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  1. Scholarship Questions 2003

  2. 2003

  3. Question 4 • An oil company operates two refineries. The refineries produce three types of fuel – aviation grade, regular grade petrol and super grade petrol. • Refinery 1 costs $160 000 per day to operate and Refinery 2 costs $175 000 per day to operate. • The oil company has contracts to produce at least 120 000 litres of aviation fuel, 300 000 litres of regular grade petrol and 108 000 litres of super grade petrol per month. • The following table gives the daily production statistics in litres:

  4. Let the number of days per month that Refinery 1 operates be x and the number of days per month that Refinery 2 operates be y. • (a) Using graph paper where necessary, determine the optimal number of days that each refinery should operate in order to minimise the total cost of operating them both. Calculate this minimum cost.

  5. Constraints: • 10 000x + 10 000y ≥ 120 000 Aviation • 20 000x + 30 000y ≥ 300 000 Regular • 6 000x + 18 000y ≥ 108 000 Super • 0 ≤ x ≤ 31 Restriction on days • 0 ≤ y ≤ 31

  6. Simplify the inequalities: • Constraints: • x+ y≥ 12 Aviation • 2x + 3y ≥ 30 Regular • x+ 3y≥ 18 Super • 0 ≤ x ≤ 31 Restriction on days • 0 ≤ y ≤ 31

  7. Objective function is • C = 160 000x + 175 000y • Gradient: -160/175 = -32/35 • Difficult to use this so look for intersection points.

  8. (0,12), (6,6), (12,2), (18,0),

  9. Costs at each vertex ($000) Optimum number of days for both refineries is 6 Minimum cost = $2 010 000

  10. The company is investigating possible changes at Refinery 2 that may alter the daily cost of operating the refinery. Assuming that the daily cost of operating Refinery 1 remains fixed at $160 000, determine the range of values that the daily cost of operating Refinery 2 could lie within and still give the production combination in (a).

  11. The gradient needs to stay between the gradients of the 2 lines that form the intersection (6,6) i.e. between -1 and -2/3

  12. 2004 Question 4

  13. Question 4 • To produce one bottle of POW takes 30 minutes and to produce one bottle of ZAP takes 20 minutes, and there are 25 hours available in total for this production-run operation. • x = bottles of POW • y = bottles of ZAP

  14. Question 4 • To produce one bottle of POW takes 30 minutes and to produce one bottle of ZAP takes 20 minutes, and there are 25 hours available in total for this production-run operation.

  15. Question 4 • Availability of a particular additive means that the chain cannot produce more than 35 bottles of POW and a combined total of 65 bottles on any production run.

  16. Question 4 • Availability of a particular additive means that the chain cannot produce more than 35 bottles of POW and a combined total of 65 bottles on any production run.

  17. Question 4 • As a minimum production requirement, at least 15 bottles of each weed killer must be produced in each run.

  18. Question 4a • The management of NAILS wants to know how many bottles of both POW and ZAP should be produced. The preliminary estimates of their potential profitability are $20 per bottle of POW and $10 per bottle of ZAP. Perform an appropriate analysis and make a recommendation to management about the amount of POW and ZAP that could be produced to maximise profit.

  19. Question 4 • Profit • Gradient is -2

  20. Feasible region (15, 15) (35, 15), (35, 22.5), (20, 45), (15, 50)

  21. Feasible region Blowup of the section

  22. Feasible region (35, 22.5) = (35, 22) and (34, 24) as need integer values

  23. Question 4b • Suppose that the profit of $20 per bottle of POW was overestimated and it was in reality only $15 per bottle. How does that change your production recommendation in part (a)?

  24. Profit = 15x + 10y

  25. Profit = 15x + 10y • This has the same gradient of our first constraint • Can have any of (20, 45), (22, 42), (24, 39), (26, 36), (28, 33), (30, 30), (32, 27), (34, 24)

  26. It is found that the sales of POW and ZAP are closely related by the function y = 9 ln(x), • where x = number of bottles of POW and y = number of bottles of ZAP. • Management therefore requests that the ratio “number of bottles of POW produced : number of bottles of ZAP produced be x : 9 ln(x)”. The constraints need to be satisfied and the estimated profit needs to be maximised, based on a profit of $15 per bottle of POW and $10 per bottle of ZAP, so as to mirror demand.

  27. y = 9lnx

  28. Equation

  29. Solve the equation derived in part (c) and make a recommendation to management about the amount of POW and ZAP that could be produced to maximise profit.

  30. Solve the equation derived in part (c) and make a recommendation to management about the amount of POW and ZAP that could be produced to maximise profit. • Use your graphics calculator: • This gives 29.661… • Integer values • 29 bottles POW and 31 bottles ZAP

  31. 2005 Question 1

  32. Question 1 • The refreshment bar sells two types of pie, steak and mince. The number of pies ordered each day needs • to satisfy the following daily constraints.

  33. At least 42 pies will be sold in total

  34. A minimum of 24 mince pies will be sold

  35. No more than 3 mince pies will be sold for every steak pie sold

  36. At most, 12 more steak pies will be sold than mince pies.

  37. Feasible region

  38. The refreshment bar pays $1.30 for each steak pie and $1.00 for each mince pie.

  39. Objective function

  40. Minimum cost • (24, 36) next point will be less • (24, 18) $47.40 • (10.5, 31.5) $45.15 • Cannot buy half a pie

  41. Minimum cost • Investigate to get minimum at • (31, 11) • 31 mince and 11 steak pies

  42. Assume the cost of steak pies remains at $1.30 each. The cost of mince pies increases so that the optimal solution is no longer that obtained in (a). State the possible changes in the cost of mince pies that give rise to these different optimal solution(s), and give the new optimal solution(s) for each of these changes.