Case Studies - Operations Research and Management [PDF]

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Operations Research and Management Case Study Q 1. Five men are available to do five different jobs. From past records, the time (in hours) that each man takes to do a job is known and is given in the following matrix. Men Jobs A B C D E

I

II

III

IV

V

2 6 4 4 5

9 8 6 2 3

2 7 5 7 9

7 6 3 3 5

1 1 1 1 1

Find the assignment of men to jobs that will minimise the total time taken. Solution: Check 1. whether minimisation problem—Yes 2. whether square matrix—Yes Then proceed to apply Hungarian Assignment Method to solve it. Step 1. Identify the smallest element of each row and subtract it from all elements of that row. Revised matrix is as given below. (Row minima for row A, B, C, D, E = 1) Men/Jobs

I

II

III

IV

V

A B C D E

1 5 3 3 4

8 7 5 1 2

1 6 4 6 8

6 5 2 2 4

0 0 0 0 0

Revised matrix I

Step 2. Identify the smallest element of each column and subtract it from all the elements of the column. Column minimal are Column I ----------> 1 Column II ----------> 2 Column III ----------> 3 Column V ----------> 4 Column VI ----------> 5

Hence, the reduced matrix works out to: Men/Jobs

I

II

III

IV

V

A B C D E

0 4 2 2 3

7 6 4 0 1

0 5 3 5 7

4 3 0 0 2

0 0 0 0 0

Revised matrix II

Step 3. Draw minimum number of horizontal and vertical lines to cover all zeros. Men/Jobs

I

II

III

IV

V

A B C D E

0 4 2 2 3

7 6 4 0 1

0 5 3 5 7

4 3 0 0 2

0 0 0 0 0

Since we have been able to over all zeros with 4 lines, which is less than the number of rows/columns i.e. 5, the solution is not optimal. Hence, proceed to step 4.

Step 4. We identify the smallest uncovered element as 2. Subtracting 2 from all uncovered elements and adding it to the elements at the point of intersection of lines, we get the revised matrix as Men/Jobs

I

II

III

IV

V

A B C D E

0 2 0 0 1

9 6 4 0 1

0 3 1 3 5

6 3 0 0 2

2 0 0 0 0

Revised matrix III

Step 5. Repeating step 3, lines drawn are as follows Men/Jobs

I

II

III

IV

V

A B C D E

0 2 0 0 1

9 6 4 0 1

0 3 1 3 5

6 3 0 0 2

2 0 0 0 0

Revised matrix IV

Since number of lines drawn are again 4 1 number of rows/columns, repeat step 4. Minimum uncovered element is 1. Subtracting 1 from all uncovered elements and adding it to the elements at point of intersection of lines, revised matrix will be

A B C D E

I

II

III

IV

V

0 1 0 0 0

9 5 4 0 0

0 2 1 3 4

6 2 0 0 3

3 0 1 1 0

Revised matrix V

Repeating step 3, number of lines drawn are 5 = no. of rows/columns. Hence, we arrive at the optimal solution.Proceed to step 6. Step 6. Making assignments on zero elements I II III IV V A Φ 9 B 1 5 C Φ 4 D Φ 0 E 0 Φ

0 2 1 3 4

6 2 0 Φ 3

3 0 1 1 Φ

Final matrix for assignments

Hence, minimum total time = 2 + 1+ 3 + 2 + 5 = 13 units (hours). This Is a case of multiple assignments, where assignment A III and B V are unique. But even after cancelling zeros at A 1 and E V, we are left with 3 zeros in column I, 2 in column II and 2 in column IV. Thus if we assign C IV, the zeros at C I and D IV get eliminated. Even now, we have 2 zeros in column I and II. These can be assigned alternatively as E I, D II or D I, E II, etc.

Q 2 A marketing manager has five salesmen and five sales districts. Considering the capabilities of the salesmen and nature of districts, the marketing manager estimates that sales per month (in hundred rupees) for each salesman in each district would be as follows— Salesmen

1 2 3 4 5

Districts A

B

C

D

E

32 40 41 22 29

38 24 27 38 33

40 28 33 41 40

28 21 30 36 35

40 36 37 36 39

Solution: Check 1. Is it the balanced problem—Yes. 2. Is it the minimization problem — No Then we convert it to standard minimisation problem by subtracting all the elements from the highest value element of the matrix i.e., 41. The minimisation matrix is thus as given below: 9 3 1 13 1 1 17 13 20 5 0 14 8 11 4 Revised matrix (minimisation type) 19 3 0 5 5 12 8 1 6 2 Using HAM for solving the revised matrix, we proceed as follows. Step 1. Row minima being subtracted from all elements of the row, the revised matrix will become 8 2 0 12 0 0 16 12 19 4 0 14 8 11 4 Matrix I 19 3 0 5 5 11 7 0 5 1 Step 2. Selecting column minima and subtracting it from all the elements of that column in the matrix, it is converted to, 8 0 0 7 0 0 14 12 14 4 0 12 8 6 4 Matrix II 19 1 0 0 5 11 5 0 0 1

Step 3. Drawing minimum number of horizontal and vertical lines across all zeros 8 0 0 7 0 0 14 12 14 4 0 12 8 6 4 19 1 0 0 5 11 5 0 0 1 Since there are only 4 lines (less than the number of rows/columns i.e., 5), the solution is not optimal. Proceed to step 4. Step 4. Identifying minimum uncovered element (4), subtracting it from all the uncovered elements and adding the same at the point of intersection of the lines, the new matrix works out to be 12 0 0 7 0 0 10 8 10 0 0 8 4 2 0 23 1 0 0 5 15 5 0 0 1 Step 5. Repeating step 3, we get 12 0 0 7 0 0 10 8 10 0 0 8 4 2 0 23 1 0 0 5 15 5 0 0 1 Here, number of lines = number of rows/columns. Hence, solution is optimal. Proceed to step 6. Step 6. Assigning on zero elements A B C D E A B C D E 1 12 Φ 7 Φ 1 12 Φ 7 Φ 0 0 2 Φ 10 8 10 or 2 10 8 10 Φ 0 0 3 8 4 2 Φ 3 Φ 8 4 2 0 0 4 23 1 Φ 5 4 23 1 Φ 5 0 0 5 15 5 Φ 1 5 15 5 Φ 1 0 0 The assignment of salesmen to districts are multiple solutions. 1→B 1→B 1→B 2→E Or, 2→A Or, 2→A 3→A 3→E 3→E 4→C 4→C 4→D 5→D 5→D 5→C

Picking the values of sales from the original matrix, Maximum sales = 38 + 36 + 41 + 41 + 35 = 191 in hundred rupees for ail the assignments. Thus, there are multiple solution for the problem, the sales remaining as Rs 19,100.

Q 3. Find the sequence that minimizes the total elapsed time to complete the following tasks (clearly state any algorithm that you may use) and calculate the minimum elapsed time. Time on Machine 1 Time on Machine 2 Time on Machine 3

3 4 6

8 3 7

7 2 5

4 5 11

9 1 5

8 4 6

7 3 12

Jobs

A

B

C

D

E

F

G

Solution: We use the sequencing technique here: Bellman-Johnson Algorithm. Given table:

Machines

A

B

C

Jobs D

M1 M2 M3

3 4 6

8 3 7

7 2 5

4 5 11

E

F

G

9 1 5

8 4 6

7 3 12

As per the above table of processing time, Minimum ti1 = 3, Maximum ti2 = 5, Minimum ti3 = 5 Hence, Min ti3 = max ti2 = 5, hence solution Is possible. Now we should add the processing time of different jobs on machines A and B and machines B and C respectively as follows: Processing Time Job Gi (=Ai + bi) Hi (= Bi + Ci) A 3+4 = 7 4+6 = 10 B 8+3 = 11 3+7 = 10 C 7+2 = 9 2+5 = 7 D 4+5 = 9 5+11 = 16 E 9+1 = 10 1+5 = 6 F 8+4 = 12 4+6 = 10 G 7+3 = 10 3+12 = 15

The minimum elapsed time among the two combinations of machines is 6 hours for Job 6 in the 2nd combination. Hence Job E will be processed last. Next minimum elapsed time is 7 hours for Job A on machine 1 (1st combination) and for Job C In 2nd combination. Hence Job A will be processed first and Job C will be scheduled at the latest remaining position i.e., position 2 before Job E. Thus, while Job E is the last job to be scheduled, Job C will be scheduled Just prior to it. Next minimum elasped time is 9 hours for Job D in Combination 1, hence Job D will be processed Immediately after Job A. Next minimum elapsed time is 10 hours for Job G in combination 1 and for Job F and Job B in 2nd combination. Hence, Job G will be processed after Job D followed by either Job F or Job B. We thus get the following sequences: A D G B F C E OR A D G F B C F For determination of total elapsed time, a table can be drawn, Case1: A → D → G → B → F → C → E Jobs M1 M2 M3 A 0 →3 3 →7 7 →13 D 3 →7 7 →12 13 →24 G 7 →14 14 →17 24 →36 B 14 →22 22 →25 36 →43 F 22 →30 30 →34 43 →49 C 30 →37 37 →39 49 →54 E 37 →46 46 →47 54 →59 Total elapsed time = 59 hours. Case2: A → D → G → B → F → C → E Jobs M1 M2 M3 A 0 →3 3 →7 7 →13 D 3 →7 7 →12 13 →24 G 7 →14 14 →17 24 →36 F 14 →22 22 →26 36 →42 B 22 →30 30 →33 42 →49 C 30 →37 37 →39 49 →54 E 37 →46 46 →47 54 →59 In the second sequencing also, the total elapsed time works out to 59 hours only. Hence, any of the sequencing can be adopted.

Q 4. A bread vendor buys every morning loaves of bread at *0.45 each by placing his order one day In advance ( at the time of receiving his previous order) and sells them at ?0.70 each. Unsold bread can be sold the next day at t 0.20 per loaf and thereafter should be treated as of no value. The pattern of demand for bread is given below: Fresh Bread Daily Sales

One Day Old Bread of Daily Sales

Probability Probability of Demand Demand 50 0.01 0 0.70 51 0.03 1 0.20 52 0.04 2 0.08 53 0,07 3 0.02 54 0.09 55 0.11 56 0.15 57 0.21 58 0,18 59 0.09 60 0.02 The vendor adopts the following order rule. If there is no stock with him at the end of the previous day, he orders 60 units. Otherwise he orders 50 or 55 whichever is nearest the actual fresh bread sale on the previous day. Starting with zero stock and a pending order for 55 loaves* simulate for 10 days and calculate the vendor's profits. FRESH BREAD ONE DAY OLD BREAD Daily Probality Cumulative Random Daily Probability Cumulative Random Sales of demand Probability numbers Sales of demand Probability numbers of demand allocated of demand 50 0.01 0.01 00 0 0.70 0.70 00 to 69 51 0.03 0.04 01 - 03 1 0.20 0.90 70 to 89 52 0.04 0.08 04 - 07 2 0.08 0.98 90 to 97 53 0.07 0.15 08 - 14 3 0.02 1.00 98 & 99 54 0.09 0.24 15 - 23 55 0.11 0.35 24 - 34 56 0.15 0.50 35 - 49 57 0.21 0.71 50 - 70 58 0.18 0.89 71 - 88 59 0.09 0.98 60 0.02 1.00 We can construct a table to see through simulation how the stocks and sales fluctuate.

Day

1 2 3 4 5 6 7 8 9 10 Total

FRESH BREAD Receipt Random Sale at the numbers start of day 55 72 55 58 * 60 06 52 60 12 53 50 74 50 58 * 55 79 55 58 * 60 70 57 60 85 58 55 71 55 58 55 21 54 60 98 60 570 549

Closing Stock

0 8 7 0 0 3 2 0 1 0

ONE DAY OLD BREAD Order for Opening Random Sale next day stock No.

60 @ 60 50 55 60 60 55 55 60 55

0 0 8 7 0 0 3 2 0 1 21

86 84

88 58 48

0 0 1 0 0 0 1 0 0 0 2

* Represents lost sales in stock Is limited. @ Previous day's closing stock is zero Estimated profit -(549 x0.70+ 2 x 0.20)-570 x 0.45 -Rs. 128.20

Q 5. Explain the scope & structure of an OR (Operation Research) model. Give an example and structure of each of following types of problems in real life: a. Linear Programming problem b. Transportation Problem c. Assignment problem, d. Game Theory problem, e. Sequencing problem f. Simulation problem. Finally give five problems in an organization wherein the OR tools cannot be applied. Solution: Operations Research is the application of the methods of mathematical science to complex problems arising in the direction and management of large systems of men, machines, materials and money in industry, business, government and defense. The distinctive approach is to develop a scientific model of systems incorporating measurements of factors such as chance and risk with which to predict and compare the outcomes of alternative decisions, strategies or controls. The purpose is to help management determine its policy and actions scientifically. It takes into account all significant factors and evaluates them as a whole, ie; OR considers the holistic approach;

because it takes care of and validates the claims of various sub-parts of an organisation. It can be called interdisciplinary approach also for the same reason. An O.R. model is defined as representation or abstraction of an actual object or situation. It shows the relationships (direct and indirect) and the interrelationships of action and reaction in terms of cause and effect. Since a model is an abstraction of reality, it may appear to be less complex than reality itself. The model, to be complete, must be representative of those aspects of reality that are being investigated. There are large number of models used in OR. Some of the basic types are described below: Physical Models :- When we utilize all forms of drawings, sketches, diagrams, groups or charts to describe, a situation or problem specific in nature, we term them as physical models. Since relationships of the parameters are depicted in the pictorial form, it becomes easier to comprehend the problem and facilitates its analysis. There are two types of physical models:a) Iconic Models:- An icon is the depiction of an object as its image or likeness. It is a useful tool but the application in management problem areas is restrictive and narrow. Simulation techniques fall under this category, where full scale product need not be manufactured, but experiments can be conducted on miniature fully functional models under simulated conditions. b) Analog Models:- These Models are similar to Iconic Models but not the exact replica of the actual system. These are small physical systems having similar characteristics such as children toys, model cars and rail line etc. The aim is to visualize how the system should look like and how routine procedures can be demonstrated as Building Models. Symbolic Models:These models are used to represent actual problems. There are two types of symbolic models: a) Verbal Models:- When we represent the problem and parameter interrelationships written or spoken in words, these are called Verbal Models. b) Mathematical Models:- When we represent the problem and its parameters by a set of mathem; expressions, these are called Mathematical models. Though these are abstract, but have definite and precise manipulation possible under the laws of mathematics. Heuristic Models:These models use intuititive rules or guidelines to solve a particular problem. These models are not basetl on any definite mathematical expressions or relationships, but problem solving based on past experience or approach formulated on the basis of definite stepped procedure. Application and Scope of Operations Research:Some of the areas of management where techniques of Operations Research are applied are listed below:(c) Marketing and Sales :. Advertising and media planning 2. Product selection, timing and competitive strategies 3. Recruitment of salesman 4. Forecasting and decision trends 5. Pricing and competitive decisions 6. Market research decisions (c) Finance and Accounting : 1. Cash flow planning 2. Credit policy analysis 3. Investment policy for maximum return 4. Dividend policies 5. Portfolio analysis 6. Profit planning

(c) Personnel: 1. Selection of personnel, mixes of age and skills 2. Recruitment policies and assignment of jobs 3. Manpower planning, wage administration 4. -Scheduling of training programmes. (c) Construction: 1. Allocation of resources to projects 2. Determination of proper workforc 3. Deployment of workforce 4. Project scheduling, monitoring and control. (c) Facilities planning: 1. Factory size location decision 2. Location and size of warehouse, distribute centres and retail outlets 3. Transportation, loading and unloading 4. Logistics system design, layout. (c) Manufacturing: 1. Employment, training, layoffs, quality control 2. Aggregate production plannit* assembly line, blending, purchasing 3. Production scheduling, production smoothing 4. Inventory control. (c) Purchasing and Procurement: 1. Bidding policies 2. Vendor analysis, replacement policies 3. Optimal buying and reordering 4. Materials transfer. (c) Maintenance and Project scheduling: 1. Preventive maintenance and maintenance 2. Project scheduling and allocation of resources 3. Maintenance crew size and scheduling 4. management and strategic planning. (c) Research and Development:- 1. Control of R and D projects 2. Product introduction Organizational design and control 4. Decision support system and MIS. Examples for the following types of problems a) Linear programming problem: Choice of investment from a variety of shares and debentures so as to maximize return on investment; to find the number of items of each type that should be manufactured so as to maximize the profit subject to production restrictions imposed by limitations on the use of machinery and labour; to determine the minimum requirement of nutrients subject to availability of foods and their prices. b) Transportation problem: To find the least expensive way of transporting shipments from the warehouses to customers. c) Assignment problem: To assign job to workers for maximum effectiveness and optimum results subject to restrictions of wages and other costs. d) Game theory problem: To provide a rational course of action in business, military operations wherein a competitive situation exists when two or more opposing parties are making decisions involving conflicting interests and the action of one depends on the action which the opponent takes. e) Sequencing problem: For patients in hospital or cars coming in for repairs in a workshop etc. when there is a need to prioritise tasks on certain number of facilities in order to reduce idle time by making effective use of available facilities and obtaining higher productivity. f) Simulation problem: In case of capital budgeting, evaluation of market demand, selling price, market share, investment requirement, growth rates or operating cost levels can be worked out by simulation.

Non - applicability of OR tools in an organization:i) Operations research uses quantitative techniques, where we require mathematical models. These models very often simplify the reality by using several assumptions. Thus they fail to represent the actual degree of complexity present in a given situation. ii) Models cannot incorporate factors which cannot be quantified. Very often these are human factors, like a worker's skill, customers satisfaction, ability of a manager. These are important in decision making process but the mathematical model representing the given situation fails to give enough weightage to them. iii) Quantitative techniques are costly to use. Hence, if a manager wants to use a model only once to solve a problem, it is very expensive to build the model. A small firm cannot afford the expenses of running a model* a computer as computer time is also expensive. iv) There is also a danger of model builder overestimating the ability of the model he has built and lose sight of the reality. The model then is not representative of the real situation and hence not workable. i) Very often the person who uses the model and the one who builds the model are different. The user i.e. the manager is not trained adequately in the interpretation and use of the results