E160 Operations Research I Fall 2008 Homework #1 Solution Chapter 2
Section 2.3
1 0 1. 1
1 1 2
0 1 1
1 3 0 4 1 8
1 1 0 1 3 0 1 1 0 4 0 1 1 0 5
1 0 0
0 1 0
1 1 0
1 1 0 4 0 1
The last row of the last matrix indicates that the original system has no sol
E160 Operations Research I Fall 2008 Homework #2 Solution Chapter 12
Section 12.2
1. Let S = soap opera ads and F = football ads. Then we wish to min z = 50S + 100F st 5S1/2 + 17F1/2 40 (men) 20S1/2 + 7F1/2 60 (women) S 0, F 0 9.Let Wi = number of widgets
E160 Operations Research I Fall 2008 Homework #3 Solution Chapter 12
Section 12.4
1. Let f(x) = profit if $x is spent on advertising. Then f(0) = 0 and for x>0, f(x) = 300x1/2 - 100x1/2 - 5000 - x. Since f(x) has no derivative at x = 0, maximum profit occ
E160 Operations Research I Fall 2008 Homework #4 Solution Chapter 12
Section 12.5
1. a = -3 b = 5, so b - a = 8. x1 = 5 - .618(8) = .056 x2 = -3 + .618(8) = 1.944. f(x1) = .115, f(x2) = 7.67 f(x2)>f(x1) so interval of uncertainty is now (.056,5]. Then x3
E160 Operations Research I Fall 2008 Homework #5 Solution Chapter 12
Section 12.3
1. f'(x) = 6x 0, (for x 0) so f(x) is convex on S. 3. f'(x) = 2x-3>0 (for x>0). Thus f(x) is convex on S. 5. f'(x) = -x-2<0, so f(x) is a concave function on S. 7. f(x1, x2)
E160 Operations Research I Fall 2008 Homework #7 Solution Chapter 12
Section 12.8
2. We wish to maximize L2/3K1/3 subject to 2L + K = 10. It is easier to maximize ln L2/3K1/3 = (2/3)ln L + (1/3)ln K (this is a concave function so we know that Lagrange mul
E160 Operations Research I Fall 2008 Homework #10 Solution Chapter 8
Section 8.6
1. We begin at Gary and include the Gary-South Bend arc. Then we add the South Bend-Fort Wayne arc. Next we add the Gary-Terre Haute arc. Finally we add the Terre Haute-Evans