 =600
4. Computation of base row. Having prepared the initial simplex tableau, the next step in the preparation
of base row values from the initial tableau by the following procedure to get tableau I:
(i) Multiply each row entry by the corresponding entry of unit profit column.
(ii) Find the sum of the products obtained in step for each column and denote it by Zj’s.
(iii) Subtract the corresponding objective factor Cj of top row from Zj. The base row values can be
positive, zero or negative.
The Zj’s and the base row values for the above illustration from the initial simplex tableau are calculated
as below:
Z1 = (1000×0) + (600×0) = 0 and (Z1- C1) = 0-0 = 0
Z2 = (1×0) + (0×0) = 0 and (Z2 - C2) = 0 - 0 = 0
Z3 = (0 × 0) + (l× 0) = 0 and (Z3 – C3) = 0 - 0 = 0
Z4 = (10 ×0) + (5 × 0) = 0 and (Z4–C4) = 0 -30 = -30
Z5 = (6 × 0) + (4 × 0) = 0 and (Z5–C5) = 0 - 20 = -20
The simplex tableau now will be
Tableau I
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5. Examining base row values. Examine all base row values in Tableau I. If they (all values) are positive,
it is the optimum solution and the solution can be read from constant and programme columns. If all
values (Zj-Cj) are negative, the largest numerically negative value among the base row values should
be searched. The column in which the largest value lies is known as key column. In the above case,
column 4 is the key column. And now we should proceed to step (6) and (7) below.
6. Computation of ratio column. Having located the key column, the next step is to compute the ratio
column. This may be done by dividing the constant column value of each row by key column entry of
that row. So, in Tableau I, the ratio column comes to 1000/10 and 600/5.
Now examine the ratio-column of tableau to locate the key row.
The row for which entry in ratio column is non-negative and numerically smallest is taken as the key
row. For example, in Tableau I the least non-negative entry in the ratio column is 100 hence Row I is
the key row. Negative values in the ratio column are not considered because solution values of real
variables cannot be negative.
The entry lying in the cell at the intersection of key row and key column is known as key number, e.g.,
in Tableau I, 10 is the key number.
7. Preparation of Tableau II and Tableau III. After identifying key row, key column and key number in
Tableau I in step (6), Tableau II is prepared having some columns but with the following changes :
(i) In Tableau II, the slack variables lying in the programme column of key row of tableau I is
replaced by variable of the key column in tableau I, e.g., in tableau II of the above example, the
entry in programme column of key row is obtained by replacing the slack variable S1 in the
programme column of the key row in tableau I by the variable Y.
(ii) The value of Cj lying in the key column of Tableau I is entered in profit column of key row in
tableau II, e.g., 30 is written in profit column of Row I.
(iii) The remaining entries of the row in tableau II corresponding to key row in tableau I are obtained
by dividing the corresponding entries of key row in tableau I by key number, e.g., the entry in
first row and constant column of tableau II shall be 100 (100/10). Similarly, other entries in Row;
I are (1/10, 0/10, 10/10 and 6/10). Now the tableau II will be as under:
Tableau II
j 1 2 3 4 5 6
Cj 0 0 0 30 20
Unit
profit
Programme
Constant S1 S2 X Y Ratio
1 30 X 100 0.5 0 1 0.6 167
key number
2 0 S2 100 -0.5 1 0 1 100
Key row
Zj 3000 3 0 30 18
Zj-Cj 3000 3 0 0 -2
Key column
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(iv) The entries in Row 2 of tableau II are made in the following way : The entry for any column in
Row II of Tableau II is obtained by multiplying the entry in the row of Tableau II obtained in step
(iii) above by the value lying in key column of row in tableau I and then subtracting the product
from the corresponding entry of the column in Row II of Tableau I, e.g., 100 in constant column
of Row II in tableau II is equal to 600 - (100 × 5) = 100.
Similarly other entries in Row II in Tableau II are
[(0 - (0.1 × 5) = - (0.5)], [(1- (0 × 5) - 1)], [(5 - (1 × 5) =0)], [(4 - (0.6 ×5) =1)].
The base row in Tableau II is obtained by applying the operations in step (4) on the values of
Tableau II and it is examined whether the optimal solution is obtained or not. If optimal solution
is reached, we shall stop at this point. If not, operations in step (6) and (7) will be repeated on the
entries in Tableau II to get the tableau III and so on the procedure is repeated till we get the
optimal solution.
It is observed from Tableau II of the given problem that only Zj-Cj is negative. Hence the column
No. 5 of tableau II is the, key column. Similarly from the ratio column of Tableau II we find the
entry 100 in row II is minimum hence Row II become the key row and I is the key number.
The Tableau III will now be prepared as
Tableau III
In Tableau III all Zj-Cj are non-negative. Hence the optimal solution is reached. From programme and
constant columns of Tableau III, the result is read as X = 40, Y = 100 with minimum profit of Rs. 3,200.
2. Minimisation of objective function. The second problem of simplex method is minimisation of objective
function. For this purpose, the problem is first changed to maximisation problems by writing Z* = -Z = -
(aX + bY)
Thus, the constraint remains unchanged. The maximum value of Z* is determined on the same lines as
described in maximisation problem. Maximisation of Z* implies minimisation of Z.
Degeneracy Method. Sometimes, it happens that during the course of simplex procedure, we get two or
more entries in the ratio column of any tableau to be identical and minimum. Now, the question arises
which row should be taken as key row? The selection of key row determines the variable to be deleted.
This is known as degeneracy. The problem of degeneracy may also occur when one of the constraints on
the right hand side of the equation is zero.
The degeneracy is resolved by the following procedure:
(a) Each element of the row which have identical entries in the ratio column is divided by the key column
number of the respective row.
(b) The values so obtained are compared step by step from left to right. Priority is to be given to identify
Unit
profit
Cj
Programme
0
Constant
0
S1
0
S2
30
X
20
Y
30 X 40 0.4 -0.6 1 0
20 Y 100 -0.5 1 0 0
Zj 3,200 2 2 30 20
Zj-Cj 3,200 42 2 0 0
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columns corresponding to slack and artificial variables. The remaining columns are considered from
left to right.
(c) The comparison is stopped as soon as the rows yield unequal ratios. The row having algebraically
smaller ratio is taken to be the key row.
After selecting the key row, regular simplex procedure is resumed.
Slack and Artificial Variables
Slack variables.
A slack variable represents costless process whose function is to ‘use up otherwise unemployed capacity,
say, machine hour or warehouse capacity. Effectually the slack variable represents unused capacity and it
will be zero only, if production facilities are fully utilisesd. There are always non-negative or positive and
explains the unallocated portion the given limited resources. The main purpose of introducing slack variables
in the simplex method of linear programming is to convert the inequalities(i.e., constraints); into equalities
or, say, equations.
Artificial variables: The artificial variables are fictitious and do not have any physical meaning. These
variables are assigned a very large per unit penalty in the objective function. The penalty is designated as
-- M for maximisation problem and +M in minimisation problem where M is always greater than zero.
There may be situations where the constraints involve mixture of > = and < signs. In such cases, the slack
variables cannot provide a starting feasible solution and we have to introduce both slack and artificial
variables to have a starting basis. The artificial variables are denoted as A1,A2………….An in the programme
column of initial simplex tableau. These artificial variables should be removed first by using simplex
criterion. If at any stage of simplex procedure the basic feasible solution contains artificial variable but the
variable to be deleted at that stage is not the artificial variable, it means we are departing from the simplex
criterion for selecting the variable.
The iterative procedure does not work under following circumstances -
(i) One or more artificial variables appeals in the program column at a stage when the base row values
are all positive.
(ii) No artificial variables remain in the programme column.
(iii) The problem is redundant if one or more artificial variables occur in programme column and the
corresponding values in the constant column are zero at a stage when all Zj-Cj are positive or nonnegative.
Problems and Solutions
Problem 1. Find the non-negative values of X1, X2 and X3 that maximise the expression
Z=3X1+5X2+4X2
subject to the following restraints
2X1 + 3X2< 8
2X2 + 5X3<10
3X1+ 2X2+4X3<15
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Solution :
By introducing the slack variables to change the restrictions into equalities, we get;
2X1+ 3X2 + S1 = 8
2X2 + 5X3 + S2 =10
3X1+ 2X2+4X3 + S3= 1
Now, objective function becomes
Z = 3X1 + 5X2 +4X3 + 0S1 + 0S2 + 0S3
Now various tables are prepared to get the non-negative values of X1, X2 and X3 to maximise Z in the
following way:
Tableau I
Cj 3 5 4 0 0 0
Profit Programme Constant X1 X2 X3 S1 S2 S3 Ratio
0 S1 8 2 3 0 1 0 0 2.67
0 S2 10 0 2 5 0 1 0 5
0 S3 15 3 2 4 0 0 1 7.5
Zj 0 0 0 0 0 0 0
Cj-Zj - 3 5 4 0 0 0
Tableau II
Cj 3 5 4 0 0 0
Profit Programme Constant X1 X2 X3 S1 S2 S3 Ratio
5 X2 8/3 2/3 1 0 1/3 0 0
0 S2 14/3 - 4/3 0 5 -2/3 1 0
0 S3 20/3 5/3 0 4 -2/3 1 1 14/15
Zj 40/3 10/3 5 0 5/3 0 0 29/12
Cj-Zj - - 1/3 0 4 -5/3 0 0
Tableau III
Cj 3 5 4 0 0 0
Profit Programme Constant X1 X2 X3 S1 S2 S3 Ratio
5 X2 8/3 2/3 1 0 1/3 0 0 4
4 X3 14/15 -4/15 0 1 -2/15 1/5 0 -7/2
0 S3 89/15 41/15 0 0 -2/15 -4/5 0 89/41
Zj 34/15 5 4 17/15 4/5 0 0
Ci-Zj - 11/15 0 0 -17/15 -4/5 0
Tableau IV
Cj 3 5 4 0 0 0
Profit Programme Constant X1 X2 X3 S1 S2 S3 Ratio
5 X2 150/123 0 1 0 45/123 24/123 0
4 X3 930/615 0 0 1 -90/615 75/615 0
3 X1 89/41 1 0 0 -2/41 -12/14 0
Zj 765/41 3 5 4 45/41 24/41 0
Ci-Zj - 0 0 0 -45/41 -24/41 0
≥
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Thus, in Table IV all Cj-Zj are zero or negative hence it is the optimal solution.
In this, X1=89/41; X2 =150/123 and X3 =930 /615 or 62/41
and maximum profit Z = 765/41
Problem 2. Find the maximum value of Z = 3X1+ 5X2 + 4X3. Where X1, X2, X3 ≥ 0 subject lo the following
constraints.
2X1 + 3X2 < 18
2X2+ 5X3 <18
3X1 + 2X2 + 4X3 < 25
Solution:
As here the objective function is to be maximised, we shall introduce the slack variables in the
constraints and the objective function and we get the following equations:
2X1 + 3X2+S1 + 0.S2 + 0.S3 =18
2X2+5X3 + 0.SI + S2 + 0.S3=18
3X1 + 2X2 +4X3 + 0.S1 +0.S2 +S3 = 25
So, Z = 3X1+ 5X2 + 4X3 +0S1+ 0S2 + 0S2
Now, various simplex tableaus are prepared to get the optimal solution.
Tableau I
Cj 0 0 0 3 5 4
Profit Programme Constant S1 S2 S3 X1 X2 X3
Ratio
0 S1 18 1 0 0 2 3 0 6
0 S2 18 0 1 0 0 2 5 9
0 S3 25 0 0 1 3 2 4 12.5
Zj 0 0 0 0 0 0 0
Zj – Cj — 0 0 0 –3 –5 –4
Tableau II
Cj 0 0 0 3 5 3
Profit Programme Constant S1 S2 S3 X1 X2 X3
Ratio
5 X2 6 1/3 0 0 2/3 1 0
0 S2 6 -2/3 1 0 4/3 0 5 6/5
0 S3 13 -2/3 0 1 5/3 0 4 13/4
Zj 30 5/3 0 0 10/3 0 0
Zj – Cj - 5/3 0 0 1/3 0 -4
Tableau III
Cj 0 0 0 3 5 4
Profit Programme Constant S1 S2 S3 X1 X2 X3
Ratio
5 X2 6 1/3 0 0 2/3 1 0 9
4 X3 6/5 -2/15 1/5 0 -4/15 0 1 Neg.
0 S3 41/5 -2/15 -4/5 1 41/15 0 0 3
Zj 74/5 17/15 4/5 0 34/15 5 4
Zj – Cj – 17/15 4/5 0 11/15 0 0
≥
α
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In Tableau IV all Zj - Cj are non-negative or positive hence it is the optimal solution. The optimal solution
is X1 = 3; X2 = 4; X3 = 2 and the maximum profit Zj = 37.
Problem 3. Maximise 3X1 + 2X2 under the following restrictions:
X1 > 0, X2 > 0
2X1+X2 < 40
X1+X2 < 24
2X1+3X2 < 60
Solution. By introducing slack variables in constraints, we get
2X1 + X2 + S1 +0S2 + 0S3 = 40
X1 + X2 + 0S1+ S2 + 0S3 = 24
2X1 + 3X2 + 0S1 + 0S2 + S3 = 60
and the objective function (Z) now can be represented by
Z = 3X1 + 2X2 + 0S1+ 0S2 +0S3
Now, the following tableaus are to be prepared to get the optimal solution:
Tableau IV
Programme Cj 0 0 0 3 5 4
Profit Constant S1 S2 S3 X1 X2 X3 Ratio
5 X2 4 15/41 8/41 -10/41 0 1 0
4 X3 2 -6/41 5/41 4/41 0 0 1
3 X4 3 -2/41 -12/41 15/41 1 0 0
Zj 37 45/41 24/41 15/41 3 5 4
Zj - Cj 45/41 24/41 15/41 0 0 0
Tableau I
C1 3 2 0 0 0
Profit Programme Constant X1 X2 S1 S2 S3 Ratio
0 S1 40 2 1 1 0 0 20
0 S2 24 1 1 0 2 0 24
0 S3 60 2 3 0 0 1 30
Zj 0 0 0 0 0 0
Cj – Zj 3 2 0 0 0
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Since all Cj-Zj are either zero or negative, it is the optimal solution of the problem. Here X1,
= 16; X2 = 8 and maximum profit Zj = 64.
Problem 4. A company manufactures two items X1 and X2. They are sold at a profit of Rs. 30 per unit of X1
and Rs. 20 per unit of X2. X1 requires 2kgs of materials, 3 man-hours and 1 machine hour per unit. X2
requires 1 kg of material, 2 man hours and 3 machine hours per unit.
During each production run there are 280 kgs of material available, 500 labour hours and 420 hours of
machines used. How much of the two items should the company produce to maximize profits?
Maximise 30X1 + 20X2 subject to
2x1+x2 < 280
3xl + 2x2 < 500
x1 + 3x2 < 420
x1, x2, x3 > 0
Solution.
The objective function to be maximised in this problem may be expressed in the following way:
Z = 30X1+ 20X2
Subject to constraints
2x1 + x2 < 280
C1 3 2 0 0 0
Profit Programme Constant X1 X2 S1 S2 S3 Ratio
3 X1 20 1 1/2 1/2 0 0 40
0 S2 4 0 1/2 –1/2 1 0 8
0 S3 20 0 2 –1 0 1 10
Zj 60 3 3/2 3/2 0 0
Cj – Zj — 0 1/2 –3/2 0 0
Tableau II
C1 3 2 0 0 0
Profit Programme Constant X1 X2 S1 S2 S3 Ratio
3 X1 16 1 0 1 –1 0
2 X2 8 0 1 –1 2 0
0 S3 4 0 0 1 –4 1
Zj 64 3 2 1 1 0
Cj – Zj — 0 0 –1 –1 0
Tableau III
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3x1 + 2x2 < 500
x1 + 3x2 < 420
x1, x2, x3 > 0
Now, by introducing slack variables, we get
2X1 + X2 +S1 + 0S2 + 0S3 =280
3X1 + 2X2 + 0S1 + S2 + 0S2 = 500
X1 + 3X2 + 0S1 + 0S2 +S3 = 420
and Z = 30X1 + 20X2 + 0S1+ 0S2 +S3
Now, various tables will be prepared as under:
Tableau I
Cj 30 20 0 0 0
Profit Programme Constant X1 X2 S1 S2 S3 Ratio
0 S1 280 2 1 1 0 0 140
0 S2 500 3 2 0 1 0 166.67
0 S3 420 1 3 0 0 1 420
Zj 0 0 0 0 0 0
Cj-Zj 30 20 0 0 0
Tableau II
Cj 30 20 0 0 0
Profit Programme Constant X1 X2 S1 S2 S3 Ratio
30 X1 140 1 1/2 1/2 0 0 280
0 S2 80 0 1/2 -3/2 1 0 160
0 S3 280 0 5/2 -1/2 0 1 112
Zj 4200 30 15 15 0 0
Cj-Zj 0 0 5 -15 0 0
Tableau III
Cj 30 20 0 0 0
Profit Programme Constant X1 X2 S1 S2 S3 Ratio
30 X1 84 1 0 3/5 0 -1/5
0 S2 24 0 0 -7/5 1 -1/5
20 X2 112 0 1 -1/5 0 2/5
Zj 4760 30 20 14 0 25/2
Cj-Zj — 0 0 -14 0 -25/2
In the above Tableau III, all Cj-Zj are zero or negative, hence it is optimal solution X1 = 84 ; S2
= 112 and maximum profit (Zj) = 4,760.
Problem 5. An animal feed company must produce 200 lbs. of a mixture containing
the ingredients X1 and X2. X1costs Rs. 3 per Ib. and X2 costs Rs. 8 per Ib. Not more than 80 lbs of X1can be
used and minimum quantity to be used for X2 is 60 lbs. Find how much of each ingredient should be used
if the company wants to minimise the cost.
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Solution:
In the above problem, the objective function to be minimised is—
Z = 3X1 + 8X2, subject to the following constraints—
X1 < 80; X1 + X2 = 200; X2 > 60.
As, in this problem, the constraints contain all the symbols, < , = , > , so slack and artificial variables to
change the restrictions into equalities and we get—
X1 + S1 = 80;
X1 + X2 +A1 = 200 and
X2 – S2 + A2 = 60
Hence the objective function becomes—
Z* = -Z = -3X1 -8X2 + 0S1 + 0S2 - M.A1 -M.A2
Here Z * is to be minimised, hence penalty for A1 and A2 in the objective function will be -M. The
various tableaus under simplex method will be prepared in the following manner -
Tableau I
Cj 0 0 -M -M -3 -8
Profit Programme Constant S1 S2 A1 A2 X1 X2 Ratio
0 S1 80 1 0 0 0 1 0
-M A1 200 0 0 1 0 1 1 200
-M A2 60 0 -1 0 1 0 1 60
Zj 0 M -M -M -M -2M
Zj - Cj 0 M 0 0 3-M 8-2M
Tableau II
Cj 0 0 -M -M -3 -8
Profit Programme Constant S1 S2 A1 A2 X1 X2 Ratio
0 S1 80 1 0 0 0 1 0 80
-M A1 140 0 0 1 -1 1 0 100
0 X2 60 0 1 0 -1 0 1
Zj 0 -M+8 -M M-8 -M -8
Zj - Cj 0 -M+8 0 2M-8 -M+8 0
Tableau III
Cj 0 0 -M -M -3 -8
Profit Programme Constant S1 S2 A1 A2 X1 X2 Ratio
-3 X1 80 1 0 0 0 1 0
-M A1 60 -1 1 1 -1 0 0 60
-8 X2 60 0 -1 0 1 0 1 -60
Zj -1200 5 0 0 0 -3 -8
Zj - Cj 5 0 M M 0 0
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Here all ZJ-CJ in Tableau IV are positive, hence it is the optimal solution. This is given X1= 80,
X2 = 120, the minimum value being Rs. 1200.
Problem: 6
The annual hand made furniture show and sales occurs next month and the school of vocational studies is
playing to make furniture for the sale. There are three wood working classes - I year, II year, III year at the
school and they have decided to make three styles of chairs A, B and C. Each chair must receive work in
each class and the time in hours for each chair in each class is given.
Chair I year II year III year
A 2 4 3
B 3 3 2
C 2 1 4
In the next month there will be 120 available in I year class, 160 in the second year class and 100 hours in
third year class to produce chairs. The teacher of the wood working class feels that a maximum of 40 chairs
can be sold at the show. The teacher has determined that the profit from each type of chair will be A — Rs.
40, B — Rs. 35 and C — Rs. 30.
Formulate a linear programming model to determine how many chairs should be produced to maximise
profit.
Solution:
Let x1, are the chairs produced of A type
x2 are the chairs produced of B type
x3 are the chairs produced of C type
Formulation
2x1 + 3x2 +2x3< 120
4 x1 + 3x2+ x3 < 160
3x1 + 2x2 + 4x3< 100
x1, x2, x3 > 0
Objective function
Maximise Z = 40 x1+35 x2+30 x3
The problem is solved by simple method, convert inequalities into equalities by adding slack variables
S.T.
2x1 + 3x2 + 2x3+ xa = 120
4x1 + 3x2 + x3 + xb= 160
3x1 + 2x2 + 4x3 + xc = 100
Rewriting objective function as Z – 40 x1 - 35x2 - 30x3 = 0
Selection of non-basic variables n – m = 6 - 3 = 3
where n = No. of variables
m = No. of constraint equation, where xa, xb and xc are slack variables.
∴
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Non basic variables xl = x2 = x3 = 0 gives.
∴ xa = 120
∴ xb = 160
∴ xc= 100
The initial solution is represented in starting table.
Starting table
∴
Basics Z x1 x2 x3 xa xb xc RHS Ratio
Z 1 -40 -35 -30 0 0 0 0 -
xa 0 2 3 2 1 0 0 120 60
xb 0 4 3 1 0 1 0 160 40
xc 0 3 2 4 0 0 1 100 33.33
Entering variable x1 leaving variable = xc literation no. 1
Basics Z x1 x2 x3 xa xb xc RHS Ratio
Z 1 0 - 25/3 70/3 0 0 40/3 4000/3 -
xa 0 0 5/3 - 2/3 1 0 - 2/3 160/3 32
xb 0 0 1/3 - 13/3 0 1 - 4/3 80/3 80
xc 0 1 2/3 4/3 0 0 1/3 100/3 50
Entering variable x2 leaving variable = xa literation no.2
Basics Z x1 x2 x3 xa xb xc RHS Ratio
Z 1 0 0 20 5 0 10 1600 -
xa 0 0 1 - 2/5 3/5 0 - 2/5 32
xb 0 0 0 42/10 - 3/25 1 - 94/75 20
xc 0 1 0 19/9 - 2/5 0 17/15 12
As Z row has all positive values hence optimal solution has reached
∴ Z = 1600
x1= 12 units
x2= 32 units
x3= 0
Problem 7.
A company has three plants F1, F2 F3 from which it supplies to 4 markets: ABCE. Determine the optimal
transportation plan from the following table giving the plant to market shipping costs, quantities available
Plant
Markets
A B C D Available at plant
F1 13 11 15 20 2
F2 17 14 12 13 6
F3 18 18 15 12 7
Requirement 3 3 4 5 15
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at each plant and quantities required at each market.
Solution:
The problem is balanced transportation problem because the demand and supply both are equal to
15. Now obtain the initial feasible solution by Vogel’s method. The cost matrix and the penalties for
given problem are as under:
Now, penalty of first column is maximum and the lowest cost in column (1,1), i.e., Rs. 13, so allocate 2
units (maximum quantities available in F,) to cell (1,1) and delete row (1) as its requirements is complete.
Now penalty of column (2) is maximum and the lowest cost is 14 so, we can allocate 3 units to cell (2,2)
as 3 units are the maximum requirement for market B. The requirement of column (2) is exhausted
hence column (2) is deleted. The shrunken matrix becomes as—
Factory Warehouses Markets Row
A B C D Penalty
(1) (2) (3) (4)
Avail-
Ability
F1
13 11 16 20 2 (13 –11) =
2
F2 17 14 12 13 6 (13 – 12) =
1
F3 18 18 15 12 7 (15 – 12) =
3
Amount
required
3 3 4 5 15
Column
penalty
(17–13)
= 4
(14–11)
=3
(15 - 12)
=3
(13 – 12)
=1
— —
Factory Warehouse Availability Penalty
(1) (3) (4)
F2 17 12 13 3 1
F3 18 15 12 7 3
Requirement 1 4 5 10
Penalty 1 3 1
Now, we consider earlier row(3)or column(3)because both have minimum cost of Rs. 12. Let us consider
column (3) and allocate 3 units to cell (2,3) and delete row F2. Now the balance will be allocated to row
F3. Thus, the initial solution by Vogel’s method is given in the following table:
Factory Warehouse Availability
A B C D
F1 2 – – – 2
F2 – 3 3 – 6
F3 1 – 1 5 15
Requirement 3 3 4 5 7
Here m+n-1 =6 equal to the occupied cells hence solution is feasible. Now, the optimality can be
tested by calculating the implicit cost for each cell. The calculations are given below:
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In the above table in cell 1,2, the implicit cost is greater than the actual cost hence 8 units’ are allocated to
(we get the following solution):
Factory Warehouse Availability
A B C D
13 11 15 20
F1
13 11 9 7 2 –5
17 14 12 13
F2 2 4 6 –2
16 14 12 10
18 18 15 12
F3
2 29 5 6
12 15 14 12
Requirement 3 3 4 5 15
Vj 18 16 14 12
In the above table, the minimum of negative θ is one hence take θ =1 and get the revised solution and test
the optimality again with the help of the following table:
Factory Warehouse Availability
A B C D
F1 2– θ θ – – 2
F2 – 3– θ 3+ θ – 6
F3 1– θ – 1– θ 5 7
Required 3 3 4 5 15
Factory Warehouse
A B C D
Availability
13 11 15 20
F1 2 –5
13 12 10 17
2
17 14 13
F2 3 –3
15 14 9
6
16 18 15 12
F3 5 0
18 17 15 12
7
Vj 18 17 15 12
Required 3 3 4 5 15
Thus, it can be observed from the above table that implicit cost for all cells is less than the corresponding
actual costs. Hence the improve solution is optimal and the minimum cost is Rs. 156 as under
Transportation cost
F1A = 11 × l = 13
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F1B = ll × 1 = 11
F2B = 14 × 2 = 28
F2C = 12 × 4 = 48
F3A = 18 × 2 = 36
F3D = 12 × 5 = 60
Rs. 196
Problem 8. The Raja Company has two factories A and B located at some distance and three regional
warehouse R, S, T. The transportation manager must schedule shipments for the coming week according
to the following:
Warehouse R requires 70 tonnes
Warehouse S requires 60 tonnes
Warehouse T requires 50 tonnes
Capacity of factory A—100 tonnes
Capacity of factory B—200 tonnes
Transportation costs are as follows:
From factory A to warehouse R — Rs. 30 per tonne
From factory A to warehouse S — Rs. 10 per tonne
From factory A to warehouse T — Rs.50 per tonne
From factory B to warehouse R — Rs. 20 per tonne
From factory B to warehouse S — Rs. 40 per tonne
From factory B to warehouse T — Rs. 60 per tonne
Find the least cost shipping schedule.
Solution:
In the given problem, the total quantities available are 300 tonnes (100 tonnes in factory A and 200 tonnes
in factory B) whereas the total requirements for all the three warehouses are only 180 tonnes (70 tonnes +
60 tonnes + 50 tonnes). So, it is a problem of unbalanced transportation. The total supply is larger than the
surplus supply available. We can prepare the initial feasible table as follows by lowest cost entry method:
Initial feasible table by lowest cost entry method
Factory Warehouse
R S T X
Availability
30 10 50 0
A 60 40
10 10 50 -10
100
20 40 60 0
B 70 120
20 20 60 4
200
Required 70 60 50 (120) 300
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In the above table it can be observed that m + n- 1 =5. It means 5 cells are occupied and the rule m + n - I is
satisfied. Hence the solution is feasible.
In the above table, all implicit costs are either less or equal to the actual transportation cost hence this
represents the optimal solution.
The least cost transportation schedule will be as follows:
Rs.
From factory A to warehouse S : 40 tonnes @ 10 per tonne = 4,00
From factory A to warehouse T : 40 tonnes @ 50 per tonne = 2,000
From factory B to warehouse R : 70 tonnes @ 20 per tonne = 1,400
From factory B to warehouse T : 10 tonnes @ 60 per tonne = 600
Minimum total transportation cost = 4,400
4.2 Transportation
Transportation applications relate to a LPP where goods are to be transported from “m” production locations
(factories) to “n” sales locations (warehouses). The objectives are (a) To meet the differing availabilities
and requirements of these locations and (b) To minimise the total transportation costs.
The transportation application can be solved in three stages:
(a) Preliminary Check,
(b) Initial Basic Feasible Solution (IBFS),
(c) Optimality Test.
Methods of finding the initial basic feasible solution to a transportation problem:
IBFS can be determined using any of the following methods:
(a) Northwest Corner Rule,
(b) Least Cost Cell Method,
(c) Vogel’s Approximation Method (VAM).
Sages involved in determining the solution to the Transportation Problem.
(1) Stage 1: Preliminary check involves the following:
(a) Verify Objective = Minimisation. In case of profit matrix, convert the same into an Opportunity
Loss Matrix, by subtracting each number from the highest number in the matrix.
(b) Verify Nature of data = Balanced.
Data is said to be balanced if Total Availability = Total Requirement.
In case of unbalanced data, a dummy column or row should be introduced with zero transportation costs.
(2) Stage 2: IBFS can be determined using any of the following methods:
(a) Northwest Corner Rule,
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(b) Least Cost Cell Method,
(c) Vogel’s Approximation Method (VAM).
(3) Stage 3: Optimality Test: It consists of the following steps
(a) Computation of margin numbers, ‘U’ and ‘V for all rows and columns such that U
+ V = Cost of allocated cells,
(b) Computation of U + V for unallocated cells,
(c) Computation of Cost Less (U + V) for unallocated cells, i.e. Step 1 minus Step 2 above.
Northwest Corner Rule in determining the IBFS.
Step Procedures
1 Ensure Availability = Requirement, by inserting dummy row or column, if
required.
2 Go to top left hand corner cell of the matrix
Compare availability and requirement.
Allocate availability or requirement, whichever is less, to that cell.
Cancel the row or column where availability or requirement is exhausted.
3 Go to top left hand corner cell of the resultant matrix (after cancellation of
row or column in Step 2)
Repeat Step 2 procedure till all the row availability and column
requirements are satisfied.
Areas of Application:
Northwest Corner Rule may be used in transportation applications
• Within the campus of an organisation as cost differences are not significant.
• Obligations where cost is not the criteria for decision-making.
Least Cost Cell Method of determining the IBFS.
Advantage: The cost associated with each route is taken into consideration. So, this method leads to a
better allocation than North West Corner Rule Method.
Step Procedures
1 • Ensure Availability = Requirement, by inserting dummy row or column, if
required.
2 • Identify the cell with the lowest cost. In case of a tie, arbitrary selection may
be made.
• Compare availability and requirement.
• Allocate availability or requirement, whichever is less, to that cell.
• Cancel the row or column where availability or requirement is exhausted.
3 • Identify the next lowest cell in the matrix.
• Repeat Step 2 procedure till all the row availability and column requirements
are satisfied.
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Vogel’s Approximation Method (VAM) for obtaining IBFS
Steps involved in Optimality Test:
Optimality test involves the following steps:
(1) Table 1: Ui + Vj for allocated cells:
(a) Select the row/column with maximum number of allocations.
(b) For that row/column, Ui/Vj is equal to zero. [Ui for Rows; Vj for Columns]
(c) The other set of numbers Ui/Vj are computed in such a way that Ui + Vj = Cost of allocated cells.
Note : The Ui + Vj table can be completed only if the IBFS is non-degenerate.
IBFS is said to be degenerate if number of allocations < (No. of rows + No. of Columns - 1).
In case of degeneracy, a dummy allocation “e” (a number very close to zero) is made in the least cost
unallocated cell falling in a non-dummy row or column.
(2) Table 2: Ui + Vj for unallocated cells:
(a) Draw a matrix for the given rows and columns.
(b) Block out the allocated cells.
(c) Compute Ui + Vj (total of margin numbers) for all unallocated cells.
(3) Table 3: Net Evaluation Table = Δij
(a) Draw a matrix for the given rows and columns.
(b) Block out the allocated cells.
(c) Compute Cost Difference for all unallocated cells. Cost Difference = Number in Table (1) Less
Number in Table (2).
Decision:
(a) If all numbers in the Net Evaluation Table are non-negative (i.e. > 0), IBFS is optimal and unique.
Steps Procedures
1 Compute Cost Differences for each row and column.
Cost Difference is the difference between the least cost and the next least
cost in that row/column.
In case of tie in least cost, Cost Difference = 0.
2 Ascertain the maximum of cost differences and select that row or column for
allocation
3 Choose the least cost cell in the selected row or column for allocation.
4 Compare availability and requirement for that cell.
Allocate availability or requirement, whichever is less, to that cell.
Cancel the row or column where availability or requirement is exhausted.
5 Compute Cost Differences for the resultant matrix and repeat the above
procedure till all row availability and column requirements are satisfied.
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(b) If all numbers are positive and one of the cells contains zero, IBFS is optimal but not unique. Alternative
solution exists. Total number of solutions = Number of Zeroes + 1
(c) If any one number in Table 3 is negative, solution is not optimal. Reallocation should be made to
determine Alternative Basic Feasible Solution (ABFS).
(d) ABFS is tested for optimality by adopting the same procedure.
Concept of “loop” diagram in Transportation
(a) Identify the worst negative in Table 3. If there is a tie, choose the least cost cell.
(b) Draw a loop with the following principles:
• Loop should commence from and end in the selected worst negative cell.
• It should have only allocated cells as its other corners.
• Only horizontal and vertical lines (not diagonal) shall be permitted.
• Loop should result in a even sided figure.
(c) Identify the selected cell as having scope for allocation. It is marked with plus (+) sign.
(d) Other corners of the loop are identified with (-) and (+) signs alternatively.
(e) ABFS is determined by reference to reallocation as specified by the corners of the loop.
(f) Reallocation is done as under
• Identify the allocations to the negative cells (corners marked with (-) sign),
• Select the minimum out of the above allocations,
• Add this minimum to (+) corners and subtract this minimum from (-) corners,
Other allocations remain unaffected.
(g) This ABFS is tested for optimality using the procedure outlined in the earlier question. In case negatives
still arise in the Net Evaluation Table, the reallocation procedure should be continued further.
Treatment of different situations in the context of Transportation Problems:
Situation Treatment
Total availability not
equal to total
requirement
• It is called unbalanced transportation application.
• Introduce dummy row or column to balance availability and
requirement.
• All entries (costs) of the dummy row/column shall be zeroes.
Maximisation
objective
• Select the highest element of profit in the matrix.
• Subtract each element from the maximum element.
• Take the resultant opportunity cost matrix with minimisation
objective, for allocation procedure.
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Problems and Solutions
Problem 1.
A Company has four factories Fl, F2, F3 and F4 manufacturing the same product. Production and raw
material costs differ from factory to factory and are given in the following table. The transportation costs
from the factories to sales depots S1, S2 and S3 are also given. The sales price and the total requirement at
each depot as also the product capacity at each factory is also stated. Determine the most profitable
production and distribution schedule and the corresponding profit. The surplus production should be
taken to yield zero profit.
Degeneracy • Optimality test can be done only if number of allocations =
[No. of rows + No. of Columns - 1]
• If number of allocations is less, all Ui and Vj numbers cannot
be computed. This situation is called degeneracy.
• If IBFS is degenerate, introduce a dummy allocation (e), a
small number very close to zero, in the least cost unallocated
cell falling in a regular row or column. (i.e. non-dummy row
or column)
Existence of zeroes in
he Net Evaluation
Table
• Zero in the Net Evaluation Table (Table 3) indicates the
availability of alternative solutions. (provided all other
numbers are positive)
• The alternative optimal solutions can be determined by
drawing a loop commencing from such cell having zero as net
evaluation.
• Reallocation will be based on the loop drawing procedure
outlined above.
Prohibited routes • Sometimes some routes may not be available due to reasons
like bad road conditions, strike etc.
• Such prohibited routes are identified with a high cost.
represented by or “M”, a very large cost close to infinity.
• Due to assignment of very large cost, such routes would
automatically be eliminated in the final solution.
Situation Treatment
Particulars Fl F2 F3 F4
Sales price
per unit
Requirements
per unit
Production Cost per unit 15
18
14 13
Raw material Cost per unit 10 9 12 9
Transportation cost per unit to
S1 3 9 5 4 34 80
S2 1 7 4 5 32 120
S3 5 8 3 6 31 150
Production capacity (units) 10 150 50 100
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Solution:
From the data given, the profit matrix is constructed as follows:
Profit = Selling Price - Production Cost - Material Cost - Transportation Cost
• In the above matrix, the objective is maximization of profit. This is converted into a opportunity cost
minimization objective by subtracting each element from the highest element i.e. 8.
• Also, the above matrix is unbalanced data since total Capacity is 310 units and total Requirement is
350 units. This is converted into balanced data by introducing the dummy factory F- 5 with zero as its
entries.
• The revised matrix (balanced minimization) is given below.
Place F -1 F - 2 F - 3 F - 4 Requirement
S - 1 6 -2 3 8 80
S - 2 6 -2 2 5 120
S - 3 1 - 4 2 3 150
Capacity 10 150 50 100
Place F -1 F - 2 F - 3 F - 4 F - 5 Requirement
S - 1 2 10 5 0 0 80
S - 2 2 10 6 3 0 120
S -3 7 12 6 5 0 150
Capacity 10 150 50 100 40 350
The above matrix is now amenable for applying VAM.
Initial Basic Feasible Solution (IBFS) is determined as under:
Cost Diff:
Note: Upon full allocation, the relevant row/column may be cancelled in pencil.
I 0 0 1 3 0 In the above IBFS,
II 0 0 1 3 _ • Number of allocated cells is 7.
III
5
2 0 2 − • m + n − 1
(i.e. Rows + Columns −1)= 3+ 5-1=7.
IV _ 2 0 2 _
V _ 2 0 _ _
VI − 2 _ − −
Hence, there is no degeneracy. This can
be tested for optimality.
Place F-1 F-2 F-3 F-4 F-5 Requirement Cost Differences
I II III IV V VI
80 80/0 0 2 — — — —
S−1 2 10 5 0 0
10 90 20 120/110/90/0 2 1 1 3 4 10
S−2
2 10 6 3 0
60 50 40 150/110/60 5 1 1 1 6 12
S−3
7 12 6 5 0
Capacity 10/0 150/60/0 50/0 100/20/0 40/0 350
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OPTIMALITY TEST:
Table 1 = Ui + Vj for allocated cells computed as below:
Ui + Vj 2 − 0 = 2 10 − 0 = 10 6 − 2 = 4 3 − 0 = 3 0 − 2 = −2
80
0 −3 = −3 2 10 5 0 0
10 90 20
(base) = 0
2 10 6 3 0
60 50 40 12 – 10 = 2
7 12 6 5 0
Table 2 = Ui + Vj for unallocated cells computed as below:
−3+2 = −1 −3+10 = 7 −3+ 4 = 1 −3+(−2) = −5
0 + 4 = 4 0 + (−2)= −2
2+2 = 4 2+3 = 5
Table 3 = Net Evaluation Table (NET) = Table 1 - Table 2 for unallocated cells is computed below:
2− (−1) = 3 10 −7= 3 5 − 1= 4 0 − (−5) = 5
6 − 4 = 2 0 − (−2)= 2
7 − 4 = 3 5 − 5 = 0
Conclusion: Since all entries in NET are non-negative, the IBFS is optimal. Since one entry in NET is zero,
the above optimal solution is not unique. One alternative solution exists.
Computation of Maximum Profit: (from the profit matrix formulated first)
Place F − l F −2 F −3 F−4 F−5
S- l 80 × 8 = 640
S-2 10×6 = 60 90×(−2) = −180 20 × 5 = 100
S-3 60×(−4) = −240 50 × 2 = 100 40 × 0 = 0
Determination of Alternative Optimal Solution: Since alternative solution is available, (one entry in NET
= 0), the alternative optimal solution is determined by drawing a loop from the “Zero” entry in the NET.
The loop is shown below in the NET of the IBFS.
80
2 10 5 0 0
10 90 + 20=110 20
2 10 6 3 0
60 – 20 = 40 50 40
7 12 6 5 0
2− (−1) = 3 10 −7= 3 5 − 1= 4 0 − (−5) = 5
+ ve 6 − 4 = 2 − ve 20 0 − (−2)= 2
7 − 4 = 3 -ve 60 5 − 5 = 0 + ve
Since the least allocation of the negative corners of the loop is 20, the alternative solution (optimal solution)
determined by adding 20 to the positive corners, subtracting 20 from the negative corners and leaving the
other cells undisturbed.
The alternative allocation is shown below:
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Resource Management
The profit form the above alternative allocation is:
Place F − 1 F − 2 F − 3 F − 4 F − 5
S−1 80 × 8 = 640
S−2 10 × 6 = 60 110 × (−2) = – 220
S−3 40 × (−4) = −160 50 × 2 = 100 20 × 3 = 60 40 × 0 = 0
Maximum Profit = Total of above = Rs.480
Note: The alternative solution need not be tested further for optimality, since the IBFS is already optimal.
Problem 2.
A company has three factories and four customers. It furnishes the following schedule of profit per unit on
transportation of goods to customers in rupees. You are required to solve the transportation problem to
maximize the profit. Determine the resultant optimal profit.
Factory /
Customer
A B C D E Supply
P 4 19 22 11 0 100
Q 0 9 14 14 0 30
R 6 6 16 14 0 70
Demand 40 20 60 30 50
Initial Basic Feasible Solution is determined as under:
Factory /
Customer
A B C D E Supply Cost Differences
I II III IV
10 60 30 4 7 7 7
A
4 19 22 11 0 100/90/0
30 0 9 B − −
0 9 14 14 0 30/0
20 50 6 0 0 − C
6 6 16 14 0 70/20/0
Demand 40/10/0 20/0 60/0 30/0 50/0 200
I 4 3 2 3 0 In the above IBFS,
II 4 3 2 3 − • Number of allocated cells is 6.
III 2 13 6 3 − • m + n- 1 (i.e. Rows + Columns- 1)
IV 4 − 2 11 − = 3 + 5-1 = 7.
Hence, there is a degeneracy.
Cost Diff:
Factory /
Customer
A B C D Supply
P 40 25 22 33 100
Q 44 35 30 30 30
R 38 38 28 30 70
Demand 40 20 60 30
Solution: The above problem is profit maximization. Hence it should be converted into an opportunity
loss matrix by subtracting the highest element i.e. 44 from the rest. Also the problem is unbalanced as
demand = 150 and Supply = 200. Hence it should be balanced by introducing a dummy column with all
entries as 0. Performing the above operations, the resultant balanced minimization matrix is -
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259
To overcome degeneracy, a dummy allocation represented by ‘e’ (where ‘e’ = 0.00000.......1) is made in the
least cost unallocated cell falling in a regular row or column i.e. non dummy row or column. [It should be
noted that degeneracy does not make the IBFS as wrong answer; it is only a situation not a defect.]
OPTIMALITY TEST:
Table 1 = Ui + Vj for allocated cells computed as below:
+ ve 19+4 = 15 − ve 60 0−(−2) = 2
9 + 0 = 9 14 − 18 = −4 14 − 7 = 7 0−(−6) = 6
− ve ‘e’ 16−24 = −8 14 − 13 = 1
Ui & Vj 4 −0 = 4 6 − 2 = 4 22 − 0 = 22 11 − 0= 11 0 − 2 = − 2
(base) 0 10 60 30
4 19 22 11 0
0 − 4= −4 30
0 9 14 14 0
6 − 4 = 2 e 20 50
6 6 16 14 0
0 + 4 = 4 0+(−2) = −2
− 4 + 4 = 0 − 4 + 22 = 18 − 4 + 11=7 − 4+ (−2) = − 6
2 + 22 =24 2 + 11 = 13
19−4 = 15 0−(−2) = 2
9−0 = 9 14 − 18 = − 4 14 − 7 =7 0−(−6) = 6
16 − 24 = − 8 14 −13 =1
Table 2 = Ui + Vj for unallocated cells computed as below:
Table 3 = Net Evaluation Table (NET) = Table 1- Table 2 for unallocated cells is Table computed below:
There are two negative elements in the NET, hence allocation is not optimal. The loop is created as below-
ABFS 1: The new Ui + Vj for allocated cells is computed from the above.
Table 1 = Ui + Vj for allocated cells computed as below:
Ui & Vj 4 −0 = 4 6−(−6) =12 22− 0 = 22 11− 0 =11 0 − (−6) = 6
(base) 0 10 60 30
4 19 22 11 0
0 − 4 = − 4 30
0 9 14 14 0
16− 22 = − 6 20 e 50
6 6 16 14 0
‘e’ is subtracted / added.
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Resource Management
Table 2 = Ui + Vj for unallocated cells computed as below:
19−12 = 7 0 − 6 = −6
9−8= 1 14−18 =−4 14−7 =7 0 − 2 = −2
6 −(−2) = 8 14−5 =9
Ui & Vj 4 − 0 = 4 6 − (−6) =12 22 − 0 = 22 11− 0 =11 0 − (0) = 0
(base) 0 10 60 30 50
4 19 22 11 0
0 − 4 = −4 30
0 9 14 14 0
16−22= −6 20 50
6 6 16 14 0
19 − 12 = 7
9 −8 = 1 14 − 18 = −4 14 −7 = 7 0 − (−4) = 4
6 −(−2) = 8 14 −5 = 9 0 − (−6) = 6
0 + 12 = 12
− 4 + 12 = 8 −4 + 22 = 18 −4 + 11 = 7 − 4 + 0 = −4
− 6 + 4 = −2 −6 + 11 =5 − 6 + 0 = − 6
Table 3 = Net Evaluation Table (NET) = Table 1 - Table 2 for unallocated cells is computed below -
There are negative elements in the NET; hence allocation ABFS 1 is not optimal. The loop is created as
below -
ABFS 2 : The new Ui + Vj for allocated cells is computed from the above.
Table 1 = Ui + Vj for allocated cells computed as below:
Table 2 = Ui + Vj for unallocated cells computed as below:
Table 3 = Net Evaluation Table (NET) = Table 1 - Table 2 for unallocated cells is computed below:
There are negative elements in the NET, hence allocation is not optimal. The loop is created as below:
+ ve 19 − 12 = 7
− ve 30 9 −8 = 1 14 − 18 = −4 14 −7 = 7 0 − (−4) = 4
6 − (−2) = 8 14 −5 = 9 0 − (−6) = 6
0+ 12= 12 0 + 6 = 6
− 4+12 = 8 −4 + 22 = 18 −4+11=7 −4 + 6 = 2
−6 + 4 = −2 −6+11=5
19−12 = 7 − ve 60 0 − 6 = −6
9−8= 1 14−18 = −4 14−7 =7 0 − 2 = −2
6 −(−2) = 8 + ve 14−5 =9 − ve 50
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261
ABFS 3: The new Ui + Vj for allocated cells is computed from the above.
Table 1 = Ui + Vj for allocated cells computed as below:
Ui & Vj 4 − 0 = 4 6− (−2) = 8 14 (−4) = 18 11− 0 =11 0 − 0 = 0
(base) 0 20 30 50
4 19 22 11 0
0 − 4 = − 4 20 10
0 9 14 14 0
16 − 18 = -2 20 50
6 6 16 14 0
Table 2 = Ui + Vj for unallocated cells computed as below:
Table 3 = Net Evaluation Table (NET) = Table 1- Table 2 for unallocated cells is computed below.
Since there are no non-negative elements in the NET, ABFS 3 is optimal and unique. Computation of
Maximum Profit: (from the profit matrix formulated first)
Maximum Profit = Total of above = Rs.5130
Problem 3.
The information on the available supply to each warehouse, requirement of each market and the unit
transportation cost from each warehouse to each market is given below:
0 + 8 = 8 0+18 = 18
− 4 + 8 = 4 − 4 + 11 = 7 − 4 + 0 = − 4
− 2 + 4 = 2 − 2 + 11 = 9 − 2 + 0 = − 2
19−8 = 11 22−18 = 4
9−4 = 5 14−7 =7 0 − (−4) = 4
6−2 = 4 14−9 = 5 0 − (−2) = 2
Place A B C D E
P 20 × 40 = 800 30×33=990 50×0 = 0
Q 20 × 44 = 880 10×30 = 300
R 20 × 38 = 760 50×28= 1400
Market Supply
Warehouse
Ml M2 M3 M4
A 5 2 4 3 22
B 4 8 1 6 15
C 4 6 7 5 8
Demand 7 12 17 9
The shipping clerk has worked out the following schedule from his experience
Units 12 1 9 15 7 1
From warehouse A A A B C C
To – market M2 M3 M4 M3 Ml M3
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Resource Management
You are required to-
(i) Check and see if the clerk has the optimal schedule
(ii) Find the optimal schedule and minimum total shipping cost and
If the clerk is approached by a carrier of route C to M2, who offers to reduce his rate in the hope of getting
some business, by how much should the rate be reduced before the clerk considers giving him an order.
Solution:
Initial Basic Feasible Solution is determined as under from the data given above
Place M-l M-2 M-3 M-4 Requirement Cost Differences
I II III IV V VI
A 12 1 9
5 2 4 3
B 15
4 8 1 6
C 7 1
4 6 7 5
Demand
I In the above IBFS,
II • Number of allocated cells is 6.
III • m + n-1 (i.e. Rows + Columns-1)
IV = 3 + 4-1=6.
V
VI
Hence, there is no degeneracy. This can be tested for
optimality.
Cost Diff :
Cost Differences have not been computed since the Clerk’s allocation is taken as the IBFS.
OPTIMALITY TEST:
Table 1 = Ui + Vj for allocated cells computed as below:
Ui & Vj 4 − 7 = − 3 2 − 4= −2 0(base) 3 − 4= −1
4 − 0 = 4 12 1 9
5 2 4 3
1 − 0= 1 15
4 8 1 6
7 −0 = 7 7 1
4 6 7 5
4 + (−3) = 1
1 + (−3) = −2 l + (−2) = −l l + (−l) = 0
7 + (−2) = 5 7 + (−1) = 6
Table 2 = Ui + Vj for unallocated cells computed as below:
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263
Table 3 = Net Evaluation Table (NET) = Table1-Table 2 for unallocated cells is computed below:
5 − 1=4
4 −(−2) = 6 8 − (−l) = 9 6−0 = 6
6 − 5 = 1 5−6 = −1
5 − 1 = 4 + ve −ve
4 − (−2) = 6 8 −(−l) = 9 6 − 0 = 6
6 − 5 = 1 −ve + ve 5 − 6 = −l
Ui & Vj 4−2 = 2 2−0 = 2 4−0 = 4 3−0 = 3
(base) 0 12 1+1 = 2 9−1=8
5 2 4 3
1 − 4 = −3 15
4 8 1 6
5 − 3 = 2 7 0+1=1
4 6 7 5
There is one negative element in the NET, hence scheduling by the clerk is not optimal. 1 is the least
allocated to the negative corner. The loop is created as below -
ABFS 1: The new Ui + Vj for allocated cells is computed from the above.
Table 1 = Ui + Vj for allocated cells computed as below:
Table 2 = Ui + Vj for unallocated cells computed as below:
0 + 2 = 2
− 3+2 = −1 −3 + 2 = −1 −3 + 3 = 0
2 + 2 = 4 2 + 4 = 6
5 − 2 = 3
4 − (−l) = 5 8 − (−l) = 9 6 − 0 = 6
6 − 4 = 2 7 − 6=1
Place M−l M−2 M−3 M−4
A 12×2 = 24 2×4 = 8 8 × 3 = 24
B 15× 1 = 15
C 7×4 = 28 1 ×5 = 5
Table 3 = Net Evaluation Table (NET) = Table 1 - Table 2 for unallocated cells is computed below:
All elements in the NET are non-negative; hence the revised allocation is optimal and unique.
The optimal allocation and cost is -
Minimum Cost - Total of above = Rs.104
If the clerk wants to consider the carrier of route C to M - 2 for giving an order, then this transportation cost
should be less than the carrier of routes C to M - 1 and C to M - 4 (which are presently allocated cells /
routes.) Hence the carrier of route C to M - 4 has to bring down his rate to Rs.3 from the present Rs.6.
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Resource Management
Problem 4.
X Company is interested in taking loans from banks for its projects — P, Q, R, S, T. The rates of interest and
the lending capacity differ from bank to bank. All these projects are to be completed. The relevant details
are provided below. Assuming the role of a consultant, advice the Company as to how it should take the
loans so that the total interest payable is least. Find out alternate optimum solutions, if any.
Interest rate in % for projects Max Credit
Source Bank
P Q R S T (in 000s)
Private Bank 20 18 18 17 17 Any
amount
Nationalised Bank 16 16 16 15 16 400
Co-operative Bank 15 15 15 13 14 250
Amount required
(in 000s)
200 150 200 125 75
Solution.
Total amount required as loan = Rs.750 (000s). The private bank can give any amount. The data is
made balanced by putting 100 against the private bank.
Initial Basic Feasible Solution is determined as under:
P Q R S T Amount Cost Differences
Part.
I II III IV V VI
100
Private 20 18 18 17 17 100/0 0 1 0 0 0 18
150 50 200
National 16 16 16 15 16 400/250/50/
0
1 0 0 0 0 16
50 125 75
Co-op 15 15 15 13 14 250/125/50/
0
1 1 0 − − −
Required
200/150/
0
150/0 200/0 125/0 75/0 750
Cost
Diff:
I 1 1 1 2 2 In the above IBFS,
II 1 1 1 − 2 • Number of allocated cells is 7.
III 1 1 1 _ − • m + n - 1 (i.e. Rows + Columns - 1)
IV 4 2 2 − − = 3 + 5−1=7.
V − 2 2 − −
VI − 2 − − −
Hence, there is no degeneracy.
This can be tested for optimality.
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265
Optimality Test:
Table 1 = Ui + Vj for allocated cells computed as below:
3 + 15 = 18 3 + 15 = 18 3 + 13 = 16 3+ 14=17
1 + 13 = 14 1 + 14= 15
0 + 15 = 15 0 + 15=15
Table 2 = Ui + Vj for unallocated cells computed as below:
Ui & Vj 15−0=15 16−1 = 15 16−1 = 15 13−0= 13 14−0=14
18−15=3 100
20 18 18 17 17
16−15 = 1 150 50 200
16 16 16 15 16
(base) 0 50 125 75
15 15 15 13 14
Table 3 = Net Evaluation Table (NET) = Table 1 - Table 2 for unallocated cells is computed below:
Solution is optimal since all elements in NET are non-negative. However there are four zeroes and so the
solution is not unique. There are four alternate solutions.
Computation of Minimum Cost: (amounts in ‘000s and interest rate in %)
20−18 = 2 18−18 = 0 17−16= 1 17−17 = 0
15−14= 1 16−15 = 1
15−15 = 0 15−15 = 0
Particulars P Q R S T
Private 100× 18= 18
National 150×16 =24.00 50× 16 = 8 200×16= 32
Co-op 50×15 = 7.50 125×13= 16.25 75×14= 10.50
20 − 18 = 2 − ve 100 18 − 18 = 0 17 − 16 = 1 17 − 17 = 0
+ ve − ve 15 − 14 = 1 16 − 15 = 1
15 − 15 = 0 15 − 15 = 0
Minimum Cost = Total of above = Rs.l,16,250
Determination of Alternative Optimal Solution -1:
The alternative optimal solution is determined by drawing a loop from the “Zero” entry in the NET. The
loop is shown below in the NET of the IBFS.
Since the least allocation of the negative corners of the loop is 100, the alternative solution (optimal solution)
is determined by adding 100 to the positive corners, subtracting 100 from the negative corners and leaving
the other cells undisturbed.
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Resource Management
The alternative allocation is shown below:
Minimum Cost = Total of above = Rs.1,16,250
Determination of Alternative Optimal Solution 3: The alternative optimal solution is determined by
drawing a loop from the “Zero” entry in the NET. The loop is shown below in the NET of the IBFS.
Since the least allocation of the negative corners of the loop is 50, the alternative solution (optimal solution)
is determined by adding 50 to the positive corners, subtracting 50 from the negative corners and leaving
the other cells undisturbed.
The alternative allocation is shown below:
Minimum Cost = Total of above = Rs.l, 16,250
Determination of Alternative Optimal Solution - 3: The alternative optimal solution is determined by
drawing a loop from the “Zero” entry in the NET. The loop is shown below in the NET of the IBFS
100 – 100 = 0 0 + 100 = 100
20 18 18 17 17
150 50 + 100 = 150 200 - 100 = 100
16 16 16 15 16
50 125 75
15 15 15 13 14
Particulars P Q R S T
Private 100 x 18 =18.00
National 150 x 16 = 24.00 150 x 16 = 24.00 200 x 16=32.00
Co- op 50 x 15 = 7.50 125 x 13 = 16.25 75 x 14 = 10.50
The cost from above alternative re-allocation is :
17 – 16 = 1 17 – 17 = 0
15 – 14 = 1 16 – 15 = 1
20 – 18 = 2 18 – 18 = 0
15 – 15 = 0 15 – 15 = 0
+ ve
20
150 + 50 = 200
16
50 – 50 = 0
18
16 16 15 16
18 17 17
100
50–50=0
0 + 50 = 50
200
125
15 15 15 13 14
Particulars
Private
National
Co-op
P Q R S T
100 x 18 = 18.00
200 x 16 = 32.00
50 x 15 = 7.50
200 x 16 = 32.00
125 x 13 = 16.25 75 x 14 = 10.50
The cost from the above alternative re-allocation is :
75
+ ve
-- ve50
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267
Since the least allocation of the negative corners of the loop is 50, the alternative solution (optimal solution)
is determined by adding 50 to the positive corners, subtracting 50 from the negative corners and leaving
the other cells undisturbed.
Minimum Cost = Total of above = Rs.l,16,250
Determination of Alternative Optimal Solution - 4: ]
The alternative optimal solution is determined by drawing a loop from the “Zero” entry in the NET. The
loop is shown below in the NET of the IBFS.
Since the least allocation of the negative corners of the loop is 75, the alternative solution (optimal solution)
is determined by adding 75 to the positive corners, subtracting 75 from the negative corners and leaving
the other cells undisturbed.
Minimum Cost = Total of above = Rs. 1,16,250
20 – 18 = 2 18 – 18 = 0 17 – 16 = 1 17 – 17 = 0
15 – 14 = 1 16 – 15 = 1
15 – 15 = 0 15 – 15 = 0
+ve
– ve50
– ve200
Particulars P Q R S T
Private 100 x 18 = 18.00
National 200 x 16 = 32.00 50 x 16 = 8.00 150 x 16=24.00
Co- op 50 x 15 = 7.50 125 x 13 = 16.25 75 x 14 = 10.50
20
150 + 50 = 200
16
50 – 50 = 0
18
16 16 15 16
18 17 17
100
50 200 – 50 = 150
125
15 15 15 13 14
0 + 50 = 50 75
The cost from above alternative re-allocation is :
17 – 16 = 1 17 – 17 = 0
15 – 14 = 1 16 – 15 = 1
20 – 18 =2 18 – 18 = 0
+ ve 15 – 15 = 0 15 – 15 = 0
– ve 150
The cost from the above alternative re-allocation is :
Particulars
Private
National
Co-op
P Q R S T
25 x 18 = 4.50
75 x 16 = 12.00 125 x 16 = 20.00 200 x 16 = 32.00
125 x 13 = 16.25
75 x 17 = 12.75
20
150 - 75 = 75
16
50 + 75 = 125
18
16 16 15 16
18 17 17
100 - 75 = 25
200
125
15 15 15 13 14
50 + 75 = 125
- ve = 75
– ve 100
+ ve
0 + 75 = 75
75 - 75 = 0
125 x 15 = 18.75
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Resource Management
4.3 Replacement of Machine and other Relevant Concept
Replacement of Machines and Equipment
Machines are purchased and replacement of old machines-are made mainly for two reasons:
1. To increase the productive capacity and
2. To reduce cost of production.
Various other reasons for replacement are the following,
1. To get rid of worn out, broken down or obsolete machines,
2. To accommodate larger sizes of work and increase the machine capacity,
3. To reduce labour cost by introducing semi-automatic machines or machines more than one of which
can be operated by a single operator,
4. To simplify operations by using automatic machines which are capable of performing variety of work
usually performed by a number of different machines,
5. To minimise repair cost and reduce idle time.
An analysis of the above six reasons will lead to either increase in capacity or reduction in cost or both.
Factors on which equipment is replaced: The replacement plan depends on evaluation of present and
replacement machines from the point of view of technical suitability and cost saving features.
The points to check for replacement studies vary from industry, to industry on management conditions
and management policies. But some factors are common to practically all cases. These are:
Technical Factors:
(i) Inadequacy from the stand paint of range, speed, accuracy, strength, rigidity, output and capacity,
(ii) Obsolescence and equipment worn out condition,
(iii) Special advantage of the new machine as to easiness of set ups convenience of operation, safety,
reliability performance, control panels and special features,
(iv) Flexibility and versatility of the machine.
Cost Factors:
(i) High repair cost of existing machine,
(ii) High remodelling cost of existing machine,
(iii) Less chance of spoilage and rejection work causing: saving in cost,
(iv) Faster rate of production causes lower cost,
(v) Replacement of skilled -workers by semi-skilled and: unskilled workers leading to lesser labour cost,
(vi) Compactness of the machine leading to a saving in-space which means saving of overhead costs,
(vii) Machine pay back period i.e. how soon the cost of the equipment is recovered,
(viii)Life of the new machine giving effective service,
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(ix) Flexibility and versatility of the machine tending to reduce idle time cost with changes in methods of
production-which might occur in future,
(x) Availability of funds for the acquisition of the equipment or possibility of special arrangement like
hire-purchase or government loans or other accommodations.
Replacement Programmes: It is prudent to have phased .policies of machine replacement plans than to
wait until breakdowns occur causing production hold ups. There are different forms of the programme.
1. A definite amount of money or a certain percentage of earning of the company is used each year to
replace existing machines which are either superseded by improved models or are not in tip top
condition or are having insufficient capacity.
2. Replacement is made of the oldest or most inadequate machine each year by upto date machines of
greater accuracy or higher capacity. Some times automatic machines are gradually introduced in this
way which is capable of doing several operations with lesser number of operators.
3. The economy of working on various machines are studied .and replacement of machines are made
only to have a definite cost reduction.
Whatever may be the programme, replacement question is to be carefully considered in prosperous and
normal years only. In slum or dull periods, replacement should not be done unless it is unavoidable.
The basic concept of ‘Replacement Theory’ is to take decision about the time when an item of ‘Capital
Asset’ should be replaced by another of the same type or by a different one. In other words ‘Replacement
Theory’ concerns about ‘optimum period of replacement’.
For the purpose of ‘Replacement Theory’ Capital equipment are basically divided under two broad
categories:
1. Replacement policy for Equipment which detoriates with time gradually;
2. Items which fail suddenly and cannot be made workable by incurring repairing costs.
Again, replacement of Capital equipment which gradually detoriates with time, can be worked-out under
two different ways:
(i) Ignoring the concept of time value of money and considering the time value of
money.
(ii) When time value of money is not considered: Determination of optimum period of
‘Replacement’ Let us, consider the following formula:
T(r) = C - S + Σ
=
r
t 1
Mt
where, T(r) denotes owning and maintaining cost of keeping an equipment for ‘r years’
C = Capital cost of the equipment
S = Salvage value of the equipment at the end of ‘r years’
Mt = Cost of maintenance in year t
Accordingly, the average cost, A(r) can be calculated as under:
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Resource Management
A(r) =
r
1
⎥⎦
⎤
⎢⎣
⎡
+ − Σ=
r
T
C S Mt
1
Decision: When to replace? The optimal replacement period would be the year, in which A(r) = Average
cost is minimum.
(ii) When time value of money is considered: In this case the concept of present value of money is
considered. Considering the interest rate, which is also termed as discounting factor or discounting
rate with that the present values of maintenance costs are to be evaluated for each year and final
annualised cost, for different years are also evaluated. Decision: When to replace? The decision will
be taken by considering the following rules:
A. Replacement will not be made if annualised cost > next period’s cost.
B. Replacement will be made in that year only, when annualised cost < next period’s cost. In other
words, replacement will be made every r years.
when A(r) = the annualized cost of replacing every r years is the minimum.
Important Assumptions
(a) Salvage/scrap value of the equipment to be replaced should be considered as ‘nil’
(b) The maintenance costs are assumed to be incurred at the beginning of the year.
Replacement of items which fail suddenly
Sometimes, the failure of an item may cause a complete breakdown of the system. The costs of failure, in
such a case will be quite higher than the cost of the item itself. So, it is important to know in advance as to
when the failure is likely to take place so that item can be replaced before it actually fails and the time of
failure can be predicted from the probability distribution of failure time obtained from past experience and
then, to find the optimal value of time t which minimises the total cost involved in the system. Mainly two
types of replacement policies are there:
(1) Individual replacement policy in which an item is replaced immediately after it fails.
(2) Group replacement policy in which all items are replaced, irrespective of whether they have failed or
not, with a provision that if any item fails before the optimal time, it may be individually replaced.
Let us, explain through example how ‘Replacement of items which fail suddenly’ is considered: Following
failure rates have been observed for a certain type of light bulbs:
Week: 1 2 3 4 5
Per cent failing by the end of week: 10 25 50 80 100
There are 1,000 bulbs in use, and it costs Rs. 2 to replace an individual bulb which has burnt out. If all bulbs
were replaced simultaneously it would cost 50 paisa per bulb. It is proposed to replace all bulbs at fixed
intervals of time, whether or not they have burnt out, and to continue replacing burnt out bulbs as and
when they fail. At what intervals all the bulbs should be replaced? At what group replacement price per
bulb would a policy of strictly individual replacement become preferable to the adopted policy?
Solution: Let, Pi be the probability that a light bulb, which was new when placed in position for use, fails
during the ith week of its life.
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Thus, following frequency distribution is obtained assuming to replace burnt out bulbs as and when they
fail,
P1 = the prob. of failure in Ist week = (10/100) = 0.10
P2 = the prob. of failure in IInd week = (25 - 10/100 = 0.15
P3 = the prob. of failure in IIIrd week = (50 - 25)/l00 = 0.25
P4 = the prob. of failure in IVth week = (80 - 50)/100 = 0.30
P5 = the prob. of failure in Vth week = (100 - 80)/100 = 0.20
Total = 1.00
Since the sum of all probabilities can never be greater than unity, therefore all further probabilities P6, P7
and P8 so on, will be zero. Thus, a bulb that has already lasted four weeks is sure to fail during the fifth
week. Furthermore, assume that
(i) Bulbs that fail during a week are replaced just before the end of that week, and
(ii) The actual percentage of failures during a week for a subpopulation of bulbs with the same age is the
same as the expected percentage of failures during the week for that subpopulation.
Let Ni be the number of replacements made at the end of the ith week, if all 1000 bulbs are new initially.
Thus,
N0 = N0 = 1,000
N1 = N0pl = 1,000 × 0.10 = 100
N2 = N0p2 + N1p1 = 1,000×0.15+ 100×0.10 = 160
N3 = N0P3 + N1p2 + N2p1 = 1,000×0.25+ 100×0.15 + 160×0.10 = 281
N4 = N0p4 + N1p3 + N2P2 + N3p1 = 377
N5 = N0p5 + N1p4 + N2P3 + N3p2 + N4p1 = 350
N6 = 0 + N1p5 + N2p4 + N3p3 + N4p2 + N5p1 = 230
N7 = 0 + 0 + N2p5 + N3p4 + N4p3 + N5p2 + N6p1 = 286
and so on.
It has been observed that the expected number of bulbs burnt out in each week increases until 4th week and
then decreases until 6th week and again starts increasing. Thus, the number will continue to oscillate and
ultimately the system settles down to a steady state in which the proportion of bulbs failing in each week
is the reciprocal of their average life.
Since the mean age of bulbs
= l×p1+2×p2 + 3×p3 + 4×p4 + 5 ×p5
= 1 × 0.10 + 2 × 0.15 + 3 × 0.25 + 4 × 0.30 + 5 × 0.20 = 3.35 weeks.
the number of failures in each week in steady state becomes
= 1,000/3.35 = 299 and the cost of replacing bulbs only on failure = 2 × 299 (at the rate of Rs. 2 per bulb) =
Rs. 598.
Since the replacement of all 1,000 bulbs simultaneously costs 50 paise per bulb and replacement of an
individual bulb on failure costs Rs. 2, therefore cost of replacement of all bulbs simultaneously is as given
in the following table:
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Resource Management
The cost of individual replacement in the third week exceeds the average cost for two weeks.
Thus it would be optimal to replace all the bulbs after every two weeks; otherwise the average cost will
start increasing.
Further, since the group replacement after one week costs Rs. 700 and the individual replacement after one
week costs Rs. 598, the individual replacement will be preferable.
Replacement—Staffing Problem
Problems concerning recruitment and promotion of staff can sometimes be analysed in a manner similar
to that used in replacement problems in industry.
A faculty in a college is planned to rise to strength of 50 staff members and then to remain at that level. The
wastage of recruits depends upon their length of service and is as follows:
End
of week
Cost of individual
replacement
Total cost of group replacement Average cost
per week
1 100 × 2 = 200 1,000 × 0.50 +100×2 = Rs.700 Rs. 700.00
2 160× 2 = 320 700+ 160×2 = Rs. 1,020 Rs.510.00
3 281 ×2 = 562 1,020 + 281 ×2 = Rs. 1,582 Rs. 527.33
the end of year
(i) Find the number of staff members to be recruited every year.
(ii) If there are seven posts of Head of Deptt. for which length of service is the only creterion of promotion,
what will be average length of service after which a new entrant should expect promotion?
Solution:
Let us assume that the recruitment per year is 100. From above it is clear that the 100 who join in the
first year will become zero in 10th year, the 100 who join in the 2nd year will (serve for 9 years and)
become 5 at the end of the 10th year and the 100 who join in the 3rd year will (serve or 8 years and)
become 14 at the end of the 10th year and so on. Thus, when the equilibrium is attained, the distribution
length of service of staff members will be as under:
Year: 1 2 3 4 5 6 7 8 9 10
Total % who left up to: 5 35 36 65 70 76 80 86 95 100
Year No. of staff members
0 100
1 95
2 65
3 44
4 35
5 30
6 24
7 20
8 14
9 5
10 0
Total 432
∴
Operation Management
273
(i) Thus if 100 staff members are recruited every year, the total number of staff members after 10 years of
service = 432
100 To maintain a strength of 50, the number to be recruited every year =
432
100
x 50 = 11.6
It is assumed that those staff members who completed x years’ service but left before x + 1 years’
service, actually left immediately before completing x + 1 years.
If it is assumed that they left immediately after completing x years’ service, the total number will
become 432 - 100 = 332 and 100 the required intake will be 50 x
332
100
= 15
In actual practice they may leave at any time in the year so that reasonable number of recruitments
per year =
2
11.6 +15
= 13 (approx)
(ii) If we recruit 13 persons every year then we want 7 seniors. Hence if we recruit 100 every year, we
shall require
13
7
x 100 = 54 (approx.) seniors.
It can be seen that 54 seniors will be available if we promote them during 6th year of their service ( ∵
0 + 5 + 14 + 20 + 24 = 63 > 54 ).
∴ The promotion of a newly recruited staff member will be due after completing 5 years and before
putting in 6 years of service.
Problems and Solutions
Problem : 1. The following table gives the running costs per year and resale values of a certain equipment
whose purchase price is Rs. 6,500. At what year is the replacement due optimally.
∴
Year 1 2 3 4 5 6 7 8
Running costs (Rs.) 1,400 1,500 1,700 2,000 2,400 2,800 3,300 3,900
Resale value (Rs.) 4,000 3,000 2,200 1,700 1,300 1,000 1,000 1,000
Year 1 2 3 4 5 6
Replacement cost at the
beginning of the year 100 110 125 140 160 190
Salvage value at the
end of the year 60 50 40 25 10 0
Operating costs 25 30 40 50 65 80
Problem: 2. A truck-owner finds from his past experience that the maintenance costs are Rs. 200 for the
first year and then increase by Rs. 2,000 every year. The cost of the Truck Type A is Rs. 9,000. Determine the
best age at which to replace, i.e. truck. If the optimum replacement is followed what will be the average
yearly cost of owing and operating the Truck? Truck Type B cost Rs. 10,000. Annual operating costs are Rs.
400 for the first year and then increase by Rs. 800 every year. The Truck owner have now the Truck Type A
which is one year old. Should it be replaced with B type, and if so, when?
Problem:3. A Plant Manager is considering replacement policy to a new machine. He estimates the following
costs:
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Resource Management
Find the year when replacement is to be made.
Problem : 4. A fleet owner finds from his past records that the costs per year of running a vehicle whose
purchase price is Rs 50,000 are as under:
Year 1 2 3 4 5 6 7
Running cost Rs.) 5,000 6,000 7,000 9,000 11,500 16,000 18,000
Resale value (Rs.) 30,000 15,000 7,500 3,750 2,000 2,000 2,000
Thereafter, running cost increases by Rs. 2,000, but resale value remains constant at Rs. 2,000. At what age
is a replacement due?
Problem : 5. The following mortality rates have been observed for a certain type of light bulbs:
There are 1,000 bulbs in sue, and it costs Rs. 2 to replace an individual bulb which has burnt out. If all bulbs
were replaced simultaneously it would cost 50 paise per bulb. It is proposed to replace all bulbs at fixed
intervals, whether or not they have burnt out, and to continue replacing burnt out bulbs as they fail. At
what intervals should all the bulbs be replaced?
Problem : 6. An electric company which generates and distributes electricity conducted a study on the life
of poles. The repatriate life data are given in the following table:
Life data of electric poles
Week 1 2 3 4 5
Per cent failing by end of week 10 25 50 80 100
(i) If the Company now installs 5,000 poles and follows a policy of replacing poles only when they fail,
how many poles are expected to be replaced each year during the next ten years?
To simplify the computation assume that failures occur and replacements are made only at the end of
a year.
(ii) If the cost of replacing individually is Rs. 160 per pole and if we have a common group replacement
policy it costs Rs. 80 per pole, find out the optimal period for group replacement.
Problem : 7. A manufacturer is offered two machines A and B. A is priced at Rs. 5,000 and running costs are
estimated Rs. 800 for each of the first five years, increasing by Rs. 200 per year in the sixth and subsequent
year. Machine B, which has the same capacity as A, costs Rs. 2,500 but will have running costs of Rs. 1,200
per year for six years, increasing by Rs. 200 per year thereafter.
If money is worth 10% per year, which machine should be purchased? (Assume that the machines will
eventually be sold for scrap at a negotiable price).
Years after
installation:
1 2 3 4 5 6 7 8 9 10
Percentage poles
failing:
1 2 3 5 7 12 20 30 16 4
∴
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275
Problem: 8. Suppose that a special purpose type of light bulb never lasts longer than 2 weeks. There is a
chance of 0.3 that a bulb will fail at the end of the first week. There are 100 new bulbs initially. The cost per
bulb for individual replacement is Re. 1 and the cost per bulb for a group replacement is Re. 0.50. It is
cheapest to replace all bulbs: (i) individually, (ii) every week, (iii) every second week, (iv) every third
week?
Problem: 9. A manufacturing firm is considering a policy of replacing certain key electrical components
belonging to one group of machines on a group replacement basis instead of making repairs as needed.
There are approximately 100 parts of one type that have the mortality distribution shown below. The cost
of replacing parts on individual basis is estimated to be Rs. 10 per part whereas that on group basis comes
to Rs. 3 per part. Compare the average weekly cost of the two replacement alternatives:
Week Probability of failure during week
1 0.3
2 0.1
3 0.1
4 0.2
5 0.3
Year Net capital
cost (Rs.) (C–
S)
Running
Cost
(Rs.)
Cumulative
running
cost (Rs.)
Total cost
Rs.
(2) + (4)
Average
Annual cost
(Rs.) (5)÷(1)
(1) (2) (3) (4) (5) (6)
1 2,500 1,400 1,400 3,900 3,900
2 3,500 1,500 2,900 6,400 3,200
3 4,300 1,700 4,600 8,900 2,967
4 4,800 2,000 6,600 11,400 2,850
5 5,200 2,400 9,000 14,200 2,840
6 5,500 2,800 11,800 17,300 2,882
Year
(1)
Capital cost
(Rs.)
(2)
Maintenance
cost (Rs.)
(3)
Cumulative
maintenance cost
(Rs.)
(4)
Total cost
(Rs.) (5)
(2) + (4)
Avg. annual
cost (Rs.) (6)
(5) ÷ (l)
1 9,000 200 200 9,200 9,200
2 9,000 2,200 2,400 11,400 5,700
3 9,000 4,200 6,600 15,600 5,200
4 9,000 6,200 12,800 21,800 5,450
Solution 1: Chart showing optimal Replacement period
∴ Optimal replacement period at the end of 5th year.
Solution 2 Truck A Chart Showing Optimal Replacement Period
∴
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Resource Management
Year Capital cost
(Rs.)
Maintenance
cost (Rs.)
Cumulative
maintenance
cost (Rs.)
Total cost
(Rs.)
Avg. annual
cost
(Rs.)
(1) (2) (3) (4) (2) + (4) (5) ÷ (1)
1 10,000 400 400 10,400 10,400
2 10,000 1,200 1,600 11,600 5,800
3 10,000 2,000 3,600 13,600 4,533
4 10,000 2,800 6,400 16,400 4,100
5 10,000 3,600 10,000 20,000 4,000
6 10,000 4,400 14,400 24,400 4,067
∴Optimal replacement period end of 3rd year.
Truck B Chart Showing Optimal Replacement Period
Comparing the average cost of Trucks A and B when the current truck A is one year old.
Average Annual Maintenance Cost
At the year third, the Truck A can be replaced by Truck B and subsequently it can be replaced at the end of
5 years of its use.
Solution 3.
Chart Showing Optimal Replacement Period
Year A B
1 5,700 10,400
2 5,200 5,800
3 5,450 4,533
The asset should be replaced by the end of 2nd year where the average annual cost is lowest.
Year Net capital cost
(Rs.) (Cost -
Scrap)
Operation
cost (Rs.)
Cumulative
operation
Costs (Rs.)
Total cost (Rs.) Average annual
cost (Rs.)
1 40 25 25 65 65
2 60 30 55 115 57.5*
3 85 40 95 180 60
4 115 50 145 260 65
5 150 65 210 360 72
6 190 80 290 480 80
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277
Solution 4.
Chart Showing Optimal Replacement Period
Year
(1)
Net capital
cost (Rs.)
(2)
Annual
maintenance cost
(Rs.)
(3)
Cumulative
operation costs
(Rs.)
(4)
Total cost
(Rs.)
(5)
Average
annual cost
(Rs.) (6)
(5) ÷ (l)
1 20,000 5,000 5,000 25,000 25,000
2 35,000 6,000 11,000 46,000 23,000
3 42,500 7,000 18,000 60,500 20,167
4 46,250 9,000 27,000 73,250 18,313
5 48,000 11,500 38,500 86,500 17,300
6 48,000 16,000 54,500 1,02,500 17,083*
7 48,000 18,000 72,500 1,20,500 17,214
Otimal replacement at the end of 6th year.
Solution 5.
Chart Showing Optimal Replacement Period
It should be replaced by the end of Second Week.
Solution 6.
Chart Showing Optimal Replacement Period
Average life of the pole 1 × 0.01 + 2 × 0.02 + 3 × 0.03 + 4 × 0.05 + 5 × 0.07 + 6 × 0.12 + 7 ×0.20 + 8 ×0.3 + 9 ×
0.16 + 10 × 0.04 = 7.05
No. of poles to be replaced every year =
7.05
5000
= 709
Average yearly cost on individual replacement = 709×160 = Rs. 1,13,440.
Group Replacement: Initial Cost = 5,000 × 80 = 4,00,000
1 2 3 4 5
% failing during the week 10 15 25 30 20
Week Bulbs to be replaced cost Weekly
cost
Cumulative
cost
Total
cost
Avg.
Weekly
cost
1 1,000 × 0.1 = 100 200 200 700 700
2 1,000 ×0.15 + 100 × 0.1 = 160 320 520 1,020 510
3 1,000×0.25+100×0.15+160 ×0.10 =
281
562 1,082 1,582 527
Initial cost = 1000 x .5 = 500
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Resource Management
Optimal replacement at the end of the 5th year.
Solution 7.
Machine A: Chart Showing Optimal Replacement Period
Year No. of poles to be replaced Yearly
cost
Cumulative
cost
Total cost Avg.annual
cost
1 5000 × 0.01 = 50 8,000 8,000 40,8000 40,8000
2 5000 × 0.02 +50 ×.01 = 101 16,160 24,160 42,4160 21,2080
3 5000 × 0.03+50 × 0.02
+101× 0.01= 152
24,320 48,480 44,8480 14,9493
4 5000 × 0.05 + 50 × 0.03+
101× 0.02+152 × 0.01 = 256
40,960 89,440 48,9440 12,2360
5 5,000 × 0.07 + 50 × 0.05 + 101
× 0.03 + 152 × 0.02 + 265
× 0.01 = 362
57,920 1,47,360 5,47,360 1,09,472
6 5,000 × 1.2 + 50 × 0.07 + 101
× 0.05 + 152 × 0.03 + 256 ×
0.02 + 362 × 0.01
= 6023
9,63,680 11,11,040 15,11,040 2,51,840
Year Running
cost (Rs.)
Discount
factor
Discounted
running
Cost (Rs.)
Cumulative
running Cost
(Rs.)
Total
cost
(Rs.)
Cumulative
discounted
factor
Annualized
cost
(Rs.)
(1) (2) (3) (4) (5) (6) (7) (8) = (6) ÷7
1 800 1.00 800 800 5,800 1.00 5,800
2 800 0.9091 727 1,527 6,527 1.9091 3,419
3 800 0.8264 661 2,188 7,188 2.7355 2,628
4 800 0.7513 601 2,789 7,789 3.4868 2,234
5 800 0.6830 546 3,335 8,335 4.1698 1,999
6 1,000 0.6209 621 3,956 8,956 4.7907 1,869
7 1,200 0.5645 677 4,633 9,633 5.3552 1,799
8 1,400 0.5132 718 5,351 10,351 5.8684 1,764
9 1,600 0.4665 746 6,097 11,097 6.3349 1,752
10 1,800 0.4241 763 6,860 11,860 6.7590 1,755
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279
Year
Running
cost (Rs.)
Discount
factor
Discounted
running
Cost (Rs.)
Cumulative
running
Cost (Rs.)
Total
cost
(Rs.)
Cumulative
discounted
factor
Annualized
cost (Rs.)
(1) (2) (3) (4) (5) (6) (7) (8)=6÷7
1 1,200 1.00 1,200 1,200 3,700 1.0000 3,700
2 1,200 0.9091 1,091 2,291 4,791 1.9091 2,510
3 1,200 0.8264 992 3,283 5,783 2.7353 2,114
4 1,200 0.7513 902 4,184 6,684 3.4868 1,917
5 1,200 0.6830 820 5,004 7,504 4.1698 1,800
6 1,200 0.6209 745 5,750 8,250 4.7907 1,722
7 1,400 0.5645 790 6,540 9,040 5.3552 1,688
8 1,600 0.5132 821 7,361 9,861 5.8684 1,680
9 1,800 0,4665 840 8,201 10,701 6.3349 1,689
Machine B:
Chart showing optimal Replacement period
Machine B should be purchased as its annual cost is lowest i.e. Rs. 1,680.
Solution 8.
Chart Showing Optimal Replacement Period
Probability 0.3; 0.7
Individual replacement: Average life of bulb = 1 × 0.3 + 2 × 0.7 = 1.7
Weekly replacement =
100
1.7 = 59
Group replacement: 100 × 0.5 = 50
Week Items to be replaced Weekly
cost
Cumulative
cost
total cost Avg. weekly cost
1 100 × 0.3 = 30 30 30 80 80
2 100 × 0.7 + 30 × 0.3 = 79 79 109 159 79.50
It should be replaced by the end of second week. It cannot be replaced every third week as the life of the
bulb is only 2 weeks.
Solution: 9 Replacement cost on individual basis
5
100×10 = Rs.1,000
= Rs. 200 per week. Replacement
cost on group basis 3(100 × 0.3) + 3(100 × 0.1) + 3(100 × 0.1) + 3(100 × 0.2) + 3(100 × 0.3) = 300/5 = Rs.60/-
4.4 Change of Technology and its Implication
Technology is a resource of profound importance not only in production but also to profitability and
growth of the entire business organisation. Technology drives productivity and also drives change in this
world. Technological change is a major factor in gaining competitive advantage. The development and
Innovative use of technology can provide a firm a distinctive competence. Competitive advantage can be
achieved not just from creating new technology but also by applying and integrating existing technologies.
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Resource Management
Technology is a significant ingredient in virtually all production and operations management decisions.
Advances in computer technologies (both hardware and software) automation, robotics, lasers, and
information and communication technologies have had broad-reaching impact across all industries. To
stay competitive, manufacturing and service organisations must adopt new technologies.
Firms that have used technology as a competitive weapon have effectively integrated their technology
strategy and business strategy. As these firms invent and develop new technologies, they offer new products
and services.
Technology is defined to be know-how, physical things and procedures used to produce products and
services. Know-how means knowledge and judgement of how, when and why to employ equipments,
processes and procedures. Knowledge includes craftsmanship and experience, physical things are
equipments and tools, procedures are the rules and techniques for operating the equipment and performing
the work.
Technologies require a support network to be implemented. A support network consists of physical,
informational and organisational relationships that make a technology complete and allow it to function
as intended.
Advanced technology refers to the application of the latest scientific or engineering discoveries to the
design of production and operations processes.
Technology and technique do not mean the same. Technique is the totality of methods rationally arrived at
and having absolute efficiency whereas technology is the organisation and application of knowledge for
the achievement of practical purposes. Some of the examples of technology are manufacturing or production
technology, design technology, computer technology, communication technology, nuclear technology,
satellite; communication technology, space technology, missile technology, laser technology and the like.
Advances in technologies create new products and services and reshape processes. Technology takes many
forms, beginning with ideas, knowledge, and experience and then utilising them to create new sad better
ways of doing things. Technology provides distinctive competency and competitive advantage to a firm
over others. The impact of technology is pervasive.
A vital factor in production organisations is whether the technology involves capital intensive operation
(i.e., large investment in plant and machinery) or labour-intensive operations. Technology will affect: (i)
the organisation of production, (ii) the capital investment in plant and equipments buildings etc., (iii) the
scale (or volume) of production operations, (iv) the influence of labour relations in-production operations.
The management of production is vitally concerned with the technology of the production processes. It
must organise according to the technology adopted and by the adoption of productive system which is
either capital intensive or labour intensive. Also, it must design a highly sophisticated production control
system for batch production, or a material management system for high capacity assembly line operations.
The organisation is not simply a technical or social system. It requires structuring and integrating human
activities around various technologies. The technical system is determined by the task requirements and
shaped by the specialisation of knowledge and skills required, the types of machinery and equipments
involved, the information processing requirements and the layout of facilities.
Any change in the technical system affects the organisational elements. The impact of technology on the
organisation-its goals, structure, psychological system and managerial system will be quite significant.
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281
Technology can be classified as: (i) manual technology, (ii) mechanised technology (or mechanisation), (iii)
Automated technology (or automation), (iv) current technology, (v) appropriate technology, (vi) state-ofthe-
art technology, (vii)advanced technology, (viii) obsolete technology (or out-dated technology), (ix)
capital-intensive technology, (x) labour intensive technology, etc.
Manual technology is the use of muscular power to do work which was prevalent before industrial
revolution. Mechanised technology or mechanisation uses machine power in place of muscle power and is
the first step towards automation. Examples of mechnisation are power operated tools, tool changing
devices, powered materials handling equipments such as conveyors, jib cranes for loading and unloading
heavy jobs etc. Automated technology or automation is any form of equipment or machine which will
carryout a preset program or sequence of operations and at the same time measure and correct its actual
performance in relation to that program. Current technology is any technology currently used by a firm in
its operations. Appropriate technology is the technology which meets the requirements of time, place and
objectives of a firm at a particular point of time. Appropriateness is an inherent quality in technology.
Depending on the social, political, economic and other conditions prevailing in a country at a particular
point of time and also depending on the priorities of the country, a technology becomes appropriate even
though it may not be an advanced technology or state-or-the-art technology. (For example, whether capitalintensive
or labour-intensive technology is the appropriate technology to be adopted by a firm or industry
or a country).
State-of-the-art technology is a modern technology which has been adopted by many developed countries
in the world. This technology will enable the firm to produce state-of the art products using state-of-the art
design. It is also known as proven technology. Advanced technology is the latest technology based on the
latest scientific or engineering discoveries and used in the design and production processes. It is also
known as new technology which is developed by a firm which is a technology leader in its field of operation.
Examples of advanced technologies are space technology, missile technology, information technology,
laser technology, bio-technology and so on.
Obsolete technology is an out-dated technology which has been replaced by a superior technology, thereby
resulting in obsolescence of the old technology which has been replaced.
Capital-intensive technology involves huge investments in capital assets such as equipment and machinery,
materials handling and storage systems, information handling (storage and retrieval) systems,
communication systems and office automation equipments.
Labour-intensive technology does not involve investment in huge capital intensive systems but make use
of abundant labour (manpower) available in the country. For example, in India, textile and mining Industries
adopt labour-intensive technologies.
The Choice of Technology: The choice of technology depends on several factors, both internal and external
to the organisation choosing the technology. The various internal factors are : (i) availability of funds for
investment, (ii) product life cycle and technology-life cycle position, (iii) present plant capacity and
technology adopted (i.e., current technology).
Technology can be quite capital intensive and require high investment in equipments, machines and
processes. The question is whether the firm can afford to invest in a new technology which may be highly
expensive. Also, the change to a new technology is not advisable when the product is in the saturation or
decline stage in its life cycle. New technologies for processing may be best suited for developing and
manufacturing new products. The new technology chosen should be capable of matching with the existing
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Resource Management
technology and plant capacity so that there will be a synergy effect on the plant capacity when a new
technology is adopted.
The external factors involved in the choice of technology are: (i) Government polices and regulations,
availability of resources such as raw materials, energy, skilled labour etc., required for using the new
technology, (ii) market scenario (market demand, customer requirement of product quality etc.).
Technology Life-Cycle: Like a product has its life cycle, technology also has a life cycle. The various phases
or stages in a technology life cycle are:
(i) Innovation in which stage a new technology (product or process technology) is developed, (ii)
Syndication during which stage, the technology is demonstrated and slowly commercialised, (iii)
Diffusion stage in which a new technology gradually replaces the current technology and (iv)
Substitution stage in which the current technology becomes obsolete and is completely replaced by
the new technology.
The recent development in the technology of product design comprises certain tools provided by the
information sciences that contribute to better, cheaper and faster design of products. They are:
(i) Computer-aided design (CAD)
(ii) Computer-aided design and manufacture (CAD/CAM).
(i) Computer Aided Design (CAD): It is an electronic system using computers for designing new parts
or products or modifying existing ones, replacing the traditional drafting work done by a draftsman
on a drafting board. The CAD consists of a powerful desktop computer and graphics software that
enables a designer to manipulate geometric shapes. The designer can create drawings and view them
from any angle on a display monitor. CAD softwares have been developed for designing electronic
circuits, printed-circuit-board design, designing and drafting three dimensional drawings and also
for analysis of heat and stress in mechanical designs.
Advantages of CAD are:
(i) Allows designers to save time and money by shortening design and development cycle time, (ii) Eliminates
prototype model building to prove the design, (iii) Allows designers to determine costs and test such
variable as stress, tolerance, product variability, interchangeability and serviceability, (iv) Low cost of
design even for a custom-built, low volume product,(v) Eliminates manual drafting completely,(vi) Makes
review of numerous options in design possible before final commitments are made because of the speed
and ease with which sophisticated designs can be manipulated, (vii) Faster development, better products
and accurate flow of information to other departments and (viii) Product cost can be determined at the
design stage itself.
Production technology
In addition to the developments in design technology, a number of advances are made in technology used
to enhance production. Some these advancements in production technology are:
(i) Numerical control and Computer Numerical Control (NC and CNC), (ii) Automated process controls,
(in) Vision systems (automated inspection systems), (iv) Robots, (v) Automated identification systems
(AIS), (vi) Automated storage and retrieval system (ASRS), (vii) Automated guided vehicles (AGV),
(viii) Automated Sow lines, (ix) Automated assembly systems, (x) Flexible manufacturing systems
(FMS), (xi) Computer integrated manufacturing (CIM), (xii) Enterprise resource planning.
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283
The above aspects of production technologies are discussed in detail in the following section.
(i) Numerical Control and Computer Numerical Control: Many machines such as lathe, milling, drilling
and boring machines are now designed for electronic control called numerical control (NC). The
numerically controlled (NC) machines have control systems which read instructions and translate
them into machine operations. When these machines are programmed through their own
minicomputers and have memories to store these programs, they are called computer numerical control
(CNC) machines.
Advantages of N/C Machines: (i) Smaller machine setup time, (ii) Machine motions and tool changing
are controlled by instructions on a control system, (iii) Increased productivity and higher quality, (iv)
Suitable for low volume production.
Advantages of CNC Machines: (i) Instructions may be stored and handled more efficiently, ii) Micro
computer system controls the machine settings and operations rather than human beings, (iii) Real
time and off-line diagnostic possibilities may be built into the CNC system, (iv) Machining data and
operator instructions may be displayed on the screen of computer.
Application of NC/CNC Machines : These machines are used to machine parts (i) with complex
machining requirements, (ii) requiring high precision, (iii) in developmental stages where many changes
may be needed, (iv) normally requiring extensive tooling, (v) requiring fast or slow machining speeds,
(vi) made from expensive raw materials and (vii) required in small quantities of repetitive batches.
(ii) Automated Process Control: Automated process control makes use of information technology to
monitor and control a physical process. It is also used to determine and control temperatures, pressures
and quantities in petroleum refineries, cement plants, chemical plants, steel mills, nuclear reactors
etc., process control systems operate in a number of ways : (a) Sensors collect data, (b) Analog devices
read data periodically (say once a minute or once a second), (c) Measurements are translated into
digital signals which are transmitted to a digital computer, (d) Computer programs read the digital
data and analyse the same, (e) The resulting output may take numerical forms which include messages
on computer consoles or printers, signal to motors to change valve settings, warning lights or sirens
etc.
(iii) Vision Systems (Automated Inspection Systems): These are machines or equipments that have been
integrated into the inspection of products for controlling quality. They combine video camera and
computing technology to take physical dimensions of parts, compare the measurements to standards
and determine whether the parts meet quality specifications. Vision systems are also used for visual
inspection in food processing organisations. Automated inspection systems facilitate 100 percent
inspection which will lead to improved product quality and reduced inspection costs.
(iv) Robots: Robotics or robotry is a fast developing field of technology in which human like machines
perform production tasks. A robot is a reprogrammable, multifunctional manipulator designed to
move materials, parts, tools or specialised devices through variable programmed motions for the
performance of a variety of tasks. Robots are machines which are flexible, have the ability to hold,
more and grab items. They are controlled by micro computers which when programmed guide the
machines through their predetermined operations.
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Advantages:
(i) do not strike work, (ii) do not mind hot, dirty, dusty working conditions, (iii) can work at high
speeds, (iv) will not sue if injured, (v) can work long hours without breaks, (vi) can be used for welding,
painting, assembly work, loading, unloading, material handling and other repetitive, monotonous
work.
(v) Automated Identification Systems (AIS): These use bar codes, radio frequencies, magnetic stripes,
optical character recognition and machine vision to sense and input data into computers. These systems
replace human beings to read data from products, documents, parts and containers and interpret the
data. An example is the system used to identify and read the bar code on an item in the check-out
counters at grocery stores. A scanner reads the identification number from the bar code on the item
accesses a computer data base, and sends the price of the item to the cash register and updates the ;
inventory data in the inventory system.
(vi) Automated Storage and Retrieval Systems (ASRS): Computer controlled warehouses use ASRS;
which provide for the automatic placement and withdrawal of parts and products into and from
designated storage places in the warehouse. Such systems are commonly used in distribution facilities
of retailers.
(vii) Automated Guided Vehicles (AGVs): These are automated materials handling and delivery systems
which can take the form of mono-rails, conveyors, driverless trains, pallet trucks and unit load carriers.
AGVs are electronically guided and controlled vehicles used to move parts and equipments. AGVs
usually follow either embedded guide wires or paint strips through operations until their destinations
are reached.
(viii) Automated Flow Lines: An automated flow line includes several automated machines which are
linked by automated transfer machines and handling machines. The raw material feeders automatically
feed the individual machines and operations are carried out without human intervention. After an
item is machined on one machine on the line, the partially completed item is automatically transferred
to the next machine on the line in a predetermined sequence, until the job is completed. Major
components such as automobile rear axle housings are produced using automated flow lines. These
systems are also known as fixed automation or hard automation because the flow lines are designed
to produce only one type of component or product. These systems are suitable for products with high
and stable demand because of very high initial investment required and the difficulty of changing
over to other products. But production systems which provide greater flexibility (For example, Flexible
Manufacturing Systems) are more favoured than fixed automation.
(ix) Automated Assembly Systems: In this system automated assembly machines or equipments are linked
together by automated materials handling equipments. Examples of automated assembly equipments
are robotic welders or component insertion unit which are used to join one or more parts or components
or assemblies. The partially assembled product is moved to the next assembly equipment automatically
by the automated materials handling equipments and this process is repeated until the whole assembly
is completed. Automated assembly systems require unique product design which is suitable to these
systems unlike product designs suitable for manual assembly operations. Some of the principles that
are applied while designing products for automated assembly are:
(i) The amount of assembly required must be reduced: For example, use of castings or plastic
mouldings (single part) instead of making the part by assembling two or more sheet metal parts
fastened together by bolts and nuts or by rivets.
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(ii) The number of fasteners required must be reduced: Joining two or more parts by welding is
preferred to joining by fasteners such as screws, rivets, bolts and nuts. Product designs should
facilitate this.
(iii) Components must be designed to be automatically delivered and positioned: Hoppers, slotted
chutes, vibratory bowls etc. are used to feed the part and orient them to deliver to the assembly
equipment and position in the proper place for assembly to be done automatically. Product
designs should facilitate this.
(iv) Products must be designed for layered assembly and vertical insertion of parts: Product design
must facilitate build up of assembly from a base upward in layers and the components inserted
vertically into the assembly.
(v) Parts must be designed such that they are self aligning: Design should facilitate aligning of parts
as and when they are inserted into assemblies.
(vi) Products must be designed into major modules for production (modular design): Each module
is assembled in an automated assembly system and them the modules are assembled into the
final product.
(vii) Quality of components should be high: High quality components avoid jams in feeding and
assembly mechanisms. The advantages of automated assembly units are low production cost
per unit, high product quality and higher product flexibility.
(x) Flexible Manufacturing System (FMS): A flexible manufacturing system (FMS) is a configuration of
a group of production machines (or workstations) connected by automated material handling and
transferring machines and integrated by computer system which can give instructions to produce
hundreds of different parts in whatever order specified.
An FMS is a type of flexible automation system which builds on the programmable automation of NC
and CNC machines. Materials are automatically handled and loaded and unloaded for machining
operations. Programs and tooling set up can be quickly changed and production can be quickly
switched on from one job to another with no loss of change over- time.
Key components of an FMS are:
(i) Several computers controlled machining centres or workstations having CNC machines and robots
for loading and unloading.
(ii) Computer controlled transport system (AGVs) for moving materials and parts from one machine to
another and in and out of the system.
(iii) Computer controlled robots for loading and unloading stations.
(iv) An automated storing and retrieval system.
All the above subsystems of FMS are controlled by a central computer with the needed software. Raw
materials are loaded on the AGVs which bring them to the work centres as per the sequence of
operations unique to each part. The route is determined by the central computer. The robots lift the
materials from the AGV and place on the workstation, where the required operations are carried out.
After the completion of operations, the robots unload the job and place it on the AGV to move the job
to the next workstation as per the sequence of operations.
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Resource Management
The FMS is suitable for intermediate flow strategies with medium level of product varieties and volumes
(40 to 2000 units per part). Also, FMS can produce low variety high volume products in the same way
as fixed automation systems.
Advantages: (i) Improved capital utilisation, (ii) Lower direct labour cost, (iii) Reduced inventory,
(iv) Consistent quality, (v) Improved productivity.
Disadvantage: (i) High initial capital investment, (ii) Limited ability to adopt to product changes, (iii)
Substantial preplanning, tooling and fixture requirements, (iv) Standardisation of part designs needed
to reduce number of tools required, (v) Requires long planning and development cycle to install FMS.
(xi) Computer Integrated Manufacturing (CIM): Computer Integrated Manufacturing is a system which
acts as a bridge or umbrella to integrate product design and engineering, process planning and
manufacturing using complex computer systems. It integrates CAD, CAM, FMS, inventory control,
warehousing and shipping. A computer-aided drafting generates the necessary electronic instructions
to run a direct numerically controlled (DNC) machine and any design change initiated at a CAD
terminal can incorporate that change in the part produced on the shop floor within a few minutes.
When this kind of capability is integrated with inventory control, warehousing and shipping as a part
of a flexible manufacturing system, the entire system is called computer-integrated-manufacturing.
The CIM systems work as follows:
1. Top Management decides to make a product based on market opportunities, the company’s strength
and weaknesses and formulates its strategic plan based on competitive advantage.
2. Production/Operations Management implements production process, supplier coordination, material
planning, scheduling operations, delivery schedules and cost controlling functions.
3. Computer-Aided-Design facilitates the design of the product, its quality analysis, planning the
manufacturing process, designing tools and. fixtures and machine loading programs.
4. Computer-Aided-Manufacturing allows fabrication of raw materials into parts to be sent to the
assembly lines.
5. Automated Storage and Retrieval System (ASRS) and automated guided vehicles (AGVs) facilitate
storage/retrieval, movement of incoming materials and parts, work-in-process and finished products
Computer Integrated Manufacturing System
Top Management
Management
Information
System (MIS)
Productional
Operations
Management
(PQM)
Computer
Aided
Manufacturing
(CAM)
Computer
Aided
Design
(CAD)
Robots
Automated Storage
and Retrieval System
(ASRS) and
Automated Guided
vehicles (AGVS)
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287
6. Robots assemble the subunits into final products, test them with automated equipments and put the
finished products into containers for shipment.
7. Management Information System (MIS) integrates all the elements of the computer integrated systems
among themselves and also with the top management of the organisation.
(xii) Enterprise Resource Planning (ERP) System: ERP systems comprise latest comprehensive software
packages to automate a number of business processes. This software integrates most of the business
functions in an organisation. ERP systems have automated manufacturing processes, organised account
books, streamlined corporate departments such as human resources and facilitate business process
re-engineering.
ERP systems need complex set of software programs and heavy investment to implement them. Several
ERP software packages have been developed by leading software companies such as SAP, Oracle,
J.D. Edwards, People Soft and Baan. The latest development in ERP system has been the integration
of e-business capabilities.
E-business uses the internet to conduct or facilitate business transactions such as sales, purchasing,
communication, inventory management, customer service, placing purchase orders and checking the
status of purchase orders etc. ERP software packages were modified with additional features to facilitate
e-business.
Automation Issues
Some of the issues to be deliberated in the use of automation are:
(i) Not all automation projects are successful.
(ii) Automation is not a substitute for poor management.
(iii) Some automation projects may not be worth-while based on economic analysis.
(iv) Some operations are not technically feasible to be automated.
(v) Small and start-up business may not be able to invest for automation projects.
Advantages and Disadvantages of Automation
Advantages: (i) Increased output and higher productivity, (ii) Improved and uniform quality, (iii) Reduced
costs, (iv) Fewer accidents, (v) Better production control, (vi) Dangerous and unpleasant taste can be
performed by robots.
Disadvantages: (i) Heavy capital investment, (ii) Displacement of labour, (iii) Loss of suggestions from
employees, (iv) Design specifications for raw materials can not be relaxed, (v) Cost of shutdown of automated
plant due to shortage of materials is quite high, (vi) Dehumanisation.
Factories of the Future
The features of factories of the future are:
1. Stress on high product quality.
2. Greater emphasis on flexibility.
3. Faster execution and delivery of customer orders.
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Resource Management
4. Change in production economics - fixed costs will become variable and variable costs will change to
fixed costs. Product will be designed using CAD/CAM which forms the basis for process planning
also. CIM systems will be extensively used.
5. Organisational structure changes with line personnel becoming staff personnel and vice versa. The
mainstream activities will be maintenance, quality assurance, product design and engineering
managing technology change, software development etc.
6. The factories of future will be driven by computers used in CIM systems.