Assn 2
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3-28.
Let A = number of shares of stock A B = number of shares of stock B C = number of shares of stock C D = number of shares of stock D
a. To get data on a per share basis multiply price by rate of return or risk measure value. Min 10A s.t. 100A 12A
+ 3.5B
+
4C
+ 3.2D
+ 50B + 4B
+ 80C + 4.8C
+ 40D + 4D
= 200,000 ≥ 18,000
≤ ≤ 80C ≤ 40D ≤ A, B, C, D ≥ 0
100A 50B
(9% of 200,00)
100,000 100,000 100,000 100,000
Solution: A = 333.3, B = 0, C = 833.3, D = 2500 Risk: 14,666.7 Return: 18,000 (9%) from constraint 2 b. Max
12A +
4B
+
4.8 C
+
4D
100 + A 100 A
50 B
+
80 C
+
40 D
s.t.
50 B 80 C A, B, C, D ≥ 0
40 D
=
200,000
≤
100,000
≤
100,000
≤
100,000
≤
100,000
Solution: A = 1000, B = 0, C = 0, D = 2500 Risk: 10A + 3.5B + 4C + 3.2D = 18,000 Return: 22,000 (11%) c. The return in part (b) is $4,000 or 2% greater, but the risk index has increased by 3,333. Obtaining a reasonable return with a lower risk is a preferred strategy in many
financial firms. The more speculative, higher return investments are not always preferred because of their associated higher risk. 3-35.
a.
min
Let B = number of copies done by Benson Printing J = number of copies done by Johnson Printing L = number of copies done by Lakeside Litho 2.45 + B
2.5J +
2.75 L
s.t. B J 0.9 + B B -
0.99 + J 0.1J
L 0.995 L
≤ ≤ ≤ =
30,000 50,000 50,000 75,000
≥ 0 L ≥ 30,000 B, J, L ≥ 0
Benson Johnson Lakeside # useful reports Benson - Johnson % Minimum Lakeside
Optimal Solution: B = 4,181, J = 41,806, L = 30,000 b. Suppose that Benson printing has a defective rate of 2% instead of 10%. The new optimal solution would increase the copies assigned to Benson printing to 30,000. In this case, the additional copies assigned to Benson Printing would reduce on a one-forone basis the number assigned to Johnson Printing. c. If the Lakeside Litho requirement is reduced by 1 unit, total cost will decrease by $0.2210. 4-5.
Let
x1 = amount of ingredient A x2 = amount of ingredient B x3 = amount of ingredient C Mi 0.10x n 1 s.t. 1x1 1x1 1x1 or 1x1 or -1 /2x1
+ 0.03x 2
+ 0.09x 3
+ +
1x2 + 1x2 +
-
1x2 +
1x3 ≥ 10 1x3 ≤ 15 ≥ 1x2 ≥0 1x3 ≥ 1/2x1 1x3 ≥ 0
x1, x2, x3 ≥ 0 Solution: x1 = 4, x2 = 4, x3 = 2 Cost = $0.70 per gallon.
[1] [2] [3] [4]
Let A = number of shares of stock A B = number of shares of stock B C = number of shares of stock C D = number of shares of stock D
a. To get data on a per share basis multiply price by rate of return or risk measure value. Min 10A s.t. 100A 12A
+ 3.5B
+
4C
+ 3.2D
+ 50B + 4B
+ 80C + 4.8C
+ 40D + 4D
= 200,000 ≥ 18,000
≤ ≤ 80C ≤ 40D ≤ A, B, C, D ≥ 0
100A 50B
(9% of 200,00)
100,000 100,000 100,000 100,000
Solution: A = 333.3, B = 0, C = 833.3, D = 2500 Risk: 14,666.7 Return: 18,000 (9%) from constraint 2 b. Max
12A +
4B
+
4.8 C
+
4D
100 + A 100 A
50 B
+
80 C
+
40 D
s.t.
50 B 80 C A, B, C, D ≥ 0
40 D
=
200,000
≤
100,000
≤
100,000
≤
100,000
≤
100,000
Solution: A = 1000, B = 0, C = 0, D = 2500 Risk: 10A + 3.5B + 4C + 3.2D = 18,000 Return: 22,000 (11%) c. The return in part (b) is $4,000 or 2% greater, but the risk index has increased by 3,333. Obtaining a reasonable return with a lower risk is a preferred strategy in many
financial firms. The more speculative, higher return investments are not always preferred because of their associated higher risk. 3-35.
a.
min
Let B = number of copies done by Benson Printing J = number of copies done by Johnson Printing L = number of copies done by Lakeside Litho 2.45 + B
2.5J +
2.75 L
s.t. B J 0.9 + B B -
0.99 + J 0.1J
L 0.995 L
≤ ≤ ≤ =
30,000 50,000 50,000 75,000
≥ 0 L ≥ 30,000 B, J, L ≥ 0
Benson Johnson Lakeside # useful reports Benson - Johnson % Minimum Lakeside
Optimal Solution: B = 4,181, J = 41,806, L = 30,000 b. Suppose that Benson printing has a defective rate of 2% instead of 10%. The new optimal solution would increase the copies assigned to Benson printing to 30,000. In this case, the additional copies assigned to Benson Printing would reduce on a one-forone basis the number assigned to Johnson Printing. c. If the Lakeside Litho requirement is reduced by 1 unit, total cost will decrease by $0.2210. 4-5.
Let
x1 = amount of ingredient A x2 = amount of ingredient B x3 = amount of ingredient C Mi 0.10x n 1 s.t. 1x1 1x1 1x1 or 1x1 or -1 /2x1
+ 0.03x 2
+ 0.09x 3
+ +
1x2 + 1x2 +
-
1x2 +
1x3 ≥ 10 1x3 ≤ 15 ≥ 1x2 ≥0 1x3 ≥ 1/2x1 1x3 ≥ 0
x1, x2, x3 ≥ 0 Solution: x1 = 4, x2 = 4, x3 = 2 Cost = $0.70 per gallon.
[1] [2] [3] [4]
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