Solutions to Review Problems
1. maximize z = 5x1 +4x2
subject to
x1 + 2 x2  6
–2x1 + x2  5
5x1 + 3 x2  15
x1  0, x2  0
a) Optimal solution
12
15
120
x1  , x 2  , z 
7
7
7
b) Dual Problem
min z '  6 w1  5w2  15w3
subject to
w1  2 w2  5w3  5
2 w1  w2  3w3  4
w1 , w2 , w3  0
c) Complementary Slackness conditions
x1 ( w1  2w2  5w3  5)  0
x 2 (2w1  w2  3w3  4)  0
w1 (6  x1  2 x 2 )  0
w2 (5  2 x1  x 2 )  0
w1 (15  5 x1  3x 2 )  0
Substituting x1 
12
15
, x 2  , we obtain
7
7
12
( w1  2 w2  5w3  5)  0  w1  2w2  5w3  5
7
15
(2 w1  w2  3w3  4)  0  2w1  w2  3w3  4
7
w1  0  0
44
0
7
w1  0  0
w2 
 w2  0
The optimal solution to the dual problem is w1 




5


 w1  7


   w2  0


6

w3  7





5
6
120
, w2  0, w3  , z ' 
.
7
7
7
2. Consider the linear programming problem
Maximize z= 2x1+3x2-x3
Subject to
x1+2x2-x3+x4 =6
x1-3x2-3x3+ x5=10
xj>=0, j=1,2,3
a) x  [6,0,0,0,4]T is a feasible solution for this problem, i.e. it satisfies all the constraints. Also,
1
1
0 
1 
columns   and   corresponding to variables x1 , x5 are linearly independent. So
x  [6,0,0,0,4]T is a basic feasible solution to the problem above.
b) Dual Problem
min z '  6 w1  10w2
subject to
w1  w2  2
2w1  3w2  3
 w1  3w2  1
w1 , w2  0
Complementary Slackness conditions:
x1 ( w1  w2  2)  0
x 2 (2w1  3w2  3)  0
x3 ( w1  3w2  1)  0
x 4  w1  0 or w1 (6  x1  2 x 2  x3 )  0
x5  w2  0 or w2 (10  x1  3x 2  3x3 )  0
Substituting x1  6, x2  0, x3  0, x4  0, x5  4, we obtain
6  ( w1  w2  2)  0
 w1  w2  2
0  (2 w1  3w2  3)  0
0  ( w1  3w2  1)  0
0  w1  0
4  w2  0
 w2  0



 w1  2
  
 w2  0



Now w*  [2,0]T is a feasible solution to the dual problem with z’=12.
x *  [6,0,0,0,4]T is a feasible solution to the primal problem with z=12.
According to the Duality Theorem, x * , w* are optimal solutions in the respective problems.
Ex. 1 (3.2)
An optimal solution is [0.27, 1.83, 0.94, 0.5, 0, 0]T with objective function value 117.81.
The dual problem has optimal solution with objective function value 117.81. The 1st, 2nd, 3rd and
4th constraints in the dual problem are satisfied as strict equations at any optimal solution, i.e. the
slacks in these constraints are equal to 0 in every optimal solution to the dual problem.
Ex. 3 (3.2)
There are no feasible solutions to the problem.
The dual problem is either infeasible or has no finite optimum.
Ex. 4 (3.2)
There are solutions to the dual problem with arbitrary large dual objective function values.
The primal problem is infeasible.
Ex. 5 (3.2)
max z  c T x
s.t.
Ax  b
x0
b  [12, 21, 8, 2, 5]T , w  [0, 4, 5, 0, 3]T is an optimal solution to the dual problem
The optimal value of the primal objective function is
0 
 4
 
z  z '  b T w  12 21 8 2 55  139
 
0 
3
Ex.11 (3.2)
max z  4 x1  2 x 2  3 x3
s.t.
2 x1  3 x 2 
x3  12
x1  4 x 2  2 x3  10
3 x1 
x 2  x3  10
x1 , x 2 , x3  0
min z '  12 w1  10 w2  10 w3
s.t.
2 w1  w2  3w3  4
3w1  4 w2  w3  2
w1  2 w2  w3  3
w1 , w2 , w3  0
Optimal solution to the primal problem x1  2, x2  0, x3  4, z  20.
Complementary slackness conditions:
x1 (2 w1  w2  3w3  4)  0  2  (2 w1  w2  3w3  4)  0
x 2 ( 3w1  4 w2  w3  2)  0  0  ( 3w1  4 w2  w3  2)  0 
w1  0
w1  0

x3 ( w1  2 w2  w3  3)  0  4  ( w1  2 w2  w3  3)  0 



  w2  3w3  4  w2  1
w1 (2 x1  3 x 2  x3  12)  0 w1  4  0

w3  1 
2 w2  w3  3

w2 ( x1  4 x 2  2 x3  10)  0  w2  0  0


w3 (3 x1  x 2  x3  10)  0  w3  0  0

Optimal solution to the dual problem w1  0, w2  1, w3  1, z '  20.
Ex. 7 (3.3)
max z  x1  x 2  x3  x 4
s.t.
x1  2 x 2  x3  3x 4  12
x1  3x 2  x3  2 x 4  8
2 x1  3x 2  x3  2 x 4  7
x1 , x 2 , x3 , x 4  0
a) Solution using the Revised Simples method
1 0 0
 x5 
12
0 






1
B  0 1 0, x B   x 6   B b   8 , c B  0, z  c BT x B  0
0 0 1
 x7 
 7 
0
z1  c1  1, z 2  c 2  1, z 3  c3  1, z 4  c 4  1, z 5  c5  0, z 6  c6  0, z 7  c7  0,
x3  entering var .
1
 1
1 1 0


t 3  B A3   1 , 2nd basic var . departing ( x6 ). E  0 1 0
 1
0 1 1
1 1 0
 x5 
20
0 






1
1
B  0 1 0, x B   x3   B b   8 , c B  1, z  c BT x B  8, c BT B 1  [0,1,0]
0 1 1
 x7 
 7 
0
1 
2
 1




z1  c1  [0,1,0]1  1  0, z 2  c 2  [0,1,0] 3   1  2, z 3  c3  [0,1,0] 1   1  0
2
 3
 1
1
 3
1
0 




z 4  c 4  [0,1,0]2  1  1, z 5  c5  [0,1,0]0  0  0, z 6  c6  [0,1,0]1  0  1,
2
0
0
0 
z 7  c7  [0,1,0]0  0  0
1
Optimal solution to the Primal Problem: x1  0, x2  0, x3  8, x4  0, z  8
b) Dual Problem
min z '  12w1  8w2  7 w3
s.t.
w1  w2  2w3  1
2w1  3w2  3w3  1
 w1  w2  w3  1
3w1  2w2  2w3  1
w1 , w2 , w3  0
Optimal solution to the Dual Problem:
1 1 0
T
T
1
w  c B B  [0,1,0]0 1 0  [0,1,0]
0 1 1
or
w1  ( z 5  c5 )  c5  0  0  0, w2  ( z 6  c6 )  c6  1  0  1, w3  ( z 7  c7 )  c7  0  0  0
and z’=8
#8
Primal Problem
max z  x1  3 x3
Dual Problem
min z '  4 w1  5w2
s.t.
s.t.
x1  2 x 2  7 x3  4
x1  3 x 2  x3  5
x1 , x 2 , x3  0
Optimal Solution to the Dual problem:
w1  3, w2  2, z'  2
w1  w2  1
2 w1  3w2  0
7 w1  w2  3
w1 , w2  unrestrict ed
Complementary Slackness conditions:
x1 ( w1  w2  1)  0
x1  0  0




x 2 (2 w1  3w2  0)  0
x2  0  0
x1  2 



x3 (7 w1  w2  3)  0
  x3  16  0
  x2  1 
x3  0
w1 ( x1  2 x 2  7 x3  4)  0
3( x1  2 x 2  7 x3  4)  0 


w2 ( x1  3 x 2  x3  5)  0 
 2( x1  3 x 2  x3  5)  0
Optimal solution to the Primal problem:
x1  2, x2  1, x3  0, z  2
#10.
maximize z = 12x1 + 20x2 +21x3 + 18x4
subject to
24x1 + 40x2 + 46x3 +44x4  1200
x1 + x2 + x3 + x4  30
3x1 + 6x2 + 6x3 + 6x4 150
x1  0, x2  0, x3  0, x4  0
Someone claims that the final simplex tableau has the objective row
12
X1
20
x2
21
x3
18
x4
0
x5
0
X6
0
x7
0
1
0
3
0
4
3
540
The solution to the dual problem from this tableau is
w1  0, w2  4, w3  3
z '  1200w1  30w2  150w3  570  540
so there must be a mistake in the tableau.