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Dr.T.Sivakami
Assistant Professor
Department of Management Studies
Bon Secours College for Women
Thanjavur
Operations Research
Operations Research
The British/Europeans refer to "operational research", the Americans to "operations research" -
but both are often shortened to just "OR" (which is the term we will use). Another term which is
used for this field is "management science" ("MS"). The Americans sometimes combine the
terms OR and MS together and say "OR/MS" or "ORMS". Yet other terms sometimes used are
"industrial engineering"("IE"), "decision science" ("DS"), and “problem solving”. In recent years
there has been a move towards a standardization upon a single term for the field, namely the term
"OR".
The Origins of Operations Research.
The formal activities of Operations Research (OR) were initiated in England during World War
II when a team of British scientists set out to make decisions regarding the best utilization of war
material. Following the end of the war, the ideas advanced in military operations were adapted to
improve efficiency and productivity in the civilian sector. Today, OR is a dominant decision
making tool.
What is operations research?
Operations
The activities carried out in an organization.
Research
The process of observation and testing characterized by the scientific method. Situation, problem
statement, model construction, validation, experimentation, candidate solutions.
Definition of OR
Defining OR is difficult task as its boundaries and content are not yet fixed. It can be regarded as
use of mathematical and quantitative techniques to substantiate the decision being taken. Further,
it is multidisciplinary which takes tools from subjects like mathematics, statistics, engineering,
economics, psychology etc. and uses them to score the consequences of possible alternative
actions. Today it has become professional discipline that deals with the application of scientific
methods to decision-making.
Few other definitions of OR are as follows:
 “OR is concerned with scientifically deciding how to best design and operate man-machine
system usually requiring the allocation of scare resources.”
– Operations Research Society,
America
 “OR is essentially a collection of mathematical techniques and tools which in conjunction
with system approach, are applied to solve practical decision problems of an economic or
engineering nature’’.
– Daellenbach and George
 “OR utilizes the planned approach (updated scientific method) and an interdisciplinary team
in order to represent complex functional relationships as mathematical models for the purpose of
providing a quantitative analysis’’.
– Thieraub and Klekamp
 “OR is a scientific knowledge through interdisciplinary team effort for the purpose of
determining the best utilization of limited resources.”
 “OR is a scientific approach to problem solving for executive
management”.
– H.A. Taha
– H.M. Wagner
Scope of Operations Research
The scope of OR is not only confined to any specific agency like defence services but today it is
widely used in all industrial organisations. It can be used to find the best solution to any problem
be it simple or complex. It is useful in every field of human activities, where optimisation of
resources is required in the best way. Thus, it attempts to resolve the conflicts of interest among
the components of organization in a way that is best for the organisation as a whole. The main
fields where OR is extensively used are given below, however, this list is not exhaustive but only
illustrative.
(i) National Planning and Budgeting
OR is used for the preparation of Five Year Plans, annual budgets, forecasting of income and
expenditure, scheduling of major projects of national importance, estimation of GNP, GDP,
population, employment and generation of agriculture yields etc.
(ii) Defence Services
Basically formulation of OR started from USA army, so it has wide application in the areas such
as: development of new technology, optimization of cost and time, tender evaluation, setting and
layouts
of defence projects, assessment of “Threat analysis”, strategy of battle, effective maintenance
and replacement of equipment, inventory control, transportation and supply depots etc.
(iii) Industrial Establishment and Private Sector Units
OR can be effectively used in plant location and setting finance planning, product and process
planning, facility planning and construction, production planning and control, purchasing,
maintenance management and personnel management etc. to name a few.
(iv) R & D and Engineering
Research and development being the heart of technological growth, OR has wide scope for and
can be applied in technology forecasting and evaluation, technology and project management,
preparation of tender and negotiation, value engineering, work/method study and so on.
(v) Business Management and Competition
OR can help in taking business decisions under risk and uncertainty, capital investment and
returns, business strategy formation, optimum advertisement outlay, optimum sales force and
their distribution, market survey and analysis and market research techniques etc.
(vi) Agriculture and Irrigation
In the area of agriculture and irrigation also OR can be useful for project management,
construction of major dams at minimum cost, optimum allocation of supply and collection points
for fertilizer/seeds and agriculture outputs and optimum mix of fertilizers for better yield.
(vii) Education and Training
OR can be used for obtaining optimum number of schools with their locations, optimum mix of
Students/teacher student ratio, optimum financial outlay and other relevant information in training
of graduates to meet out the national requirements.
(viii) Transportation
Transportation models of OR can be applied to real life problems to forecast public transport
requirements, optimum routing, forecasting of income and expenses, project management for
railways, railway network distribution, etc. In the same way it can be useful in the field of
communication.
(ix) Home Management and Budgeting
OR can be effectively used for control of expenses to maximize savings, time management, work
study methods for all related works. Investment of surplus budget, appropriate insurance of life
and properties and estimate of depreciation and optimum premium of insurance etc.
Phases of OR Study
OR is a logical and systematic approach to provide a rational basis for decision-making. The
phases of OR must be logical and systematic. The various steps required for the analysis of a
problem under OR are as follows:
Step I. Observe the Problem Environment
The first step of OR study is the observation of the environment in which the problem exists. The
activities that constitute this step are visits, conferences, observations, research etc. with the help
of such activities, the OR analyst gets sufficient information and support to proceed and is better
prepared to formulate the problem.
Step II. Analyse and Define the Problem
In this step not only the problem is defined but also uses, objectives and limitations of the study
that are stressed in the light of the problem. The end results of this step are clear grasp of need
for a solution and understanding of its nature.
Step III. Develop a Model
The next step is to develop model, which is representation of same real or abstract situation. OR
models are basically mathematical models representing systems, process or environment in form
of equations, relationships or formulae. The activities in this step is to defining interrelationships
among variables, formulating equations, using known OR models or searching suitable alternate
models. The proposed model may be field tested and modified in order to work under stated
environmental constraints. A model may also be modified if the management is not satisfied with
the answer that it gives.
Step IV. Selection of Data Input
It is a established fact that without authentic and appropriate data the results of the OR models
cannot be trusted. Hence, taping right kind of data is a vital step in OR process. Important
activities in this step are analysing internal-external data and facts, collecting opinions and using
computer data banks. The purpose of this step is to have sufficient input to operate and test the
model.
Step V. Solution and Testing
In this step the solution of the problems is obtained with the help of model and data input. Such a
solution is not implemented immediately and this solution is used to test the model and to find its
limitations if any. If the solution is not reasonable or if the model is not behaving properly,
updating and modification of the model is considered at this stage. The end result of this step is
solution that is desirable and supports current organisational objectives.
Step VI. Implementation of the Solution
This is the last phase of the OR study. In OR the decision-making is scientific but
implementation of decision involves many behavioural issues. Therefore, implementation
authority has to resolve the behavioural issues, involving the workers and supervisors to avoid
further conflicts. The gap between management and OR scientist may offer some resistance but
must be eliminated before solution is accepted in totality. Both the parties should play positive
role, since the implementation will help the organisation as a whole. A properly implemented
solution obtained through OR techniques results in improved working conditions and wins
management support.
Features / Characteristics of OR
The significant features of operations research include the followings:
(i) Decision-making. Every industrial organisation faces multifacet problems to identify best
possible solution to their problems. OR aims to help the executives to obtain optimal solution
with the use of OR techniques. It also helps the decision maker to improve his creative and
judicious capabilities, analyse and understand the problem situation leading to better control,
better co-ordination, better systems and finally better decisions.
(ii) Scientific Approach. OR applies scientific methods, techniques and tools for the purpose of
analysis and solution of the complex problems. In this approach there is no place for guess work
and the person bias of the decision maker.
(iii) Inter-disciplinary Team Approach. Basically the industrial problems are of complex
nature and therefore require a team effort to handle it. This team comprises of
scientist/mathematician and technocrates. Who jointly use the OR tools to obtain a optimal
solution of the problem. The tries to analyse the cause and effect relationship between various
parameters of the problem and evaluates the outcome of various alternative strategies.
(iv) System Approach. The main aim of the system approach is to trace for each proposal all
significant and indirect effects on all sub-system on a system and to evaluate each action in terms
of effects for the system as a whole. The interrelationship and interaction of each sub-system can
be handled with the help of mathematical/analytical models of OR to obtain acceptable solution.
(v) Use of Computers. The models of OR need lot of computation and therefore, the use of
computers becomes necessary. With the use of computers it is possible to handle complex
problems requiring large amount of calculations.
Limitations of Operations Research
OR has some limitations however, these are related to the problem of model building and the
time and money factors involved in application rather than its practical utility. Some of them
are as follows:
(i) Magnitude of Computation. Operations research models try to find out optimal solution
taking into account all the factors. These factors are enormous and expressing them in quantity
and establishing relationships among these require voluminous calculations which can be
handled by computers.
(ii) Non-Quantifiable Factors. OR provides solution only when all elements related to a
problem can be quantified. All relevant variables do not lend themselves to quantification.
Factors which cannot be quantified, find no place in OR study. Models in OR do not take into
account qualititative factors or emotional factors which may be quite important.
(iii) Distance between User and Analyst. OR being specialist’s job requires a mathematician
or statistician, who might not be aware of the business problems. Similarly, a manager fails to
understand the complex working of OR. Thus there is a gap between the two. Management
itself may offer a lot of resistance due to conventional thinking.
(iv) Time and Money Costs. When basic data are subjected to frequent changes, incorporating
them into the OR models is a costly proposition. Moreover, a fairly good solution at present
may be more desirable than a perfect OR solution available after sometime. The computational
time increases depending upon the size of the problem and accuracy of results desired.
(v) Implementation. Implementation of any decision is a delicate task. It must take into account
the complexities of human relations and behaviour. Sometimes, resistance is offered due to
psychological factors which may not have any bearing on the problem as well as its solution.
LPP .
Linear Programming Problems in maths is a system process of finding a maximum or minimum value of
any variable in a function, it is also known by the name of optimization problem. LPP is helpful in
developing and solving a decision making problem by mathematical techniques. The problem is generally
given in a linear function which needs to be optimized subject to a set of different constraints. Major usage
of LPP is in advising the management to to make the most efficient & effective use of the scarce resources
Different Parts of the LP Model
1. Decision Variable: Variables which are changeable & going to impact the decision function. Like the
profit function is effected by both sales and price, now which one of these two is changeable, will be our
decision variable. 2. Objective Function: Linear function of the objective, either to Maximize or minimize,
like Maximize Profit, sales,production etc. and Minimize Cost, Loss, energy, consumption, wastage etc. 3.
Constraints: Any kind of limitation or scarcity explained through a function like Limitations of raw
materials, time, funds, equipment's etc. Non-Negative Constraints will also be there which will remain
non-negative all the time.
Formulation of an LP Problem:
1. Recognize the decision variables and assign symbols to them like X, Y, Z & so on). Now these are
the quantities we wish to find out. 2. Express all the constraints in terms of inequalities in relation
the decision variable. 3. Formulate the objective function in terms of the decision variables. 4. Add
the non-negativity condition/constraints
Let us look at this diet problem, A house wife wishes to mix two types of food F1 and F2 in such a way
that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B.
Food F1 costs E60/Kg and Food F2 costs E80/kg. Food F1 contains 3 units/kg of vitamin A and 5 units/kg
of vitamin B while Food F2 contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Formulate this
problem as a linear programming problem to minimize the cost of the mixtures,
Let’s follow the steps given to formulate and LP: 1. The kgs of the F1 and F2 in the mixture are our
decision variables. Suppose the mixture has X Kg of Food F1 and Y Kg of food F2. 2. In this example, the
constraints are the minimum requirements of the vitamins. The minimum requirement of vitamin A is 8
units. Therefore 3X + 4Y ≥ 8. Similarly, the minimum requirement of vitamin B is 11 units. Therefore,5X
+ 2Y ≥ 11 3. The cost of purchasing 1 Kg of food F1 is E60. The cost of purchasing 1 Kg of food F2 is
E80. The total cost of purchasing X Kg of food F1 and Y Kg of food F2 is C = 60X + 80Y, which is the
objective function. 4. The non-negativity conditions are X ≥ 0, Y ≥ 0
Therefore the mathematical formulation of the LPP is:
Minimize: C = 60X + 80Y
Subject to: 3X + 4Y ≥ 8
5X + 2Y ≥ 11
X ≥ 0 , Y ≥ 0
Steps for solving LPP through Graphical Method :
Step 1. Formulate the LPP problems and develop objective function along with all the constraints function.
Step 2. Graph the feasible region and find the corner points. The coordinates of the corner points can be
obtained by either inspection or by solving the two equations of the lines intersecting at that point. Step 3.
Make a table listing the value of the objective function at each corner point. Step 4. Determine the optimal
solution from the table in step 3. If the problem is of maximization (minimization) type, the solution
corresponding to the largest (smallest) value of the objective function is the optimal solution of the LPP.
Ex: A furniture company produces inexpensive tables and chairs. The production process for each is
similar in that both require a certain number of hours of carpentry work and a certain number of labour
hours in the painting department. Each table takes 4 hours of carpentry and 2 hours in the painting
department. Each chair requires 3 hours of carpentry and 1 hour in the painting department. During the
current production period, 240 hours of carpentry time are available and 100 hours in painting is available.
Each table sold yields a profit of E7; each chair produced is sold for a E5 profit. Find the best combination
of tables and chairs to manufacture in order to reach the maximum profit
The decision variables can be defined as X = number of tables to be produced & Y = number of chairs to
be produced. Now linear programming (LP) problem can be formulated in terms of X and Y and Profit
(P). maximize P = 7X + 5Y (Objective function) subject to 4X + 3Y ≤ 240 (hours of carpentry constraint)
2X + Y ≤ 100 (hours of painting constraint) X ≥ 0, Y ≥ 0 (Non-negativity constraint). Therefore the
mathematical formulation of the LPP is: Maximize: P = 7X + 5Y Subject to: 4X + 3Y ≤ 240 2X + Y ≤ 100
X ≥ 0 , Y ≥ 0 To find the optimal solution to this LP using the graphical method we first identify the
region of feasible solutions and the corner points of the of the feasible region. The graph for this example
is plotted in the next slide
The graphical method is one of the easiest way to solve a small LP problem. However this is useful only
when the decision variables are not more than two. It is not possible to plot the solution on a two-
dimensional graph when there are more than two variables and we must turn to more complex methods.
Another limitation of graphical method is that, an incorrect or inconsistent graph will produce inaccurate
answers,so one need to be very carefulwhile drawing and plotting the graph. A very useful method of
solving linear programming problems of any size is the so called Simplex method. We will discuss
simplex method once you are done with the practice on graphical method.
Transportation Problem | Set 1
(Introduction)
 Difficulty Level : Basic
 Last Updated : 25 Nov, 2019
Transportation problem is a special kind of Linear Programming Problem (LPP) in
which goods are transported from a set of sources to a set of destinations subject to the
supply and demand of the sources and destination respectively such that the total cost of
transportation is minimized. It is also sometimes called as Hitchcock problem.
Types of Transportation problems:
Balanced: When both supplies and demands are equal then the problem is said to be a
balanced transportation problem.
Unbalanced: When the supply and demand are not equal then it is said to be an
unbalanced transportation problem. In this type of problem, either a dummy row or a
dummy column is added according to the requirement to make it a balanced problem.
Then it can be solved similar to the balanced problem.
Methods to Solve:
To find the initial basic feasible solution there are three methods:
1. NorthWest Corner Cell Method.
2. Least Call Cell Method.
3. Vogel’s Approximation Method (VAM).
Basic structure of transportation problem:
In the above table D1, D2, D3 and D4 are the destinations where the products/goods are to
be delivered from different sources S1, S2, S3 and S4. Si is the supply from the
source Oi. dj is the demand of the destination Dj. Cij is the cost when the product is
delivered from source Si to destination Dj.
.
Transportation Problem | Set 2 (NorthWest
Corner Method)
 Last Updated : 28 Oct, 2020
An introduction to Transportation problem has been discussed in the previous article, in
this article, finding the initial basic feasible solution using the NorthWest Corner Cell
Method will be discussed.
Explanation: Given three sources O1, O2 and O3 and four
destinations D1, D2, D3 and D4. For the sources O1, O2 and O3, the supply
is 300, 400 and 500 respectively. The destinations D1, D2, D3 and D4 have
demands 250, 350, 400 and 200 respectively.
Solution: According to North West Corner method, (O1, D1) has to be the starting point
i.e. the north-west corner of the table. Each and every value in the cell is considered as the
cost per transportation. Compare the demand for column D1 and supply from the
source O1 and allocate the minimum of two to the cell (O1, D1) as shown in the figure.
The demand for Column D1 is completed so the entire column D1 will be canceled. The
supply from the source O1 remains 300 – 250 = 50.
Now from the remaining table i.e. excluding column D1, check the north-west corner
i.e. (O1, D2) and allocate the minimum among the supply for the respective column and
the rows. The supply from O1 is 50 which is less than the demand for D2 (i.e. 350), so
allocate 50 to the cell (O1, D2). Since the supply from row O1 is completed cancel the
row O1. The demand for column D2 remain 350 – 50 = 300.
From the remaining table the north-west corner cell is (O2, D2). The minimum among the
supply from source O2 (i.e 400) and demand for column D2 (i.e 300) is 300, so
allocate 300 to the cell (O2, D2). The demand for the column D2 is completed so cancel
the column and the remaining supply from source O2 is 400 – 300 = 100.
Now from remainig table find the north-west corner i.e. (O2, D3) and compare
the O2 supply (i.e. 100) and the demand for D2 (i.e. 400) and allocate the smaller (i.e.
100) to the cell (O2, D2). The supply from O2 is completed so cancel the row O2. The
remaining demand for column D3 remains 400 – 100 = 300.
Proceeding in the same way, the final values of the cells will be:
Note: In the last remaining cell the demand for the respective columns and rows are equal
which was cell (O3, D4). In this case, the supply from O3 and the demand
for D4 was 200 which was allocated to this cell. At last, nothing remained for any row or
column.
Now just multiply the allocated value with the respective cell value (i.e. the cost) and add
all of them to get the basic solution i.e. (250 * 3) + (50 * 1) + (300 * 6) + (100 * 5) + (300
* 3) + (200 * 2) = 4400
Transportation Problem | Set 3 (Least Cost
Cell Method)
 Last Updated : 04 Nov, 2020
The North-West Corner method has been discussed in the previous article. In this article,
the Least Cost Cell method will be discussed.
Solution: According to the Least Cost Cell method, the least cost among all the cells in
the table has to be found which is 1 (i.e. cell (O1, D2)).
Now check the supply from the row O1 and demand for column D2 and allocate the
smaller value to the cell. The smaller value is 300 so allocate this to the cell. The supply
from O1 is completed so cancel this row and the remaining demand for the
column D2 is 350 – 300 = 50.
Now find the cell with the least cost among the remaining cells. There are two cells with
the least cost i.e. (O2, D1) and (O3, D4) with cost 2. Lets select (O2, D1). Now find the
demand and supply for the respective cell and allocate the minimum among them to the
cell and cancel the row or column whose supply or demand becomes 0 after allocation.
Now the cell with the least cost is (O3, D4) with cost 2. Allocate this cell with 200 as the
demand is smaller than the supply. So the column gets cancelled.
There are two cells among the unallocated cells that have the least cost. Choose any at
random say (O3, D2). Allocate this cell with a minimum among the supply from the
respective row and the demand of the respective column. Cancel the row or column with
zero value.
Now the cell with the least cost is (O3, D3). Allocate the minimum of supply and demand
and cancel the row or column with zero value.
The only remaining cell is (O2, D3) with cost 5 and its supply is 150 and demand
is 150 i.e. demand and supply both are equal. Allocate it to this cell.
Now just multiply the cost of the cell with their respective allocated values and add all of
them to get the basic solution i.e. (300 * 1) + (250 * 2) + (150 * 5) + (50 * 3) + (250 * 3)
+ (200 * 2) = 2850
Transportation Problem (Vogel’s
Approximation Method)
 Difficulty Level : Hard
 Last Updated : 25 Nov, 2019
The North-West Corner method and the Least Cost Cell method has been discussed in
the previous articles. In this article, the Vogel’s Approximation method will be
discussed.
Solution:
 For each row find the least value and then the second least value and take the absolute
difference of these two least values and write it in the corresponding row difference as
shown in the image below. In row O1, 1 is the least value and 3 is the second least
value and their absolute difference is 2. Similarly, for row O2 and O3, the absolute
differences are 3 and 1 respectively.
 For each column find the least value and then the second least value and take the
absolute difference of these two least values then write it in the corresponding column
difference as shown in the figure. In column D1, 2 is the least value and 3 is the
second least value and their absolute difference is 1. Similarly, for
column D2, D3 and D3, the absolute differences are 2, 2 and 2 respectively.
 These value of row difference and column difference are also called as penalty. Now
select the maximum penalty. The maximum penalty is 3 i.e. row O2. Now find the cell
with the least cost in row O2 and allocate the minimum among the supply of the
respective row and the demand of the respective column. Demand is smaller than the
supply so allocate the column’s demand i.e. 250 to the cell. Then cancel the
column D1.
 From the remaining cells, find out the row difference and column difference.
 Again select the maximum penalty which is 3 corresponding to row O1. The least-cost
cell in row O1 is (O1, D2) with cost 1. Allocate the minimum among supply and
demand from the respective row and column to the cell. Cancel the row or column
with zero value.
 Now find the row difference and column difference from the remaining cells.
 Now select the maximum penalty which is 7 corresponding to column D4. The least
cost cell in column D4 is (O3, D4) with cost 2. The demand is smaller than the supply
for cell (O3, D4). Allocate 200 to the cell and cancel the column.
 Find the row difference and the column difference from the remaining cells.
 Now the maximum penalty is 3 corresponding to the column D2. The cell with the
least value in D2 is (O3, D2). Allocate the minimum of supply and demand and cancel
the column.
 Now there is only one column so select the cell with the least cost and allocate the
value.
 Now there is only one cell so allocate the remaining demand or supply to the cell
 No balance remains. So multiply the allocated value of the cells with their
corresponding cell cost and add all to get the final cost i.e. (300 * 1) + (250 * 2) + (50
* 3) + (250 * 3) + (200 * 2) + (150 * 5) = 2850
Transportation Problem | Set 5 (
Unbalanced )
 Last Updated : 27 Nov, 2019
An introduction to the transportation problem has been discussed in this article. In this
article, the method to solve the unbalanced transportation problem will be discussed.
Below transportation problem is an unbalanced transportation problem.
The problem is unbalanced because the sum of all the supplies i.e. O1, O2, O3 and O4 is
not equal to the sum of all the demands i.e. D1, D2, D3, D4 and D5.
Solution:
In this type of problem, the concept of a dummy row or a dummy column will be used. As
in this case, since the supply is more than the demand so a dummy demand column will
be added and a demand of (total supply – total demand) will be given to that column
i.e. 117 – 95 = 22 as shown in the image below. If demand were more than the supply
then a dummy supply row would have been added.
Now that the problem has been updated to a balanced transportation problem, it can be
solved using any one of the following methods to solve a balanced transportation problem
as discussed in the earlier posts:
1. NorthWest Corner Method
2. Least Cost Cell Method
3. Vogel’s Approximation Method
UNIT 3 Inventory Control Problems
Introduction
The word ‘inventory’ means simply a stock of idle resources of any kind having an economic
value. In other words, inventory means a physical stock of goods, which is kept in hand for
smooth and efficient running of future aff airs of an organization. It may consist of raw materials,
work-in-progress, spare parts/consumables, finished goods, human resources such as unutilized
labor, financial resources such as working capital, etc. It is not necessary that an organization has
all these inventory classes but whatever may be the inventory items, they need efficient
management as generally a substantial amount of money is invested in them. The basic inventory
decisions in- clude: 1) How much to order? 2) When to order? 3) How much safety stock should be
kept? The problems faced by diff erent organizations have necessitated the use of scientific
techniques in the management of inventories known as inventory control. Inventory control is
concerned with the acquisition, storage, and handling of inventories so that the inventory is
1.1
available whenever needed and the associated total cost is minimized.
Reasons for Carrying Inventory
Inventories are carried by organisations because of the following major reasons :
1. Improve customer service- An inventory policy is designed to respond to indi- vidual
customer’s or organization’s request for products and services.
2. Reduce costs- Inventory holding or carrying costs are the expenses that are in- curred
for storage of items. However, holding inventory items in the warehouse can indirectly
reduce operating costs such as loss of goodwill and/or loss of po- tential sale due to
shortage of items. It may also encourage economies of pro- duction by allowing larger,
longer and more production runs.
3. Maintenance of operational capability- Inventories of raw materials and work- in-
progress items act as buff er between successive production stages so that downtime
in one stage does not aff ect the entire production process.
4. Irregular supply and demand- Inventories provide protection against irregular supply
and demand; an unexpected change in production and delivery sched- ule of a product
or a service can adversely affect operating costs and customer service level.
5. Quantity discount- Large size orders help to take advantage of price-quantity
discount. However, such an advantage must keep a balance between the storage cost
and costs due to obsolescence, damage, theft, insurance, etc.
6. Avoiding stockouts (shortages)- Under situations like, labor strikes, natural disasters,
variations in demand and delays in supplies, etc., inventories act as buffer against stock
out as well as loss of goodwill.
Costs Associated with Inventories
Various costs associated with inventory control are often classified as follows :
1. Purchase (or production) cost: It is the cost at which an item is purchased, or if an item is
produced.
2. Carrying (or holding) cost: The cost associated with maintaining inventory is known as
holding cost. It is directly proportional to the quantity kept in stock and the time for
which an item is held in stock. It includes handling cost, main- tenance cost,
depreciation, insurance, warehouse rent, taxes, etc.
3. Shortage (or stock out) cost: It is the cost which arises due to running out of stock. It
1.2
1.3
includes the cost of production stoppage, loss of goodwill, loss of profitability, special
orders at higher price, overtime/idle time payments, loss of opportunity to sell, etc.
4.Ordering (or set up) cost: The cost incurred in replenishing the inventory is known as ordering
cost. It includes all the costs relating to administration (such as salaries of the persons working
for purchasing, telephone calls, computer costs, postage, etc.), transportation, receiving and
inspection of goods, processing pay- ments, etc. If a firm produces its own goods instead of
purchasing the same from an outside source, then it is the cost of resetting the equipment for
production.
Basic Terminologies
The followings are some basic terminologies which are used in inventory theory:
1. Demand
It is an eff ective desire which is related to particular time, price, and quantity. The demand
pattern of a commodity may be either deterministic or probabilistic. In case of deterministic
demand, the quantities needed in future are known with certainty. This can be fixed (static)
or can vary (dynamic) from time to time. On the contrary, probabilistic demand is uncertain
over a certain period of time but its pattern can be described by a known probability
distribution.
2. Ordering cycle
An ordering cycle is defined as the time period between two successive replen- ishments.
The order may be placed on the basis of the following two types of inventory review
systems:
• Continuous review: In this case, the inventory level is monitored continu- ously
until a specified point (known as reorder point) is reached. At this point, a new
order is placed.
• Periodic review: In this case, the orders are placed at equally spaced intervals of
time. The quantity ordered each time depends on the available inventory level at
the time of review.
3. Planning period
This is also known as time horizon over which the inventory level is to be con- trolled. This
can be finite or infinite depending on the nature of demand.
4. Lead time or delivery lag
The time gap between the moment of placing an order and actually receiving it is referred
to as lead time. Lead time can be deterministic (constant or variable) or probabilistic.
1.4
5. Buff er (or safety) stock
Normally, demand and lead time are uncertain and cannot be predetermined completely.
So, to absorb the variation in demand and supply, some extra stock is kept. This extra stock
is known as buff er stock.
6. Re-order level
The level between maximum and minimum stocks at which purchasing activity must start
for replenishment is known as re-order level.
Economic Order Quantity (EOQ)
The concept of economic ordering quantity (EOQ) was first developed by F. Harris in 1916. The
concept is as follows: Management of inventory is confronted with a set of opposing costs. As the
lot size increases, the carrying cost increases while the ordering cost decreases. On the other hand,
as the lot size decreases, the carrying cost decreases but the ordering cost increases. The two
opposite costs can be shown graphically by plotting them against the order size as shown in Fig.
1.1 below :
Fig. 1.1: Graph of EOQ
Economic ordering quantity(EOQ) is that size of order which minimizes the average total cost of carrying
inventory and ordering under the assumed conditions of cer- tainty and the total demand during a given
period of time is known.
List of Symbols
The following symbols are used in connection with the inventory models presented in this chapter
:
1.5
1.6
c = purchase (or manufacturing) cost of an item
c1 = holding cost per quantity unit per unit time
c2 = shortage cost per quantity unit per unit item
c3 = ordering (set up) cost per order (set up)
R = demand rate
P = production rate
t = scheduling period which is variable
tp = prescribe scheduling period
D = total demand or annual demand
q = lot (order) size
L = lead time
x = random demand
f (x) = probability density function for demand x.
z = order level
Deterministic Inventory Models
1.7.1 Model I(a): EOQ model without shortage
The basic assumptions of the model are as follows:
• Demand rate R is known and uniform.
• Lead time is zero or a known constant.
• Replenishment rate is infinite, i.e., replenishments are instantaneous.
• Shortages are not permitted.
• Inventory holding cost is c1 per unit per unit time.
• Ordering cost is c3 per order.
Our objective is to determine the economic order quantity q∗ which minimizes the average total
cost of the inventory system. An inventory-time diagram with inventory level on the vertical axis
and time on the horizontal axis is shown in Fig. 1.2. Since the actual consumption of inventory
varies constantly, the concept of average inventory is applicable here. Average Inventory =
1/2[maximum level + minimum level] = (q + 0)/2 = q/2.
1.7
2
Fig. 1.2: Inventory-time diagram when lead time is a known constant
Thus, the average inventory carrying cost is = average inventory × holding cost =
1
qc1.
The average ordering cost is (R/q)c3. Therefore, the average total cost of the inventory system is
given by
C(q) =
1
c1q +
c3R
. (1.1)
2 q
Since the minimum average total cost occurs at a point when average ordering cost
and average inventory carrying cost are equal, therefore, we have 1
c1q = c3R
which
2 q
gives the optimal order quantity
√
q∗ =
2c3R
. (1.2)
c1
This result was derived independently by F.W. Harris and R.H. Wilson in the year 1915. Thats
why the model is called Harris-Wilson model.
Characteristics of Model I(a)
(i) Optimal ordering interval t∗ = q∗/R =
√
2c3
c1R√
(ii) Minimum average total cost Cmin = C(q∗) = 2c1c3R
If in Model I(a), the ordering cost is taken as (c3 + kq) where k is the ordering cost per unit item
ordered then there will be no change in the optimal order quantity q∗.
In this case, the average total cost is
C(q) =
1
c1q +
c3R
+ kR. (1.3)
2 q
Example 1.1: A manufacturing company purchases 9000 parts of a machine for its annual
requirements, ordering one month usage at a time. Each part costs Rs.20.
c1
c1
The ordering cost per order is Rs. 15 and the carrying charges are 15% of the average inventory
per year. Suggest a more economic purchasing policy for the company. How much would it save
the company per year ?
Solution: Given that R = 9000 parts/year, c1 = (15/100)×20 = Rs.3 per part/year, c3 =
Rs.15 per order. Using Harris-Wilson formula,
√
q∗ = 2c3R/c1 = 300 units
t∗ = q∗/R = 1/30 year = 12 days (approx.)
√
Cmin = 2c1c3R = Rs.900
If the company follows the policy of ordering every month, then lot size
of inventory each month q = 9000/12 = 750 parts, annual storage cost =
c1(q/2) = Rs. 1125,
annual ordering cost = 15 × 12 = Rs. 180.
The total cost per year = 1125 + 180 = Rs. 1305.
Therefore, the company should purchase 300 parts at time interval of 12 days instead of ordering
750 parts each month. Then there will be a net saving of Rs. 405.
Lot Size in Discrete Units
Let the lot size q be constrained to values u, 2u, 3u, · · · Then the necessary conditions for
optimal q, i.e., q∗ are
C(q∗) ≤C(q∗ +u) (1.4)
C(q∗) ≤ C(q∗ − u) (1.5)
From equations (1.4) and (1.5), we get by simplifying
q∗(q∗ − u) ≤
2Rc3
≤ q∗(q∗ + u) (1.6)
Example 1.2: Demand in an inventory system is at a constant and uniform rate of 2400 kg. per
year. The carrying cost is Rs. 5 per kg. per year. No shortage is allowed. The replenishment cost
is Rs. 22 per order. The lot size can only be in 100 kg. unit. What is the optimal lot size of the
system ?
Solution: Given that R = 2400kg/year, c1 = Rs.5/kg/year, c3 = Rs.22 per order, u = 100
kg. The optimal lot size q∗ has to satisfy the inequality
q∗(q∗ − u) ≤
2c3R
≤ q∗(q∗ + u)
i.e., q∗(q∗ − 100) ≤ 21120 ≤ q∗(q∗ + 100) By
trial and error, we find the optimal lot size q∗ = 200 kg.
Sensitivity of Lot Size System
The average cost C(q) of the lot size system is a func√
tion of the controllable variable
q where C(q) = 1 c1q+c3R
. The optimal results are q∗ = 2c3R
and C∗ = C(q∗) =
√
2c1c3R.
2 q c1
Suppose that instead of the optimal lot size q∗, the decision maker uses another lot size q′
which
is related to q∗ by the relation q′
= bq∗, b > 0.
Let C′
designate the average total cost of the system then. We use the ratio C′
/C∗
as a measure of sensitivity. It can be shown that C′
/C∗ = (1 + b2
)/(2b).
So, the measure of sensitivity is a function of b and is independent of the other pa- rameters c1, c3
and R.
1.7.2 Model I(b): EOQ Model with Diff erent Rates of Demand
This inventory system operates on the assumptions of Model I(a) except that the de- mand rates
are diff erent in diff erent cycles but order quantity is fixed in each cycle. The objective is to
determine the order size in each reorder cycle that will minimize the total inventory cost. Suppose
that the total demand D is specified over the plan- ning period T . If t1, t2, ..., tn denote the lengths
of successive n inventory cycles and D1, D2, ..., Dn are the demand rates in these cycles,
respectively, then the total period T is given by T = t1+t2+...+tn. Fig. 1.3 depicts the inventory
system under consideration.
Fig. 1.3: Inventory-time diagram for diff erent cycles
Suppose that each time a fixed quantity q is ordered. Then the number of orders in
q
the time period T is n = D/q. Thus, the inventory carrying for the time period T is
1 1 1 1 1
2
qt1c1 +
2
qt2c1 + ... +
2
qtnc1 =
2
qc1(t1 + t2 + ... + tn) =
2
qc1T
Total ordering cost = (Number of orders) × c3 = D
c3
Hence, the total inventory cost is C(q) = 1
c1qT + c3D
2 q
The optimal ordering quantity (q∗) is then determined by the first order condition as
√
q∗ =
2c3(D/T )
c1
UNIT 4
Hungarian Algorithm for Assignment
Problem | Set 1 (Introduction)
 Difficulty Level : Hard
 Last Updated : 04 Mar, 2020
Let there be n agents and n tasks. Any agent can be assigned to perform any task, incurring
some cost that may vary depending on the agent-task assignment. It is required to perform
all tasks by assigning exactly one agent to each task and exactly one task to each agent in
such a way that the total cost of the assignment is minimized.
Example: You work as a manager for a chip manufacturer, and you currently have 3
people on the road meeting clients. Your salespeople are in Jaipur, Pune and Bangalore,
and you want them to fly to three other cities: Delhi, Mumbai and Kerala. The table below
shows the cost of airline tickets in INR between the cities:
The question: where would you send each of your salespeople in order to minimize fair?
Possible assignment: Cost = 11000 INR
Other Possible assignment: Cost = 9500 INR and this is the best of the 3! possible
assignments.
Brute force solution is to consider every possible assignment implies a complexity
of Ω(n!).
The Hungarian algorithm, aka Munkres assignment algorithm, utilizes the following
theorem for polynomial runtime complexity (worst case O(n3
)) and guaranteed optimality:
If a number is added to or subtracted from all of the entries of any one row or column of a
cost matrix, then an optimal assignment for the resulting cost matrix is also an optimal
assignment for the original cost matrix.
We reduce our original weight matrix to contain zeros, by using the above theorem. We try
to assign tasks to agents such that each agent is doing only one task and the penalty
incurred in each case is zero.
Core of the algorithm (assuming square matrix):
1. For each row of the matrix, find the smallest element and subtract it from every element
in its row.
2. Do the same (as step 1) for all columns.
3. Cover all zeros in the matrix using minimum number of horizontal and vertical lines.
4. Test for Optimality: If the minimum number of covering lines is n, an optimal
assignment is possible and we are finished. Else if lines are lesser than n, we haven’t
found the optimal assignment, and must proceed to step 5.
5. Determine the smallest entry not covered by any line. Subtract this entry from each
uncovered row, and then add it to each covered column. Return to step 3.
Explanation for above simple example:
Below is the cost matrix of example given in above diagrams.
2500 4000 3500
4000 6000 3500
2000 4000 2500
Step 1: Subtract minimum of every row.
2500, 3500 and 2000 are subtracted from rows 1, 2 and
3 respectively.
0 1500 1000
500 2500 0
0 2000 500
Step 2: Subtract minimum of every column.
0, 1500 and 0 are subtracted from columns 1, 2 and 3
respectively.
0 0 1000
500 1000 0
0 500 500
Step 3: Cover all zeroes with minimum number of
horizontal and vertical lines.
Step 4: Since we need 3 lines to cover all zeroes,
we have found the optimal assignment.
2500 4000 3500
4000 6000 3500
2000 4000 2500
So the optimal cost is 4000 + 3500 + 2000 = 9500
An example that doesn’t lead to optimal value in first attempt:
In the above example, the first check for optimality did give us solution. What if we the
number covering lines is less than n.
cost matrix:
1500 4000 4500
2000 6000 3500
2000 4000 2500
Step 1: Subtract minimum of every row.
1500, 2000 and 2000 are subtracted from rows 1, 2 and
3 respectively.
0 2500 3000
0 4000 1500
0 2000 500
Step 2: Subtract minimum of every column.
0, 2000 and 500 are subtracted from columns 1, 2 and 3
respectively.
0 500 2500
0 2000 1000
0 0 0
Step 3: Cover all zeroes with minimum number of
horizontal and vertical lines.
Step 4: Since we only need 2 lines to cover all zeroes,
we have NOT found the optimal assignment.
Step 5: We subtract the smallest uncovered entry
from all uncovered rows. Smallest entry is 500.
-500 0 2000
-500 1500 500
0 0 0
Then we add the smallest entry to all covered columns, we get
0 0 2000
0 1500 500
500 0 0
Now we return to Step 3:. Here we cover again using
lines. and go to Step 4:. Since we need 3 lines to
cover, we found the optimal solution.
1500 4000 4500
2000 6000 3500
2000 4000 2500
So the optimal cost is 4000 + 2000 + 2500 = 8500
Travelling Salesman Problem
A travelling salesman, named Rover plans to visit each of n cities. He wishes to visit each city
once and only once, arriving back to city from where he started. The distance between City i and
City j is cij.
If there are n cities, there are (n - 1)! possible ways for his tour. For example, if the
number of cities to be visited is 5, then there are 4! different combinations. Such type of
problems can be solved by the assignment method.
Let cij be the distance (or cost or time) between City i to City j and
xij =
1 if a tour includes travelling from city i to city j (for i ≠ j)
0 otherwise
The following example will help you in understanding the travelling salesman problem of
operation research.
Example: Travelling Salesman Problem
A travelling salesman, named Rolling Stone plans to visit five cities 1, 2, 3, 4 & 5. The
travel time (in hours) between these cities is shown below:
To
From 1 2 3 4 5
1 ∞ 5 8 4 5
2 5 ∞ 7 4 5
3 8 7 ∞ 8 6
4 4 4 8 ∞ 8
5 5 5 6 8 ∞
How should Mr. Rolling Stone schedule his touring plan in order to minimize the total
travel time, if he visits each city once a week?
"As every thread of gold is valuable, so is every minute of time." - John Mason
Solution
After applying steps 1 to 3 of the Hungarian method, we get the following assignments.
Table
To
From 1 2 3 4 5
1 ∞ 1 3 1
2 1 ∞ 2 1
3 2 1 ∞ 2
4 3 ∞ 4
5 3 ∞
Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros
in the reduced matrix.
Table
Select the smallest element from all the uncovered elements. Subtract this smallest
element from all the uncovered elements and add it to the elements, which lie at the
intersection of two lines. Thus, we obtain another reduced matrix for fresh assignment.
Repeating step 3 on the reduced matrix, we get the following assignments.
Table
To
From 1 2 3 4 5
1 ∞ 2 1
2 ∞ 1 1
3 1 ∞ 2
4 3 ∞ 5
5 4 ∞
The above solution suggests that the salesman should go from city 1 to city 4, city 4 to city
2, and then city 2 to 1 (original starting point). The above solution is not a solution to the
travelling salesman problem as he visits city 1 twice.
The next best solution can be obtained by bringing the minimum non-zero element, i.e., 1
into the solution. Please note that the value 1 occurs at four places. We will consider all
the cases separately until the acceptable solution is obtained. To make the assignment in
the cell (2, 3), delete the row & the column containing this cell so that no other assignment
can be made in the second row and third column.
Now, make the assignments in the usual manner as shown in the following table.
Table
He starts from city 1 and goes to city 4; from city 4 to city 2; from city 2 to city 3; from city
3 to city 5; from city 5 to city 1.
Substituting values from original table:
4 + 7 + 6+ 4 + 5 = 26 hours.
UNIT V
REPLACEMENT OF ITEMS DETERIORATING WITH TIME
13.1 Introduction
In any establishment, sooner or later equipment needs to be replaced, particularly when new
equipment gives more efficient or economical service than the old one. In some cases, the old
equipment might fail and work no more or is worn out. In such situations it needs more expenditure
on its maintenance than before. The problem in such situation is to determine the best policy to be
adopted with respect to replacement of the equipment. The replacement theory provides answer to
this question in terms of optimal replacement period. Replacement theory deals with the analysis of
materials and machines which deteriorate with time and fix the optimal time of their replacement
so that total cost is the minimum.
13.2 Replacement Decisions
The problem is to decide the best policy to adopt with regard to replacement. The need for
replacement arises in a number of different following situations so that different types of decisions
may have to be taken.
 It may be necessary to decide whether to wait for a certain item to fail which might cause
some loss or to replace earlier at the expense of higher cost of the item.
 The item can be considered individually to decide whether to replace now or if not when to
reconsider the item in question.
 It is necessary to decide whether to replace by the same item or by a different type of item.
13.3 Types of Replacement Problems
i) Replacement policy for items, efficiency of which declines gradually with time without
change in money value.
ii) Replacement policy for items, efficiency of which declines gradually with time but with
change in money value.
iii) Replacement policy of items breaking down suddenly
a) Individual replacement policy
b) Group replacement policy
iv) Staff replacement
In this lesson we confine ourselves to first two situations only
13.4 Replacement of Items that Deteriorate with Time
There are certain items which deteriorate gradually with usage and such items decline in efficiency
over a period of time. Generally, the maintenance cost of certain items always increase gradually
with time and a stage comes when the maintenance cost becomes so large that it is better and
economical to replace the item with a new one. There may be number of alternatives and we may
have a comparison between various alternatives by considering the costs due to waste, scrap, loss
of output, damage to equipment and safety risks etc.
13.4.1 Replacement of items whose maintenance cost increases with time and the value of
money remains same during the period
The following costs are considered in such decisions:
C: Capital cost of a certain item say a machine
s(t): The selling or scrap value of the item after t years
f(t): Operating (or maintenance) cost of the item at time t
n: Optimal replacement period of the item
The operating cost function f(t) is assumed to be strictly positive. It may be continuous or discrete.
13.4.2 When t is a continuous variable
Now the annual cost of the machine at time t is given by C � S(t) + f(t) and since the total
maintenance cost incurred on the machine during n years is .
Total cost T incurred on machine during n years is given by
���������������
Thus the average annual cost incurred on the machine per year during n years is given by
�����������
To determine the value of optimal period (n), the principle of minima will be employed.
�����������
���������������
�����������
Clearly
�����������
Therefore, it can be seen that A(n) or TA = f(n) is a minimum for T provided that f(t) is
non�decreasing and f(0) = 0. Hence, if time is measured continuously, then the average annual
cost will be minimized by replacing the item when the average cost becomes equal to the current
maintenance cost.
13.4.3 When time is a discrete variable
Here the period of time is considered as fixed and n takes values 1,2,3, � then
�����������
By using finite differences, A(n) will be a minimum for that value of n for which
���������������
or
����������� �
For this, we write
�����������
�����������
�����������
�����������
�����������
Similarly, it can be shown
�����������
This suggests the replacement policy that if time is measured in discrete units, then the average
annual cost will be minimized by replacing item when the next period�s maintenance cost become
greater than the current average cost. Hence, replace the equipment at the end of n years, if the
maintenance cost in the (n+1)th year is more than the average total cost in the (n)th year and the
(n)th year�s maintenance cost is less than the previous year�s average total cost. The following
examples will illustrate this methodology
Example 1 A milk plant is considering replacement of a machine whose cost price is Rs. 12,200
and the scrap value Rs. 200. The running (maintenance and operating) costs in Rs. are found from
experience to be as follows:
Year: 1 2 3 4 5 6 7 8
Running Cost: 200 500 800 1200 1800 2500 3200 4000
When should the machine be replaced?
Solution The computations can be summarized in the following tabular form:
Table 13.1 Calculations for average cost of machine
(In Rupees)
Year
(1)
Running
Cost
(2)
Cumulative
Running
Cost
(3)
Depreciation Cost
(4)
Total Cost TC
(5) = (3) + (4)
Average
Cost
(6) = (5)/(1)
1 200 200 12000 12200 12200
2 500 700 12000 12700 6350
3 800 1500 12000 13500 4500
4 1200 2700 12000 14700 3675
5 1800 4500 12000 16500 3300
6 2500 7000 12000 19000 3167
7 3200 10200 12000 22200 3171
8 4000 14200 12000 26200 3275
From the table it is noted that the average total cost per year, A(n) is minimum in the 6th year (Rs.
3167). Also the average cost in 7th year (Rs.3171) is more than the cost in 6th year. Hence the
machine should be replaced after every 6 years.
Example 2
A Machine owner finds from his past records that the maintenance costs per year of a machine
whose purchase price is Rs. 8000 are as given below:
Year: 1 2 3 4 5 6 7 8
Maintenance Cost: 1000 1300 1700 2200 2900 3800 4800 6000
Resale Price: 4000 2000 1200 600 500 400 400 400
Determine at which time it is profitable to replace the machine.
Solution C = Rs. 8000. Table 13.2 shows the average cost per year during the life of machine.
Here, The computations can be summarized in the following tabular form:
Table 13.2 Calculations for average cost of machine
Year
f(t) Cumulative
maintenance
cost
Scrap
value
Total cost
1 1000 1000 4000 5000 5000
2 1300 2300 2000 8300 4150
3 1700 4000 1200 10800 3600
4 2200 6200 600 13600 3400
5 2900 9100 500 16600
6 3800 12900 400 20500 3417
7 4800 17700 400 25300 3614
8 6000 23700 400 31300 3913
The above table shows that the value of TA during fifth year is minimum. Hence the machine
should be replaced after every fifth year.
Example 3
The cost of a machine is Rs. 6100 and its scrap value is only Rs.100. The maintenance costs are
found to be
Year: 1 2 3 4 5 6 7 8
Maintenance Cost (in Rs.): 100 250 400 600 900 1250 1600 2000
When should the Machine be replaced?
Solution
C = 6100, s(t) = 100 The computations can be summarized in the following tabular form:
Table 13.3 Calculations for average cost of machine
Replace
at the end
of year
Cumulative
maintenance cost
Total cost
1 100 100 6100 6100
2 250 350 6350 3175
3 400 750 6750 2250
4 600 1350 7350 1737.50
5 900 2250 8250 1650
6 1250 3500 9500
7 1600 5100 11100 1585.7
8 2000 7100 13100 1637.50
It is now observed that the machine should be replaced at the end of sixth year otherwise the
average cost per year will start to increase.
13.4.4 Replacement of items whose maintenance cost increases with time and the money value
changes at a constant rate
To understand this let us define the following terms:
Money Value
Since money has a value over time, therefore the explanation of the statement: �Money is worth
10% per year� can be given in two ways:
(a) In one way, spending Rs.100 today would be equivalent to spend Rs.110 in year�s time.
In other words if we plan to spend Rs.110 after a year from now, we could spend Rs.100
today and an investment which would be worth Rs.110 next year.
(b) Alternatively if we borrow Rs.100 at the interest of 10% per year and spend Rs.100 today,
we have to pay Rs.100 after one year (next year).
Thus, we conclude that Rs.100 is equal to Rs.110 a year from now. Consequently Rs. 1 from a year
now is equal to (1+0.1)-1 rupee today.
Present Worth Factor
As we have seen, a rupee a year from now will be equivalent to (1+0.1)-1 rupee today at the
discount rate of 10% per year. So, one rupee in n years from now will be equal to (1+0.1)-n.
Therefore, the quantity (1+0.1)-n is called the Present Worth Factor (PWF) or Present Value (PV)
of one rupee spent in n years from now. In general, if r is the rate of interest, then (1+r)-n is called
PWF or PV of one rupee spent in n years from now onwards. The expression (1+r)-n is known as
compound amount factor of one rupee spent in n years.
Discount Rate
Let r be the rate of interest. Therefore present worth factor of unit amount to be spent after one
year is .
Then v is known as the discount rate. The optimum replacement policy for replacement of item
where maintenance costs increase with time and money value changes with constant rate can be
determined by following method:
Suppose that the item (which may be machine or equipment) is available for use over a series of
time periods of equal intervals (say one year). Let
C = Purchase price of the item to be replaceds
Rt = Running (or maintenance) cost in the tth year
R = Rate of interest
�is the present worth of a rupee to be spent in a year hence.
Let the item be replaced at the end of every nth year. The year wise present worth of expenditure on
the item in the successive cycles of n years can be calculated as follows:
Year 1 2---- n n+1 n+2 ---- 2n 2n+1
Present
worth
C+R1 R2v Rnvn-1 (C+R1)vn R2vn+1 Rnv2n-1 (C+R1)v2n
Assuming that machines has no resale value at the time of replacement, the present worth of the
machine in n years will be given by
���������������
Summing up the right-hand side, column-wise
�����������
���������������
����������� ,
using sum of an infinite G.P.
�����������
�����������
f(n) and f(n+1) given above at n = 0,1,2�, are called the weighted average cost of previous n years
with weights 1,v, v2, ----vn-1respectively. P(n) is the amount of money required now to pay all the
future costs of acquiring and operating the equipment when it is renewed every n years. However,
if P (n) is less than P (n+1) then replacing the equipment each n year is preferable to replacing
each n years is preferable to replacing each (n+1) years. Further, if the best policy is replacing
every n years, then the two inequalities P (n+1) � P (n) > 0 and P (n-1) - P (n) < 0 must hold,
without giving the proof we shall state the following two inequalities which holds good at n, the
optimal replacement interval.
�����������
�����������
As a result of these two inequalities, rules for minimizing costs may be stated as follows:
1. Do not replace if the operating cost of next period is less than the weighted average of
previous costs.
2. Replace if the operating cost of the next period is greater than the weighted average of the
previous costs.
Working Procedure
The step-by-step procedure for solving the problem is stated as under:
1. Write in a column the running/maintenance costs of machine or equipment for different
years, Rn.
2. In the next column write the discount factor indicating the present value of a rupee
received after (i-1) years,
3. The two column values are multiplied to get present value of the maintenance costs, i.e.,
.
4. These discounted maintenance costs are then cumulated to the ith year to get .
5. The cost of machine or equipment is added to the values obtained in Step 4 above to
Obtain C+ .
6. The discount factors are then cumulated to get .
7. The total costs obtained in (Step 5) are divided by the corresponding value of the
accumulated discount factor for each of the years.
8. Now compare the column of maintenance costs which is constantly increasing with the
last column. Replace the machine in the latest year that the last column exceeds the column
of maintenance costs.
Example 4
A milk plant is offered an equipment A which is priced at Rs.60,000 and the costs of operation and
maintenance are estimated to be Rs.10,000 for each of the first 5 years, increasing every year by
Rs. 3000 per year in the sixth and subsequent years. If money carries the rate of interest 10% per
annum what would the optimal replacement period?
Solution
Table 13.4 Determination of optimal replacement period
At the end
of year
(n)
Operating &
maintenance
cost
Rn
Discounted
factor
Discounted
operation &
maintenance
cost
Cumulative
Discounted
operation &
maintenance
cost
Discounted total
cost
Cumulative
discounted
factor
Weighted
average
annual cost
(1) (2) (3) (4)=(2)x(3) (5) (6)=(5)+60000 (7) (8)=(6)+(7)
1 10000 1.0000 10000.00 10000.00 70000.00 1.00 70000.00
2 10000 0.9091 9091.00 19091.00 79091.00 1.91 41428.42
3 10000 0.8264 8264.00 27355.00 87355.00 2.74 31933.83
4 10000 0.7513 7513.00 34868.00 94868.00 3.49 27207.75
5 10000 0.6830 6830.00 41698.00 101698.00 4.17 24389.18
6 13000 0.6209 8071.70 49769.70 109769.70 4.79 22913.08
7 16000 0.5645 9032.00 58801.70 118801.70 5.36 22184.36
8 19000 0.5132 9750.80 68552.50 128552.50 5.87 21905.89
9 22000 0.4665 10263.00 78815.50 138815.50 6.33 21912.82
10 25000 0.4241 10602.50 89418.00 149418.00 6.76 22106.52
From Table 13.4 we find the weighted cost is minimum at the end of 8th year, hence the equipment
should be replaced at the end of 8th year.
Example 5
A Manufacturer is offered two machines A and B. Machine A is priced at Rs. 5000 and running
cost is estimated at Rs. 800 for each of the first five years, increasing by Rs. 200 per year in the
sixth and subsequent years. Machine B, with the same capacity as A, costs Rs. 2500, but has
running cost of Rs. 1200 per year for six years, thereafter increasing by Rs. 200 per year. If money
is worth 10% per year, which machine should be purchased? (Assume that the machines will
eventually be sold for scrap at a negligible price).
Solution
Since money is worth 10% per year, therefore discount rate is
Table 13.5 Computation of weighted average cost for machine A
At the
end of
year
(n)
Operating &
maintenance
cost
Rn
Discounted
factor
Discounted
operation &
maintenance
cost
Cumulative
Discounted
operation &
maintenance
cost
Discounted
total cost
Cumulative
discounted
factor
Weighted
average
annual cost
(1) (2) (3) (4)=(2)x(3) (5) (6)=(5)+
5000
(7) (8)=(6)+(7)
1 800 1.0000 800 800 5800 1 5800
2 800 0.9091 727 1527 6527 1.9091 3419.035
3 800 0.8264 661 2188 7188 2.7355 2627.819
4 800 0.7513 601 2789 7789 3.4868 2233.98
5 800 0.6830 546 3336 8336 4.1698 1999.098
6 1000 0.6209 621 3957 8957 4.7907 1869.61
7 1200 0.5645 677 4634 9634 5.3552 1799.025
8 1400 0.5132 718 5353 10353 5.8684 1764.13
9 1600 0.4665 746 6099 11099 6.3349 1752.043
10 1800 0.4241 763 6862 11862 6.759 1755.053
From table 13.5 we conclude that for machine A 1600<1752.043<1800. Since the running cost of
9th year is 1600and that of 10th year is 1800 and 1800>1752.043, it would be economical to replace
machine A at the end of nine years.
Table 13.6 Computation of weighted average cost for machine B
At the
end of
year
(n)
Operating &
maintenance
cost
Rn
Discounted
factor
Discounted
operation &
maintenance
cost
Cumulative
Discounted
operation &
maintenance
cost
Discounted total
cost
Cumulative
discounted
factor
Weighted
average
annual cost
(1) (2) (3) (4)=(2)x(3) (5) (6)=(5)+
2500
(7) (8)=(6)+(7)
1 1200 1.0000 1200.00 1200.00 3700.00 1.00 3700.00
2 1200 0.9091 1090.92 2290.92 4790.92 1.91 2509.52
3 1200 0.8264 991.68 3282.60 5782.60 2.74 2113.91
4 1200 0.7513 901.56 4184.16 6684.16 3.49 1916.99
5 1200 0.6830 819.60 5003.76 7503.76 4.17 1799.55
6 1200 0.6209 745.08 5748.84 8248.84 4.79 1721.84
7 1400 0.5645 790.30 6539.14 9039.14 5.36 1687.92
8 1600 0.5132 821.12 7360.26 9860.26 5.87 1680.23
9 1800 0.4665 839.70 8199.96 10699.96 6.33 1689.05
10 2000 0.4241 848.20 9048.16 11548.16 6.76 1708.56
In table13.6 we find that 1800<1689<2300 so it is better to replace the machine B after 8th year.
The equivalent yearly average discounted value of future costs is Rs. 1748.60 for machine A and it
is 1680.23for machine B. Hence, it is more economical to buy machine B rather than machine A.

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Operations Research

  • 1. Dr.T.Sivakami Assistant Professor Department of Management Studies Bon Secours College for Women Thanjavur Operations Research
  • 2. Operations Research The British/Europeans refer to "operational research", the Americans to "operations research" - but both are often shortened to just "OR" (which is the term we will use). Another term which is used for this field is "management science" ("MS"). The Americans sometimes combine the terms OR and MS together and say "OR/MS" or "ORMS". Yet other terms sometimes used are "industrial engineering"("IE"), "decision science" ("DS"), and “problem solving”. In recent years there has been a move towards a standardization upon a single term for the field, namely the term "OR". The Origins of Operations Research. The formal activities of Operations Research (OR) were initiated in England during World War II when a team of British scientists set out to make decisions regarding the best utilization of war material. Following the end of the war, the ideas advanced in military operations were adapted to improve efficiency and productivity in the civilian sector. Today, OR is a dominant decision making tool. What is operations research? Operations The activities carried out in an organization. Research The process of observation and testing characterized by the scientific method. Situation, problem statement, model construction, validation, experimentation, candidate solutions. Definition of OR Defining OR is difficult task as its boundaries and content are not yet fixed. It can be regarded as use of mathematical and quantitative techniques to substantiate the decision being taken. Further, it is multidisciplinary which takes tools from subjects like mathematics, statistics, engineering, economics, psychology etc. and uses them to score the consequences of possible alternative actions. Today it has become professional discipline that deals with the application of scientific methods to decision-making.
  • 3. Few other definitions of OR are as follows:  “OR is concerned with scientifically deciding how to best design and operate man-machine system usually requiring the allocation of scare resources.” – Operations Research Society, America  “OR is essentially a collection of mathematical techniques and tools which in conjunction with system approach, are applied to solve practical decision problems of an economic or engineering nature’’. – Daellenbach and George  “OR utilizes the planned approach (updated scientific method) and an interdisciplinary team in order to represent complex functional relationships as mathematical models for the purpose of providing a quantitative analysis’’. – Thieraub and Klekamp  “OR is a scientific knowledge through interdisciplinary team effort for the purpose of determining the best utilization of limited resources.”  “OR is a scientific approach to problem solving for executive management”. – H.A. Taha – H.M. Wagner Scope of Operations Research The scope of OR is not only confined to any specific agency like defence services but today it is widely used in all industrial organisations. It can be used to find the best solution to any problem be it simple or complex. It is useful in every field of human activities, where optimisation of resources is required in the best way. Thus, it attempts to resolve the conflicts of interest among the components of organization in a way that is best for the organisation as a whole. The main fields where OR is extensively used are given below, however, this list is not exhaustive but only illustrative. (i) National Planning and Budgeting OR is used for the preparation of Five Year Plans, annual budgets, forecasting of income and expenditure, scheduling of major projects of national importance, estimation of GNP, GDP, population, employment and generation of agriculture yields etc. (ii) Defence Services Basically formulation of OR started from USA army, so it has wide application in the areas such as: development of new technology, optimization of cost and time, tender evaluation, setting and
  • 5. of defence projects, assessment of “Threat analysis”, strategy of battle, effective maintenance and replacement of equipment, inventory control, transportation and supply depots etc. (iii) Industrial Establishment and Private Sector Units OR can be effectively used in plant location and setting finance planning, product and process planning, facility planning and construction, production planning and control, purchasing, maintenance management and personnel management etc. to name a few. (iv) R & D and Engineering Research and development being the heart of technological growth, OR has wide scope for and can be applied in technology forecasting and evaluation, technology and project management, preparation of tender and negotiation, value engineering, work/method study and so on. (v) Business Management and Competition OR can help in taking business decisions under risk and uncertainty, capital investment and returns, business strategy formation, optimum advertisement outlay, optimum sales force and their distribution, market survey and analysis and market research techniques etc. (vi) Agriculture and Irrigation In the area of agriculture and irrigation also OR can be useful for project management, construction of major dams at minimum cost, optimum allocation of supply and collection points for fertilizer/seeds and agriculture outputs and optimum mix of fertilizers for better yield. (vii) Education and Training OR can be used for obtaining optimum number of schools with their locations, optimum mix of Students/teacher student ratio, optimum financial outlay and other relevant information in training of graduates to meet out the national requirements. (viii) Transportation Transportation models of OR can be applied to real life problems to forecast public transport requirements, optimum routing, forecasting of income and expenses, project management for railways, railway network distribution, etc. In the same way it can be useful in the field of communication. (ix) Home Management and Budgeting OR can be effectively used for control of expenses to maximize savings, time management, work study methods for all related works. Investment of surplus budget, appropriate insurance of life and properties and estimate of depreciation and optimum premium of insurance etc.
  • 6. Phases of OR Study OR is a logical and systematic approach to provide a rational basis for decision-making. The phases of OR must be logical and systematic. The various steps required for the analysis of a problem under OR are as follows: Step I. Observe the Problem Environment The first step of OR study is the observation of the environment in which the problem exists. The activities that constitute this step are visits, conferences, observations, research etc. with the help of such activities, the OR analyst gets sufficient information and support to proceed and is better prepared to formulate the problem. Step II. Analyse and Define the Problem In this step not only the problem is defined but also uses, objectives and limitations of the study that are stressed in the light of the problem. The end results of this step are clear grasp of need for a solution and understanding of its nature. Step III. Develop a Model The next step is to develop model, which is representation of same real or abstract situation. OR models are basically mathematical models representing systems, process or environment in form of equations, relationships or formulae. The activities in this step is to defining interrelationships among variables, formulating equations, using known OR models or searching suitable alternate models. The proposed model may be field tested and modified in order to work under stated environmental constraints. A model may also be modified if the management is not satisfied with the answer that it gives. Step IV. Selection of Data Input It is a established fact that without authentic and appropriate data the results of the OR models cannot be trusted. Hence, taping right kind of data is a vital step in OR process. Important activities in this step are analysing internal-external data and facts, collecting opinions and using computer data banks. The purpose of this step is to have sufficient input to operate and test the model. Step V. Solution and Testing In this step the solution of the problems is obtained with the help of model and data input. Such a solution is not implemented immediately and this solution is used to test the model and to find its limitations if any. If the solution is not reasonable or if the model is not behaving properly, updating and modification of the model is considered at this stage. The end result of this step is solution that is desirable and supports current organisational objectives.
  • 7. Step VI. Implementation of the Solution This is the last phase of the OR study. In OR the decision-making is scientific but implementation of decision involves many behavioural issues. Therefore, implementation authority has to resolve the behavioural issues, involving the workers and supervisors to avoid further conflicts. The gap between management and OR scientist may offer some resistance but must be eliminated before solution is accepted in totality. Both the parties should play positive role, since the implementation will help the organisation as a whole. A properly implemented solution obtained through OR techniques results in improved working conditions and wins management support. Features / Characteristics of OR The significant features of operations research include the followings: (i) Decision-making. Every industrial organisation faces multifacet problems to identify best possible solution to their problems. OR aims to help the executives to obtain optimal solution with the use of OR techniques. It also helps the decision maker to improve his creative and judicious capabilities, analyse and understand the problem situation leading to better control, better co-ordination, better systems and finally better decisions. (ii) Scientific Approach. OR applies scientific methods, techniques and tools for the purpose of analysis and solution of the complex problems. In this approach there is no place for guess work and the person bias of the decision maker. (iii) Inter-disciplinary Team Approach. Basically the industrial problems are of complex nature and therefore require a team effort to handle it. This team comprises of scientist/mathematician and technocrates. Who jointly use the OR tools to obtain a optimal solution of the problem. The tries to analyse the cause and effect relationship between various parameters of the problem and evaluates the outcome of various alternative strategies. (iv) System Approach. The main aim of the system approach is to trace for each proposal all significant and indirect effects on all sub-system on a system and to evaluate each action in terms of effects for the system as a whole. The interrelationship and interaction of each sub-system can be handled with the help of mathematical/analytical models of OR to obtain acceptable solution. (v) Use of Computers. The models of OR need lot of computation and therefore, the use of computers becomes necessary. With the use of computers it is possible to handle complex problems requiring large amount of calculations.
  • 8. Limitations of Operations Research OR has some limitations however, these are related to the problem of model building and the time and money factors involved in application rather than its practical utility. Some of them are as follows: (i) Magnitude of Computation. Operations research models try to find out optimal solution taking into account all the factors. These factors are enormous and expressing them in quantity and establishing relationships among these require voluminous calculations which can be handled by computers. (ii) Non-Quantifiable Factors. OR provides solution only when all elements related to a problem can be quantified. All relevant variables do not lend themselves to quantification. Factors which cannot be quantified, find no place in OR study. Models in OR do not take into account qualititative factors or emotional factors which may be quite important. (iii) Distance between User and Analyst. OR being specialist’s job requires a mathematician or statistician, who might not be aware of the business problems. Similarly, a manager fails to understand the complex working of OR. Thus there is a gap between the two. Management itself may offer a lot of resistance due to conventional thinking. (iv) Time and Money Costs. When basic data are subjected to frequent changes, incorporating them into the OR models is a costly proposition. Moreover, a fairly good solution at present may be more desirable than a perfect OR solution available after sometime. The computational time increases depending upon the size of the problem and accuracy of results desired. (v) Implementation. Implementation of any decision is a delicate task. It must take into account the complexities of human relations and behaviour. Sometimes, resistance is offered due to psychological factors which may not have any bearing on the problem as well as its solution. LPP . Linear Programming Problems in maths is a system process of finding a maximum or minimum value of
  • 9. any variable in a function, it is also known by the name of optimization problem. LPP is helpful in developing and solving a decision making problem by mathematical techniques. The problem is generally given in a linear function which needs to be optimized subject to a set of different constraints. Major usage of LPP is in advising the management to to make the most efficient & effective use of the scarce resources Different Parts of the LP Model 1. Decision Variable: Variables which are changeable & going to impact the decision function. Like the profit function is effected by both sales and price, now which one of these two is changeable, will be our decision variable. 2. Objective Function: Linear function of the objective, either to Maximize or minimize, like Maximize Profit, sales,production etc. and Minimize Cost, Loss, energy, consumption, wastage etc. 3. Constraints: Any kind of limitation or scarcity explained through a function like Limitations of raw materials, time, funds, equipment's etc. Non-Negative Constraints will also be there which will remain non-negative all the time. Formulation of an LP Problem: 1. Recognize the decision variables and assign symbols to them like X, Y, Z & so on). Now these are the quantities we wish to find out. 2. Express all the constraints in terms of inequalities in relation the decision variable. 3. Formulate the objective function in terms of the decision variables. 4. Add the non-negativity condition/constraints Let us look at this diet problem, A house wife wishes to mix two types of food F1 and F2 in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food F1 costs E60/Kg and Food F2 costs E80/kg. Food F1 contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while Food F2 contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Formulate this problem as a linear programming problem to minimize the cost of the mixtures, Let’s follow the steps given to formulate and LP: 1. The kgs of the F1 and F2 in the mixture are our decision variables. Suppose the mixture has X Kg of Food F1 and Y Kg of food F2. 2. In this example, the constraints are the minimum requirements of the vitamins. The minimum requirement of vitamin A is 8 units. Therefore 3X + 4Y ≥ 8. Similarly, the minimum requirement of vitamin B is 11 units. Therefore,5X + 2Y ≥ 11 3. The cost of purchasing 1 Kg of food F1 is E60. The cost of purchasing 1 Kg of food F2 is E80. The total cost of purchasing X Kg of food F1 and Y Kg of food F2 is C = 60X + 80Y, which is the
  • 10. objective function. 4. The non-negativity conditions are X ≥ 0, Y ≥ 0 Therefore the mathematical formulation of the LPP is: Minimize: C = 60X + 80Y Subject to: 3X + 4Y ≥ 8 5X + 2Y ≥ 11 X ≥ 0 , Y ≥ 0 Steps for solving LPP through Graphical Method : Step 1. Formulate the LPP problems and develop objective function along with all the constraints function. Step 2. Graph the feasible region and find the corner points. The coordinates of the corner points can be obtained by either inspection or by solving the two equations of the lines intersecting at that point. Step 3. Make a table listing the value of the objective function at each corner point. Step 4. Determine the optimal solution from the table in step 3. If the problem is of maximization (minimization) type, the solution corresponding to the largest (smallest) value of the objective function is the optimal solution of the LPP. Ex: A furniture company produces inexpensive tables and chairs. The production process for each is similar in that both require a certain number of hours of carpentry work and a certain number of labour hours in the painting department. Each table takes 4 hours of carpentry and 2 hours in the painting department. Each chair requires 3 hours of carpentry and 1 hour in the painting department. During the current production period, 240 hours of carpentry time are available and 100 hours in painting is available. Each table sold yields a profit of E7; each chair produced is sold for a E5 profit. Find the best combination of tables and chairs to manufacture in order to reach the maximum profit The decision variables can be defined as X = number of tables to be produced & Y = number of chairs to be produced. Now linear programming (LP) problem can be formulated in terms of X and Y and Profit (P). maximize P = 7X + 5Y (Objective function) subject to 4X + 3Y ≤ 240 (hours of carpentry constraint) 2X + Y ≤ 100 (hours of painting constraint) X ≥ 0, Y ≥ 0 (Non-negativity constraint). Therefore the mathematical formulation of the LPP is: Maximize: P = 7X + 5Y Subject to: 4X + 3Y ≤ 240 2X + Y ≤ 100 X ≥ 0 , Y ≥ 0 To find the optimal solution to this LP using the graphical method we first identify the region of feasible solutions and the corner points of the of the feasible region. The graph for this example
  • 11. is plotted in the next slide The graphical method is one of the easiest way to solve a small LP problem. However this is useful only when the decision variables are not more than two. It is not possible to plot the solution on a two- dimensional graph when there are more than two variables and we must turn to more complex methods. Another limitation of graphical method is that, an incorrect or inconsistent graph will produce inaccurate answers,so one need to be very carefulwhile drawing and plotting the graph. A very useful method of solving linear programming problems of any size is the so called Simplex method. We will discuss simplex method once you are done with the practice on graphical method. Transportation Problem | Set 1 (Introduction)  Difficulty Level : Basic  Last Updated : 25 Nov, 2019 Transportation problem is a special kind of Linear Programming Problem (LPP) in which goods are transported from a set of sources to a set of destinations subject to the supply and demand of the sources and destination respectively such that the total cost of transportation is minimized. It is also sometimes called as Hitchcock problem. Types of Transportation problems: Balanced: When both supplies and demands are equal then the problem is said to be a balanced transportation problem. Unbalanced: When the supply and demand are not equal then it is said to be an unbalanced transportation problem. In this type of problem, either a dummy row or a
  • 12. dummy column is added according to the requirement to make it a balanced problem. Then it can be solved similar to the balanced problem. Methods to Solve: To find the initial basic feasible solution there are three methods: 1. NorthWest Corner Cell Method. 2. Least Call Cell Method. 3. Vogel’s Approximation Method (VAM). Basic structure of transportation problem: In the above table D1, D2, D3 and D4 are the destinations where the products/goods are to be delivered from different sources S1, S2, S3 and S4. Si is the supply from the source Oi. dj is the demand of the destination Dj. Cij is the cost when the product is delivered from source Si to destination Dj. . Transportation Problem | Set 2 (NorthWest Corner Method)  Last Updated : 28 Oct, 2020 An introduction to Transportation problem has been discussed in the previous article, in this article, finding the initial basic feasible solution using the NorthWest Corner Cell Method will be discussed.
  • 13. Explanation: Given three sources O1, O2 and O3 and four destinations D1, D2, D3 and D4. For the sources O1, O2 and O3, the supply is 300, 400 and 500 respectively. The destinations D1, D2, D3 and D4 have demands 250, 350, 400 and 200 respectively. Solution: According to North West Corner method, (O1, D1) has to be the starting point i.e. the north-west corner of the table. Each and every value in the cell is considered as the cost per transportation. Compare the demand for column D1 and supply from the source O1 and allocate the minimum of two to the cell (O1, D1) as shown in the figure. The demand for Column D1 is completed so the entire column D1 will be canceled. The supply from the source O1 remains 300 – 250 = 50. Now from the remaining table i.e. excluding column D1, check the north-west corner i.e. (O1, D2) and allocate the minimum among the supply for the respective column and the rows. The supply from O1 is 50 which is less than the demand for D2 (i.e. 350), so allocate 50 to the cell (O1, D2). Since the supply from row O1 is completed cancel the row O1. The demand for column D2 remain 350 – 50 = 300.
  • 14. From the remaining table the north-west corner cell is (O2, D2). The minimum among the supply from source O2 (i.e 400) and demand for column D2 (i.e 300) is 300, so allocate 300 to the cell (O2, D2). The demand for the column D2 is completed so cancel the column and the remaining supply from source O2 is 400 – 300 = 100. Now from remainig table find the north-west corner i.e. (O2, D3) and compare the O2 supply (i.e. 100) and the demand for D2 (i.e. 400) and allocate the smaller (i.e. 100) to the cell (O2, D2). The supply from O2 is completed so cancel the row O2. The remaining demand for column D3 remains 400 – 100 = 300.
  • 15. Proceeding in the same way, the final values of the cells will be: Note: In the last remaining cell the demand for the respective columns and rows are equal which was cell (O3, D4). In this case, the supply from O3 and the demand for D4 was 200 which was allocated to this cell. At last, nothing remained for any row or column. Now just multiply the allocated value with the respective cell value (i.e. the cost) and add all of them to get the basic solution i.e. (250 * 3) + (50 * 1) + (300 * 6) + (100 * 5) + (300 * 3) + (200 * 2) = 4400 Transportation Problem | Set 3 (Least Cost Cell Method)  Last Updated : 04 Nov, 2020 The North-West Corner method has been discussed in the previous article. In this article, the Least Cost Cell method will be discussed.
  • 16. Solution: According to the Least Cost Cell method, the least cost among all the cells in the table has to be found which is 1 (i.e. cell (O1, D2)). Now check the supply from the row O1 and demand for column D2 and allocate the smaller value to the cell. The smaller value is 300 so allocate this to the cell. The supply from O1 is completed so cancel this row and the remaining demand for the column D2 is 350 – 300 = 50. Now find the cell with the least cost among the remaining cells. There are two cells with the least cost i.e. (O2, D1) and (O3, D4) with cost 2. Lets select (O2, D1). Now find the demand and supply for the respective cell and allocate the minimum among them to the cell and cancel the row or column whose supply or demand becomes 0 after allocation.
  • 17. Now the cell with the least cost is (O3, D4) with cost 2. Allocate this cell with 200 as the demand is smaller than the supply. So the column gets cancelled. There are two cells among the unallocated cells that have the least cost. Choose any at random say (O3, D2). Allocate this cell with a minimum among the supply from the respective row and the demand of the respective column. Cancel the row or column with zero value.
  • 18. Now the cell with the least cost is (O3, D3). Allocate the minimum of supply and demand and cancel the row or column with zero value. The only remaining cell is (O2, D3) with cost 5 and its supply is 150 and demand is 150 i.e. demand and supply both are equal. Allocate it to this cell.
  • 19. Now just multiply the cost of the cell with their respective allocated values and add all of them to get the basic solution i.e. (300 * 1) + (250 * 2) + (150 * 5) + (50 * 3) + (250 * 3) + (200 * 2) = 2850 Transportation Problem (Vogel’s Approximation Method)  Difficulty Level : Hard  Last Updated : 25 Nov, 2019 The North-West Corner method and the Least Cost Cell method has been discussed in the previous articles. In this article, the Vogel’s Approximation method will be discussed. Solution:  For each row find the least value and then the second least value and take the absolute difference of these two least values and write it in the corresponding row difference as shown in the image below. In row O1, 1 is the least value and 3 is the second least value and their absolute difference is 2. Similarly, for row O2 and O3, the absolute differences are 3 and 1 respectively.  For each column find the least value and then the second least value and take the absolute difference of these two least values then write it in the corresponding column difference as shown in the figure. In column D1, 2 is the least value and 3 is the second least value and their absolute difference is 1. Similarly, for
  • 20. column D2, D3 and D3, the absolute differences are 2, 2 and 2 respectively.  These value of row difference and column difference are also called as penalty. Now select the maximum penalty. The maximum penalty is 3 i.e. row O2. Now find the cell with the least cost in row O2 and allocate the minimum among the supply of the respective row and the demand of the respective column. Demand is smaller than the supply so allocate the column’s demand i.e. 250 to the cell. Then cancel the column D1.
  • 21.  From the remaining cells, find out the row difference and column difference.  Again select the maximum penalty which is 3 corresponding to row O1. The least-cost cell in row O1 is (O1, D2) with cost 1. Allocate the minimum among supply and demand from the respective row and column to the cell. Cancel the row or column with zero value.
  • 22.  Now find the row difference and column difference from the remaining cells.  Now select the maximum penalty which is 7 corresponding to column D4. The least cost cell in column D4 is (O3, D4) with cost 2. The demand is smaller than the supply for cell (O3, D4). Allocate 200 to the cell and cancel the column.
  • 23.  Find the row difference and the column difference from the remaining cells.  Now the maximum penalty is 3 corresponding to the column D2. The cell with the least value in D2 is (O3, D2). Allocate the minimum of supply and demand and cancel the column.
  • 24.  Now there is only one column so select the cell with the least cost and allocate the value.  Now there is only one cell so allocate the remaining demand or supply to the cell  No balance remains. So multiply the allocated value of the cells with their corresponding cell cost and add all to get the final cost i.e. (300 * 1) + (250 * 2) + (50 * 3) + (250 * 3) + (200 * 2) + (150 * 5) = 2850
  • 25. Transportation Problem | Set 5 ( Unbalanced )  Last Updated : 27 Nov, 2019 An introduction to the transportation problem has been discussed in this article. In this article, the method to solve the unbalanced transportation problem will be discussed. Below transportation problem is an unbalanced transportation problem. The problem is unbalanced because the sum of all the supplies i.e. O1, O2, O3 and O4 is not equal to the sum of all the demands i.e. D1, D2, D3, D4 and D5. Solution: In this type of problem, the concept of a dummy row or a dummy column will be used. As in this case, since the supply is more than the demand so a dummy demand column will be added and a demand of (total supply – total demand) will be given to that column i.e. 117 – 95 = 22 as shown in the image below. If demand were more than the supply
  • 26. then a dummy supply row would have been added. Now that the problem has been updated to a balanced transportation problem, it can be solved using any one of the following methods to solve a balanced transportation problem as discussed in the earlier posts: 1. NorthWest Corner Method 2. Least Cost Cell Method 3. Vogel’s Approximation Method UNIT 3 Inventory Control Problems Introduction The word ‘inventory’ means simply a stock of idle resources of any kind having an economic value. In other words, inventory means a physical stock of goods, which is kept in hand for smooth and efficient running of future aff airs of an organization. It may consist of raw materials, work-in-progress, spare parts/consumables, finished goods, human resources such as unutilized labor, financial resources such as working capital, etc. It is not necessary that an organization has all these inventory classes but whatever may be the inventory items, they need efficient management as generally a substantial amount of money is invested in them. The basic inventory decisions in- clude: 1) How much to order? 2) When to order? 3) How much safety stock should be kept? The problems faced by diff erent organizations have necessitated the use of scientific techniques in the management of inventories known as inventory control. Inventory control is concerned with the acquisition, storage, and handling of inventories so that the inventory is 1.1
  • 27. available whenever needed and the associated total cost is minimized. Reasons for Carrying Inventory Inventories are carried by organisations because of the following major reasons : 1. Improve customer service- An inventory policy is designed to respond to indi- vidual customer’s or organization’s request for products and services. 2. Reduce costs- Inventory holding or carrying costs are the expenses that are in- curred for storage of items. However, holding inventory items in the warehouse can indirectly reduce operating costs such as loss of goodwill and/or loss of po- tential sale due to shortage of items. It may also encourage economies of pro- duction by allowing larger, longer and more production runs. 3. Maintenance of operational capability- Inventories of raw materials and work- in- progress items act as buff er between successive production stages so that downtime in one stage does not aff ect the entire production process. 4. Irregular supply and demand- Inventories provide protection against irregular supply and demand; an unexpected change in production and delivery sched- ule of a product or a service can adversely affect operating costs and customer service level. 5. Quantity discount- Large size orders help to take advantage of price-quantity discount. However, such an advantage must keep a balance between the storage cost and costs due to obsolescence, damage, theft, insurance, etc. 6. Avoiding stockouts (shortages)- Under situations like, labor strikes, natural disasters, variations in demand and delays in supplies, etc., inventories act as buffer against stock out as well as loss of goodwill. Costs Associated with Inventories Various costs associated with inventory control are often classified as follows : 1. Purchase (or production) cost: It is the cost at which an item is purchased, or if an item is produced. 2. Carrying (or holding) cost: The cost associated with maintaining inventory is known as holding cost. It is directly proportional to the quantity kept in stock and the time for which an item is held in stock. It includes handling cost, main- tenance cost, depreciation, insurance, warehouse rent, taxes, etc. 3. Shortage (or stock out) cost: It is the cost which arises due to running out of stock. It 1.2 1.3
  • 28. includes the cost of production stoppage, loss of goodwill, loss of profitability, special orders at higher price, overtime/idle time payments, loss of opportunity to sell, etc. 4.Ordering (or set up) cost: The cost incurred in replenishing the inventory is known as ordering cost. It includes all the costs relating to administration (such as salaries of the persons working for purchasing, telephone calls, computer costs, postage, etc.), transportation, receiving and inspection of goods, processing pay- ments, etc. If a firm produces its own goods instead of purchasing the same from an outside source, then it is the cost of resetting the equipment for production. Basic Terminologies The followings are some basic terminologies which are used in inventory theory: 1. Demand It is an eff ective desire which is related to particular time, price, and quantity. The demand pattern of a commodity may be either deterministic or probabilistic. In case of deterministic demand, the quantities needed in future are known with certainty. This can be fixed (static) or can vary (dynamic) from time to time. On the contrary, probabilistic demand is uncertain over a certain period of time but its pattern can be described by a known probability distribution. 2. Ordering cycle An ordering cycle is defined as the time period between two successive replen- ishments. The order may be placed on the basis of the following two types of inventory review systems: • Continuous review: In this case, the inventory level is monitored continu- ously until a specified point (known as reorder point) is reached. At this point, a new order is placed. • Periodic review: In this case, the orders are placed at equally spaced intervals of time. The quantity ordered each time depends on the available inventory level at the time of review. 3. Planning period This is also known as time horizon over which the inventory level is to be con- trolled. This can be finite or infinite depending on the nature of demand. 4. Lead time or delivery lag The time gap between the moment of placing an order and actually receiving it is referred to as lead time. Lead time can be deterministic (constant or variable) or probabilistic. 1.4
  • 29. 5. Buff er (or safety) stock Normally, demand and lead time are uncertain and cannot be predetermined completely. So, to absorb the variation in demand and supply, some extra stock is kept. This extra stock is known as buff er stock. 6. Re-order level The level between maximum and minimum stocks at which purchasing activity must start for replenishment is known as re-order level. Economic Order Quantity (EOQ) The concept of economic ordering quantity (EOQ) was first developed by F. Harris in 1916. The concept is as follows: Management of inventory is confronted with a set of opposing costs. As the lot size increases, the carrying cost increases while the ordering cost decreases. On the other hand, as the lot size decreases, the carrying cost decreases but the ordering cost increases. The two opposite costs can be shown graphically by plotting them against the order size as shown in Fig. 1.1 below : Fig. 1.1: Graph of EOQ Economic ordering quantity(EOQ) is that size of order which minimizes the average total cost of carrying inventory and ordering under the assumed conditions of cer- tainty and the total demand during a given period of time is known. List of Symbols The following symbols are used in connection with the inventory models presented in this chapter : 1.5 1.6
  • 30. c = purchase (or manufacturing) cost of an item c1 = holding cost per quantity unit per unit time c2 = shortage cost per quantity unit per unit item c3 = ordering (set up) cost per order (set up) R = demand rate P = production rate t = scheduling period which is variable tp = prescribe scheduling period D = total demand or annual demand q = lot (order) size L = lead time x = random demand f (x) = probability density function for demand x. z = order level Deterministic Inventory Models 1.7.1 Model I(a): EOQ model without shortage The basic assumptions of the model are as follows: • Demand rate R is known and uniform. • Lead time is zero or a known constant. • Replenishment rate is infinite, i.e., replenishments are instantaneous. • Shortages are not permitted. • Inventory holding cost is c1 per unit per unit time. • Ordering cost is c3 per order. Our objective is to determine the economic order quantity q∗ which minimizes the average total cost of the inventory system. An inventory-time diagram with inventory level on the vertical axis and time on the horizontal axis is shown in Fig. 1.2. Since the actual consumption of inventory varies constantly, the concept of average inventory is applicable here. Average Inventory = 1/2[maximum level + minimum level] = (q + 0)/2 = q/2. 1.7
  • 31. 2 Fig. 1.2: Inventory-time diagram when lead time is a known constant Thus, the average inventory carrying cost is = average inventory × holding cost = 1 qc1. The average ordering cost is (R/q)c3. Therefore, the average total cost of the inventory system is given by C(q) = 1 c1q + c3R . (1.1) 2 q Since the minimum average total cost occurs at a point when average ordering cost and average inventory carrying cost are equal, therefore, we have 1 c1q = c3R which 2 q gives the optimal order quantity √ q∗ = 2c3R . (1.2) c1 This result was derived independently by F.W. Harris and R.H. Wilson in the year 1915. Thats why the model is called Harris-Wilson model. Characteristics of Model I(a) (i) Optimal ordering interval t∗ = q∗/R = √ 2c3 c1R√ (ii) Minimum average total cost Cmin = C(q∗) = 2c1c3R If in Model I(a), the ordering cost is taken as (c3 + kq) where k is the ordering cost per unit item ordered then there will be no change in the optimal order quantity q∗. In this case, the average total cost is C(q) = 1 c1q + c3R + kR. (1.3) 2 q Example 1.1: A manufacturing company purchases 9000 parts of a machine for its annual requirements, ordering one month usage at a time. Each part costs Rs.20.
  • 32. c1 c1 The ordering cost per order is Rs. 15 and the carrying charges are 15% of the average inventory per year. Suggest a more economic purchasing policy for the company. How much would it save the company per year ? Solution: Given that R = 9000 parts/year, c1 = (15/100)×20 = Rs.3 per part/year, c3 = Rs.15 per order. Using Harris-Wilson formula, √ q∗ = 2c3R/c1 = 300 units t∗ = q∗/R = 1/30 year = 12 days (approx.) √ Cmin = 2c1c3R = Rs.900 If the company follows the policy of ordering every month, then lot size of inventory each month q = 9000/12 = 750 parts, annual storage cost = c1(q/2) = Rs. 1125, annual ordering cost = 15 × 12 = Rs. 180. The total cost per year = 1125 + 180 = Rs. 1305. Therefore, the company should purchase 300 parts at time interval of 12 days instead of ordering 750 parts each month. Then there will be a net saving of Rs. 405. Lot Size in Discrete Units Let the lot size q be constrained to values u, 2u, 3u, · · · Then the necessary conditions for optimal q, i.e., q∗ are C(q∗) ≤C(q∗ +u) (1.4) C(q∗) ≤ C(q∗ − u) (1.5) From equations (1.4) and (1.5), we get by simplifying q∗(q∗ − u) ≤ 2Rc3 ≤ q∗(q∗ + u) (1.6) Example 1.2: Demand in an inventory system is at a constant and uniform rate of 2400 kg. per year. The carrying cost is Rs. 5 per kg. per year. No shortage is allowed. The replenishment cost is Rs. 22 per order. The lot size can only be in 100 kg. unit. What is the optimal lot size of the system ? Solution: Given that R = 2400kg/year, c1 = Rs.5/kg/year, c3 = Rs.22 per order, u = 100 kg. The optimal lot size q∗ has to satisfy the inequality q∗(q∗ − u) ≤ 2c3R ≤ q∗(q∗ + u)
  • 33. i.e., q∗(q∗ − 100) ≤ 21120 ≤ q∗(q∗ + 100) By trial and error, we find the optimal lot size q∗ = 200 kg. Sensitivity of Lot Size System The average cost C(q) of the lot size system is a func√ tion of the controllable variable q where C(q) = 1 c1q+c3R . The optimal results are q∗ = 2c3R and C∗ = C(q∗) = √ 2c1c3R. 2 q c1 Suppose that instead of the optimal lot size q∗, the decision maker uses another lot size q′ which is related to q∗ by the relation q′ = bq∗, b > 0. Let C′ designate the average total cost of the system then. We use the ratio C′ /C∗ as a measure of sensitivity. It can be shown that C′ /C∗ = (1 + b2 )/(2b). So, the measure of sensitivity is a function of b and is independent of the other pa- rameters c1, c3 and R. 1.7.2 Model I(b): EOQ Model with Diff erent Rates of Demand This inventory system operates on the assumptions of Model I(a) except that the de- mand rates are diff erent in diff erent cycles but order quantity is fixed in each cycle. The objective is to determine the order size in each reorder cycle that will minimize the total inventory cost. Suppose that the total demand D is specified over the plan- ning period T . If t1, t2, ..., tn denote the lengths of successive n inventory cycles and D1, D2, ..., Dn are the demand rates in these cycles, respectively, then the total period T is given by T = t1+t2+...+tn. Fig. 1.3 depicts the inventory system under consideration. Fig. 1.3: Inventory-time diagram for diff erent cycles Suppose that each time a fixed quantity q is ordered. Then the number of orders in
  • 34. q the time period T is n = D/q. Thus, the inventory carrying for the time period T is 1 1 1 1 1 2 qt1c1 + 2 qt2c1 + ... + 2 qtnc1 = 2 qc1(t1 + t2 + ... + tn) = 2 qc1T Total ordering cost = (Number of orders) × c3 = D c3 Hence, the total inventory cost is C(q) = 1 c1qT + c3D 2 q The optimal ordering quantity (q∗) is then determined by the first order condition as √ q∗ = 2c3(D/T ) c1
  • 35. UNIT 4 Hungarian Algorithm for Assignment Problem | Set 1 (Introduction)  Difficulty Level : Hard  Last Updated : 04 Mar, 2020 Let there be n agents and n tasks. Any agent can be assigned to perform any task, incurring some cost that may vary depending on the agent-task assignment. It is required to perform all tasks by assigning exactly one agent to each task and exactly one task to each agent in such a way that the total cost of the assignment is minimized. Example: You work as a manager for a chip manufacturer, and you currently have 3 people on the road meeting clients. Your salespeople are in Jaipur, Pune and Bangalore, and you want them to fly to three other cities: Delhi, Mumbai and Kerala. The table below shows the cost of airline tickets in INR between the cities: The question: where would you send each of your salespeople in order to minimize fair? Possible assignment: Cost = 11000 INR Other Possible assignment: Cost = 9500 INR and this is the best of the 3! possible assignments.
  • 36. Brute force solution is to consider every possible assignment implies a complexity of Ω(n!). The Hungarian algorithm, aka Munkres assignment algorithm, utilizes the following theorem for polynomial runtime complexity (worst case O(n3 )) and guaranteed optimality: If a number is added to or subtracted from all of the entries of any one row or column of a cost matrix, then an optimal assignment for the resulting cost matrix is also an optimal assignment for the original cost matrix. We reduce our original weight matrix to contain zeros, by using the above theorem. We try to assign tasks to agents such that each agent is doing only one task and the penalty incurred in each case is zero. Core of the algorithm (assuming square matrix): 1. For each row of the matrix, find the smallest element and subtract it from every element in its row. 2. Do the same (as step 1) for all columns. 3. Cover all zeros in the matrix using minimum number of horizontal and vertical lines. 4. Test for Optimality: If the minimum number of covering lines is n, an optimal assignment is possible and we are finished. Else if lines are lesser than n, we haven’t found the optimal assignment, and must proceed to step 5. 5. Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to step 3. Explanation for above simple example: Below is the cost matrix of example given in above diagrams. 2500 4000 3500 4000 6000 3500 2000 4000 2500 Step 1: Subtract minimum of every row. 2500, 3500 and 2000 are subtracted from rows 1, 2 and 3 respectively. 0 1500 1000 500 2500 0 0 2000 500 Step 2: Subtract minimum of every column. 0, 1500 and 0 are subtracted from columns 1, 2 and 3
  • 37. respectively. 0 0 1000 500 1000 0 0 500 500 Step 3: Cover all zeroes with minimum number of horizontal and vertical lines. Step 4: Since we need 3 lines to cover all zeroes, we have found the optimal assignment. 2500 4000 3500 4000 6000 3500 2000 4000 2500 So the optimal cost is 4000 + 3500 + 2000 = 9500 An example that doesn’t lead to optimal value in first attempt: In the above example, the first check for optimality did give us solution. What if we the number covering lines is less than n. cost matrix: 1500 4000 4500 2000 6000 3500 2000 4000 2500 Step 1: Subtract minimum of every row. 1500, 2000 and 2000 are subtracted from rows 1, 2 and 3 respectively. 0 2500 3000 0 4000 1500 0 2000 500 Step 2: Subtract minimum of every column. 0, 2000 and 500 are subtracted from columns 1, 2 and 3 respectively. 0 500 2500 0 2000 1000 0 0 0
  • 38. Step 3: Cover all zeroes with minimum number of horizontal and vertical lines. Step 4: Since we only need 2 lines to cover all zeroes, we have NOT found the optimal assignment. Step 5: We subtract the smallest uncovered entry from all uncovered rows. Smallest entry is 500. -500 0 2000 -500 1500 500 0 0 0 Then we add the smallest entry to all covered columns, we get 0 0 2000 0 1500 500 500 0 0 Now we return to Step 3:. Here we cover again using lines. and go to Step 4:. Since we need 3 lines to cover, we found the optimal solution. 1500 4000 4500 2000 6000 3500 2000 4000 2500 So the optimal cost is 4000 + 2000 + 2500 = 8500 Travelling Salesman Problem A travelling salesman, named Rover plans to visit each of n cities. He wishes to visit each city once and only once, arriving back to city from where he started. The distance between City i and City j is cij. If there are n cities, there are (n - 1)! possible ways for his tour. For example, if the number of cities to be visited is 5, then there are 4! different combinations. Such type of problems can be solved by the assignment method. Let cij be the distance (or cost or time) between City i to City j and
  • 39. xij = 1 if a tour includes travelling from city i to city j (for i ≠ j) 0 otherwise The following example will help you in understanding the travelling salesman problem of operation research. Example: Travelling Salesman Problem A travelling salesman, named Rolling Stone plans to visit five cities 1, 2, 3, 4 & 5. The travel time (in hours) between these cities is shown below: To From 1 2 3 4 5 1 ∞ 5 8 4 5 2 5 ∞ 7 4 5 3 8 7 ∞ 8 6 4 4 4 8 ∞ 8 5 5 5 6 8 ∞ How should Mr. Rolling Stone schedule his touring plan in order to minimize the total travel time, if he visits each city once a week? "As every thread of gold is valuable, so is every minute of time." - John Mason Solution After applying steps 1 to 3 of the Hungarian method, we get the following assignments. Table To From 1 2 3 4 5 1 ∞ 1 3 1 2 1 ∞ 2 1 3 2 1 ∞ 2 4 3 ∞ 4
  • 40. 5 3 ∞ Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix. Table Select the smallest element from all the uncovered elements. Subtract this smallest element from all the uncovered elements and add it to the elements, which lie at the intersection of two lines. Thus, we obtain another reduced matrix for fresh assignment. Repeating step 3 on the reduced matrix, we get the following assignments. Table To From 1 2 3 4 5 1 ∞ 2 1 2 ∞ 1 1 3 1 ∞ 2 4 3 ∞ 5 5 4 ∞ The above solution suggests that the salesman should go from city 1 to city 4, city 4 to city 2, and then city 2 to 1 (original starting point). The above solution is not a solution to the travelling salesman problem as he visits city 1 twice. The next best solution can be obtained by bringing the minimum non-zero element, i.e., 1 into the solution. Please note that the value 1 occurs at four places. We will consider all the cases separately until the acceptable solution is obtained. To make the assignment in the cell (2, 3), delete the row & the column containing this cell so that no other assignment can be made in the second row and third column. Now, make the assignments in the usual manner as shown in the following table.
  • 41. Table He starts from city 1 and goes to city 4; from city 4 to city 2; from city 2 to city 3; from city 3 to city 5; from city 5 to city 1. Substituting values from original table: 4 + 7 + 6+ 4 + 5 = 26 hours. UNIT V REPLACEMENT OF ITEMS DETERIORATING WITH TIME 13.1 Introduction In any establishment, sooner or later equipment needs to be replaced, particularly when new equipment gives more efficient or economical service than the old one. In some cases, the old equipment might fail and work no more or is worn out. In such situations it needs more expenditure on its maintenance than before. The problem in such situation is to determine the best policy to be adopted with respect to replacement of the equipment. The replacement theory provides answer to this question in terms of optimal replacement period. Replacement theory deals with the analysis of materials and machines which deteriorate with time and fix the optimal time of their replacement so that total cost is the minimum. 13.2 Replacement Decisions The problem is to decide the best policy to adopt with regard to replacement. The need for replacement arises in a number of different following situations so that different types of decisions may have to be taken.  It may be necessary to decide whether to wait for a certain item to fail which might cause some loss or to replace earlier at the expense of higher cost of the item.  The item can be considered individually to decide whether to replace now or if not when to reconsider the item in question.  It is necessary to decide whether to replace by the same item or by a different type of item. 13.3 Types of Replacement Problems i) Replacement policy for items, efficiency of which declines gradually with time without change in money value.
  • 42. ii) Replacement policy for items, efficiency of which declines gradually with time but with change in money value. iii) Replacement policy of items breaking down suddenly a) Individual replacement policy b) Group replacement policy iv) Staff replacement In this lesson we confine ourselves to first two situations only 13.4 Replacement of Items that Deteriorate with Time There are certain items which deteriorate gradually with usage and such items decline in efficiency over a period of time. Generally, the maintenance cost of certain items always increase gradually with time and a stage comes when the maintenance cost becomes so large that it is better and economical to replace the item with a new one. There may be number of alternatives and we may have a comparison between various alternatives by considering the costs due to waste, scrap, loss of output, damage to equipment and safety risks etc. 13.4.1 Replacement of items whose maintenance cost increases with time and the value of money remains same during the period The following costs are considered in such decisions: C: Capital cost of a certain item say a machine s(t): The selling or scrap value of the item after t years f(t): Operating (or maintenance) cost of the item at time t n: Optimal replacement period of the item The operating cost function f(t) is assumed to be strictly positive. It may be continuous or discrete. 13.4.2 When t is a continuous variable Now the annual cost of the machine at time t is given by C � S(t) + f(t) and since the total maintenance cost incurred on the machine during n years is . Total cost T incurred on machine during n years is given by ��������������� Thus the average annual cost incurred on the machine per year during n years is given by ����������� To determine the value of optimal period (n), the principle of minima will be employed. ����������� ���������������
  • 43. ����������� Clearly ����������� Therefore, it can be seen that A(n) or TA = f(n) is a minimum for T provided that f(t) is non�decreasing and f(0) = 0. Hence, if time is measured continuously, then the average annual cost will be minimized by replacing the item when the average cost becomes equal to the current maintenance cost. 13.4.3 When time is a discrete variable Here the period of time is considered as fixed and n takes values 1,2,3, � then ����������� By using finite differences, A(n) will be a minimum for that value of n for which ��������������� or ����������� � For this, we write ����������� ����������� ����������� ����������� ����������� Similarly, it can be shown ����������� This suggests the replacement policy that if time is measured in discrete units, then the average annual cost will be minimized by replacing item when the next period�s maintenance cost become greater than the current average cost. Hence, replace the equipment at the end of n years, if the maintenance cost in the (n+1)th year is more than the average total cost in the (n)th year and the (n)th year�s maintenance cost is less than the previous year�s average total cost. The following examples will illustrate this methodology
  • 44. Example 1 A milk plant is considering replacement of a machine whose cost price is Rs. 12,200 and the scrap value Rs. 200. The running (maintenance and operating) costs in Rs. are found from experience to be as follows: Year: 1 2 3 4 5 6 7 8 Running Cost: 200 500 800 1200 1800 2500 3200 4000 When should the machine be replaced? Solution The computations can be summarized in the following tabular form: Table 13.1 Calculations for average cost of machine (In Rupees) Year (1) Running Cost (2) Cumulative Running Cost (3) Depreciation Cost (4) Total Cost TC (5) = (3) + (4) Average Cost (6) = (5)/(1) 1 200 200 12000 12200 12200 2 500 700 12000 12700 6350 3 800 1500 12000 13500 4500 4 1200 2700 12000 14700 3675 5 1800 4500 12000 16500 3300 6 2500 7000 12000 19000 3167 7 3200 10200 12000 22200 3171 8 4000 14200 12000 26200 3275 From the table it is noted that the average total cost per year, A(n) is minimum in the 6th year (Rs. 3167). Also the average cost in 7th year (Rs.3171) is more than the cost in 6th year. Hence the machine should be replaced after every 6 years. Example 2 A Machine owner finds from his past records that the maintenance costs per year of a machine whose purchase price is Rs. 8000 are as given below: Year: 1 2 3 4 5 6 7 8 Maintenance Cost: 1000 1300 1700 2200 2900 3800 4800 6000 Resale Price: 4000 2000 1200 600 500 400 400 400 Determine at which time it is profitable to replace the machine. Solution C = Rs. 8000. Table 13.2 shows the average cost per year during the life of machine. Here, The computations can be summarized in the following tabular form: Table 13.2 Calculations for average cost of machine
  • 45. Year f(t) Cumulative maintenance cost Scrap value Total cost 1 1000 1000 4000 5000 5000 2 1300 2300 2000 8300 4150 3 1700 4000 1200 10800 3600 4 2200 6200 600 13600 3400 5 2900 9100 500 16600 6 3800 12900 400 20500 3417 7 4800 17700 400 25300 3614 8 6000 23700 400 31300 3913 The above table shows that the value of TA during fifth year is minimum. Hence the machine should be replaced after every fifth year. Example 3 The cost of a machine is Rs. 6100 and its scrap value is only Rs.100. The maintenance costs are found to be Year: 1 2 3 4 5 6 7 8 Maintenance Cost (in Rs.): 100 250 400 600 900 1250 1600 2000 When should the Machine be replaced? Solution C = 6100, s(t) = 100 The computations can be summarized in the following tabular form: Table 13.3 Calculations for average cost of machine Replace at the end of year Cumulative maintenance cost Total cost 1 100 100 6100 6100 2 250 350 6350 3175 3 400 750 6750 2250 4 600 1350 7350 1737.50 5 900 2250 8250 1650 6 1250 3500 9500 7 1600 5100 11100 1585.7 8 2000 7100 13100 1637.50 It is now observed that the machine should be replaced at the end of sixth year otherwise the average cost per year will start to increase.
  • 46. 13.4.4 Replacement of items whose maintenance cost increases with time and the money value changes at a constant rate To understand this let us define the following terms: Money Value Since money has a value over time, therefore the explanation of the statement: �Money is worth 10% per year� can be given in two ways: (a) In one way, spending Rs.100 today would be equivalent to spend Rs.110 in year�s time. In other words if we plan to spend Rs.110 after a year from now, we could spend Rs.100 today and an investment which would be worth Rs.110 next year. (b) Alternatively if we borrow Rs.100 at the interest of 10% per year and spend Rs.100 today, we have to pay Rs.100 after one year (next year). Thus, we conclude that Rs.100 is equal to Rs.110 a year from now. Consequently Rs. 1 from a year now is equal to (1+0.1)-1 rupee today. Present Worth Factor As we have seen, a rupee a year from now will be equivalent to (1+0.1)-1 rupee today at the discount rate of 10% per year. So, one rupee in n years from now will be equal to (1+0.1)-n. Therefore, the quantity (1+0.1)-n is called the Present Worth Factor (PWF) or Present Value (PV) of one rupee spent in n years from now. In general, if r is the rate of interest, then (1+r)-n is called PWF or PV of one rupee spent in n years from now onwards. The expression (1+r)-n is known as compound amount factor of one rupee spent in n years. Discount Rate Let r be the rate of interest. Therefore present worth factor of unit amount to be spent after one year is . Then v is known as the discount rate. The optimum replacement policy for replacement of item where maintenance costs increase with time and money value changes with constant rate can be determined by following method: Suppose that the item (which may be machine or equipment) is available for use over a series of time periods of equal intervals (say one year). Let C = Purchase price of the item to be replaceds Rt = Running (or maintenance) cost in the tth year R = Rate of interest �is the present worth of a rupee to be spent in a year hence. Let the item be replaced at the end of every nth year. The year wise present worth of expenditure on the item in the successive cycles of n years can be calculated as follows: Year 1 2---- n n+1 n+2 ---- 2n 2n+1
  • 47. Present worth C+R1 R2v Rnvn-1 (C+R1)vn R2vn+1 Rnv2n-1 (C+R1)v2n Assuming that machines has no resale value at the time of replacement, the present worth of the machine in n years will be given by ��������������� Summing up the right-hand side, column-wise ����������� ��������������� ����������� , using sum of an infinite G.P. ����������� ����������� f(n) and f(n+1) given above at n = 0,1,2�, are called the weighted average cost of previous n years with weights 1,v, v2, ----vn-1respectively. P(n) is the amount of money required now to pay all the future costs of acquiring and operating the equipment when it is renewed every n years. However, if P (n) is less than P (n+1) then replacing the equipment each n year is preferable to replacing each n years is preferable to replacing each (n+1) years. Further, if the best policy is replacing every n years, then the two inequalities P (n+1) � P (n) > 0 and P (n-1) - P (n) < 0 must hold, without giving the proof we shall state the following two inequalities which holds good at n, the optimal replacement interval. ����������� ����������� As a result of these two inequalities, rules for minimizing costs may be stated as follows: 1. Do not replace if the operating cost of next period is less than the weighted average of previous costs.
  • 48. 2. Replace if the operating cost of the next period is greater than the weighted average of the previous costs. Working Procedure The step-by-step procedure for solving the problem is stated as under: 1. Write in a column the running/maintenance costs of machine or equipment for different years, Rn. 2. In the next column write the discount factor indicating the present value of a rupee received after (i-1) years, 3. The two column values are multiplied to get present value of the maintenance costs, i.e., . 4. These discounted maintenance costs are then cumulated to the ith year to get . 5. The cost of machine or equipment is added to the values obtained in Step 4 above to Obtain C+ . 6. The discount factors are then cumulated to get . 7. The total costs obtained in (Step 5) are divided by the corresponding value of the accumulated discount factor for each of the years. 8. Now compare the column of maintenance costs which is constantly increasing with the last column. Replace the machine in the latest year that the last column exceeds the column of maintenance costs. Example 4 A milk plant is offered an equipment A which is priced at Rs.60,000 and the costs of operation and maintenance are estimated to be Rs.10,000 for each of the first 5 years, increasing every year by Rs. 3000 per year in the sixth and subsequent years. If money carries the rate of interest 10% per annum what would the optimal replacement period? Solution Table 13.4 Determination of optimal replacement period At the end of year (n) Operating & maintenance cost Rn Discounted factor Discounted operation & maintenance cost Cumulative Discounted operation & maintenance cost Discounted total cost Cumulative discounted factor Weighted average annual cost (1) (2) (3) (4)=(2)x(3) (5) (6)=(5)+60000 (7) (8)=(6)+(7) 1 10000 1.0000 10000.00 10000.00 70000.00 1.00 70000.00 2 10000 0.9091 9091.00 19091.00 79091.00 1.91 41428.42
  • 49. 3 10000 0.8264 8264.00 27355.00 87355.00 2.74 31933.83 4 10000 0.7513 7513.00 34868.00 94868.00 3.49 27207.75 5 10000 0.6830 6830.00 41698.00 101698.00 4.17 24389.18 6 13000 0.6209 8071.70 49769.70 109769.70 4.79 22913.08 7 16000 0.5645 9032.00 58801.70 118801.70 5.36 22184.36 8 19000 0.5132 9750.80 68552.50 128552.50 5.87 21905.89 9 22000 0.4665 10263.00 78815.50 138815.50 6.33 21912.82 10 25000 0.4241 10602.50 89418.00 149418.00 6.76 22106.52 From Table 13.4 we find the weighted cost is minimum at the end of 8th year, hence the equipment should be replaced at the end of 8th year. Example 5 A Manufacturer is offered two machines A and B. Machine A is priced at Rs. 5000 and running cost is estimated at Rs. 800 for each of the first five years, increasing by Rs. 200 per year in the sixth and subsequent years. Machine B, with the same capacity as A, costs Rs. 2500, but has running cost of Rs. 1200 per year for six years, thereafter increasing by Rs. 200 per year. If money is worth 10% per year, which machine should be purchased? (Assume that the machines will eventually be sold for scrap at a negligible price). Solution Since money is worth 10% per year, therefore discount rate is Table 13.5 Computation of weighted average cost for machine A At the end of year (n) Operating & maintenance cost Rn Discounted factor Discounted operation & maintenance cost Cumulative Discounted operation & maintenance cost Discounted total cost Cumulative discounted factor Weighted average annual cost (1) (2) (3) (4)=(2)x(3) (5) (6)=(5)+ 5000 (7) (8)=(6)+(7) 1 800 1.0000 800 800 5800 1 5800 2 800 0.9091 727 1527 6527 1.9091 3419.035 3 800 0.8264 661 2188 7188 2.7355 2627.819 4 800 0.7513 601 2789 7789 3.4868 2233.98 5 800 0.6830 546 3336 8336 4.1698 1999.098 6 1000 0.6209 621 3957 8957 4.7907 1869.61 7 1200 0.5645 677 4634 9634 5.3552 1799.025 8 1400 0.5132 718 5353 10353 5.8684 1764.13 9 1600 0.4665 746 6099 11099 6.3349 1752.043 10 1800 0.4241 763 6862 11862 6.759 1755.053
  • 50. From table 13.5 we conclude that for machine A 1600<1752.043<1800. Since the running cost of 9th year is 1600and that of 10th year is 1800 and 1800>1752.043, it would be economical to replace machine A at the end of nine years. Table 13.6 Computation of weighted average cost for machine B At the end of year (n) Operating & maintenance cost Rn Discounted factor Discounted operation & maintenance cost Cumulative Discounted operation & maintenance cost Discounted total cost Cumulative discounted factor Weighted average annual cost (1) (2) (3) (4)=(2)x(3) (5) (6)=(5)+ 2500 (7) (8)=(6)+(7) 1 1200 1.0000 1200.00 1200.00 3700.00 1.00 3700.00 2 1200 0.9091 1090.92 2290.92 4790.92 1.91 2509.52 3 1200 0.8264 991.68 3282.60 5782.60 2.74 2113.91 4 1200 0.7513 901.56 4184.16 6684.16 3.49 1916.99 5 1200 0.6830 819.60 5003.76 7503.76 4.17 1799.55 6 1200 0.6209 745.08 5748.84 8248.84 4.79 1721.84 7 1400 0.5645 790.30 6539.14 9039.14 5.36 1687.92 8 1600 0.5132 821.12 7360.26 9860.26 5.87 1680.23 9 1800 0.4665 839.70 8199.96 10699.96 6.33 1689.05 10 2000 0.4241 848.20 9048.16 11548.16 6.76 1708.56 In table13.6 we find that 1800<1689<2300 so it is better to replace the machine B after 8th year. The equivalent yearly average discounted value of future costs is Rs. 1748.60 for machine A and it is 1680.23for machine B. Hence, it is more economical to buy machine B rather than machine A.