FAQ(Frequently Asked Questions )
on
Operations Research
(Quantitative Techniques for Managerial Decision Making)
by Mahesh P.Kumar
INTRODUCTION
A person who wants to be a successful manager must possess the art
and skills of rational and logical decision making. It ,however, has been
experienced that many managers suffer from what is known as decidophobia.
Such managers are not able to make the right type of decision at proper
time as they keep on postponing the decisions . Under such situation the
manager feels the need of certain techniques that can help him in the decision
making.Here quantitative techniques comes to the rescue of decision maker
. In fact operations research techniques create information base that helps
the management in decision making.
Q. WHAT IS OPERATIONS RESEARCH
Ans. Operations Research is collection
of various optimizing techniques.Operations research is a branch
of quantitative techniques ,which are concerned with creation of information
base that helps in decision making.
Operations Research is set of certain quantitative techniques like linear
programming, assignment technique, transportation technique, integer programming,
etc. All these techniques aim at optimizing i.e. maximizing the profits
or minimizing the cost. In operations research the practitioner first converts
the real life problem in a mathematical model and then a suitable technique
is applied to solve the problem . During the solution process various alternatives
are developed and finally decision is made. The decision making is concerned
with selection of best alternative among the various alternatives so developed.Infact
OR/MS generates information and provides logic that helps management in
decision making.
OR/MS involves rational approach . It does so by making optimum utilization
of the scarce resources. In management resources are classified as
six M's i.e. Men,Money,Material, Machines, Methods,and Markets.Operations
research involves the use of techniques that ensure the best utilization
of these resources.
OR/MS Techniques can be used to solve the problems faced during various
'operations'.In 1940's operations research techniques were basically applied
in defense operations .But today OR/MS Techniques are applied in many operations
belonging to business, commerce, industry, and also various areas of management
namely production, finance, personnel, and marketing .
OR/MS is a dynamic field of study.OR/MS methods are being improved
over the period of time.
Operations Research is a set of quantitative methods of scientific
nature in the sense that it involves the observation of a problem, then
defining it and converting it in a operations research model, solving it
manually or through software help and finally implementing it and ensuring
follow up.
The study and solution of problem through operations research requires
team work. It may involve help from statistician, mathematician, economist,
management expert and computer expert. In fact many softwares (like TORA,
LINDO, etc.) have been developed by various authors , publishers and software
development agencies. These software help the user to step through the
iterative computations and thereby saves time and effort.
The term is known as operational research in the UK, where as
is better known as operations research in US. however in any case
it is being abbreviated as OR. Some experts also name it as Management
Science ,which implies use of scientific methodology to solve the problems
of business management .In other words Management Science is concerned
with application of quantitative techniques to solve management problems.
To sum up the meaning ,OR/MS is collection of advanced quantitative/statistical
techniques that ensure better decision to management problems than the
solutions just based on past experience or manager's personal institution.
.Q. WHY TO STUDY * OR/MS*?
Ans. OR/MS provide our decision
making process a sense of direction. Decisions based on OR/MS techniques
aim at the following.
1)To make optimum utilization of the resources. We know that
every firm ,company
, or even country has limited amount of resources. OR/MS techniques
ensure effective utilization of the scarce resources
2)To minimize cost: OR/MS aim at minimization of costs. In the
broader sense minimization of costs includes minimization of time , wastage,
and manpower requirements.
3)To maximize profits: OR/MS techniques include maximization
of profits, and also maximization of efficiency and productivity.
4)To provide information base for decision making: OR/MS
is concerned with creation of quantitative information base for selection
of a particular course of action from a set of alternatives.
5)To provide solutions for management problems :OR/MS helps
management in determination of policies and actions scientifically.
Q. BRIEF HISTORY OF* OR/MS*
Ans. Operations
Research was a subject of discussion in the days of world war II
,When many countries involved were facing the problem of scarce resources,
available to them in limited quantity. At this point of time operations
research originated as a tool of military research.However when the war
was over the experts in this field were invited to deal with the problems
of limited resource utilization in the shattered industry.
Q.
MEANING OF LINEAR PROGRAMMING
Ans
We know the resources are limited but these have alternative uses.That is why a rational person is concerned with allocation of limited resources among alternative uses in such a way that he is able to minimize the cost of operation under his decision.This is ensured through optimum utilization of the resources.Infact the problems that maximization/minimization of an objective function,subject to certain structural constaints are known as optimizing problems.Since world war II,many optimizing techniques have been evolved and applied in the fields of economics, business management, commerce, trade ,and administration.The optimizing techniques help in solving a great deal of problems better known as programming problems.The programming problems involve allocation of limited resource among competing activities.Infact these problems deal with the situation when a number of resources(like Men, Money, Materials,Machines,Methods etc. )are available manufacture certain products.As the total amount of resources available is restricted/limited ,therefore the decision maker faces the problem of "How to allocate the scarce resource among various competing activities(say, products) such that cost of manufacture is minimized of profits resulting from them are maximized?".In other words there exist many feasible allocations within these constraints or restrictions,out of which optimum allocation is to be selected that maximizes profits or minimizes the cost.Linear Progamming is makes decisions under such situations.
Linear Programming is concerned with allocation of available limited resources among alternative uses thereof in such a manner that decision maker is able to maximize the profits or minimize the costs.
Q.MEANING OF TRANSPORTATION
Ans.
Transportation is one of the various techniques of operations research ,and is based on fundamental assumption of linear programming.
Transportation is concerned with movement of goods from one place to another place.In transportation we do not study modes of transportation rather we study few transportation situations ,such as transporting goods from plant/ factory to a particular warehouse,from warehouse to a retailer , or from retailer to a consumer.The place from where the goods start moving is called source/origin and the place where these are supposed to reach is called destination.
The transportation problem is to transport the goods that are stored at various origins to different destinations in such a manner that total supply is exhausted to meet the total demand , and at the same time total transportation cost is minimized.
Q. HOW A GENERAL TRANSPORTATION PROBLEM IS EXPRESSED.
Ans.A general transportation problem is expressed by way of a table as given below.
General Transportation Table(3 by3type)
.
|
D-I
|
D-II
|
D-III
|
Supply
|
O-I
|
C11(x11)
|
C12(x12)
|
C13(x13)
|
S1
|
O-II
|
C21(x21)
|
C22(x22)
|
C23(x23)
|
S2
|
O-III
|
C31(x31)
|
C32(x32)
|
C33(x33)
|
S3
|
Dem.
|
D1
|
D2
|
D3
|
TD/TS
|
EXPLANATION:
The model implies
C11 is cost of transporting goods from origin-I to destination-I
C12 is cost of transporting goods from origin-I to destination-II
C13 is cost of transporting goods from origin-I to destination-III
C21 is cost of transporting goods from origin-II to destination-I
C22 is cost of transporting goods from origin-II to destination-II
.....and so on.
S1 is supply of source -I
S2 is supply of source -II
S3 is supply of source -III
D1 is demand of destination-I
D2 is demand of destination-II
D3 is demand of destination-III
X11 is quantity to be allocated from origin-I to destination-I
X12 is quantity to be allocated from origin-I to destination-II
X 13 is quantity to be allocated from origin-I to destination-III
X21 is quantity to be allocated from origin-II to destination-I
X22 is quantity to be allocated from origin-II to destination-II
.... and so on.
Q.WHAT ARE VARIOUS * ASSUMPTIONS * OF TRANSPORTATION TECHNIQUE
Ans.
Transportation model is based on following set of assumptions:
§ Total supply is equal to total demand:
it is also known as balanced transportation problem.
§ Per unit cost of transportation from each source to each destination is known and certain.
§ The aim is to minimize total transportation cost for the organization as a whole and not for individual plant.
Q.Explain THE INITIAL BASIC FEASIBLE SOLUTION
Ans.
The initial feasible solution is obtained through certain methods, namely
NORTH WEST CORNER METHOD ,
LEAST COST ENTRY METHOD,
and
VOGEL’S APPROXIMATION METHOD.
Q. EXPAIN NORTH -WEST CORNER METHOD, by solving following problem.
A Transportation Problem:
.
|
D-I
|
D-II
|
D-III
|
Supply
|
O-I
|
7
|
5
|
2
|
25
|
O-II
|
3
|
4
|
1
|
75
|
O-III
|
9
|
8
|
6
|
100
|
Dem.
|
50
|
60
|
90
|
200
|
Ans.
Basic feasible solution of a transportation problem can be obtained through certain methods, north west corner method is one of these methods.Steps of solving a transportation problem through north -west corner methods along with the solution table are given below:
.
SOLUTION:
Transportation Table-I
(North West Corner Method)
.
| D-I
| D-II
| D-III
| Supply
|
| O-I | 7(25)| 5 | 2 | 25 |
| O-II | 3(25)4(50)| 1 | 75 |
| O-III | 9 | 8(10)6(90)| 100 |
| Dem. | 50 | 60 | 90 | 200
| | | | | | |
The above given transportation table is obtained through following procedure:
Step-1
Select the north west corner i.e. upper left corner of the transportation matrix.In the above given table cell(I,I) is towards the north west.Here corresponding supply is 25 units whereas corresponding demand is 50 units ,and we allocate 25 units i.e. the quantity which is minimum of the corresponding supply and corresponding demand.
Step-2
Adjust respective supply and demand numbers i.e. as we allocate 25 units in cell(I,I),we are left with demand balance of 25 units whereas whole of the supply is exhausted.
Step-3
(A) If the supply of the first row is exhausted we move downward the first cell in the second row and repeat the step-2 .
(B)If demand of the forst column is met we move horizontally to the next cell in the second column and repeat the step-2.
After allocating 25 units in cell(1,1)
we allocate 25 units in cell(2,1) ,
then we allocate 50 units in cell(2,2) ,
then we allocate 10 units in cell(3,2) ,
finally we allocate 90 units in cell(3,3)
Step-4
Continue the procedure till total supply is used to meet the total demand.
Total transportation cost= 1070 (175+75+200+80+540)
Q.Solve the above given problem through LEAST COST ENTRY METHOD
Ans.
Transportation Table-I
(Least Cost Entry Method)
.
| D-I
| D-II
| D-III
| Supply
|
| O-I | 7 | 5(10)2(15)| 25 |
| O-II | 3 | 4 | 1(75)| 75 |
| O-III | 9(50)8(50)| 6 | 100 |
| Dem. | 50 | 60 | 90 | 200
| | | | | |
Step1:Select the cell with lowest per unit transportation cost,among all the rows and columns of the table(if minimum cost is not uniquei.e. two or more points have the same minimum cost then the cell with lowestallocation is preferred.).In the above table cell(2,3) with costof Re.1 is selected .
Step2:Allocate as many units as possible in cell determinedin step 1. As corresponding supply is 75 whereas demand is 90 ,therefore75 units are allocated in this cell. Here Supply balance of O-II is zero,therefore second row is deleted.
Step3:Repeat step 1 and step 2 in the reduced table. The nextlowest cost is 2 in the cell(1,3) ,where allocation of 15 units is made.Afterthis demand of destination -II becomes zero that is second column is deleted.The next lowest cost is Rs.5 which lies in cell( 2,1) where 10 units areallocated. After this row one is delete because dsupply of plant1 is fully over. Next lowest cost (8) is allocated with 50 units and finally50 units are allocated in cell(3,1) having cost of Rs.9
In the above table total transportation cost as per least cost method is calculated to be Rs1005 i.e
.Rs.50 + 30 + 75 + 450 + 400
Q.EXPLAIN THE VOGELS APPROXIMATION METHOD,withthe same example.
Ans.Solution of the previously given transportation problem as obtained through Vogels Approximation Methodis given below:
Transportation Table-I
(Vogel's Approximation Method)
.
| D-I
| D-II
| D-III
| Supply
|
| O-I | 7 | 5 | 2(25)| 25 |
| O-II | 3(50)| 4 | 1(25)| 75 |
| O-III | 9 | 8(60)6(40)| 100 |
| Dem. | 50 | 60 | 90 | 200
| | | | | |
Steps involved in VAM are explained as follows:
Step1:determine penalties in each row and column
(penalty is defined as difference between least cost and next leastcost )
Step2: determine the highest penalty among them and encircleit.
Here MAXIMUM penalty is 4 in first column. In this column least costis 3 where we allocate 50 units (it is lower value out of supply (75) anddemand(50)
Notice; if there is tie among highest penalty then cell with lowerallocation is preferred.
Step 3:Reduce the supply or demand by the amount allocated tothe cell. Delete the row or column where balance comes to be zero.In theabove column third column is deleted.
Step4 : determine penalties for the reduced matrix.In the aboveproblem 3 is maximum penalty in the second round.Here we make allocationof 25 units in cell(2,3). After this row two is deleted.
In the third round maximum penalty is 4 in the third column.Here wemake allocation of 25 units in cell (1,3).Now supplyof first row is over ,therefore first row is deleted.
Now we are left with only two cost cells i.e.(3,2) and (3,3) ,herewe make allocation of 40 units in cell(3,3) and finally allocation of 60units in cell(3,2).
Step5 calculate total transportation cost
TOTAL TRANSPORTATION COST= 945 (50+150+25+480+240)
thus in the given problem
METHOD------------------------------TOTAL TRANSPORTATION COST
NORTH-WEST CORNER 1070
LEAST COST ENTRY 1050
VOGEL'S APPROXIMATION * 945*(BEST METHOD!)
thus Vogel's Method gives minimum cost Basic Feasible Solution.Q.EXPLAIN WHEN A TRANSPORTATION PROBLEM HAS A DEGENERATESOLUTION
Ans.
Degeneracy puts an obstacle in computation process. A problem is saidto be not degenerate when two conditions are met
(a) required number of allocations =m+n-1. If actually there are lessthan the required allocations then the solution is said to be degenerate. In such situation we introduce dummy allocation (s).Dummy allocation isrepresented by the symbol Ê (epsilon) .Ultimately Ê is assigned 0 value i.e. Ê tends to zero(0) .Dummy allocation isintroduced in LEAST COST EMPTY CELL,(it however should not resultinto loop .if it results into closed loop then we shift to NEXT LEASTCOST EMPTY CELL.
(b) Allocations must be independent:
Q. What is meant by UNBALANCED TRANSPORTATION PROBLEM?
Ans.when total supply is not equal to total demand
the problem is said to be unbalanced.
here we may face two situations:
(1)When total Supply exceeds the total Demand:
in this case we add a dummy column with zero cost elements .This column
has demand quantity
equal to the difference between supply and demand.
For Example:
|
D-I |
D-II |
D-III |
supply |
| O-I |
2 |
5 |
4 |
500 |
| O-II |
7 |
3 |
2 |
300 |
| demand |
200 |
100 |
150 |
|
Before solving this problem it is balanced as
|
D-I |
D-II |
D-III |
Dummy |
Supply |
| O-I |
2 |
5 |
4 |
0 |
500 |
| O-II |
7 |
3 |
2 |
0 |
300 |
| demand |
200 |
100 |
150 |
250 |
800 |
(1)When total demand exceeds the total supply:
in this case we add a dummy row with zero cost elements .This row has
supply quantity
equal to the difference between demand and supply.
For Example:
| . |
D-I |
D-II |
D-III |
SUPPLY |
| O-I |
6 |
5 |
3 |
400 |
| O-II |
1 |
7 |
2 |
300 |
| Demand |
500 |
300 |
175 |
|
Before solving this problem it is balanced as
|
D-I |
D-II |
D-III |
supply |
| O-I |
6 |
5 |
3 |
400 |
| O-II |
1 |
7 |
2 |
300 |
| Dummy |
0 |
0 |
0 |
275 |
| demand |
500 |
300 |
175 |
975 |
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Mahesh P.Kumar
[email protected]
b-24/114,KISHAN NAGAR,NEAR LOWER MALL
PATIALA-147001
PUNJAB
INDIA