1. Question: The Southern Electric Company is contemplating the future of one of its plant located in South Carolina. Three alternative decisions are being considered; 1)Expand the plant 2)Maintain the status quo at the plant 3)Sell the plant now The following payoff table describes the decision situation:

2. Payoff Table States Of Nature Good Economic Conditions Poor Economic Conditions Decision Expand Maintain status quo Sell now $ 800, 000 1, 300, 000 320, 000 $ 500, 000 - 150, 000 320, 000

3. Determine the best decision using each of the decision criteria. 1)Maxi-max 2)Maxi-min 3)Mini-max Regret 4)Hurwicz Criterion 5)Equal likelihood

4. Maxi-max: This results in the following maximum payoffs and decision accordingly: Expand$ 800, 000 Status quo 1, 300, 000 Sell 320, 000 Decision: Status quo is to be maintained Maxi-min: This results in the following minimum payoffs and decision accordingly: Maximum

5. Expand$ 500,000 Status quo - 150, 000 Sell now 320, 000 Decision: Expand the Plant Mini-max Regret Maximum Good Economic Conditions Poor Economic Conditions Decision Expand Status quo Sell now500,000 – 320,000 = 180,000 1,300,000 – 1,300,000 = 0 1,300,000 – 320,000= 980,000 $ 500,000 – 500,000 = $ 0 500,000 – (- 150,000) = 650,000 $ 1,300,000 – 800,000 = $ 500,000

6. The maximum regret for each decision is determined and the decision corresponding to the minimum of the regret is selected. Expand $ 500, 000 Status quo 650, 000 Sell now 980,000 Decision: Expand the plant Hurwicz Criterion Let us assume that α =.3, the 1 – α =.7. The following decision will result Minimum

7. Expand (800,000 x.3) + (500,000 x.7) = $ 650,000 Status quo (1,300,000 x.3) + (-150,000 x.7) = $ 575,000 Sell now (320,000 x.3) +(320,000 x.7) = $ 320,000 Decision: Expand the plant Equal Likelihood This criteria weights each state of nature equally i.e, α =.5, and 1- α =.5 Maximum

8. Expand (800,000 x.5) + (500,000 x.5) = $ 650,000 Status quo (1,300,000 x.5) + (- 150,000 x.5) = $ 575,000 Sell now (320,000 x.5) + (320,000 x.5) = $ 320,000 Decision: Expand the plant Question: Consider the following payoff table of three product decisions (A, B, & C) and three future market conditions (payoff = $ (M): Maximum)

9. Payoff Table Market Conditions 12 Decision A B C $ 1.0 0.8 0.7 $ 2.0 1.2 0.9 3 $ 0.5.9 1.7

10. Determine the best decision using the following decision criteria: i.Maximax ii.Maximin Solution: Maximax Criteria A$ 2.0 B 1.2 C 1.7 Decision: Product-A Maximum

11. Maximin Criteria A$ 0.5 B 0.8 C 0.7 Decision: Product-B Maximum

12. Decision making situation with Probabilities (under Risk) It is often possible for the decision maker to know enough about the future states of nature to assign probabilities that each will occur, which is decision making under risk. The most widely used decision making criterion under risk is expected value computation. All the payoffs for a particular alternative are multiplied by the respective probabilities associated with the future conditions, thus arriving at the expected value, by using the following formula.

13. Where = probability of outcome = outcome The managers then compare the expected values of all alternatives and use the resulting information to make a decision. Example: Assume that it is now possible for the Southern Electric Company to estimate probability of.7 that good market conditions will exist and a probability of.3 that poor economic

14. conditions will exist in future. Determine the best decision using expected value. Solution: The expected values for each decision alternative are computed as follows; EV(expand) = (800,000 x.7) + (500,000 x.3) $ 710,000 EV(status quo) = (1,300,000 x.7) + (-150,000 x.3) = $ 865,000 EV(Sell) = (320,000 x.7) + (320,000 x.3) = $ 320,000 Maximum)

15. Decision: Maintain status quo ( since it has the highest expected value). Example: An investment has the potential of providing cash returns of Rs.300, Rs.500 & Rs.750 per year through the year 2007.The returns will be determined by the state of the economy. Management places a 50% probability of normal economic conditions, 30% likely hood of recession and 20% excellent economic conditions. Based upon his experience & the information, the expected value of the cash flow from the investment can be calculated as follows:

16. Based upon the assumption of potential cash flows & the estimated likely hood of various levels of economic growth, Rs 490 is the best estimate of the annual cash flow, that will be generated by the estimate. This return can now be compared with alternative investments to determine the investment that generates the highest expected return. Cash Flow (Rs)Economic ConditionsExpected Value(Rs) 300x.3 (Recession)90 500x.5 (Normal)250 750x.2 (Boom)150 490

17. AlternativesLight Demand (Probability of.2) Moderate Demand (Probability of.3) Heavy Demand (Probability of.5) 1. Buy one truckPayoff: $ 30,000Payoff: $ 32,000Payoff: $ 20,000 2.Buy two TrucksPayoff: $ 24,000Payoff: $ 34,000Payoff: $ 31,000 3.Buy three TrucksPayoff: $ 10,000Payoff: $ 15,000Payoff: $ 40,000 States Of Nature Example

18. The owner of an organization could use a payoff matrix such as this to determine the expected value of buying one, two or three trucks. Based on experience, current trends & potential judgment, the owner has assigned each future condition a probability of occurrence. Also the owner has estimated the revenue payoffs of each of three alternatives under each of the 3 states of nature.

19. On the basis of this analysis, the owner would get the highest expected value from alternative- 2, that is from buying two Trucks. Decision Alternatives Expected Values Alternative-1 (buy one Truck).2($ 30,000)$6000.3($ 32,000)$9600.5($ 20,000)$10000$25600 Alternative-2 (buy two Trucks).2($ 24,000)$4800.3($ 34,000)$10200.5($ 31,000)$15500$30500 Alternative-3 (buy three Trucks).2(-$ 10,000)$-2000.3($ 15,000)$4500.5($ 40,000)$20000$22500

20. Northwest Electric Company wants to install a generating plant. They have two options, either to go for a large hydroelectric plant or to put smaller Coal Powered Plant. They have developed the following energy demand estimates for the next 20 years. High Demand:P=.6 Moderate Demand:P=.3 Low Demand:P=.1 Example:

21. They have also collected the following additional information regarding both the plants Large Hydroelectric Plant High demand environment will yield annual profit of $130 Million. Moderate demand environment will yield annual profit of $80 Million. Low demand environment will yield annual profit of $ 20 Million. Small Coal Powered Plant High demand environment will yield annual profit of $ 45 Million. Moderate demand environment will yield annual profit of $60 M Low demand environment will yield annual profit of $ 100 Million.

22. Decision? Build Hydroelectric Plant or Coal Powered Plant Hydroelectric Plant Coal Powered Plant High Moderate Low Demand Probabilities.6x130x20=1560.1x20x20=40.3x80x20=480.3x60x20=360.6x45x20=540.1x100x20=200 $ 1560 480 40 $ 2080 850 $ 1230 $ 540 360 200 $ 1100 420 $ 680 Expected Value Plant Cost Net Value Plant Cost Net Value

23. The net expected value of profits generated by the Hydroelectric Plant is $1.23 Billion over the next 20 years, $ 550 Million advantage over the coal powered plant. The Large Hydroelectric Plant seems to be the appropriate decision based on the analysis of the available information with the company.

24. Linear Programming This is a mathematical technique that is used to find the best solution to a given problem from a set of feasible solutions. The essence of linear programming is optimizing the allocation of scarce resources. Linear means that the relationship among variables can be expressed as directly proportional functions. Example;

25. Consider a case of an automobile dealer. The inventory is divided between Mehran Cars & Shahzore trucks. Assume that consumer demand is sufficiently great that any combination of these vehicles will be purchased & that the Mehran contributes $ 650 per unit to profits after costs have been paid and the Shahzore contributes $ 950. Following are the conditions: Objective:To maximize profits

26. Alternative: Dealer can stock any combination of Mehrans and Shahzores Resource Limitation: Limited storage sets the dealer quota at an upper limit of 15 Mehrans or 10 Shahzores (since Shahzore requires 50% more space than Mehran)

27. Mathematical equation is P=$ 650 A + $ 950 Bwhere A=Number of Mehrans B=Number of Shahzores P=Total Profit Profit is a linear function of A & B. A & B are the decision variables of the problem. Total capacity is 15 Mehrans or 10 Shahzores (15As or 10Bs). A Mehran takes one unit of space & a Shahzore takes 1.5 units of space.

28. An equation that will require solution with respect to space limitation constraint is A + 1.5B ≤ 15 It is mandatory for the dealer that he must carry both Mehran & Shahzore in inventory. This can be explained by A,B ≥ 1 The problem facing can be summarized in the following manner: To Maximize P= $ 650 A + $ 950 B Subject toA + 1.5 B ≤ 15 & A,B ≥ 1

29. MehransShahzoreSpace UnitsProfits 1914.59200.00 28148900.00 38159550.00 4714.59250.00 56148950.00 66159600.00 7514.59300.00 84149000.00 94159650.00 10314.59350.00 112149050.00 122159700.00 13114.59400.00 Effect of Alternative Inventory Combination of Mehrans & Shahzores

30. The optimal solution subject to the given constraints is an inventory composed of 12 Mehrans & 2 Shahzore trucks. This solution gives the dealer a profit contribution of $ 9700, the highest of all possible combinations of Mehrans and Shahzore trucks.

31. Fixed Cost = $ 500 Variable Cost= $ 7.5 per unit Selling Cost= $ 10 per unit Breakeven point satisfies following equation: Fixed Cost + Variable Cost = Revenue 500 + 7.5 x = 10 x, where x is the number of units 2.5 x = 500 x=500/2.5 = 200 Breakeven point = 200 units (or $ 2000.00) Example:

32. Computation No. of Units Variable CostFixed CostTotal Cost 50= 7.5 x 50 = 375500= 875 100= 7.5 x 100 = 750500= 1250 150= 7.5 x 150 = 1125500= 1625 200= 7.5 x 200 = 1500500= 2000 250= 7.5 x 250 = 1875500= 2375

33. Break-even Diagram Profit Loss Total Cost Fixed Cost A B F E D O Production quantity Dollars C Variable Cost

34. BREAK-EVEN ANALYSIS A break-even point may be defined as the level of activity at which the income equals the costs and there is no profit or loss. At break-even point; Income = Total Cost Income=Fixed Cost +Variable Cost `or

35. or where = Break-even Quantity =Sale Price per Unit =Variable Cost per Unit =Fixed Cost(total fixed cost)

36. Example: Several graduates students at Whitewater University formed a company called the New River rafting company to produce rubber rafts. The initial investment in plant and equipment is estimated to be $ 2,000. labor and raw material cost is approximately $ 5 per raft. If the rafts can sold at a price of $ 10 each, what volume of demand would be necessary to break even ? Solution; Fixed Cost = = $ 2,000

37. Variable Cost = = $ 5 per raft Price = = $ 10 per raft Hence the break even point is

38. UnitsFixed CostTotal CostTotal Revenue 020002,0000 10020002,5001,000 20020003,0002,000 30020003,5003,000 40020004,000 50020004,5005,000 60020005,0006,000 70020005,5007,000 80020006,0008,000 Break – Even of Process - A

39. Fixed Cost Line Total Cost Line Revenue Line Break even Point Loss Area Units ( Rafts) Amount ($) Break-Even of Process-A

40. The students from the New river Rafting Company believe demand for their product will far exceed the break even point. They are now contemplating a larger initial investment 0f $ 10,000 for more automated plant that would reduce the variable cost to $ 2 per raft. a.What is the break even point for this new process (Process-A)? b.Compare the process-A with the Process-B. For what volume of demand should each process be chosen?

41. Fixed Cost = $ 1,000.00 Variable Cost= $ 2.0 per unit Selling Cost= $ 10 per unit Breakeven point satisfies following equation: Fixed Cost + Variable Cost = Revenue 10,000 + 2 x = 10 x, where x is the number of units 8 x = 10,000 x=10,000/8 = 1250 Breakeven point = 1250 units (or $ 12500.00) Solution

42. When the cost of two alternatives is a function of the same variable, it is usually useful to find the value of the variable for each the alternatives incur equal cost. & Where (The point of indifference between two alternatives by equating the total cost of each alternative and solving for the demand volume) Solution

43. $ 2,000 + $ 5v = $ 10, 000 + $ 2v $ 3v = $8,000 v = 2,667 Rafts If the demand for rafts is exactly 2,667 units, we can choose either process A or process B. If the demand is less than 2,667 rafts, the alternative with the lowest fixed cost, process A should be chosen. If the demand is greater than 2,667 units, the alternative with the lowest variable cost Process B should be preferred.

44. Sl. NoUnitsTotal Cost (Process - A)Total Cost (Process - B) 102,00010,000 25004,50011,000 310007,00012,000 415009,50013,000 5200012,00014,000 6250014.50015,000 7300017,00016,000 8350019,50017,000 9400022,00018,000 Comparison between Process – A & Process - B

45. Total Cost of Process-B Point of indifference = 2667 units Units ( Rafts) Amount ($) Total Cost of Process-A Comparison of Process-A & Process-B

46. Break-Even Decision Models Make or Buy Decision A manufacturing firm can buy a required component part from a supplier for $ 96 per unit delivered. Alternatively the firm can manufacture the part for a variable cost of $ 46 per unit. It is estimated that the additional fixed cost in the plant would be $ 7500, per year, if the part is manufactured. Find the number of units per year for which the cost of two alternatives will break even.

47. When the cost of two alternatives is a function of the same variable, it is usually useful to find the value of the variable for each the alternatives incur equal cost. Where

48. Break-even occurs when Therefore

49. Total annual cost for the buy alternative is: Total annual cost for the manufacturing alternative is: 150 are the number of units per year for which the cost of two alternatives will break even.

50. If you need to supply 100 units, Cost on buy mode shall be 96 x 100 = 9600 $ Cost on make mode = (46 x 100) + 7500 = 4600 + 7500 = 12100 $ Here the buy alternative is preferred. If you need to supply 200 units, Cost on buy mode shall be 96 x 200 = 19200 $ Cost on make mode = (46 x 200) + 7500 = 9200 + 7500 = 16700 $ Here the make alternative is preferred.

51. Sl. NoUnits (component parts) Total Cost (Buying) Total Cost (Make) 1007,500 2504,8009,800 31009,60012,100 415014,400 520019,20016,700 625024,00019,000 730028,80021,300 835033,60023,600 940038,40025,900 A Make or Buy Decision

52. These cost functions and the breakeven points may be shown in the figure. For requirements in excess of 150 units, the make alternative would be more economical. If the rate of use is likely to be less than 150 units, the buy alternative should be chosen.

53. Total Cost (Buying) Break - Even Units Amount ($) Total Cost (Manufacturing) A Make or Buy Decision

54. Lease or Buy Decision Another example of breakeven analysis consider the discussion to lease or buy a piece of equipment. Assume that a computer is needed for data processing in an office. Suppose that the computer can be leased for $ 50 per day which includes the cost of maintenance. Alternatively the computer can be purchased for $ 2100.The computer is estimated to have a useful life of 12 years with a salvage value of $ 400 at the end of that time. It is estimated that the annual

55. Maintenance cost will be $ 300. It costs $50 per day to operate the computer.(Time value of money is not being considered) Annual cost if the computer is leased is; Annual cost if the computer is bought is;

56. Break-even occurs when Therefore A graphical representation of the decision shows that for all levels of use exceeding 40 days per year, it would be more economical to purchase the computer. If the level of use is anticipated to be below 40 days per year, the computer should be leased.

57. Sl. NoDaysTotal Cost (Leasing) Total Cost (Buying) 1002,000 255002,250 3101,0002,500 4151,5002,750 5202,0003,000 6252,5002,350 7303,0003,500 8353,5003,750 9404,000 10454,5004,250 11505,0004,500 A Lease or Buy Decision

58. Total Cost (Leasing) Break - Even Days Amount ($) Total Cost (Buying) A Lease or Buy Decision

59. Locational Break-even Analysis (Question) Karachi, Lahore, and Peshawar are three potential locations for producing telecommunication set expected to sell for Rs. 90 per set.

60. Find the most economical location for an expected volume of 1,850 units per year. SiteFixed Cost / Year (Rs) Var. Cost/Unit (Rs) Karachi20, 00050 Lahore40, 00030 Peshawar80, 00010

61. Solution: Total cost= Fixed cost + Variable cost Total cost at KR=20,000 +(50 x 1,850) = 1,12,500 Total cost at LH=40,000 +(30 x 1,850) = 95,500 Total cost at PR=80,000 +(10 x 1,850) = 98,500 It is evident from the graph that the most economical location for a volume of 1,850 units is Lahore. Expected profit is; Total revenue – total cost = (90 x 1,850) – 95,500 = 71,000 / year

62. Cost Volume Graph Annual Volume, Units Annual Cost, $

63. It is to be noted that for volume less than 1,000 units, Karachi would be preferred, and for volumes greater than 2,000 units, Peshawar would be preferred. Less than 1,000 units, say 800 units; Total cost at KarachiRs. 60,000 Total cost at Lahore 64,000 Total cost at Peshawar 88,000 More than 2,000 units, say 2200 units;

64. Total cost at KarachiRs. 1,30,000 Total cost at Lahore 1,06,000 Total cost at Peshawar 1,02,000 The same may be shown in the cost-volume graph.

65. Q-1)Texloy Manufacturing Company must select a process for its new product, TX 142 from among three different alternatives. The following cost data have been gathered. For what volume of demand would each process be selected. SelectionProcess-AProcess-BProcess-C Fixed Cost$ 10, 000$ 20, 000$ 50, 000 Variable Cost $ 5 per unit$ 4 per unit $ 2 per unit

66. Solution: Let x represents the number of TX 142s demand, say products, then Total Cost of Process-A$ 10,000 + $ 5x Total Cost of Process-B$ 20,000 + $ 4x Total Cost of Process-C$ 50,000 + $ 2x Next we calculate the point of indifference between these processes by equating their total cost and solving for demand volume v. Note that in this problem there are three processes to consider but we can compare only two at a time.

67. Comparison 1. Process-A versus Process-B Process-A = Process-B $ 10,000 + $ 5v=$ 20,000 + $ 4v v = 10,000 Units If the demand is 10,000 units, one can choose either process-A or process-B. But if the demand is less than 10,000 units, process-A should be chosen (the alternative with lowest fixed cost). Conversely if the demand is greater than 10,000 units, we should choose process-B (the alternative with lowest variable cost).

68. Comparison 2. Process-B versus Process-C Process-B = Process-C $ 20,000 + $ 4x=$ 50,000 + $ 2x v = 15,000 Units If the demand for units is 15,000 units, we are indifferent between process-B & process-C.If demand is greater than 15,000 units, process-C should be chosen (the alternative with lowest variable cost). If demand is less than 15,000 but greater than 10,000 units process-B should be chosen (comparison-1)

69. Comparison 3. Process-C versus Process-A Process-C = Process-A $ 50,000 + $ 2v=$ 10,000 + $ 5v v = 13,333 Units We have already concluded that process-B should be selected between 10,000 & 15,000, therefore this point of indifference can be ignored. The graph of the problem is shown here.

70. UnitsProcess-A ($10,000 + $5v) Process-B ($20,000 + $4v) Process-C ($50,000 + $2v) 010,00020,00050,000 50012,50022,00051,000 100015,00024,00052,000 150017,50026,00053,000 200020,00028,00054,000 250022,50030,00055,000

71. Question-1

72. Question-2: Computer Ease Supply Store sells computer diskettes. Diskettes are purchased for $ 17.00 a box and sold for $ 19.70 a box. To remain in the competitive market, the company must constantly reevaluate its suppliers. A survey of new supplier sources yields the following options: Supplier-1; Fixed order of 100 boxes per month, of diskettes, $ 1700 plus $ 15 per box for every box over 100 boxes.

73. Supplier-2; Fixed order of 100 boxes per month, of diskettes, $ 2000 plus $ 10 per box for every box over 100 boxes. Should the company try Supplier-1, Supplier-2, or stick with the current supplies? How does the monthly demand for diskettes affect your recommendations.

74. Solution: Current situation:Total Cost = $ 17 per Box Supplier-1Total Cost = $ 1700 + $15 (for every box over 100) Supplier-2Total Cost = $ 2000 + $10 (for every box over 100) Comparison-1 Current Situation vs Supplier-1 (x + 100) 17 = 1700 + 15x, where x is number of boxes over 100. x= 0

75. If demand is less than 100 boxes, choose current scenario. Comparison-2 supplier-1 vs Supplier-2 $ 1700 + $ 15v = $ 2000 + $ 10v v = 60 So Point of indifference is 160 units. If demand is greater than 160, Supplier-2 should be preferred, otherwise choose supplier-1

76. Current Situation; 100 Boxes=17 x 100 = 1700 60 Boxes =17 x 60 = 1020 160 Boxes=17 X 160 = 2720 200 Boxes=17 X 200 = 3400 Supplier-1 100 Boxes=1700 160 Boxes=1700 + (60 x 15) = 2600 200 boxes=1700 + (100 x 15) = 3200

77. Supplier-2 100 Boxes=2,000 160 Boxes=2,000 + (60 x 10) = 2,600 200 boxes=2,000 + (100 x 10) = 3,000

78. UnitsCurrent Situation Supplier-1Supplier-2 001,7002,000 508501,7002,000 1001,700 2,000 1502,5502,4502500 2003,4003,2003,000

79. Question-2