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Fall 2008
Homework #9 Solution
Chapter 12
Section 12.12
1.
By the method from the book:
x
0
= [1/2 1/2]
T
∇
f(x,y) = [4  4x  2y, 6  2x 4y]
∇
f(.5,.5) = [1 3] Find d
0
by solving
max z = d1 + 3d2
st
d1 + 2d2
≤
2
d1,d2
≥
0
Optimal solution is d
0
= [0 1]T Choose x
1
= [.5 .5]
T
+
t
0
[.5 .5]
T
= [.5 .5t
0
.5 + .5t
0
] where t
0
solves
max
f(.5  .5t, .5 + .5t) = 3.5 + t  t
2
/2 = g(t).
0
≤
t
≤
1
Then g'(t) = 1  t = 0 for t = 1. Since g"(t)<0, t
0
= 1 and
x
1
= [0 1]
T
. Here z = f(0, 1) = 4.
∇
f(x
1
) = [2 2]. We find d
1
by solving
max z = 2d1 + 2d2
st
d1 + 2d2
≤
2
d1,d2
≥
0
Optimal solution is d
1
= [2 0]
T
. Now x
2
= [0 1]
T
+ t
1
[2 1]
T
= [2t
1
1  t
1
] where t
1
solves
max f(2t, 1  t) = 4 + 2t  6t
2
= h(t). Then h'(t) = 2  12t
= 0
0
≤
t
≤
1
for t = 1/6. Since h"(t)<0, t
1
= 1/6 and x
2
= [1/3 5/6]. At
this point z = 4.17.
By the method during discussion section:
Iteration 1:
Step 1:
max z = d
1
+ 3d
2
st
1 ≤d
1
≤
1
1 ≤d
2
≤
1
=> d0=(1,1).
Step 2: x1=x0+t0*d0
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 Spring '08
 ADLER
 Operations Research

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