Operation Management-ICAI-Inter-Process Planning-Project Planning

2.4 Process Planning
A process is a sequence of activities that is intended to achieve some result, typically to create added value
for the customers.
Types of Processes:
(i) Conversion Processes i.e., converting the raw materials into finished products (for example, converting
iron ore into iron and then to steel). The conversion processes could be metallurgical or chemical or
manufacturing or construction processes.
(ii) Manufacturing Processes can be:
(a) Forming Processes, (b) Machining Processes and (c) Assembly Processes.
(iii) Testing Processes which involve inspection and testing of products (some times considered as part of
the manufacturing processes).
Forming Processes include foundry processes (to produce castings) and other processes such as forging,
stamping, embossing and spinning. These processes change the shape of the raw material (a metal) into
the shape of the work piece without removing or adding material.
Machining Processes comprise metal removal operations such as turning, milling, drilling, grinding,
shaping, boring etc.
Assembly Processes involve joining of parts or components to produce assemblies having specific functions.
Process Planning is defined as the systematic determination of methods by which a product is to be
manufactured economically and competitively. It consists of selecting the proper machines, determining
the sequence of operations, specifying the inspection stages, and tools, jigs and fixtures such that the product
can be manufactured as per the required specification. The detailed process planning is done at each
component level.
After the final design of the product has been approved and released for production, the Production Planning
and Control department takes the responsibility of Process Planning and Process Design for converting
the product design into a tangible product. As the process plans are firmly established, the processing time
required to carryout the production operations on the equipments and machines selected are estimated.
These processing times are compared with the available machine and labour capacities and also against
the cost of acquiring new machines and equipments required, before a final decision is made to manufacture
the product completely in house or any parts or sub assemblies must be outsourced.
In transformation of raw materials into finished products, several questions need to be answered; such as:
1. What will be the production quantity?
2. What are characteristics of the products to be manufactured?
3. The availability of equipment and what kinds of equipment are to be purchased and what will be the
investment?
4. What kinds of labour are required?
5. What should be the level of automation?
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6. Make or buy the components required?
Once these questions are answered, the process planning activity can be carried out with minute details as
to how each component can be manufactured.
Process Planning is concerned with planning the conversion processes needed to convert the raw material
into finished products. It consists of two parts – (i) Process design and (ii) Operations design.
The two terms are explained as under:
Process planning establishes the shortest route that is followed from raw material stage till it leaves as a
finished part or product.
The activities that are associated with process planning are:
• List of operations to be performed and their sequence.
• Specifications of the machines and equipment required.
• Necessary toolings, jigs and fixtures.
• Gives the manufacturing details with respect to feed, speed, and depth of cut for each operation to be
performed.
• It gives the estimated or processing time of operations.
All the above information is represented in the form of a document called process sheet or route sheet.
The information given in the process sheet can be used for variety of activities.
• It becomes the important document for costing and provides the information on the various details
like set-up and operation times for each job.
• The machine and manpower requirements can be computed from the set-up and operational times.
• Helps to carryout scheduling.
• The material movement can be traced.
• It helps in cost reduction and cost control.
• It helps to determine the efficiency of a work centre.
Factors affecting process planning
(i) Volume (quantity) of production.
(ii) Delivery dates for components or products.
(iii) Accuracy and process capability of machines.
(iv) The skill and expertise of manpower.
(v) Material specifications.
(vi) Accuracy requirements of components or parts.
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Steps in Process Planning
1. Detailed study of the component drawings to identify the salient features that influence process
selection, machine selection, inspection stages and toolings required.
2. List the surfaces to be machined.
3. The surfaces to be machined are combined into basic operations. This step helps in selection of machines
for operation.
4. Determine the work centre, tools, cutting tools, jigs and fixtures and inspection stages and equipment.
5. Determine the speed, feed and depth of cut for each operation.
6. Estimate the operation time.
7. Find the total time to complete the job taking into account the loading and unloading times, handling
times, and other allowances.
8. Represent the details on the process sheet.
Process design is concerned with the overall sequences of operations required to achieve the product
specifications. It specifies the type of work stations to be used, the machines and equipments necessary to
carryout the operations. The sequence of operations is determined by (a) The nature of the product, (b) the
materials used, (c) the quantities to be produced and (d) the existing physical layout of the plant.
Operations design is concerned with the design of the individual manufacturing operation. It examines
the man-machine relationship in the manufacturing process. Operations design must specify how much
labour and machine time is required to produce each unit of the product.
Process Design-Framework
The process design is concerned with the following: (i) Characteristics of the product or service offered to
the customers, (ii) Expected volume of output, (iii) Kinds of equipments and machines available in the
firm, (iv) Whether equipments and machines should be of special purpose or general purpose, (v) Cost of
equipments and machines needed, (vi) Kind of labour available, amount of labour available and their
wage rates, (vii) Expenditure to be incurred for manufacturing processes, (viii) Whether the process should
be capital-intensive or labour-intensive, (ix) Make or buy decision and (x) Method of handling materials
economically.
Selection of process
Process selection refers to the way production of goods or services is analysing. It is the basis for decisions
regarding capacity planning, facilities (or plant) layout, equipments and design of work systems. Process
selection is necessary when a firm takes up production of new products or services to be offered to the
customers.
Three primary questions to be addressed before deciding on process selections are:
• How much varieties of products or services will the system need to handle?
• What degree of equipment flexibility will be needed?
• What is the expected volume of output?
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Process decisions:
Major process decisions are:
Process choice :
It refers to choice of a particular process, based upon the nature of product. The operations manager has to
choose from five basic process types – (i) Job shop, (ii) Batch, (iii) Repetitive or assembly line, (iv) Continuous
and (v) Project.
Vertical integration :
Vertical integration is the degree to which a firm’s own production system handles the entire supply chain
starting from procurement of raw-materials to distribution of finished goods.
Two directions of vertical integration are:
(a) Backward integration which represents moving upstream toward the sources of raw-materials and
parts, for example, a steel mill going for backward integration by owning iron ore and coal mines and
a large fleet of transport vehicles to move these raw materials to the steel plant.
(b) Forward integration in which the firm acquires the channel of distribution (such as having its own
warehouses, and retail outlets).
Procedure for process planning and design
1. The inputs required comprise the product design information, production system information and
product strategy decisions.
2. Process planning and design starts with selection of the types of processes, determining the sequence
of operation, selection of equipment, tooling, deciding about the type of layout of facilities and
establishing the control system for efficient analysing of resources to achieve most economical
production of the product.
3. The outputs are specific process plans, route sheets, flow charts, assembly charts, installation of
equipments, machinery, material handling systems and providing trained, skilled employees to
carryout the production processes to achieve the desired results.
Process analysis and process flow design :
While analysing and designing processes to transform input resources into goods and services, certain
questions need to be asked. They are:
• Is the process designed to achieve competitive advantage in terms of differentiation, response or low
cost?
• Does the process eliminate steps that do not add value?
• Does the process analysing customer value as perceived by the customer?
• Will the process enable the firm to obtain customer orders?
A number of tools help production manager to understand the complexities of process design and redesign.
Some of such tools are: (i) Flow diagram, (ii) Assembly charts, (iii) Process charts and (iv) Operation and
Route sheet.
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These tools are discussed in the following paragraphs.
Flow Diagrams: It is a drawing used to analyse the movement of people or material or product to
understand, analyse and communicate the process to others.
Assembly Charts: Assembly charts are used to provide an overall macro view of how materials and sub
assemblies are assembled to form finished products. These charts list all major materials, components, sub
assembly operations, inspections and assembly operations.
Process Charts: A process chart is understood as a graphic representation of events and information relating
to them during a series of actions or operations.
Operation process charts are similar to assembly charts except that they include specifications for the
components as well as operating and inspection times and thereby provide more instruction on how to
produce an item.
The operation analysis and routing sheets or simply the route sheets specify precisely how to produce an
item by identifying the equipment and tools to be used, the operations to be carried out and their sequence
to be followed and the machine set up and run-time estimate.
Purpose of Process Charts
Process charts can present a picture of a given process so clearly that every step of the process can be
understood by those who study the charts.
Process charts may be effective in process analysis and may help in detecting inefficiencies of the processes
currently adopted.
Types of Process Charts
Process charts can be classified as operation process charts, flow process charts, worker-machine/manmachine
charts and activity charts or multiple activity charts.
(a) Operation Process Chart
Operation process chart, the basic process chart is a graphic representation of the points at which materials
are introduced into the process and of the sequences of inspections and all operations except those involved
in material handling. It includes information considered desirable for analysis such as time required and
location.
Flow Process Chart
Flow process chart is a graphic representation of all operations, transportations, inspections, delays and
storages during a process and includes information for analysis such as time required and distance moved.
It is especially useful in detecting hidden, non-productive costs such as delays, temporary storages and
distances travelled.
Worker-Machine Chart or Man-Machine Chart or Multiple Activity Chart
Worker-machine charts or man-machine charts are graphical representation of simultaneous activities of a
worker and the machine or equipment he or she operates. These charts help identify idle time and cost of
both workers and machines. Alternative worker-machine combinations can be analysed to determine the
most efficient arrangement of worker-machine interaction for carrying out a job.
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Worker-machine charts show the time required to complete tasks that constitute a work cycle. A cycle is
the length of time required to progress through one complete combination of work activities.
Operation Analysis and Route Sheet: An operation and route sheet specifies operations and process routing
for a particular part and assembly. It conveys such information as the type of equipment, tooling, and
operations required to complete the part, equipment setup time and operation time etc.
Process improvement
It is a systematic study of the activities and flows of each process to improve the process. Once the process
is thoroughly understood, it can be improved.
Process improvement becomes necessary because of relentless pressure to provide better quality at a lower
price. The basic techniques for analysing the processes such as flow diagrams and process charts are useful
for understanding the processes and improve them. Improvements can be made in quality, through-put
time, cost, errors, safety and on-time delivery.
Process improvement is necessary when:
(i) the process is slow in responding to the customer,
(ii) the process introduces too many quality problems or errors,
(iii) the process is costly,
(iv) the process is a bottleneck, with work accumulating and waiting to go through it, and
(v) the process involves waste, pollution and little value addition.
Application of BCA in the choice of machines or process
This analysis is the most convenient method for selecting the optimum method of manufacture or machine
amongst the competing ones. The cost estimates of the competing methods (both fixed and variable costs)
are prepared and a particular quantity N is determined at which the alternatives give the same cost.
If the quantity to be manufactured is less than N the process with lower fixed cost is selected
and if the quantity to be produced is more than N the process with lower variable cost is selected.
Let FA the Annual Fixed Cost of Machine A
FB the Annual Fixed Cost of Machine B
VA – Variable Cost per unit for Machine A
VB = Variable Cost per unit for Machine B
N = Quantity at which costs on both machines will be equal.
∴ Total cost on machine A = Total cost on Machine B for Quantity N
i.e., FA + VA.N = FB+VB.N
or, N(VA-VB) = FB- FA
or, N = B A
A B
F F
V V
−
−
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The alternative with lower fixed cost will be more economical for manufacturing up to N and once the
quantity exceeds N, it is economical to select an alternative with lower variable cost.
Problems and Solutions
Problem 1:
A component can be manufactured either on centre lathe or on a turret lathe. The cost and time information
to process a component is given below.
The tooling costs are to be recovered within a year. There are no repeat orders. The requirements are to be
met in two lots.
(i) Find the quantity at which both alternatives results in equal cost. (BEP)
(ii) Give the decision rule regarding the choice of lathes
(iii) If the quantity required is 800 Nos./year, which of the machine do you propose?
Solution:
Let F1 = Fixed cost for the centre lathe. Fixed cost consists of set-up and tooling up costs.
∴ Fixed cost for centre lathe (F1) = Set-up cost + Tooling up cost
∴ F1 = No. of set-ups/year × set-up time/set-up (hrs) × [(set-up labour rate) + (Depreciation and other
expense/hr)] + tooling up costs.
= 2 x
30
60
x (10 + 2) + 200 = 212 (Rs.)
Similarly, Fixed cost of turret lathe (F2) =
2 90
60
×
x (20 + 2) + 500= Rs. 566
Variable cost for centre lathe = V1
V1 = Processing time × [(Labour cost/hr + Depreciation and other cost/hr)]
=
10
60
(10 + 2) = Rs. 2 / piece
Variable cost for turret (V2)
V2 =
5
60 (20 + 2) = 1.83 Rs. /piece.
Let, Quantity at which both alternatives gives equal cost be ‘N’.
Particulars Centre lathe Turret lathe
Setup time 30 minutes 120 minutes
Processing time 10 minutes 5 minutes
Tooling up cost (Rs.) 200 500
Labour cost/hr Rs. 2 Rs. 2
Depreciation and other cost per hour Rs. 10 Rs. 20
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N = 2 1
1 2
F –F
V –V =
566 – 212
2 – 1.83
=
354
0.17
= 2082.3 = 2083.
Thus the break-even quantity is 2083 pieces.
2. Decision Rule
(i) If quantity is below 2083, Centre lathe is preferred because of lower fixed cost.
(ii) For quantities above 2083, turret lathe is preferred.
(iii) For quantity 2083, both are equally feasible select either of the two machines.
Problem 2:
A Company’s fixed and variable costs for manufacturing a component on three alternative machines are
given below formulate the decision rules for selecting the machines.
Solution:
The total cost = fixed cost + [variable cost / unit × no. of units]
Let, Tc1 , Tc2 and Tc3 be the total costs for engine lathe, capstan lathe and automat respectively
Let, ‘x’ be the number of units to be manufactured Then,
Total cost on Engine lathe (Tc1) = 5 + 0.2x
Total cost on Capstan lathe (Tc2) = 30 + 0.lx
Total cost on Automat (Tc3) = 70 + 0.05x
Now, comparing the Tc1 and Tc2.
∴ 5 + 0.2x = 30 + 0.lx
∴ 0.lx = 25
x = 250
Comparing the Tc2 and Tc3
At B.E.P. Tc2 = Tc3
∴ 30 + 0.1x = 70+0.05x
∴ 0.05x = 40
∴ x = 800
Decision Rules
(i) If the quantity is below 300, select engine lathe
(ii) Between 300 to 800, select capstan lathe
(iii) Above 800, Automat is to be selected.
Fixed Cost (Rs.) Variable Cost (Rs/unit)
Engine lathe 5 0.20
Capstan lathe 30 0.10
Automat 70 0.05
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Problem 3 :
A Job is performed on the milling machine. The following details are given below:
Standard time for job = 6 minutes
No. of jobs to be produced = 70,000 jobs
Machine capacity = 2000 hrs/month
Machine utilisation = 90%
Compute the number of machines required.
Solution:
Standard Time (ST) =
6
60
=
1
10
hrs.
Maximum Production (MP) = 70,000
Machine Capacity (MC) = 2000 hrs/month
Utilisation of capacity (UC) = 0.9
∴ No. of Machines required (N)
N =
ST MP
MC UC
×
× =
0.1 70,000
2,000 0.9
×
× = 3.88 Machines = 4 Machines
Problem 4:
A company wants to expand the solid propellant manufacturing plant by the addition of more 1 tonne
capacity curing furnace. Each tonne of propellant must undergo 30 minutes of furnace time including
loading and unloading operations. Furnace is used only 80 per cent of the time due to power restrictions.
The required output for the new layout is to be 16 tonnes per shift (8 hours). Plant efficiency (system) is
estimated at 50 per cent of system capacity.
(a) Determine the number of furnaces required
(b) Estimate the percentage of time the furnace will be idle
Solution
Required system capacity =
Actualoutput
SystemEfficiency
=
16tonnes/shift
0.5
= 32.0 tonnes/shift
Reared system capacity (hrs) = ×
32tonnes/shift
0.8 8 hrs/shift = 5 tonnes/hrs.
Individual furnace capacity =
1tonne
0.5hr
= 2 tonnes / hr per furnace
≈
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(i) Number of furnaces required (N) =
Requiredfurnance capacity
Individual furnace capacity
N =
5tonnes/hr
2 tonnes/ hr/ per furnace
= 2.5 furnaces (say 3)
(ii) Total Hours available per shift = 3 furnace @ 8 hours = 24 furnaces hrs
(iii) Total Hours of actual use per shift =16 tonnes × 0.5 hr/tonne = 8 furnace hr
Idle hours = 24 - 8 = 16 hours
Percentage idle time =
5hrs idle
2hrs total = 66.66% ≈ 67% Idle time.
Problem 5:
A lathe machine is used for turning operation and it takes 30 minutes to process the component. Efficiency
of the lathe is 90 per cent and scrap is 20 per cent. The desired output is 600 pieces per week. Consider 48
hours per week. Determine the number of lathes required?
Solution
Assuming 50 weeks in a year.
The output per annum = 600 × 50 = 30,000 units.
The scarp rate is 20%.
∴ The quantity to be produced (including scarp)
Requiredoutput
(1 − Scraprate) =
30,000
(1 − 0.2) = 37,500 units
Total time required for turning
= 37,500 ×
30
60
= 18,750 hours
Production time required with 90 per cent efficiency
=
18,750
0.9
=20833.3 hours
Time available per lathe per annum = 48 × 50 = 2400 hrs
∴ Number of lathes required = = = ≈ Timerequired(hrs) 20833.3
8.68 9
Timeavailable(hrs) 2400
∴ No. of lathes required = 9
≈
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Problem 6:
An article is processed on three machines A, B and C as shown below:
A
B
C
B
Machine Machine operation time Preparation Cleaning
time (min/day) (min/day)
Time Processing Total
A 2 2.5 4.5 15 10
B 3 10 13 30 10
C 2 5 5 35 10
A study revealed that if the jigs for machines B and C were to be redesigned, loading and unloading
times could be reduced to 2 minutes and 1 minute respectively.
(a) Find the number of pieces produced per day (single shift of 8 hrs).
(b) Costing has shown that unless production is increased by 20 per cent the installation of new jigs
would not be worthwhile. Would you recommend redesign of jigs.
(c) If the number to be produced is large, suggest changes in present arrangement and estimate new
production rate.
Solution:
Machines Processing time (minutes) Preparation and cleaning (min/day)
A 2 + 2.5 = 4.5 25
B 3 + 10 = 13 40
C 2 + 5 = 7 35
(a) Cycle time for the job is 13 minutes.
Total production-time available/day = 480 - preparation and cleaning time/day
For machine B (critical operation) = 480 – 40 = 440 min= 440/13
Output/day = 34 pieces
(b) Redesigning the new jigs: The redesigning of new jigs will change the cycle time, i.e., the cycle
time is reduced to 12 minutes (from 13 min as the unloading and loading time is reduced by 1
minute)
Output/day = 440/12 = 36.6 = 37 pieces
Percentage increase in output =
37 34
34
−
× 100 % = 9%
So the redesign of jigs is not justified.
(c) Suggestion if the number of pieces is large: The critical operation (bottleneck) is processing on
machine B which requires 10 minutes. If we introduce one more machine of kind B, then the
cycle time will be reduced to 8 min.
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The new output =
440
8
= 55 pieces.
Problem 7:
A component is to be processed on two machines lathe and milling machine. The sequence of operation is
first turning and then milling.
The machine times are given below:
Turning 12 minutes.
Milling 20 minutes.
1. Estimate the number of machines required to machine 2500 components per week if available machine
hours per week are 48.
2. What are the steps that you propose to reduce number of machines.
Solution:
1. No. of components required to be machined = 2500/week
The available time (in hr.) per week = 48
Assuming 50 weeks in a year, no. of components to be machined/annum = 2500 × 50 = 1,25,000
annum.
The time required to machine 1,25,000 components = 1,25,000 ×
12
60
= 25,000 hrs
Time available per annum = 48 × 50 = 2400 hrs
No. of lathes required = 10.41 ≈ 11
No. of milling machines required =
×
= ≈
20
12,000
60 17.36 18
2400
2. Steps to reduce the number of machines. Increase the number of shifts. Addition of second shift
is going to reduce the number of machines by 50 per cent.
Problem 8:
The component can be processed on either of the two machines — Turrette lathe and Center lathe. The
time and cost details are given below.
Computer the breakeven quantity and state the decision rules.
Particulars Turrette Lathe Centre Lathe
1. Setup time (hrs) 4.5 0.2
2. Operation time/piece (min) 4.0 35
3. Setup cost/hour (Rs.) 350 15
4. Machining cost/hour (Rs.) 45 25
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Production Planning and Productivity Management
Solution:
Breakeven quantity refers to the quantity at which both the alternatives are equally feasible and the
total cost of both alternatives are equal
Let ‘Q’ be the breakeven quantity
Total cost of production on Turrette lathe = Total cost of production on center lathe
= [Setup cost + operation cost] Turret = [Setup cost + operation cost]Center
= [4.5×350] +
4
45
60
⎡ ⎤ ⎢ × × ⎥ ⎣ ⎦
Q = [0.2 × 15] +
35
25
60
⎡ ⎤ ⎢ × × ⎥ ⎣ ⎦
Q
or, 1575 + 3Q = 3 + 14.58Q
or, 11.58Q = 1572
or, Q = 136 units.
If the quantity is < 136, Center lathe is preferred.
If the quantity is > 136, Turrette Lathe is preferred.
Problem 9:
A customer is processed through each of the three operations A, B and C, in sequence. The process is
designed to handle 100 customers a day. The average rate at which each operation can process customers
is shown in the figure below:
A B C
(a) For an 8 hour day identify any bottlenecks in the process.
(b) What effect will bottlenecks have on overall output and other operations?
(c) If all the processes are operated for 10 hours per day, is there still a bottleneck?
(d) How will random arrivals affect the output and processing rate of this process?
Solution:
(a) Identification of bottleneck for an 8 hour a day operation.
Operation A, Output for 8 hrs/day = 15 × 8 = 120 customers
Operation B, Output for 8 hrs/day = 10 × 8 = 80 customers
Operation C, Output for 8 hrs/day = 12 × 8 = 96 customers
Looking into the above calculations, the operations B which process 10 customer/hr and 80
customers a day becomes the bottleneck.
(b) Effect of the bottleneck on the overall output.
The bottleneck operation decides the capacity and output. Here the capacity of the unit is 80
customer/day (due to bottleneck B) even though the process A and C have the output of 120 and
96 respectively.
15 Customer /hr 10 Customer /hr 12 Customer /hr
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(i) Effect of the bottleneck on other operation. Bottleneck operation B, makes the waiting time
higher for C. Because of difference in out (C > B), facility has to wait fill the customer comes
from B.
(ii) In case A, the operation will be completed and because of difference in processing rates,
operation A will be idle for 25% time unless it is used for other purpose and the full capacity
of A if used creates an inventory between A and B.
(c) If all the processes are operated for 10 hrs, then the designed output of 100 customers/hr. are
achieved still the process line is unbalanced as the processing capacities are unequal.
(d) Random arrivals affect the output and processing of customers.
The random arrivals affect the capacity utilisation also both capacity idle time and shortage.
2.5 Project Planning
Planning begins with well-defined objectives. The project team may be drawn from several organizational
departments, e.g., engineering, production, marketing, and accounting. Project definition involves
identifying the controllable and uncontrollable variables involved, and establishing project boundaries.
Performance criteria should relate to the project objectives, which are often evaluated in terms of time,
cost, and resource utilisation
Project planning is part of project management, which relates to the use of schedules such as Gantt charts
to plan and subsequently report progress within the project environment. Project management is the
discipline of organizing and managing resources (e.g. people) in such a way that the project is completed
within defined scope, quality, time and cost constraints. A project is a temporary and one-time endeav or
undertaken to create a unique product or service, which brings about beneficial change or added value.
This property of being a temporary and one-time undertaking contrasts with processes, or operations,
which are permanent or semi-permanent ongoing functional work to create the same product or service
over and over again. The management of these two systems is often very different and requires varying
technical skills and philosophy, hence requiring the development of project managements.
The first challenge of project management is to make sure that a project is delivered within defined
constraints. The second, more ambitious challenge is the optimized allocation and integration of inputs
needed to meet predefined objectives. A project is a carefully defined set of activities that use resources
(money, people, materials, energy, space, provisions, communication, etc.) to meet the predefined objectives.
Initially, the project scope is defined and the appropriate methods for completing the project are determined.
Following this step, the durations for the various tasks necessary to complete the work are listed and
grouped into a work breakdown structure. The logical dependencies between tasks are defined using an
activity network diagram that enables identification of the critical path. Float or slack time in the schedule
can be calculated using project management software. Then the necessary resources can be estimated and
costs for each activity can be allocated to each resource, giving the total project cost. At this stage, the
project plan may be optimized to achieve the appropriate balance between resource usage and project
duration to comply with the project objectives. Once established and agreed, the plan becomes what is
known as the baseline. Progress will be measured against the baseline throughout the life of the project.
Analysing progress compared to the baseline is known as earned value management.
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Gantt Chart: Gantt Chart is a principal tool used in scheduling and also in some methods of loading. This
chart was originated by the American engineer Henry L. Gantt and consists of a simple rectangular grid,
divided by series of parallel horizontal and verticular lines. The vertical lines always divide the horizontal
scale units of time. The time units can be in years, months, weeks, days, hours, minutes or even seconds
according to the work for which it is prepared. In this chart, the time which an activity takes in completing
the task is represented by the horizontal line. The length of the line is drawn in proportion to the duration
of time. Generally, the time in the chart should flow from left to right and activities be listed from top to
bottom. The progress of the work may be shown by a bar or a line within the uprights of the activity
symbol and its length should represent the amount of work completed. Horizontal lines divide the chart
into sections which can represent various work tasks (work schedule) or work centres (load schedule).
When it shows only work tasks-products, orders, or operations to be completed, it is known as Work
Schedule. When it shows the same task opposite the work centres at which they are produced-factories,
departments, workshops, machine tools or men it is known as Load Chart.
The units scheduled or loaded on these charts are always the same because these work tasks are known as
having a known standard time. The work tasks can be represented on the chart by numbers or symbols.
The symbols used on the chart may van from company to company.
Network Analysis: Routing is the first step in production planning. In small projects, routing is very
simple. Sequence of operations is almost decided and the operations can be performed one after the other
in a given sequence. But in large project, this is rather a difficult problem. There may be more than one
route to complete a job. The function of production manager is to find out the path which takes the least
time in completing the project.
In a big project, many activities are performed simultaneously. There are many activities which can be
started only at the completion of other activities. In such cases, a thorough study is required to collect the
complete details about the project and then to find out a new, better and quicker way to get the work done
in a decent way. In such cases, the first step is to draw some suitable diagram showing various activities
and their positions in the project. It should also explain the time to be taken in completing the route from
one operation to the other. It also defines the way in which the delay in any activity can affect the entire
project in terms of both money and time. Such a diagram is called network diagram. A network is a picture
of a project, a map of requirements tracing the work from a departure points to the final completion objective.
It can be a collection of all the minute details involved or only a gross outline of general functions.
Important characteristics in a Network Analysis: The following are some important points to remember
in a network analysis:
(i) The objective is to be finished within the specified time otherwise there is a penalty.
(ii) Various activities are to be completed in an order; however, a number of activities are performed
simultaneously while there are many other activities, which can be started only when some other
activities are completed.
(iii) The cost of any activity is proportional to its time of completion.
(iv) There can be hurdles in the process and the resources to be allocated may be limited. A network
graph consists of a number of points or nodes, each of which is connected to one or more of the other
nodes by routes or edges. It is a set of operations and activities describing the time orientation of a
composite project.
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135
Procedure for drawing a network diagram: The procedure for drawing a network diagram may be
explained below.
There are three basic questions and the network depends on them.
These questions are:
Which operation must be completed before each given operation can be started?
Which activities can be carried out in parallel?
Which operation immediately succeeds other given activities?
The common practice is simply to work backward through the list of operations, generating the immediate
predecessors for each operation.
Slack and float:
Slack – Slack signifies the freedom for rescheduling or to start the job. It can be calculated by the difference
between EFT and LFT for any job. A job for which the slack time is zero is known as critical job. The critical
path can be located by all those activities or events for which slack time is either zero or float time is the
least. The abbreviations EFT and LFT given in the above line have the following explanation.
EFT (Earliest Finish Time) – this is the sum of the earliest start time plus the time of duration for any
event.
LFT (Latest Finish Time) – It is calculated from the LFT of the head event. For its calculation total project
time is required. The total project time is the shortest possible time required in completing the project.
Floats – Floats in the network analysis represent the difference between the maximum time available to
finish the activity and the time required to complete it. There are so many activities where the maximum
time available to finish the activity is more than the total time required to complete it. This difference is
known as floats.
Floats may be total, free, and independent:
Total Float: Total float is the maximum amount by which duration time of an activity can be increased
without increasing the total duration time of the project. Total float can be calculated as follows:
(i) First, the difference between Earliest Start Time (EST) of tail event and Latest Finish Time (LFT) of
head event for the activity shall be calculated.
(ii) Then, subtract the duration time of the activity from the value obtained in (i) above to get the required
float for the activity.
The total float can be helpful in drawing the following conclusions:
(a) If total float value is negative, it denotes that the resources for completing the activity are not adequate
and the activity, therefore, cannot finish in time. So, extra resources or say critical path needs crashing
in order to reduce the negative float.
(b) If the total float value is zero, it means the resources are just sufficient to complete the activity without
any delay.
(c) If the total float value is positive, it points out that total resources are in excess of the amount required
or the resources should be reallocated to avoid the delay otherwise the activity will be delayed by so
much time.
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Production Planning and Productivity Management
Free Float: It is that fraction from total float of an activity which can be used for rescheduling the activity
without affecting the succeeding activity. If both tail and head events are given their earliest times, i.e., EST
and EFT the Free Float can be calculated by deducting head slack from total float, i.e.,
Free Float = Total float - Slack time of the head event.
Independent Float: It is the time by which an activity can be rescheduled without affecting the other
activities – preceding or succeeding. It may be calculated as follows:
= Earliest Start Time (EST) - Earliest Finish Time (EFT) of the activity
or, Independent Float = Free Float -Slack Time of tail event. The basic difference between slack and float
time is that a slack is used with reference to events whereas float is used with reference to activity.
Use of Float Information in Decision Making: The float information can be used in decision-making in
the following ways:
(i) Total float can affect both the previous and the subsequent activities.
(ii) Total float can be used without affecting the subsequent activities.
(iii) Independent float can be used in allocating the resources elsewhere and increasing the time of
some non-critical activities.
(iv) Negative float signifies reduction in target time to finish the work in time.
Critical Path Method (CPM): The critical path analysis is an important tool in production planning and
scheduling.Gnatt charts are also one of the tools of scheduling but they have one disadvantage for which
they are found to be unsuitable. The problem with Gnatt Chart is that the sequence of operations of a
project or the earliest possible date for the completion of the project as a whole cannot be ascertained. This
problem is overcome by this method of Critical Path Analysis.
CPM is used for scheduling special projects where the relationship between the different parts of
projects is more complicated than that of a simple chain of task to be completed one after the other.
This method (CPM) can be used at one extreme for the very simple job and at other extreme for the
most complicated tasks.
A CPM is a route between two or more operations which minimises (or maximises) some measures of
performance. This can also be defined as the sequence of activities which will require greatest normal time
to accomplish. It means that the sequence of activities which require longest duration are singled out. It is
called at critical path because any delay in performing the activities on this path may cause delay in the
whole project. So, such critical activities should be taken up first.
One of the purposes of critical path analysis is to find the sequence of activities with the largest sum of
duration times, and thus find the minimum time necessary to complete the project. The critical series of
activities is known as the ‘Critical Path’.
Under CPM, the project is analysed into different operations or activities and their relationship are
determined and shown on the network diagram. So, first of all a network diagram is drawn. After this the
required time or some other measure of performance is posted above and to the left of each operation
circle. These times are then combined to develop a schedule which minimises or maximises the measure of
performance for each operation. Thus CPM marks critical activities in a project and concentrates on them.
It is based on the assumption that the expected time is actually the time taken to complete the object.
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137
Thus CPM technique is a very useful analysis in production planning of a very large project.
PERT (Programme Evaluation and Review Technique):
There are so many modem techniques mat have developed recently for the planning and control of large
projects in various industries especially in defence, chemical and construction industries. Perhaps, the
PERT is the best known of such techniques.
PERT is a time-event network analysis technique designed to watch how the parts of a programme fit
together during the passage of time and events. This technique was developed by the special project office
of the U.S. Navy in 1958. It involves the application of network theory to scheduling, problems. In PERT
we assume that the expected time of any operation can never be determined exactly.
Major Features of PERT or Procedure or Requirement for PERT:
The following are the main features of PERT:
(i) All individual tasks should be shown in a network. Events are shown by circles. Each circle represents
an event—an event —a subsidiary plan whose completion can be measured at a given time.
(ii) Each arrow represents an activity —the time - consuming elements of a programme, the effort that
must be made between events.
(iii) Activity time is the elapsed time required to accomplish an event. In the original PERT, three-time
values are used as follows:
• t1 (Optimistic time): It is the best estimate of time if every tiling goes exceptionally well.
• t2 (Most likely time): It is an estimated time what the project engineer believes necessary to do the
job or it is the time which most often is required if the activity is repeated a number of times.
• t3 (Pessimistic time): It is also an activity under adverse conditions. It is the longest time and
rather is more difficult to ascertain.
The experiences have shown that the best estimator of time out of several estimates made by the project
engineer is:
t = 1 2 3 4
6
t + t +t
and the variance of t is given by: V (t) =
2
3 1
6
⎛ − ⎞
⎜ ⎟
⎝ ⎠
t t
(iv) The next step is to compute the critical path and the slack time.
A critical path or critical sequence of activities is one which takes the longest time to accomplish the
work and the least slack time.
Difference in PERT and CPM - Although these techniques (PERT and CPM) use the same principles and
are based on network analysis yet they are in the following respects from each other:
(i) PERT is appropriate where time estimates are uncertain in the duration of activities as measured by
optimistic time, most likely time, and pessimistic time, whereas CPM (Critical Path Method) is good
when time estimates are found with certainty. CPM assumes that the duration of every activity is
constant and therefore every activity is critical or not.
(ii) PERT is concerned with events which are the beginning or ending points of operation while CPM is
concerned with activities.
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Production Planning and Productivity Management
(iii) PERT is suitable for non-repetitive projects while CPM is designed for repetitive projects.
(iv) PERT can be analysed statistically whereas CPM not.
(v) PERT is not concerned with the relationship between time and cost, whereas CPM establishes a
relationship between time and cost and cost is proportionate to time.
Problem and Solution
Problem: 1
A small project is composed of time activities whose time estimates are given below:
Activities A, B and C can start simultaneously. Activity D follows activity A while E follows B.
Activity D and E are followed by activity G while F is dependent on C, H depends on D and E, while I
depends on F and G.
(i) Construct the network.
(ii) Find the expected duration and variance of each activity.
(iii) Calculate the slack for each event.
(iv) What is the critical path and expected project duration of the project?
(v) The project due date is 28 days, what is the probability of meeting the due date?
(vi) What should be the project duration for the probability of completion of 95%?
Solution:
Expected time Te =
4
6
a + b + m
= and Variance V =
2
6
⎛ − ⎞
⎜ ⎟
⎝ ⎠
b a
Activity Expected time (Te) Variance (v)
A 3 1
B 5 1
C 5 1
D 2 0
E 6 4
F 7 4
G 5 1
H 8 1
I 7 2
Activity A B C D E F G H I
Optimisctic time 2 2 4 2 2 3 2 5 3
Most likely time 2 5 4 2 5 6 5 8 6
Pessimistic time 8 8 10 2 14 15 8 11 15
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139
Z = s e –
SD
T T
=
28 – 23
0.833
SD
=
... Probability of completion = 96.16% (approx)
With 95% probability Z=1.65 = s – 23
SD
T
... Schedule time = 33 days
2.6 Progressing and Follow-up
Expediting or progressing ensures that, the work is carried out as per the plan and delivery schedules are met.
Progressing including activities such as status reporting attending to bottlenecks or hold-ups in production
and removing the same, controlling variations or deviations from planned performance levels, following
up and monitoring progress of work through all stages of production, co-ordinating with purchase, stores,
tool room and maintenance departments and modifying the production plans and re-plan if necessary.
Need for expediting may arise due to the following reasons.
a. Delay in supply of materials.
b. Excessive absenteeism.
c. Changes in design specifications.
d. Change in delivery schedules initiated by customers.
e. Break down of machines or tools and fixtures.
f. Errors in design and process plans.
2.7 Dispatching
Dispatching may be defined as setting production activities in motion through the release of orders work
order, shop order and instructions in accordance with the previously planned time schedules and routings.
Construction of Network
TE = 3 TE = 5 TE =19
TL = 9 TL = 11 TL = 23
Critical path is 1 3 5 9 10
B E G I
Project duration 23 days
Standard deviation σ = ij V = 2.83
B
2
1
2
3
4
5
6
7 8
9 10
TE = 0
TL = 0
TE = 5
TL = 5
TE = 11
TL = 11
TE = 16
TL = 16
TE = 23
TL = 23
TE = 5
TL = 9
TE = 12
TL = 16
A
C
D
E
F
G
H
DI
D2
6
7
7
I
5
5
8
3
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Production Planning and Productivity Management
Fig. Dispatching
Dispatching also provides a means for comparing actual progress with planned production progress.
Dispatching functions include:
a. Providing for movement of raw materials from stores to the first operation and from one operation
to the next operation till all the operations are carried out.
b. Collecting tools and fixtures from tool stores and issuing them to the user department or workers.
c. Issuing job orders authorizing operations in accordance with dates and times as indicated in
schedules or machine loading charts.
d. Issue of drawings, specifications, route cards, material requisitions and tool requisitions to the
user department.
e. Obtaining inspection schedules and issuing them to the inspection section.
f. Internal materials handling and movement of materials to the inspection area after completing
the operation, moving the materials to the next operation centre after inspection and movement
of completed parts to holding stores.
g. Returning fixtures and tools to stores after use.
2.8 Scheduling Technique and Line Balancing Problem
Scheduling: ‘Scheduling’ is the next important function of production planning and control after ‘Routing’.
It determines the starting and the completion timings for each of the operations with a view to engage
every machine and operator of the system for the maximum possible time and; without imposing
unnecessary burden over them. Scheduling is the determination of the time that should be inquired to
perform each operation and also the time that should be required to perform the entire series as routed.
Scheduling involves establishing the amount of work to be done and the time when each element of the
work will start or the order of the work.
Scheduling technique is an important technique of determining the starting and the completion timings of
each operation and that of the total manufacturing process so that the man and machines can be utilised to
the maximum.
Dispatching
Centralised Decentralised
Ex. Central Ex. Foreman
Despatch
Section
M/c. Shop Worker
Issue of order and
Instruction to foreman
Issue of orders and
Insuruction from
foreman
(Foreman Implacement in each shop) (No. central Instructions here)
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141
Scheduling depends upon a number of factors, e.g., routing, the method of production, quantity of
production, transportation of raw materials, production capacity, the probable data of delivery specified
by customers in their orders and the past records.
Relationship between ‘Routing’ and ‘Scheduling’. ‘Routing’ and ‘Scheduling’ are independent and either
of these activities cannot be undertaken independently. It is very difficult to prepare schedules without
determining the routing of sequence of operations. Routing is the prerequisite of scheduling. Unless route
or sequence of operations, tools, equipment and plants and the persons by when operations are to be
performed, are established, the time taken by each operation, the idle time of men and machine and total
time for the whole process cannot be ascertained in a convincing manner.
Conversely, scheduling is equally important for routing. It is quite difficult to route an item efficiently
through a plant without consulting previously-designed schedules. The main aim of routing is to pass the
item through the process of manufacture by a route which is the best and the most economical. And a
route or sequence of operations may be considered best which utilises the men, materials and machines to
the maximum and which consumes the shortest time during the process of production. This information
(time schedule of each operation) can be obtained from schedules. So, scheduling is necessary for effective
routing.
Thus, we can conclude that routing and scheduling are inter-related, inter-connected and inter-dependent
activities of production planning and control.
Relationship between Routing and Scheduling: Both are interconnected as scheduling is difficult without
routing and routing is also not effective without scheduling. Routing is a prerequisite for scheduling while
time to be taken ‘may form the basis of routing and that is fixed by scheduling.
Principles of Scheduling:
The principles of scheduling are:
(a) The principle of optimum task size: Scheduling tends to achieve its maximum efficiency when the
task sizes are small and all tasks are of the same order of magnitude.
(b) The principle of the optimum Production plan: Scheduling tends to achieve its maximum efficiency
when the work is planned, so that it imposes an equal/even load on all the plant.
(c) The principle of the optimum operation sequence: Scheduling tends to achieve its maximum efficiency
when the work is planned so that the work centers are normally used in the same sequence.
The first principle has a tendency when applied, not only give good results but also to be self-correcting if
it is ignored. For example, if in a functional batch production machine shop the loads imposed by different
operations vary greatly in length it is possible that it will be necessary to break many of the long operations
into one or more small batches, in order to get the other orders completed by due date. In effect, this
principle only repeats the known advantage of maintaining a high rate of stock turn over, and of single
phase ordering. The second principle merely states that the obvious fact that there will be less idle time
and waiting time, if all the plant is evenly loaded by the production planners, then if some of the machines
are over loaded perhaps because direct labour cost on them are lower and others are idle for part of the
time due to shortage of work. The third principle says about principle of flow. Some times it is also true if
we sequence some jobs, which need the same machine set up, at a time, this avoids machine ancillary time
needed, in case, the jobs of the above type are done at different times. For example, consider drilling a 10
mm hole in five different jobs may be done at a time so that the set up time required for five jobs
independently at different times are avoided.
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Production Planning and Productivity Management
Forms of Schedules:
Here we shall discuss the presentation of production schedules. Depending on the need and use, the
Schedules can be prepared in different forms.
A Production Flow Program:
If a number of components or assemblies have to be manufactured for the final assembly line and those
components are to be made concurrently, the production master flow program is prepared taking into
account the sequence of operations and the time of starting and ending each component in order to comply
with the required date of completion of the product. The necessary document for this is Operation Process
Chart and the Sequence of Operation.
Scheduling Systems:
Scheduling Systems may be classified into four groups as shown below:
(i) Unit scheduling system: This is used for scheduling when jobs are produced one by and are of different
a type that is for job production.
(ii) Batch scheduling system: When jobs are produced to order, in batches, this is used.
(iii) Mass scheduling system: When large number of items of similar type are produced that is in mass
production, this is used.
Unit Scheduling System:
Here we have two types of scheduling, one is Project scheduling and the other is Job Scheduling.
Project Scheduling: Generally, a project consists of number of activities managed by defferent Apartments
or individual supervisors. It can also be said as a complex output made up of many interdependent jobs.
Examples are: Railway coach building, Shipbuilding etc. The scheduling methods used are:
(i) Project Evaluation and Review Technique (PERT),
(ii) Critical Path Method (CPM),
(iii) Graphical Evaluation and Review Technique (GERT).
We can also use Bar charts, GANTT charts, Milestone chart, but these are less superior to the above.
Job Shop Scheduling: In Job shop scheduling, we come across varieties of jobs to be processed on different
types of machines. Separate records are to be maintained for each order. Only after receiving the order, one
has to plan for production of the job. The routing is to be specified only after taking the order. Scheduling
is done to see that the available resources are used optimally. The following are some of the methods used
for scheduling. (i) Arrival pattern of the job, (ii) Processing pattern of the job, (iii) Depending on the type of
machine used, (iv) Number of workers available in the shop, (v) Order of sequencing.
Arrival pattern of the job: This is done in two ways. Firstly, as and when the order is received, it is processed
on the principle First in First Out (FIFO). Otherwise, if the orders are received from single customer at different
point of time in a week/month, then the production manager pile up all orders and starts production depending
on the delivery date and convenience (This situation is generally known as static situation).
Processing Pattern of the Job: As the layout of Job shops of Process type and there may be duplication of
certain machines, the production planner, after receiving the order thinks of the various methods of
Operation Management
143
converting the requirement of customer / order into a production plan to suit the available facilities.
Depending on the process required, there may be backtracking, which is unavoidable. When facilities are
busily engaged, in process inventory may be a common problem.
Machine varieties available: Facilities available in the production shop will affect the scheduling. Here
the size, capacity, precession and other factors of machines will have their influence on the scheduling.
Number of Men in the production shop: Many a time we see that the number of workers available in the
job shops are very much limited, that is sometimes they are less in number than the machines in the shop
(these shops are known as labour limited shop). Depending the availability of labour, the scheduling is to
be done. In case the machines available are limited and have more men (known as machine limited shops),
then availability of machine dictates the scheduling.
Sequencing rules for single facility: When we have a single facility, and the orders are in queue, then they
are processed depending on the rules mentioned below:
(a) First in first served or first in first out (FIFS/FIFO): Here the jobs are processed as they come in. This
is commonly observed queue discipline.
(b) Shortest processing time (SPT): The jobs having shortest processing time are processed first. This is
just to avoid formation of queue. For example, when you go for Xeroxing a document, and other
person comes for Xeroxing a book, then document is Xeroxed and then the book is taken for Xeroxing.
(c) Minimum due date (MDD): Here jobs are processed in ascending order of their available time before
delivery date. By doing so, we can keep up the delivery promises. To meet the delivery promises, if
necessary, overtime, sub contracting etc., may be used.
(d) Last come first served or last in first out (LCFS/LIFO): This generally happens in case of inventory
stocking and using. When material piles up, the material at the top i.e., material last arrived is used
first.
(e) Static slack for remaining operations (SSRO): Static slack is given by: (Due date - Remaining
processing time/number of remaining operations). Here jobs are processed in ascending order of the
operations.
(f) Dynamic slack for remaining operations (DSRO): Dynamic slack is given by: (Due date - expected
time of remaining operations / number of remaining operations). Here the jobs are done in ascending
order of the ratio dynamic slack.
Basic Scheduling Problems:
The production planner may face certain problems while preparing production plans or Schedules. Some
important problems are discussed below:
(a) Flow production scheduling for fluctuating demand (known smoothening problem),
(b) Batch production scheduling, when products are manufactured consecutively,
(c) The assignment problem,
(d) Scheduling orders with random arrivals and
(e) Product sequencing.
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Production Planning and Productivity Management
Problems and Solutions
Problem: 1
A company has two plants A and B with fixed costs of Rs. 50,000 and Rs. 70,000 respectively. Both the
plants are designed to produce up to 10,000 units each. The variable costs of two plants at different of
production are as follows:
Production (Units) Plant A (Rs.) Plant B (Rs.)
2,500 36,000 29,000
5,000 45,000 39,000
7,500 77,000 51,000
10,000 1,10,000 1,15,000
Find the most economic loading schedule.
Solution:
The fixed costs are irrelevant and only the incremental costs should be considered. Incremental costs
will be as follows:
Plant A (Rs.) Plant B (Rs.)
Total incremental costs 36,000 29,000
9,000 10,000
32,000 12,000
33,000 64,000
Per unit incremental costs 14.40 11.60
3.60 4.00
12.80 4.80
13.20 25.60
Ans.
First 2500 Units B
Next 2500 Units A
Next 2500 Units B
Next 2500 Units A
Problem: 2
A company is setting an assembly line to produce 192 units per eight hour shift. The information regarding
work elements in terms of times and immediate predecessors are given
Work element Time (Sec) Immediate predecessors
A 40 None
B 80 A
C 30 D, E, F
D 25 B
E 20 B
F 15 B
G 120 A
H 145 G
I 130 H
J 115 C, I
Total 720
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145
(i) What is the desired cycle time?
(ii) What is the theoretical number of stations?
(iii) Use largest work element time rule to workout a solution on a precedence diagram.
(iv) What are the efficiency and balance delay of the solution obtained?
Solution:
The precedence diagram is represented as shown below:
(a) Cycle time =
1 8hours
r 192units
= = 150 sec/unit, where r = output.
(b) Sum of the work elements is 720 seconds, so minimum number of work stations
= = =
Σt 720second/unit
4.8 5
cycle time 150sec/unit station stations
Assignment of work elements to work stations.
(d) Efficiency =
t 720
100 100 96%
nCt 5 150
× = × =
×
Σ
Thus, the balance delay is [100 - 96] = 4 percent only.
Station Elements Work element time Cumulative time Idle time for
in sec (Sec) station
S1 A 40 40
B 80 120 05
D 25 145
S2 G 120 120
E 20 140 10
S3 H 145 145
S4 I 130 130
F 15 145 05
S5 C 30 30
J 115 145 05
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Production Planning and Productivity Management
Problem: 3
The company is engaged in the assembly of a wagon on a conveyor. 500 wagons are required per day.
Production time available per day is 420 minutes. The other information is given below regarding assembly
steps and precedence relationships. Find the minimum number of work stations, balance delay and line
efficiency.
Solution:
The element times and precedence relationships
(i) Precedence diagram is constructed as per the given details.
(ii) Determination of cycle time
Ct =
= × = = Production time/day 420 3 25, 200
50.4
Output /day 500 500
(iii) Theoretical number of work stations required
N =
Total time 195
3.87
Cycle time 50.4
= = ≈ 4 workstations
(iv) Assign the elements to work stations based on the largest element time and the work stations to
be required are 5.
Balance made according to largest number of followers task rule.
Task Time (sec) Task that must Precede
A 45 –
B 11 A
C 09 B
D 50 –
E 15 D
F 12 C
G 12 C
H 12 E
I 12 E
J 08 F, G, H, I
K 09 –
Total 195 –
sec
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147
(iv) Efficiency = = × =
× ×
T 195
100 77.38%
No.of stations C 5 50.4
(v) Balance delay = 100 - 77.38 = 22.62%
Problem: 4
The processing times (tj) in hrs for the five jobs of a single machine scheduling is given. Find the optimal
sequence which will minimise the mean flow time and find the mean flow time.
Determine the sequence which will minimise the weighted mean flow time and also find the mean flow
time
Station Task Task time (sec) Idle time (sec)
S1 A 45 5.4
S2 D 50 0.4
S3 B 11 3.4
E 15
C 09
F 12
S4 G 12 6.4
H 12
I 12
J 8
S5 K 9 41.4
Job (j) 1 2 3 4 5
Processing time (tj) hrs 30 8 10 28 16
Solution:
(a) First arrange the jobs as per the shortest processing time (SPT) sequence.
Job (j) 2 3 5 4 1
Processing time (tj) hrs 8 10 16 28 30
Job (j) 2 3 5 4 1
Processing time (tj) hrs 8 10 16 28 30
Completion time (Cj) 8 18 34 62 92
Therefore, the job sequence that minimises the mean flow time is 2-3-5-4-1.
Computation of minimum flow time (F min)
The flow time is the amount of time the job ‘f’ spends in the system. It is a measure which indicates the
waiting of jobs in the system. It is the difference between the completion time (Cj) and ready time (Rj)
for job j.
Fj = Cj - Rj
Since the ready time (Rj) = 0 for all j, the mean flow time ( j F )is equal to Cj for all j. Mean flow time
( F ) =
n
j
j 1
1 1 1
F [8 18 34 62 92] [214] 42.8
n = 5 5
Σ = + + + + = = hours
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Production Planning and Productivity Management
(b) The weights are given as follows:
Job (j) 1 2 3 4 5
Processing time (tj) hrs 30 8 10 28 16
Weight (Wj) 1 2 1 2 3
Job (j) 1 2 3 4 5
Processing time (tj hrs) 30 8 10 28 16
Weight (Wj) 1 2 1 2 3
tj/Wj 30 4 10 14 5.31
Job (j) 2 5 3 4 1
Processing time (tj) hrs 8 16 10 28 30
Fj = (Cj – Rj) 8 24 34 62 92
Wj 2 3 1 2 1
Fj x Wj 16 72 34 124 92
The weighted processing time =
j
J
Proces sing time(t )
Weight(W)
The weighted processing time is represented as
Thus, arranging the jobs in the increasing order of tj/Wj (weighted shortest processing time WSPT).
We have the optimal sequence that minimises the weighted mean flow time is 2-5-3-4 -1
flow time( w F ) :
w F =
Σ
Σ
n
j j
j=1
n
j
j=1
w F
w
The weighted mean flow time is computed as follows for optimal sequence.
Weighted mean flow time ( w F ) is computed as
w F =
(16 72 34 124 92)
(2 3 1 2 1)
+ + + +
+ + + + = 37.55 hrs.
Problem: 5
The processing times and due dates of jobs for a single machine scheduling is given.
Determine the sequence which will minimise the maximum lateness and also determine the maximum
lateness with respect to the optimal sequence.
Job (j) 1 2 3 4 5 6 7
Processing time (tj) 10 8 8 7 12 15 18
Due date (dj) 15 10 12 11 18 25 30
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Solution:
Due date (dj) is the time at which job ‘j’ is to be completed and Lateness (Lj) is the amount of time by
which the completion time of the job j differs from the due date (Lj = Cj - dj). Lateness is a measure
which gives set of due dates of times. Lateness can be positive or negative. Positive lateness indicates
the completion of job after its due date. It is therefore often desirable to optimize positive lateness.
Arranging the jobs as per Earliest Due Date (EDD), i.e., increasing order of their due dates. The sequence
is 2-4-3-1-5-6-7. This sequence gives the minimum value of maximum lateness (Lmax)
Computation of Lmax
Job (j) 1 2 3 4 5
Processing time (tj) hrs 9 7 5 11 6
Due date (in days) (dj) 16 20 25 15 40
From the above table, the maximum Lj is 48. This is the (optimised) value of Lmax.
Problem: 6
The processing times for five jobs and their due dates are given for a single machine scheduling below.
Job (j) 2 4 3 1 5 6 7
as per EDD sequence
Processing time (tj) 8 7 8 10 12 15 18
Completion time (Cj) 8 15 23 33 45 60 78
Due date (dj) 10 11 12 15 18 25 30
Lateness (Lj) (-2) 4 11 18 27 35 48
(a) Determine the sequence
(b) Total completion time
(c) Average completion time
(d) Average number of jobs in the system and average job lateness using the following priority sequencing
rules
(i) Shortest Processing Time (SPT)
(ii) Earliest Due Date (EDD)
(iii) Longest Processing Time (LPT)
(e) Compare the above characteristics for the three sequencing rules.
Solution:
(i) Shortest Processing Time (SPT) sequence.
As per this rule, the job with the shortest processing time is scheduled first and immediately followed
by next lowest processing time and so on.
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Production Planning and Productivity Management
Job Sequence Processing time Flow time Due Date Job lateness
(j) (tj) days (Fj) days (dj) days (Days)
3 5 5 25 0
5 6 11 40 0
2 7 18 20 0
1 9 27 16 11
4 11 38 15 23
Total 38 99 34
The various characteristics are:
Total completion/ time (flow time)= 38 days
Average completion time =
Total flowtime
No. of jobs =
99
5
= 19.8 days
Average number of jobs in the system =
Total flow time 99
2.61 jobs
Totalprocess time(completion) 38
= =
Average job lateness =
Total Job lateness 34
No.of jobs 5
= = 6.8 days
(ii) Earliest Due Date(EDD) rule
As per this rule priority is given to the job with earliest due date. Arranging the jobs as per EDD
sequence gives the sequence as 4-1-2-3-5.
Characteristics:
• Total completion time = 38 days
• Average completion time =
Flowtime
No. of jobs =
128
5
= 25.6 days
• Average number of jobs in the system =
Flowtime
Completion time =
128
38
= 3.37 jobs
• Average job lateness =
18
3.6days
5
=
Job sequence Processing time Flow time Due Date Job lateness
(j) (Cj) (Fj) (Dj) (Days)
4 11 11 15 0
1 9 20 16 4
2 7 27 20 7
3 5 32 25 7
5 6 38 40 0
Total 38 128 18
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(iii) Longest Processing Time (LPT) Rule
The job sequence and computations as per this priority sequencing rule is given as follows:
Job Sequence Processing time Flow time Due Date Job lateness
(j) (Cj) (Fj) (Dj) (Days)
4 11 11 15 0
1 9 20 16 4
2 7 27 20 7
5 6 33 40 0
3 5 38 25 13
Total 38 129 24
Characteristics:
• Total completion time = 38 days
• Average completion time =
Flowtime
No. of jobs =
129
5
= 25.8days
• Average number of jobs in the system =
Flowtime
No. of jobs =
129
38
= 3.39 jobs
• Average job lateness =
24
5
= 4.8 days
(b) Comparison of Priority rules
Problem: 7
The following jobs are waiting to be processed in a turning shop today (July, 23). The estimates of the time
needed to complete the jobs are as follows:
Jobs (j) Due date Processing time (tj) in days
1. July, 31 9
2. August, 2 6
3. August, 16 24
4. July, 29 5
5. August, 30 30
Sequence the jobs based on the minimum critical ratio.
Priority Total completion Avg. completion Avg. No. of jobs Avg. Job
rule time (days) time (days) in the system lateness
SPT 38 19.8 2.61 6.8
EDD 38 25.6 3.37 3.6
LPT 38 25.8 3.39 4.8
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Production Planning and Productivity Management
Solution:
The critical ratio is computed as
Critical Ratio (CR) =
Timeneededforduedate of the job Timeremaining
Timeneededto complete the job Work remaining
=
Critical Ratio (CR) =
As per the critical ratio rule, a job with the minimum critical ratio is given the first preference, i.e., the
lower is the critical ratio, higher is its priority.
The denominator of CR, i.e., the time needed to complete the job includes the processing time remaining
plus the transfer times plus the estimated waiting times remaining for the job to go through before it
is completely processed.
The table below gives the calculations of time remaining and time needed and the critical ratio.
Jobs Due date Processing time Time needed to Critical ratio
(j) (tj) in days complete the job in CR = Tr/Tn
days
1. July, 31 8 9 8/9 = 0.89
2. August, 2 10 6 10/6 = 1.167
3. August, 16 24 24 24/24 = 1.00
4. July, 29 6 5 6/5 = 1.20
5. August, 30 38 30 38/30 = 1.27
Note: Critical Ratio (CR) of less than one means that the job is already late.
The CR value of one indicates that the job is on schedule and greater than one indicates that the job
has some slack available to it. From the table, job 1 has the lowest critical ratio and has to be processed
first and job 2 has the highest critical ratio and it is scheduled last. The sequence is:
1 3 4 5 2
Problem: 8
A company has 8 large machines which receive preventive maintenance. The maintenance team is divided
into 2 crews A and B. Crew A takes the machine power and replaces parts as per given maintenance
schedule. The second crew resets the machine and puts back into operation.
At all times the no passing rule is considered to be in effect. The servicing time for each machine is given
below.
Machine a b c d e f g h
Crew A 5 4 22 16 15 11 9 4
Crew B 6 10 12 8 20 7 2 21
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Determine the optimal sequence of scheduling the factory maintenance crews to minimise their idle
times and represent it on the Gantt chart.
Solution:
Step I : The minimum processing time on crew A is kept first in sequence while machine with processing
times minimum on crew B is kept last in sequence.
Optimal sequence:
b h a e c d f g
Step II: Calculation of total elapsed time
Week
Product 1 2 3 4
A 2000 4000 1000 2000
B 3000 1000 4000 3000
Optimal Crew A Crew B Idle time on
Sequence S F S F Crew B
b 0 4 4 4 4
h 4 8 14 35 0
a 8 13 35 41 0
e 13 28 41 61 0
c 28 50 61 73 0
d 50 66 73 81 0
f 66 77 81 88 0
g 77 86 88 90 0
Crew B b h a e d f g
idle time
10 20 30 40 50 60 70 80 90 100
Time in hrs
Total elapsed time is 90 hrs.
Gantt Chart
Problem: 9
The following is a tentative master schedule for 4 weeks at a small company
Crew a b h a e c d f g
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Production Planning and Productivity Management
The labour in the key work centers for the company’s two major products is as follows
Product
Dept. A B
4 0.21 0.07
11 0.06 0.1
14 0.11 0.08
Determine the load on department 4 over the next 4 weeks.
Solution:
Hours required in each week:
Week 1: For product A = 0.21(2000) = 420
For product B = 0.07(3000) = 210
Total load for the week = 630
Week 2:
For product A = 0.21(4000) = 840
For product B = 0.07(1000) = 70
Total load for the week = 910
Week 3:
For product A = 0.21 (1000) = 210
For product B = 0.07(4000) = 280
Total load for the week = 490
Week 4:
For product A = 0.21(2000) = 420
For product B = 0.07(3000) = 210
Total load for the week 630
Note that this load profile is not very uniform, even though 5000 units of products are produced each
week.
Line Balancing
Assembly line balancing is associated with a product layout in which products are processed as they pass
through a line of work centres. An assembly line can be considered as a “production sequence” where
parts are assembled together to form an end product. The operations are carried out at different workstations
situated along the line.
The Problem of Line Balancing:
It arises due to the following factors
1. The finished product is the result of many sequential operations.
2. There is a difference in production capacities of different machines Line balancing is the apportionment
of sequential work activities into workstations in order to gain a high utilisation of labour and
equipment so as to minimise the idle time. For example, the production capacities of two machines A
and B are as under for a particular job: A 50 pieces/hour; B 25 pieces/hour.
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155
Now, if only one machine of each is provided, then machine B will produce 25 units/hour where as
the machine A can produce 50 units. But because of the sequence, only 25 units are produced per
hour, i.e., machine A will work only 50 per cent of its capacity and the remaining 30 minutes in one
hour, it is idle. This idle time can be minimised by introducing one more machine of kind B in the
production line.
Steps in Solving Line Balancing Problems
1. Define task.
2. Identify precedence requirements.
3. Calculate minimum number of workstations required to produce desired output.
4. Apply heuristics to assign task to each station.
5. Evaluate effectiveness and efficiency.
6. Seek further improvement.
Three important parameters in line balancing
Total station time
1. Line efficiency (LE) = ×
×
Total station tim
100
Cycle time no of workstations
2. Balance delay (BD) =
Total idle time for allworkstation
100
Total availableworking timeon all station
×
BD = (1-LE)
3. Smoothness Index (SI) =
= Σ k
i l
(Max. station time - station times of station i)2
SI = 0, means a perfect balance
K = total number of workstations < total number of elements
Also, CT > maximum time of any work element n
Terms used in the context of Line Balancing:
1. Workstation: A work station is a location on assembly line where given amount of work is performed.
2. Cycle time: It is the amount of time for which a unit that is assembled is available to any operator on
the line or it is the time the product spends at each work station.
Cycle Time (CT) =
Available timeperiod AT
Output units required/period Output
=
3. Task: The smallest grouping of work that can be assigned to a workstation.
4. Predecessor task: A task that must be performed before performing another (successor) task.
5. Task time: Standard time to perform element task.
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Production Planning and Productivity Management
6. Station time (sk): Total standard work content of specific workstation.
7. Balance Delay (BD): Percentage of total idle time on the line to total time spent by the product from
beginning to end of line.
B.D =
n
k l
n.CT sk 100
n CT
=
− ×
×
Σ
B.D= Balance delay
n = number of work stations, CT = Cycle time
sk = Station time
Sequencing Model
This model is used for sequencing the jobs on the machines so as to minimize the total processing time. We
have different sequencing models. They are:
(i) Sequencing ‘n’ jobs on 2 machines,
(ii) Sequencing ‘n’ jobs on 3 machines,
(iii) Sequencing ‘n’ jobs on ‘m’ machines,
(iv) Sequencing 2 jobs on ‘w’ machines,
(v) Sequencing of ‘n’ jobs on two machines.
The order in which jobs pass through machines or work-stations is a sequencing problem and analysis of
this problem is sequence analysis. A production manager or a person who has been assigned the job of
sequencing the manufacturing of a product must know about the necessary operations and the possible
order in which these can be performed so that die idle time can be minimised. The effectiveness of sequencing
may be measured by time and cost factors. Saving in time also affects the cost.
The main problem of sequencing is the assignment of various jobs to different machines. When there are
very few different types of jobs or machines, the problems is not much serious and it is solved informally
by sketching the flow mentally or on a time chart. Suppose, if these are two jobs and two machines (2x 2
problem), there is no problem of assignment of jobs to machines. There are two possible sequence in this
case —job 1 first and job 2 second or vice versa.
It becomes more tedious as the number of jobs and machines increases. We shall discuss or analyse the
following two cases:
(i) n jobs are to be processed on two machines A and B in the order AB.
(ii) n jobs are to be processed on three machines A, B and C in the order ABC.
(1) Assignment of jobs to two machines: Assuming that there are a number of jobs to be processed
on two machines and the processing time for each job on each machine is known and that each
job is first processed on Machine A and then on Machine B, the sequence of different jobs on
machine A and B may be assigned as follows:
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157
(i) First, we should find out the least processing time. If it is on Machine A, it should be ranked first on
Machine A. If it is on Machine B, it should be placed at the last in the sequence.
(ii) The next step would be to delete the job selected in step (i) above either to Machine A or B from the list
of jobs to be assigned. This process should be repeated to the remaining jobs. This process will be
continued until a complete sequence of all the jobs is obtained. If two jobs have the same time on the
same machine (if it is smaller than the time on the other machine), the order of assignment for these
two jobs is arbitrary.
This technique applies to single unit or single type batch jobs where jobs have no priority for completion.
In this technique it is also presumed that there is sufficient in-process storage space and the cost of inprocess
inventor is the same or varies very insignificantly for all units. These assumptions are valid only
for short processes. For extended process, oilier criterion such as closer inventor, cost control or expediting
priorities etc. are considered other than minimising elapsed time. Other complicating variables may be
variable transportation time between machines, improving the defective work, breakdowns of machine
and variable time caused by operator proficiencies or working conditions.
(2) Assignment of jobs to three machines: Assuming that there are a number of jobs to be processed on
three machines A, B and C in the order of ABC, i.e., first it is processed on machine A, then on
Machine B and then on machine C. The problem can be solved if either of the two conditions is
satisfied.
(i) The least processing time of any job on Machine A is greater than or equal to the maximum
processing time of any job on Machine B.
(ii) The smallest processing time for any job on Machine C is either equal to or greater than the
processing time for any job on Machine B.
If any of the above two conditions is satisfied, the job processing times are converted into the following
ways:
a. We shall add job processing times (as given) of different jobs on; Machine A and B; and
b. Again job processing time for various jobs (given) on Machine B and C shall be added.
After this process, the optional sequence can be obtained in the same way as it is assigned on two machines.
2.9 Economic Batch Production
Production managers often have to decide what quantity of output must be produced in a batch (known as
lot size or batch size). The products are manufactured in lot sizes against the anticipated demand for the
products. Often the quantity produced may exceed the quantity which can be sold. (i.e., production rates
exceed demand rates). The optimum lot size which is known as economic lot size or economic batch quantity
or economic manufacturing quantity is that quantity of output produced in one batch, which is most
economical to produce, i.e., which results in lowest average cost of production.
Determination of Economic Lot Size for Manufacturing:
The factors to be considered in arriving at the economic lot size are:
(i) Usage rate: The rate of production of parts should match with the rate of usage of these parts in the
assembly line.
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Production Planning and Productivity Management
(ii) Manufacturing cost: Higher the lot size, lower will be the cost per unit produced because of distribution
of set up costs for setting up production or machines and preparing paper work (production orders).
But the carrying cost (handling and storing costs) will increase with increase in lot size.
(iii) Cost of deterioration and obsolescence: Higher the lot size, higher will be the possibility of loss due
to deterioration (items deteriorating after shelf life) or obsolescence (due to change in technology or
change in product design).
Before deciding on production using economic lot sizes, the availability of production capacity to produce
the product in economic lot size must be verified. The economic lot size balances the two opposing costs
related to batch size i.e., setup cost for production and the inventory carrying costs resulting from inventory
of products produced when production rate exceeds usage rate or when the items produced are not
immediately consumed in the next stage of production. The set up cost per unit decreases with increase in
lot size whereas the inventory carrying cost increases with increase in lot size. Diagram below illustrates
the concept of economic batch quantity or economic lot size.
If S is the set up cost per set up, ‘C’ is the production cost per unit produced and I is the inventory carrying
changes (%) and A is the annual demand for the item in units, then,
Economic Batch Quantity (EBQ)
or Economic Lot Size (ELS)
or Economic Manufacturing Quantity (EMQ)
⎛ ⎞⎛ ⎞
× ⎜ ⎟× ⎜ ⎟
⎝ ⎠⎝ ⎠ = =
⎛ ⎞⎛ ⎞
⎜ ⎟× ⎜ ⎟
⎝ ⎠⎝ ⎠
Annualdemand SetupCost
2
2AC (inunits) per setup
CI Production Cost Inventory conveying
perunit charg es (percentage)
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159
Economic Run Length: When a firm is producing an item and keeping it in inventory for later use, instead
of buying it, the formula used to calculate economic order quantity (EOQ) can be used to calculate the
economic production quantity referred to as Economic Run Length (ERL).
If ‘p’ is the production rate and ‘d’ is the demand rate (or consumption rate), A is the annual demand for
the item in units, I is the inventory carrying charges (percentage), C is the production cost per unit, then,
Economic Run Length (ERL) =
2AS
d
CI 1 –
p
⎛ ⎞
⎜ ⎟
⎝ ⎠
=
⎛ ⎞⎛ ⎞
× ⎜ ⎟×⎜ ⎟
⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎜ ⎟× ⎜ ⎟ − ⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠
Annualdemand SetupCost
2
(inunits) per setup
Productioncos t Inventoryconveying DemandRate
perunit charg es (percentage) Production Rate
Problems and Solutions
Problem: 1
C Ltd. produces a product which has a monthly demand of 4,000 units. The product requires a component
X which is purchased at Rs. 20. For every finished product, one unit of component is required. The ordering
cost is Rs. 120 per order and the holding cost is 10% p.a.
You are required to calculate:
(i) Economic order quantity.
(ii) If the minimum lot size to be supplied is 4,000 units, what is the extra cost the company has to
incur?
(iii) What is the minimum carrying cost, the company has to incur.
Solution:
(i) Economic Order Quantity:
Annual consumption of component X (C0):
4,000 units per month for 12 months = 48,000 units;
Ordering cost per order (O) = Rs. 120 and Carrying cost (i.e., holding cost) of one unit of component
for one year (Cc) = 10% of Rs. 20 = Rs. 2
EOQ =
2 2 48,000units Rs.120
2, 400 units
Rs.2
× ×
O = =
O
C O
C
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Production Planning and Productivity Management
(ii) Extra cost to be incurred by the company:
Cost when order size is 4,000 units
⎛ ⎞
⎜ ⎟
⎝ ⎠
48,000
i.e., or 12 orders p.a.
4,000
Rs.
Ordering cost : 12 order p.a. at Rs. 120 per order 1,440
Carrying cost of average inventory :
4,000units
Rs.2
2
× 4,000
Total annual cost 5,440
cost when order size is 2,400 units (i.e,
48,000
2, 400 or 20 orders p.a.)
Rs.
Ordering costs : 20 orders p.a. at Rs.. 120 per order 2,400
Carrying cost of average inventory: 2,400
Total annual cost 4,800
Extra cost to be incurred : Rs. (5,440 - 4,800) = Rs. 640.
(iii) Minimum carrying cost to be incurred by the company:
Since carrying cost depends upon the size of the order, carrying cost will be the minimum at the
economic order quantity. Hence the minimum carrying cost is Rs. 2,400 as calculated in (ii) above.
Problem: 2
M/s Kobo Bearings Ltd., is committed to supply 24,000 bearings per annum to M/s Deluxe Fans on a
steady daily basis. It is estimated that it costs 10 paisa as inventory holding cost per bearing per month and
that the setup cost per run of bearing manufacture is Rs. 324.
(a) What is the optimum run size for bearing manufacture? (b) What should be the interval between the
consecutive optimum runs? (c) Find out the minimum inventory holding cost.
Solution:
(a) Optimum run size or Economic Batch Quantity (EBQ)
=
× × = × × =
×
2 2 24000 324
3600units
0.10 12
AnnualOutput Setupcost
AnnualCost of Carrying oneunit
(b) Interval between two consecutive optimum runs = × 30 EBQ
Monthly Output
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161
=
3600
30 54
24000 12
× =
÷ Calendar days
(c) Minimum inventory holding cost = Average inventory × Annual carry-ing cost of one unit of
inventory
= (3600÷2) ×0.10×12 = Rs. 2,160.
Problem :3
The demand of a certain item is random. It has been estimated that the monthly demand of the item has a
normal distribution with a mean of 680 and a standard deviation of 130 units. The unit price of the item is
Rs. 10 per unit; the ordering cost is Rs. 20. The inventory carrying cost is estimated to be 25 per cent per
year respectively. The procurement lead time is constant and is one week. Find the most economic ordering
policy and the expected total cost of controlling inventory, given that the service level is 97.5%.
Solution:
Under the given circumstances, Economic Order Policy involves deter-mination of EOQ, ROL and
Safety Stock for a year.
EOQ =
2 680 12 2
0.25 10
× × ×
× = 361 (appx.) units; ROL = Consumption during lead time + Safety stock
= 680 ÷ 4 + ZLT. Z for 97.5% service level may be taken as 2 (1.96 from Table)
LT = S.D. during lead time = Varience During Lead time =
16900
4
= 130/2 = 65 units
Safety stock = 2 × 65 = 130 units; ROL = 680/4 + 130 = 300 units
Expected total cost of inventory per annum: Procurement cost = Rs. 10 × 680 × 12 = Rs. 81,600
Ordering cost = (Annual Demand / EOQ) × (Ordering cost / Order) = (680 × 12 × 20)/361.
= Rs. 452 (apx.)
Inventory carrying cost = Average inventory × inventory carrying cost = (EOQ/2 + Safety stock) ×
0.25 × Rs. 10 = (361/2 + 130) × 0.25 × Rs. 10 = Rs. 776 (apx.) Total variable cost = Rs. 452 + Rs. 776 = Rs.
1,228 .
Problem: 4
A manufacturer requires 10,00,000 components for use during the next year which is assumed to consist of
250 working days. The cost of storing one component for one year is Rs. 4 and the cost of placing order is
Rs. 32.
There must always be a safety stock equal to two working days usage and the lead time from the supplier,
which has been guaranteed, will be five working days throughout the year. Assuming that usage takes
place steadily throughout the working days, delivery takes place at the end of the day and orders are
placed at the end of working day, you are required to(a) Calculate the EOQ (b) Calculate the Re-order
point.
Solution:
We are given that :- D = 10,00,000 ; S = Rs. 32; C = Rs.4
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Production Planning and Productivity Management
EOQ =
2DS 2(10,00,000)32
C 4
= = 4000 units.
(b) Re-order point is to be calculated by the following formula;
Re-order point = Safety stock + (Average usage × Lead time)
Safety stock = (10,00,000 / 250days) = 4,000 usage per day × 2 = 8,000.
Reordering point = 8000 + (4000 units per day × Number of days)
= 8000 + (4000 ×5) = 28,000 units.
Problem: 5
The monthly requirement of raw material for a company is 3000 units. The carrying cost is estimated to be
20% of the purchase price per unit, in addition to Rs. 2 per unit. The purchase price of raw material is Rs. 20
per unit. The ordering cost is Rs. 25 per order. (i) You are required to find EOQ.(ii) What is the total cost
when the company gets a concession of 5% on the purchase price if it orders 3000 units or more but less
than 6000 units per month. (iii) What happens when the company gets a concession of 10% on the purchase
price when it orders 6,000 units or more? (iv) Which of the above three ways of orders the company should
adopt?
Solution:
We are given that,
D = 3,000 × 12 = 36,000 units per annum ; S = Rs. 25;
C = 2 + 20% of Rs. 20 = 2 + 4 = Rs. 6
(i) EOQ =
2DS 2 36000 25
300,000
C 6
= × × = = 548 units (approx.)
Total cost= Ordering Cost + Cost of raw material + Storage cost
= (36,000 / 548) × 25 + (36,000 × 20)+(548/2) ×6
= Rs. 1642.33 + 7,20,000+ 1,644 = Rs. 7,23,286.
(ii) When the company has an option to order between 3000 and 6000 units, the EOQ should be
calculated with a reduction in price by 5% (due to concession); The purchase price = 95% of Rs.
20 = Rs. 19.
D = 36,000 units per annum; S = Rs. 25; C = 2 + 20% of 19 = 2 + 3.80 = Rs 5.80
EOQ =
2 36000 25 18,00,000
5.80 5.80
× × = = 557 units app.
Total cost = (36,000/557)× 25+(36,000 × 19)+ (557/2) ×5.80
= Rs. (1,615.79 + 6,84,000+1,615.30) = Rs. 6,87,231.09
(iii) When the company orders more than 6,000 units purchase price = 90% of Rs. 20 (because 10%
concession)
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163
= Rs. 18; D = 36,000 units per annum; S = Rs. 25; C = 2 + 20% of Rs. 18
= 2 + 3.60 = 5.60
EOQ =
2DS 2 36000 25
C 5.60
= × × = 567 units app. Total cost
= (36,000/567) ×25+ (36,000×18)+ (567/2) ×5.60
= Rs. 1,587.30 + 6,48,000 + 1,587.60 = Rs. 6,51,174.90
(iv) Comparing these costs, we notice that the cost is minimum for (iii) order. Therefore the company
should adopt a policy of ordering 567 units per order.
Problem: 6
M/s. Tubes Ltd. are the manufacturers of picture tubes of T.V. The following are the details of their operation
during 2001:
Average monthly market demand 2,000 tubes
Ordering cost Rs. 100 per order
Inventory carrying cost 20% per annum
Cost of tubes Rs. 500 per tube
Normal usage 100 tubes per week
Minimum usage 50 tubes per week
Maximum usage 200 tubes per week
Lead time to supply 6 – 8 weeks
Compute from the above:
(1) Economic order quantity. If the supplier is willing to supply quarterly 1,500 units at a discount
of 5%, is it worth accepting?
(2) Maximum level of stock.
(3) Minimum level of stock.
(4) Re-order level of stock.
Solution:
(1) Economic Order Quantity:
Annual usage of tubes (Co) = Normal usage per week × 52 weeks
= 100 tubes × 52 weeks
= 5,200 tubes.
Ordering cost per order (O) = Rs. 100.
Inventory carrying cost per unit per annum = 20% of Rs. 500 = Rs. 100.
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Production Planning and Productivity Management
EOQ =
2 5, 200 units Rs.100
Rs.100
× ×
O =
O
C O
C = 102 units (approx.).
Evaluation of order size of 1,500 units at 5% discount
No. of orders =
5, 200units
1, 500 units = 3.46 units or 4 (in case of a fraction, the next whole number is considered).
Rs.
Ordering cost order per year at Rs. 100 per order 400
Carrying cost of average inventory:
1, 500units 20
Rs.(500less5%)
2 100
× × 71,250
Total annual cost (excluding item cost) 71,650
Annual cost if EO Q (102 units) is adopted : Rs.
Ordering cost: 5,200 ÷ 102 or 51 orders per year of Rs. .100 per order 5,100
Carrying cost of average inventory
102units 20
Rs.500
2 100
× × 5,100
Total annual cost (excluding item cost) 10,200
Increase in annual cost: Rs. (71,650 – 10,200) = Rs. 61,450.
Amount of quantity discount: 5% × Rs. 500 × 5,200, units = Rs. 1,30,000.
Since the amount of quantity discount (Rs. 1,30,000) is more than the increase in total annual cost
(Rs. 61,450), it is advisable to accept the offer. This will result in a saving of Rs. (1,30,000 - 61,450) or
Rs. 68,550 p.a. in inventory cost.
(2) Maximum Level of Stock:
= Re-order level + Re-order quantity – (Minimum usage × Minimum delivery period)
= 1,600 units + 102 units – (50 units × 6 weeks) = 1,402 units.
(3) Minimum Level of Stock:
= Re-order level – (Normal usage × Normal delivery period) [see Note]
= 1,600 units – (100 units × 7 weeks) = 900 units.
Note: Normal delivery period is taken to be the average delivery period.
(4) Re-order Level of Stock:
= Maximum usage × Maximum delivery period = 200 units × 8 weeks = 1,600 units.
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2.10 Human Resource Planning
HRP is a process of stricking balance between human resources required and acquired in an organisation.
In other words, HRP is a process by which an organisation determines how it should acquire its desired
manpower to achieve the organisational goals. Thus, HRP helps an organisation have the right number
and kind of people at the right places and right times to successfully achieve its overall objectives. Human
resource planning is a process of determining and assuming that the organisation will have an adequate
number of qualified persons, available at the proper times, performing jobs which meet the needs of
enterprise and which provide satisfaction for the individuals involved.
HRP is the process- including forecasting, developing and controlling-by which a firm ensures that it has
the right number of people and the right kind of people at the right places at the right time doing work for
which they are economically most useful.
HRP can be defined as the comparisons of an organisation’s existing labour resources with forecast labour
demand, and hence the scheduling of activities for acquiring, training, redeploying and possibly discarding
labour. It seeks to ensure that an adequate supply of labour is available precisely when required.
HRP could be seen as a process, consisting of the following series of activities:
1. Forecasting future personnel requirements, either in terms of mathematical projections of trends in
the economy and developments in the industry, or of judgements estimates based upon specific future
plans of the company.
2. Inventing present manpower resources and analysing the degree to which these re-sources are
employed optimally.
3. Anticipating Manpower Problems by projecting present resources into the future and comparing
them with the forecast of requirements, to determine their adequacy, both quantitatively and
qualitatively.
4. Planning the necessary programmes of recruitment, selection, training, employment, Utilisation,
transfer, promotion, development, motivation and compensation so that future man power
requirements will be duly met.
Objectives of HRP :
The main objective of having human resource planning is to have an accurate number of employees required,
with matching skill requirements to accomplish organisational goals. In other words, the objectives of
human resource planning are to:
• Anticipate the impact of technology on jobs and requirements for human resources.
• Assess surplus or shortage, if any, of human resources available over a specified period of time.
• Control the human resources already deployed in the organisation.
• Ensure adequate supply of manpower as and when required.
• Ensure proper use of existing human resources in the organisation.
• Forecast future requirements of human resources with different levels of skills.
• Provide lead time available to select and train the required additional human resource over a specified
time period.
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Production Planning and Productivity Management
Importance of HRP:
1. Despite growing unemployment, there has been shortage of human resources with required skills,
qualification and capabilities to carry on works. Hence the need for human resource planning.
2. Human resource planning is also essential in the face of marked rise in workforce turnover which is
unavoidable and even beneficial. Voluntary quits, discharges, marriages, promotions and seasonal
fluctuations in business are the examples of factors leading to workforce turnover in organisations.
These cause a constant change and flow in the work force in many organisations.
3. Human resource planning is also needed in order to meet the needs of expansion and diversification
programmes of an organisation.
4. Large numbers of employees, who retire, die, leave organisations, or become incapacitated because of
physical or mental ailments, need to be replaced by the new employees. Human resource planning
ensures smooth supply of workers without interruption.
5. Technological changes and globalisation usher in change in the method of products and distribution
of production and services and in management techniques. These changes may also require a change
in the skills of employees, as well as change in the number of employees required. It is human resource
planning that enables organisations to cope with such changes.
6. The need for human resource planning is also felt in order to identify areas of surplus personnel or
areas in which there is shortage of personnel. Then, in case of surplus personnel, it can be redeployed
in other areas of organisation. Conversely, in case of shortage of personnel, it can be made good by
downsizing the work force.
Human resource planning is important to organisation because it benefits the organisation in several ways.
The important ones are mentioned below:
1. By maintaining a balance between demand for and supply of human resources, human resource
planning makes optimum use of human resources, on the one hand, and reduces labour cost
substantially, on the other.
2. Careful consideration of likely future events, through human resource planning might lead to the
discovery of better means for managing human resources. Thus, foreseeable pitfalls might be avoided.
3. Human resource planning compels management to assess critically the strength and weaknesses of
its employees and personnel policies on continuous basis and, in turn, take corrective measures to
improve the situation.
4. Human resource planning helps the organisation create and develop training and succes-sion planning
for employees and managers. Thus, it provides enough lead time for internal succession of employees
to higher positions through promotions.
5. Human resource planning meets the organisation need for right type of people in right number at
right times.
6. It also provides multiple gains to the employees by way of promotions, increase in emolu-ments and
other perquisites and fringe benefits.
7. Last but no means the least, with increase in skill, knowledge, potentialities, productivity and job satisfaction,
organisation becomes the main beneficiary. Organisation is benefited in terms of increase in prosperity /
production, growth, development, profit and, thus, an edge over its competitors in the market.
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8. Manpower shortfalls and surpluses may be avoided, to a large extent.
9. Some of the problems of managing change may be foreseen and their consequences mitigated.
Consultations with affected groups and individuals can take place at an early stage in the change
process. This may avoid resistance for change.
10. Through human resource planning, duplication of efforts and conflict among efforts can be avoided,
on the one hand, and coordination of worker’s efforts can be improved, on the other.
Forecast requirement for human resources in the future:
There are various techniques varying from simple to sophisticated ones employed in human resource
forecasting. These include:
1. Management Judgement,
2. Work-Study Method,
3. Ratio-Trend Analysis,
4. Delphi Technique,
5. Flow Models,
6. Mathematical Models.