Solutions to Homework Assignment #5
1.
a)
 53 13 0
 1  1 0
 x1 
 2
2 1 




1
B   2 5 0, B   3 3 0, x B   x2 , c B  3
1 1 1
 2  5 1
 x7 
0
b)
 53 13 0  2   203 
 203 


 
  82
x B  B 1b   23 13 0  5    143 , z  c BT x B  2 3 0 143  
3
1 1 1 10 15
15
 53 13 0  1  1
 53 13 0   1 0




t1  B 1 A1   23 13 0  2  0, t 2  B 1 A 2   23 13 0  5   1,
1 1 1  2  0
1 1 1  5 0
 53 13 0   1  83 
 53 13 0  0   1
t 3  B 1 A 3   23 13 0  3   53 , t 4  B 1 A 4   23 13 0  3   1,
1 1 1  0    3
1 1 1  3   0 
 53 13 0 1  53 
 53 13 0 0  13 


 


t 5  B 1 A 5   23 13 0 0   23 , t 6  B 1 A 6   23 13 0 1   13 
1 1 1 0 1
1 1 1 0 1
 53 13 0 0 0


t 7  B 1 A 7   23 13 0 0  0
1 1 1 1 1
z j  c j  c BT t j  c j
 83 
1
0 
34
 
z1  c1  2 3 00  2  0, z 2  c 2  2 3 01  3  0, z 3  c3  2 3 0 53   1   ,
3
  3
0
0
 53 
 13 
 1
16
5
 
z 4  c 4  2 3 0 1  1  6, z 5  c5  2 3 0 23   0  , z 6  c6  2 3 0 13   0  ,
3
3
1
1
 0 
0 
z 7  c7  2 3 00  0  0
1
So the corresponding simplex tableau looks like this:
x1 x2 x3
x4 x5
8
x1
1 0
 3 -1 53
x2
0 1
 53 -1 23
x7
0 0 -3
0 0
34
0 0
 3 -6 163
x6 x7
1
0 203
3
1
0 143
3
1 1 15
5
0 823
3
c) If we pick either x3 or x4 as the entering variable, the pivot column does not contain positive
entries. Therefore we stop here: the problem has no finite optimum.
Ex.2 (3.2)
T
An optimal solution to the dual problem is 0 3 15 0 5 with dual objection function
value 125.
By the Duality Theorem the primal problem has an optimal solution with the primal objective
function value 125. We cannot say anything about the values of the primal variables (we don’t
even know how many are there). However, since the 2nd, 3rd and 5th dual variables are different
from 0, the 2nd, 3rd and 5th constraints in the primal problem must be satisfied as strict equalities
at the optimal solution, i.e. the slacks in the 2nd, 3rd and 5th constraints of the primal problem
must be zero (according to the Complementary Slackness principle).
Ex.6 (3.2)
Primal problem:
min z  4 x  6 y
subject to
x  3y  5
2x  y  3
x, y  0
Optimal solution
4
7
58
x ,y ,z
5
5
5
Z=12
Dual problem
max z '  5u  3v
subject to
u  2v  4
3u  v  6
u, v  0
Optimal Solution
8
6
58
u  , v  , z' 
5
5
5
Z’=15
z  z' 
58
, so the optimal solutions to the two problems satisfy the Duality Theorem.
5
Ex. 10 (3.2)
max z  3 x1  4 x 2
subject to
x1  2 x 2  10
x1  x 2  8
3 x1  5 x 2  26
x1 , x 2  0
Z=12
The optimal solution to the primal problem is x1  7, x2  1, z  25
This solution satisfies the first constraint x1  2x2  10 as a strict inequality (9<10), i.e. with a
non-zero slack. The Complementary Slackness principle implies that the first variable in the
dual problem must be zero ( w1  0 ) in any optimal solution to the dual problem.
Ex. 9 (3.2)
max z  9 x1  14 x 2  7 x3
subject to
2 x1  x 2  3 x3  6
5 x1  4 x 2  x3  12
2 x2
5
x1 , x 2 , x3 unrestrict ed
5
5
27
x1 
, x 2  , x3 
is a feasible solution to this problem with z = 44.
26
2
26
min z '  6w1  12w2  5w3
subject to
2w1  5w2
9
w1  4w2  2w3  14
3w1  w2
7
w1 , w2 , w3  0
All points on this line have
w1  2, w2  1
3
2
1
Optimal solution w1  2, w2  1, w3  4, z '  44 (can be found by solving the system of three
linear equations with variables).
We
have z=z’=44, therefore by the Duality Theorem,
5
5
27
x1 
, x 2  , x3 
is optimal for the primal problem.
26
2
26
the
feasible
Ex.2 (3.3)
cB
-1
0
x5
x6
4
x1
0
0
4
x1
1
0
5
3
4
3
x2
x3
4
3
1
3
1
3
2
3
2
3
1
6
 53
 43
 23
x7
 13
3
x4
0
1
-1
x5
1
0
0
x6
0
1
1
2
0
0
1
6
4
-1
0
0
5
3
12
 13
xB
4
10
solution