KEY TO PALLAVA MIDTERM
PART A
1. When using a graphical solution procedure, the region bounded by the set of constraints is called the a. Solution b. Feasible region c. Infeasible region d. Maximum profit region e. None of the above 2. In an LP problem, at least one corner point must be an optimal solution, if an optimal solution exists. a. True b. False 3. An LP problem has a bounded bou nded feasible region. If this problem has an equality equal ity constraint, then a. This must be a minimization problem b. The feasible region must consist of a line segment. c. The problem must be degenerate. d. The problem must have more than one optimal solution. 4. Which of the following would cause a change in the feasible region? a. Increasing an objective function coefficient in a maximization problem. b. Adding a redundant constraint. c. Changing the right-hand-side of a nonredundant constraint. d. Increasing an objective function coefficient in a minimization problem. 5. If a non-redundant constraint is removed from an LP problem, then a. The feasible region will get larger. b. The feasible region will get smaller. c. The problem would become nonlinear. d. The problem would become infeasible. 6. In the optimal solution to a linear program, there are a re 20 units of slack for a constraint. From this we know that a. The dual price for this constraint is 20. b. The dual price for this constraint is 0. c. This constraint is redundant. d. The problem must be a maximization problem. 7. A linear program has been solved and sensitivity analysis has been performed. The ranges for the objective function coefficients have been found. For the profit on X1, the
upper bound is 80, the lower bound is 60, and the current value is 75. Which of the following must be true if the profit on this variable is lowered to 70 and the optimal solution is found? a. A new corner point will become optimal. b. The maximum possible total profit may increase. c. The values for all the decision variables will remain the same. d. All of the above are possible. 8. A graphical method should only be used to solve an LP problem when a. There are only two constraints b. There are more than two constraints. c. There are only two variables d. There are more than two variables 9. In LP, variables do not have to be integer valued and may take on any fractional value. This assumption is called a. Proportionality b. Divisibility c. Additivity d. Certainty 10. In solving a linear program, no feasible solution exists. To resolve this p roblem we might a. Add another variable b. Add another constraint c. Remove or relax a constraint d. Try a different computer program 11. If the feasible region gets larger due to a change in one of the constraints, the optimal value of the objective function a. Must increase or remain the same for a maximization problem. b. Must decrease or remain the same for a maximization problem. c. Must increase or remain the same for a minimization problem. d. Cannot change. 12. When alternate optimal solutions exist in an LP problem, then a. The objective function will be parallel to one of the constraints. b. One of the constraints will be redundant. c. Two constraints will be parallel. d. The problem will also be unbounded. 13. If a linear program is unbounded, the problem probably has not been formulated correctly. Which of the following would most likely cause this? a. A constraint was inadvertently omitted b. An unnecessary constraint was added to the problem c. The objective function coefficients are too large d. The objective function coefficients are too small
14. A feasible solution to an LP problem a. Must satisfy all of the problem’s constraints simultaneously. b. Need not satisfy all of the constraints, only some of them. c. Must be a corner point of the feasible region. d. Must give the maximum possible profit. 15. Assume that a company makes wooden picture frames. Frame style 1 takes 2 hours of skilled labor and 3 linear feet of wood. If the company had 40 hours of skilled labor and 48 linear feet of wood that can be used each week, what is the largest quantity of this item that the company will be able to produce given these resource constraints? a. 13.33 b. 16 c. 20 d. 24 e. none of the above
PART B
1. The Seymour Hayes Manufacturing Company produces a small component for an industrial product and distributes it to five wholesalers at a fixed delivered price of $2.50 per unit. Sales forecasts indicate that monthly deliveries will be 2700 units to Wholesaler 1; 2700 units to Wholesaler 2; 9000 units to Wholesaler 3; 4500 units to Wholesaler 4; and 3600 units to Wholesaler 5. The monthly production capacities are 4500 at Plant 1; 9000 at Plant 2; and 11250 at Plant 3. The direct costs of producing each unit are $2.00 at Plant 1, $1.00 at Plant 2, and $1.80 at Plant 3. The transformation costs of shipping a unit from a plant to a wholesaler are given below:
Plant 1 Plant 2 Plant 3
Whlslr. 1 $0.05 $0.08 $0.10
Whlslr. 2 $0.07 $0.06 $0.09
Whlslr. 3 $0.11 $0.10 $0.09
Whlslr. 4 $0.15 $0.12 $0.10
Whlslr. 5 $0.16 $0.15 $0.16
Formulate a linear programming model to indicate the optimal production amounts at each plant, and to show how many components each plant supplies each wholesaler. Let xi be the amount produced at Plant i Let xij be the amount shipped from Plant I to Whlslr. j Total capacity over three plants is equal to 24750. Total demand over five wholesalers is 22,500. So we have enough capacity to satisfy demand. Maximize: -2x1 - 1x2 - 1.80x3 +2.5x1+2.5x2+2.5x3 – 0.05x11 – 0.07x12 – 0.11x13 – 0.15x14 0.16x15 -0.08x21-0.06x22-0.10x23-0.12x24-0.15x25-0.10x31-0.09x32-0.09x33-0.09x34-0.16x34 Subject to: x1 <= 4500 x2 <= 9000 x3 <=11250 x1 = x11 + x12 + x13 + x14 + x15 x2 = x21 + x22 + x23 + x24 + x25 x3 = x31 + x32 + x33 + x34 + x35 x11 + x21 + x31 = 2700 x12 + x22 + x32 <= 2700 x13 + x23 + x33 <= 9000 x14 + x24 + x34 <= 4500 x15 + x25 + x35 <= 3600 All variables are non-negative
Optimal objective value is 20,367/- with x1 =2250, x2=9000 and x3=11,250 x11=2250,x21=450,x22=2700,x23=2250,x25=3600,x33=6750,x34=4500 2. A mill of the Fine-Webb Paper Company produces so-called liner board in jumbo reels having a standard width of 68 inches. (Each reel has a fixed length.) The company’s customers, however, order reels having smaller widths (and the same fixed length as the larger reel). Today’s orders are for 110 reels of 22-inch width, 120 reels of 20-inch width, and 80 reels of 12-inch width. These smaller widths are to be cut from the larger standard size reel. For example, the company can decide to slit a jumbo reel into two reels each 22 inches wide, and one reel 20 inches wide; this leaves 4 inches of trim waste from the 68 inch jumbo. The production scheduler, Manny Blanks, wants to manufacture today’s orders to as to minimize total trim waste. (a) Find every possible way to slit a 68 inch jumbo reel into combinations of 22-inch, 20inch and 12-inch reels. One such combination was already illustrated in the example. Calculate the trim waste for each combination. Label the combinations 1, 2, 3, …, and let xi be the number of jumbo reels cut into combination i. There are at least 10 worthwhile combinations. Other combinations have too much trim waste to be useful in a product decision Variable X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
Combination 22, 22, 22 22, 22, 20 22, 20, 20 20, 20, 20 22, 20, 12, 12 12,12,12,12,12 22, 22, 12, 12 12, 12, 12, 12, 20 12, 12, 12, 22 20, 20, 12, 12
Total 66 64 62 60 64 60 68 68
Trim Waste 2 4 6 8 4 8 0 0
58 64
10 4
(b) Formulate the problem as a linear programming model and solve using QM or Excel solver. (Note that if any smaller reels in excess of the customer requirements are produced, these too must be counted as waste. yi is real valued and denotes the number of reels produced in excess of customer requirements. xi is integer valued and denotes number of Jumbo reels cut to pattern i.
Minimize 2x1 + 4x2 + 6x3 + 8x4 + 4x5 + 8x6 + 0x7 + 0x8 + 10x9 + 4x10 + 22y1 + 20y2 + 12y3 Subject to: 3x1 + 2x2 + 1x3 + 0.x4 + 1x5 + 0.x6 + 2x7 + 0x8 + 1x9 + 0x10 - y1 = 110 0x1 + 1x2 +2x3 + 3x4 +1x5 + 0x6 + 0x7 + 1x8 + 0x9 + 2x10 – y2 = 120 0x1 + 0x2 + 0x3 +0x4 + 2x5 + 5x6 + 2x7 + 4x8 + 3x9 + 2x10 – y3 = 80
Optimal value is 340 with x 1 = 36, x2=1, x4=33,x8=20 c) Show that the objective function also can be written as Minimize 68(x1+x2+x3+…..). 68(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10) is the number of Jumbo reel inches used in total Which is the same as total reel inches in the order + total reel inches in the excess reels + total trim waste = 22(120) + 20(110) + 12(80) + 22y1+20y2+12y3+ 2x1+4x2+6x3+8x4+4x5+8x6+0x7+0x8+10x9+4x10
