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OPERATION RESEARCH &

DECISION MODELS

Prof. Srinivasan Kidambi

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Table of Contents

Topic No. ofProblems

PageNo.

Assignment 16 3 – 9

Transportation 17 10 – 18

Decision Trees 11 19 – 23

Statistical Decision Theory 15 24 – 28

Replacement 8 29 – 31

Game Theory 5 32 – 38

Inventory Management 19 39-42

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Assignment

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Assignment ModelAlgorithm

Step – 0 Verify whether the given problem is maximization or minimization, if it is maximization convert in to minimization by selecting the largest element in the matrix and subtracting other elements from the largest element

Step – 1 Verify whether No. of rows = no of columns, if not introduce a dummy row or column with zero elements

Step – 2 Subtract row minimum Step – 3 Subtract column minimum and assign zeros

Note – 1 – If number of assignments = number of rows, optimal assignment is made , if not proceed further Note – 2 – Assignment of zeros refer procedure A

Step – 4 Draw minimum number of lines to cover all zeros (refer procedure B ) , Always the number of lines should be equal to number of assignments

Step – 5 i - Select the smallest uncovered elementsii - Subtract the smallest uncovered element from the uncovered element iii – add the smallest uncovered element , to the element at the intersection of two lines iv – If the element is covered by a single line write as it is v – Assign the zeros as per procedure A , if optimal assignment is not made at this stage also repeat the procedure from Step no – 4

Procedure A – Assigning Zeros –Step – 1 (A)- if there is only one zero in a row assign that zero and delete the

corresponding zeros in that column (B)- If you find more than one zero in a row , skip that row and go to the next row(C)- After completing row wise assignment , go through column wise

Step – 2 (A) – if there is only one zero in a column assign that zero and delete the corresponding zeros in that row , (B) – if you find more than one zero in a column , skip that column and go to next column for assignment (C) – After completing column wise assignment , if any zeros are left out , repeat the above procedure till all the zeros are assigned

Procedure – B for drawing min number of lines

Step – 1 Select the row or column having an assigned zero with max number of deleted zeros , draw a line through assigned zero covering all the elements , corresponding to that row or column

Step – 2 Select again the assigned zero with next maximum number of deleted zeros, and draw the line covering all the elements , If only assigned zero is left out with no deleted zeros then a line can be drawn through assigned zero covering all the elements either horizontally or vertically

Step – 3 Repeat the above procedure till all zeros are covered by line

Note – verify whether number of lines = number of assignments

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REPLACEMENT

The problem of replacement is to decide when an item should be replaced economically by another of the same type or by a different one

The situation under which replacement of certain item is to be done are

Old item has failed and does not work at all or the old item is expected to fail shortly

The old item has deteriorated and requires expensive maintenance. A better design of equipment has been developed

The failure of an item can be classified under two categories

Gradual failure – all mechanical items will come under this category

Sudden failure – all electrical and electronic item will come under this category

The items that fail may be broadly classified into

Items that Deteriorate with time ( gradual failure )1. Money value is not considered 2. Money value is considered

Items that fail suddenly ( Sudden failure )

Procedure – Type – 1 Replacement of an item that deteriorate with time , Money value is not considered

Step – 0 Prepare the table as per the format given below and write down the year , maintenance cost , resale value in the respective column of the table

Year Maint . cost Cum.Maint cost Resale value

C-S Totalcost

Avg.Cost

Where C is the cost of equipment and S is the resale value /Scrap value

Total cost = ( C- S ) + Cum. Maint. Cost

Average cost = Total Cost / No. of years Step – 1 Find the cumulative maintenance cost , C-S ,total cost and

average cost

Step – 2 Select the minimum average cost and the corresponding period , replace the equipment at the end of this period

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Procedure – Type – 2 Replacement of an item that deteriorate with time , Money value is considered

Money Value

It is assumed that the maintenance cost increases with time and each cost is to be paid just at the start of the period

Let the money carry a rate of interest r per year . Thus a rupee invested now will be worth

( 1+r ) after a year (1+r)^2 after two years and so on

In this way , a rupee invested today will be worth (1+r)^n n years hence

Or in other words if we have to make a payment of one rupee in n years time , it is equivalent to make a payment of ( 1+r )^ -n rupee today

The quantity (1+r)^ -n is called the present worth factor (pwf)

Assumptions

a) The equipment has no salvage value b) The maintenance cost has incurred at the beginning of the different

time periods

ProcedureStep – 0 Prepare the table as per the format given below and write down the year, running cost ( maintenance cost) in the respective column of the table

Year Maint . cost

pwf Discounted Maint.cost

Cum disc Main cost

Total cost

Cum. pwf Weight . Avg cost / yr

Step – 1 Find the pwf for the various years if money is worth r % Note – 1 Always take first year pwf as 1 Note – 2 If money is worth 10 % then pwf is

1/1.1 = .9091 for the second year . Use the calculator for finding out the

pwf s for the various years

Step – 2 Find the discounted maintenance cost

Discounted maintenance cost = pwf * corresponding running cost

Step – 3 Find the cum.discounted maintenance cost , total cost for the various periods

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Total cost = cost of equipment + cum . disc . maint cost ,

Step – 4 Find the weighted avg cost per year

Weighted avg cost = total cost / cum pwf

Step – 5 when the weighted avg cost is less than the running cost of the subsequent period ,replace the equipment at the end of that period

Procedure – Type – 3Replacement of an item that fail completely sudden failure

Instead of gradual deterioration some item fail all of a sudden . All electrical and electronic items will come under this category , the failure of the item may result in complete break down of the system . The loss due to this break down is indirect . to minimize this the indirect cost such as loss in production , idle labour etc, the time of such failure is to be predicted . If the time of failure can be predicted , preventive replacement will be appropriate course of action

There arises a need of replacement policy of sudden failure items to avoid the possibility of complete breakdown . Two types of replacement policies are considered

- Individual replacement policy ( IRP)Under this policy an item is replaced immediately after its failure

- Group replacement policy (GRP)Replace as and when the item fails during the optimal period and also replace all the items including the good ones at the end of the optimal period

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Replacement- Problems

1. A truck owner from his past experience estimates that the maintenance cost per year of a truck whose purchase price is Rs. 1,50,000 and the resolve valve of truck will be as follows:

Year 1 2 3 4 5 6 7 8Maintenance(in Rs.) M(t):

Resale Value(in Rs.) S(t):

10,000

1,30,000

15,000

1,20,000

20,000

1,15,000

25,000

1,05,000

30,000

90,000

40,000

75,000

45,000

60,000

50,000

50,000

Determine at which time it is profitable to replace the truck

Ans: Replace at the end of 4th year Rs. 37, 500/-

2. A Taxi owner estimates from his past records that the cost per year for operating a taxi whose purchase price when new is Rs. 60,000 are as given below:

Age 1 2 3 4 5Operating Cost 10,000 12,000 15,000 18,000 20,000

After 5 years, the operating cost = 6,000 K. where k = 6, 7, 8, 9, 10 (K denoting age in years). If the resale value decreases by 10% of purchase price each year, what is the best replacement policy? Cost of money is zero.

Ans: Taxi should be replaced at the end of each year Rs. 16,000.

3. A plant manager is considering replacement policy for a new machine. He estimates the following cost (all costs in rupees),

Year 1 2 3 4 5 6Replacement Cost at Beginning of year:

100 110 125 140 160 190

Salvage valve at End of year:

60 50 40 25 10 0

Operating Costs: 25 30 40 50 65 80

Find an optimum replacement policy and corresponding minimum cost.

Ans: Replace at the end of 2nd year Rs. 56.5

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4. The initial cost of machine is Rs. 30,000 and running or operating expenditure which increases with age of the machine is given below:

Year 1 2 3 4 5 6 7Running cost (Rs.):

5,000 6,000 8,000 10,000 13,000 16,000 20,000

What is the replacement policy when this machine should be replaced? It is given that the rate of interest is 10% and scrap value is nil?

Ans: Replace at the end of 5th year, Rs. 15, 219.70/-

5. An engineering company is offered two types of material handling equipments A and B. A is priced at Rs. 60,000 including cost of installation and the cost for operation and maintenance are estimated to be Rs. 10,000 for each of the first five years, increasing every year by Rs. 3000 per year in the sixth and subsequent years. Equipment B with rated capacity same as A, requires an initial investment of Rs. 30, 000 but in terms of operation and maintenance costs more than A. These costs for B are estimated to be Rs.13,000 per year for the first six years, increasing every year by Rs. 4000 per year from seventh year onwards. The company expects a return of 10% on all its investments. Neglecting the scrap value of the equipment at the end of its economic life, determine which equipments the company should buy.

Ans: Replace machine A @ the end of 8th year, Rs. 19, 311 Replace machine B @ the end of 6th year, Rs. 16, 175

6. A Company has the option to buy one of the mini computers; MINICOMP and CHIPCOMP, MINICOMP costs Rs. 5 lakhs, and running and maintenance costs are Rs. 60,000 for each of the five years, increasing by Rs. 20, 000 per year in the 6th and subsequent years chip comp has the same capacity as MINICOMP, but costs only Rs. 2,50,000. However, its running and maintenance costs are Rs. 1, 20, 000 per year in the first five years, and increase by Rs. 20,000 per year thereafter. If the money is worth 10% per year, which computer should be purchased? What are the optimum replacement periods for each one of the computers? Assume that there is no salvage value for either computer. Explain your analysis.

Ans: Replace MINICOMP @ the end of 9th year, Rs. 1, 55,204. Replace the CHIPCOMP @ the end of the 7th year, Rs. 1, 73,219. MINICOMP computer

to be purchased.

7. The following mortality rates have been observed for a special type of light bulbs.

Month 1 2 3 4 5

Percent failing at the end of month: 10 25 50 80 100

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In an individual unit there are 1000 special type of bulbs in use, and it costs Rs.10 to replace an individual bulb which has burnt out. If all bulbs were replaced simultaneously it would cost Rs. 2.50 per bulb. It is proposed to replace all bulbs at fixed intervals whether or not they have burnt out, and to continue replacing burnt out bulbs as they fail. At what intervals of time the manager should replace all the bulbs?

Ans: IRP – Rs.2990, GRP – Rs.2550, GRP is recommended.

8. A computer contains 20,000 resistors. When any resistor fails, it is replaced. The cost of replacing a resistor individually is Re. 1. If all the resistors are replaced at the same time the cost per resistor to Rs. 0.40. The per cent surviving at the end of month t, and the possibility of failure during the month t are given below:

0 1 2 3 4 5 6Percent Surviving at the end of t

Probability of failure during the month:

100

-

96

0.04

90

0.06

65

0.25

35

0.30

20

0.15

0

0.20

What is the optimum replacement plan?

Ans: IRP – Rs.4926, GRP – Rs.5043.10; IRP is recommended

9. A) Machine a cost Rs.9000 operation are Rs.200 for the first year and then increase by Rs.2000 every year. Determine the best age at which to replace the machine if the optimal replacement is policy is followed what will be the average yearly cost of owning and operating the machine.

B) Machine B cost Rs. 10000 annual operating cost is Rs.400 for the first year and then increases by Rs.800 every year you have a machine of type A which is one year old. Should you replace it with machine B , if so when

ANS :- Machine A – Optimal period of replacement is in the end of 3rd year Machine A – Avg. Cost of owning & Operating the machine Rs.5200/- Machine B - Optimal period of replacement is in the end of 5th year Machine B - Avg. Cost of owning & Operating the machine Rs.4000/- Yes, Since the avg. yearly cost of machine B is less than avg. yearly cost of machine A Replace machine A with machine B. Replace A at the end of 2nd year.

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GAME THEORY

Theory of games is concerned with decision-making in situations where two or more rational players, each with a set of strategies, are involved in conditions of competition and conflicting interests.

With the pay-offs resulting from the play of various combinations of the strategies by different players being given and known to all of them, the solution to a game calls for determining the optimum strategies for the players.

For a game involving two-persons and zero-sum (where the gain for one is equal to the loss of the other) with a given set of a pay-offs expressed from the view point of the ‘maximizing’ player, obtaining the solution calls for determining, in the first place, whether the saddle point exists.

For the ‘maximizing’ players, the maximin strategy is determined. For this, the minimum pay-offs of all the strategies are obtained (as row minima) and then the strategy with the maximum among those minimum values is determined. Similarly, the maximum pay-offs of all the strategies open to the ‘minimizing’ player (column maxima) are obtained and the least value in them is considered. The strategy is called ‘minimax’. If the pay-off of the maximin strategy matches with the pay-off of the minimax strategy, then the saddle points exists. In such a case, the strategies so determined are the respective optimal strategies for the players and the pay-off involved is the value of the game. The strategies involved are termed pure.

If the saddle point does not exist, then the players have to employ mixed strategies. For each player, the optimal mix is obtained in such a manner that the pay-off is the same no matter what strategy the opponent chooses to play.

For a 2 x 2 game, involving two strategies for each of the players, the optimal mix can be obtained by using analytical method.

Game of the order 2 x n or m x 2 may be reduced to 2 x 2 games by using graphical method and then solved as such.

The games involving more than two strategies available to each of the players may be attempted to be reduced to the order of 2 x 2, by using the rule of dominance. If a game can be so reduced, it is solved by using the analytical method.

If a strategy is found to be better than, or at least as good as another strategy in terms of the comparable pay-offs, then the former is said to “dominate” the latter. Also, a weighted average of two (or more) strategies may be seen to dominate a given strategy. In any event, a dominated strategy can be deleted.

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If a game cannot be reduced to a 2 x 2 game, it can be formulated as a linear programming problem and solved accordingly. In general, any game can be formulated and solved as an LPP.

TEST YOUR UNDERSTANDING

Mark the following statements as T {True} or F {False}

1. In a two-person game, both the players must have an equal number of strategies.

2. The zero – sum game implies that any gain of one player is exactly matched by a loss to the other so that their sum is equal to zero.

3. The solution to a game implies determining optimal strategies for both the players and value of the game.

4. The solution to a two-person game is based on the assumption that player A will always play his strategy first and the other player, B, would play his strategy thereafter.

5. In the two-person games, both the players are assumed to be in full knowledge of the strategies available to each one and the pay-offs resulting from each combination of strategies.

6. For a two-person game with, respectively, 6 and 5 strategies available to the two players, a total of 30 conditional pay-offs would be involved.

7. Player A’s strategy is determined on the basis of maximin criterion whereas player b’s is on minimax criterion.

8. Saddle point is the point of equilibrium.

9. It is possible that multiple minimax/maximin strategies might exist in a two player game.

10. If the largest element in the pay-off matrix is negative, the game should favour player B irrespective of what the optimal strategies of the players are.

11. If the value of the game is a negative value, it implies that the game is favouring player B.

12. A game is said to be fair when both the players know about the optimal strategy of each other and the losing player pays off the amount involved to the gaining player.

13. Mixed strategy for a player can involve no more than two strategies.

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14. Mixed strategies for each of the players are determined such that same pay-off would be expected irrespective of the strategy adopted by the opponent..

15. Mixed strategies are adopted by the players when they are doubtful about the optimal strategies for them.

16. Strategies involved in mixed strategies of the player are played in a random manner in the ratio determined in the optimal mix.

17. An optimal strategy (0, 2/9, 7/9) implies that in every nine plays, the first strategy should never be played, the second strategy twice and the third strategy be adopted in the remaining seven times.

18. The graphic approach to the solution of games can be applied when one of the player has two strategies available.

19. Every game is solvable irrespective of whether the players adopt pure or mixed strategy.

20. In case of m by 2 games, the value of the game is given by the highest point in the lower envelope.

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SHEMATIC PRESENTATION OF THE METHOD OFSOLUTION OF TWO-PERSON ZERO-SUM GAME

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GAME THEORY - PROBLEMS

1. Two leading firms, Nirmala Textiles Ltd and Swati Rayons Ltd., for years have been selling shirting, which is but a small part of both firms total sales. The Marketing Director of Nirmala Textiles raised the question, “What should the firms strategies be in terms of advertising for the product in question?” The system group of Nirmala Textiles developed the following data for varying degrees of advertising:

a) No advertising, medium advertising and heavy advertising for both firms will result in equal market share.

b) Nirmala Textiles with no advertising : 40 percent of the market with medium advertising by Swati Rayons and 28 per cent of the market with heavy advertising by Swati Rayons.

c) Nirmala Textiles using medium advertising : 70 per cent of the market with no advertising by Swati Rayons and 45 per cent of the market with heavy advertising by Swati Rayons.

d) Nirmala Textiles using heavy advertising: 75 per cent of the market with no advertising by Swati Rayons and 52.5 per cent of the market with medium advertising by Swati Rayons.

2. Shruti Ltd has developed a sales forecasting function for its products and the products of its competitors, Purnima Ltd. There were four strategies S1, S2, S3, and S4 available to Shruti Ltd and the strategies P1, P2, P3 to Purnima Ltd. The pay-offs corresponding to all the above combinations of the strategies are given below. From the table we can see that, for example, if strategy, S1 is employed by Shruti Ltd and strategy P1 by Purnima Ltd, then there shall be a gain of Rs. 30,000 in quarterly sales to the former. Other entries can be similarly interpreted.

Considering this information, state what would be the optimal strategy for Shruti Ltd? Purnima Ltd? What is the value of the game? Is the game fair?

Purnima Ltd’s StrategyP1 P2 P3

Shruti’s Strategy

S1 30 16 25

S2 48 40 60

S3 19 20 18

S4 60 59 55

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3. In a small town, there are only two stores that handles sundry goods – ABC and XYZ. The total number of customers is equally divided between the two, because price and quality of goods sold are equal. Both stores have good reputation in the community, and they render equally good customer service. Assume that a gain of customer of ABC is a loss to XYZ and vice-versa. Both stores plan to run annual pre Diwali sales during the first week of November. Sales are advertised through a local newspaper, radio and television media. With the aid of an advertising firm, store ABC constructed the game matrix given below.

(Figures in the matrix represent a gain or loss of customers)

Strategy Of ABCStrategy Of XYZ

Newspaper Radio Television

Newspaper 30 40 -80

Radio 0 15 -20

Television 90 20 50

Determine optimal strategies and worth of such strategies for both ABC and XYZ.

4. Reduce the following two-person zero sum game to 2 x 2 order, and obtain the optimal strategies for each player and the value of the games:

Process BB1 B2 B3 B4

Players A

A1 3 2 4 0A2 3 4 2 4A3 4 2 4 0A4 0 4 0 8

5. A Company is currently involved in negotiation with its union on the upcoming wage contract. With the aid of an outside mediator, the table below was constructed by the management group. The pluses are to be interpreted as proposed wage increases while a minus figure indicates that a wage reduction is proposed. The mediator informs the management group that he has been in touch with the union and that they have constructed a table that is comparable to the table developed by the management. Both the company and the union must decide on an overall strategy before negotiation begin. The management group understands the relationship of company strategies to union strategies in the following table but lacks specific knowledge of game theory to select the best strategy (or Strategies) for the firm. Assist the management on this problem. What game value and strategies are available to the opposing groups?

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Conditional costs to company

(In Lac Rs)

Company Strategies

Union StrategiesU1 U2 U3 U4

C1 + 0.25 + 0.27 + 0.35 - 0.02C2 + 0.20 + 0.16 + 0.08 + 0.08C3 + 0.14 + 0.12 + 0.15 + 0.13C4 + 0.30 + 0.14 + 0.19 0.00

6. Use the graphical method for solving the following game and find the value of the game

Player A

Player B

B1 B2 B3 B4

A1 2 2 3 -2

A2 4 3 2 6

7. Solve the following game graphically

Player A

Player B

B1 B2

A1 1 2

A2 4 5

A3 9 -7

A4 -3 -4

A5 2 1

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INVENTORY MODELS

Problems

1. Alpha industry estimates that it will sell 12,000 units of its product for the forthcoming year. The ordering cost is Rs. 100 per order and the carry cost / unit / year is 20% of the purchase price. The purchase price / unit is Rs. 50/-. Find

a) EOQ (490 units)b) No. of order/year (24)c) Time bet-orders (0.04 yrs)d) Optimal total cost (Rs. 6, 04,899).

2. A company for one of the A class items placed 6 orders each of size 200 in a year. Given ordering cost Rs. 400, Holding cost 40%, cost/unit Rs.40/-. Find out the loss to the company in not operating scientific inventory policy. What are your recommendations for the future.

Ans: 80/-3. For the data given below find EOQ & Total variable cost.

a) Annual Demand = 5000 units, Unit price = 20/-, Ordering cost/order = Rs.16/- Storage rate2% p.a, Interest rate – 12% p.a, Obsolescence rate 6% p.a.

Ans: (200,800).b) Det. Total variable cost that would result for the item if an incorrect price

of Rs. 12.80 is used.Ans: (250, 820)

4. The purchase manager has decided to place an order for a minimum qty of 500 nos. of a particular item in order to get a discount of 10%. From the records, it was found out that in the last year, 8 orders each of size 200 nos. were placed.

Given: Ordering cost = Rs. 500/orderInventory carrying cost = 40%Cost/unit = Rs.400/-

Is the purchase manager justified in his decision? What is the effect of his decision on the company?

Ans: Yes, Saving over EOQ policy of ordering Rs. 42, 400, saving over present policy of ordering Rs. 46, 400/-

5. A company uses annually 50,000 units of an item. Each costing Rs.1.20. Each order costs Rs.45/- and Inventory carrying cost is 15% of the average Inventory value.

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Find:a) EOQ

Ans.5000.b) If the company operates 250 days a year, the procurement time being 10

days and safety stock is 500 units. Find Reorder Level, Max Level, Min Level and Avg. inventory.

Ans: 2500, 5500, 3000 units.

6. The average monthly consumption of an item is 200 units and the normal lead time is 1 month. If the max consumption has been upto 250 units per month and the max. Lead time is 1½ months. What would be the buffer stock for the item if the item is controlled by Fixed order quantity system (FOQ) system.

Ans: 175 units.

7. A firm has a demand distribution during a constant lead time with a std. deviation of 250 units. The firm wants to provide 98% service level.

a) How much safety stock should be carried?b) If the demand during lead time is 1200 units. What is the appropriate

Reorder level (Z value for 98% is 2.05)?

Ans: 512, 1712 units.

8. An airline has determined that 10 spare brake cylinders will give them a stock out risk of 30%, whereas 14 will reduce the risk to 15% and 16 to 10%. It takes 4 months to receive the cylinder from the supplier and the airline uses on average of 4 cylinders/ months. At what stock level should they reorder assuming they wish to maintain an 85% service level?

Ans: 30 cylinders.9. The average demand for an item is 120 units / year. The lead time is one months

and the demand during lead time follows normal distribution with average of 10 units and S.D. of 2 units. If the item is ordered once in 4 months and the policy of the company is that there should not be more than 1 stock out every 2 years. Det. Reorder level?

10. A scrutiny of past records gives the following distribution for lead time and daily demand during lead time.

Lead time distribution: LT (days) 3 4 5 6 7 8 9 10

Freq 2 3 4 4 2 2 2 1

Demand distribution: Demand/day 0 1 2 3 4 5 6 7

Freq 2 4 5 5 4 2 1 2

What should be the safety stock?

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Ans: 52/-.

11. Data on the distribution of lead time for a pump component were collected as shown below. The management would like to set safety stock that would limit the stock out risk to 10%.

Lead Time Frequency 0 – 1 10 1 – 2 20

2 – 3 70 3 – 4 40 4 – 5 30 5 – 6 10

6 – 7 10 7 – 8 10

How many weeks of safety stock are required to protect the desired service level.

Ans: 2.6 weeks.

12. Find the optimal order qty for a product for which the price break is as follows.

Qty Unit Price 0 <= Q1 < 50 Rs. 10

50 <= Q2 < 100 Rs. 9 100 <= Q3 Rs. 8

The monthly demand for the product is 200 units. The storage cost is 25% of the unit cost and the ordering cost is Rs. 20/- order.

13. Find the optimum order qty for a product for which the price breaks are as follows

Qty Unit Price 0 <= Q1 < 100 Rs. 20 per unit

100 <= Q2 < 200 Rs. 18 per unit 200 <= Q3 Rs. 20 per unit

The monthly demand for the product is 400 units. The storage cost is 20% of the unit cost of the product and the cost of ordering is Rs. 25 per order.

Ans: EOQ Q3* = 79 TC = 6770Q2* = 75 TC = 7480Q1* = 70 TC = 8283

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14. A company purchase one of its items for Rs. 2 per unit without qty discount. The ordering cost is Rs. 20/- per order and the carrying cost is 20% of its purchase price per unit per year. The annual demand is 2500 units. A new vendor offers quantity discount for the same item on per the following quantity discount scheme. Find the best order quantity.

Qty Unit Price 0 <= Q1 < 1500 C 1500 <= Q2 < 200 97% of C

2500 <= Q3 95% of C

15. An automobile factory manufactures a particular type of gear within the factory. This gear is used in the final assembly. The particulars of this gear are

D = Demand rate - 14,000 units / yearP = Production rate - 35,000 units / yearS = Setup cost - Rs. 500 / setupH = Carrying cost - Rs. 15 / unit / year

Find Economic Batch quantity & Cycle time?

Ans: 1248, t1 – 13 days, t2 – 20 days, t = 33 days.

16. A Company manufactures a low cost bearing which is used in main product line. The demand of the bearing is 10,000 units / months and the production rate of the bearing is 25,000 units per months. The carrying cost is Re. 0.02 per unit per year and the setup cost is Rs. 18 per set up. Find EBQ & cycle time.

17. Find the most economic batch qty of a product on a machine if the product rate of that item on the machine is 200 pieces / day and the demand is uniform at the rate of 100 pieces / day. The setup cost is Rs. 200 per batch and the cost of holding one item in inventory is Rs. 0.81 per day.

b) How will the batch quantity vary if the rate production rate is infinite?

18. The annual demand for a component is 7200 units. The carrying cost is Rs. 500 / unit / year. The ordering cost is Rs. 1500/- order and the shortage cost is Rs. 2000 / unit / year. Find EOQ, max inventory max shortage qty, inventory period & shortage period.

19. The demand of an item in a store is 18,000 units per year. The purchase price of the item is Rs. 5/- per unit and its carrying cost is Rs. 1.2 per unit per year and the ordering cost is Rs. 400/- per order. The shortage cost is Rs. 5/- per unit per year. Find EOQ, no of orders / year, the max inv, max, shortage qty and the total cost.

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Dynamic Programming INTRODUCTION

The decision-making process often involves several decisions to be taken at different times. The mathematical technique of optimizing a sequence of interrelated decisions over a period of time is called dynamic programming.Dynamic programming (DP) differs from linear programming in two ways:

(i) In DP, there is no set procedure (algorithm) as in LP to solve any decision-problem. The DP technique allows to break the given problem into a sequence of easier and smaller subproblems, which are then solved in a sequential order (stage).

(ii) LP approach provides one-time period (single stage) solution to a problem whereas DP approach is useful for decision-making over time and solves each subproblem optimally.

DYNAMIC PROGRAMMING TERMINOLOGY

Regardless of the type or size of a dynamic programming problem, there are certain terms and concepts that are common in every problem.

1. STAGE : The dynamic programming can be decomposed or divided into a sequence of smaller subproblems called stages. At each stage there are a number of decision alternatives (courses of action) and a decision is made by selecting the most suitable alternative. Stages very often represent different time periods in the planning period of the problem, places, people or other entities. For example, in the replacement problem each year is a stage, in the salesman allocation problem each territory represent a stage.

2. STATE : Each stage in the dynamic programming problem has a certain number of states associated with it. These states represent various conditions of the decision process at a stage. The variables that specify the condition of the decision process or describe the status of the system at a particular stage are called state variables. These variables provide information for analyzing the possible effects that the current decision could have upon future courses of action. At any stage of the decision-making process there could be a finite or infinite number of states. For example, a specific city is referred to as state variable, in any stage of the shortest route problem.

3. RETURN FUNCTION : At each stage, a decision that can affect the state of the system at the next stage and help in arriving at the optimal solution at the current stage is made. Every decision that is made has its own merit in terms of worth or benefit associated with it and can be described in an algebraic equation form. This equation is generally called a return function, in general, depends upon the state variable as well as the

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decision made at a particular stage. An optimal policy or decision at a stage yields optimal (maximum or minimum) return for the given value of the state variable.

DEVELOPING OPTIMAL DECISION POLICY

Dynamic programming is an approach in which the problem is broken down into a number of smaller subproblems called stages. These subproblems are then solved sequentially until the original problem is finally solved. A particular sequence of alternatives (course of action) adopted by the decision-maker in a multistage decision problem is called a policy. The optimal policy, therefore, is the sequence of alternatives that achieves the decision-maker’s objective. The solution of a dynamic programming problem is based upon Bellman’s principle of optimality (recursive optimalization technique), which states:

The optimal policy must be one such that, regardless of how a particular state is reached, all later decisions (choices) proceeding from that state must be optimal.

Based on this principle of optimality, we find the best policy by solving one stage at a time, and then sequentially adding a series of one-stage-problems that are solved until the overall optimum of the initial problem is obtained. The solution procedure is based on a backward induction process and forward induction process. In the first process, the problem is solved by solving the problem in the last stage and working backwards towards the first stage, making optimal decisions at each stage of the problem. In certain cases, the second process is used to solve a problem by first solving the initial stage of the problem and working towards the last stage, making an optimal decision at every stage of the problem.

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Dynamic Programminng Problems

SHORTEST ROUTE PROBLEM

1. A salesman located in a city A decided to travel to city B. He knew the distances of alternative routes from city A to city B. He then drew a highway network map as shown in the Fig. The city of origin A, is city 1. The destination city B, is city 10. Other cities through which the salesman will have to pass through are numbered 2 to 9. The arrow representing routes between cities and distances in kilometres are indicated on each route. The salesman’s problem is to find the shortest route that covers all the selected cities from A to B.

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2. A company has decided to introduce a product in three phases. Phase 1 will feature making a special offer at a greatly reduced rate to attract the first-time buyers. Phase 2 will involve intensive advertising to persuade the buyers to continue purchasing at a regular price. Phase 3 will involve a follow up advertising and promotional campaign.A total of Rs 5 million has been budgeted for this marketing campaign. If m is the market share captured in Phase 1, fraction f2 of m is retained in Phase 2, and fraction f3 of market share in Phase 2 in retained in Phase 3. The expected values of m, f2 and f3 at different levels of money expanded are given below. How should the money be allocated to the three phases in order to maximize the final share?

Money Spent Effect On Market Share(Rs millions) m per cent f2 f3

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0.030.050.700.800.850.90

0.050.070.850.900.930.95