(a) Solve the following linear programming problem by the graphical method
(not by the simplex algorithm.)
Minimize:z = 16x1 + 9x2
subject to:x1 + x2 = 34 ,9x1 + 23x2 = 320, 23x1+ 9x2 = 320 and x1,x2 = 0
(the = signs above in (subject to) are actually meant to be greater or equal to
signs).
State the minimum value of z and the values of x1, x2 at which it occurs
(b) Given that x1 and x2 are unrestricted in sign, formulate the dual of the
problem in part (a). Show that the feasible region for the dual problem is
a line segment in R^3, and find the optimal point.
(c) Now suppose the objective function in the primal problem is changed to
z = 24x1 + 9x2. Show that the dual problem now has no feasible
solution. State what this tells you about the primal problem, and explain
it by reference to the graph.
Minimize z = 16x1 + 9x2
Subject to
X1 + x2 ≥ 34
9x1+23x2 ≥ 320
23 x1 + 9x2 ≥ 320
X1 ≥0, x2≥ 0
A
B
C
D
The vertices of the feasible region are determined as follows.
Solve x1 =0
23x1 +9x2 = 320
X2 = 320/9 = 320/9
A is (0, 320/9)
x1+x2 = 34
23x1 + 9x2 = 320
23x1 + 23 x2 = 782
14x2 = 462
X2 = 462/14 = 33
X1 =1
The point B is (1, 33)
9x1 + 23x2 = 320
X1 + x2 = 34
9x1 + 9x2 = 306
14x2=14
X2=1
X1=33
The point C is (33,1)
9x1 + 23x2 = 320
x2 = 0
x1 = 320/9
The point D is 320/9
Point
A
B
C
D
X1
0
1
33
320/9
X2
320/9
33
1
0
z
320
313
537
568.89
The minimum value of z is 313.
X1 = 1
X2 = 33
The dual of
Minimize z = 16x1 + 9x2
Subject to
X1 + x2 ≥ 34
9x1+23x2 ≥ 320
23 x1 + 9x2 ≥ 320
is
Maximixe w = 34y1 + 320 y3 +320 y3
Subject to
Y1 + 9y2 + 23 y3 = 16
Y1 + 23 y2 + 9y3 = 9
Y1,y2,y3≥0
The feasible region is the line segment lying in the nonnegative octant of the line determined by the
intersection of the planes
Y1 + 9y2 + 23 y3 = 16
Y1 + 23 y2 + 9y3 = 9
9
41
The meeting point of this line with the coordinate planes are (0, 64 , 64), (9/2, 0, 1/2), and
(41/2, −1/2, 0). So the feasible region is the line segment joining the points (0,
(9/2, 0, 1/2) which are determined as follows.
Y1 = 0
9y2 + 23 y3 = 16
23 y2 + 9y3 = 9
𝑦2 =
16 23
|
|
9
9
9 23
|
|
23 9
= −448 = 64
𝑦3 =
9 16
|
|
23 9
9 23
|
|
23 9
= −448 = 64
(0,
9 41
, )
64 64
Y2 = 0
Y1 + 23 y3 = 16
Y1 + 9y3 = 9
14 y3 = 7
Y3 = ½
Y1 = 9-9/2= 9/2
(9/2, 0, 1/2)
Y3 = 0
Y1 + 9y2 = 16
Y1 + 23y2 = 9
-14y2 = 7
Y2 = -1/2
−63
9
−287
41
9 41
, ),
64 64
and
Y1 = 16 + 9/2 = 41/2
(41/2, −1/2, 0)
9
41
The feasible region is the line segment joining (0, 64 , 64) and (9/2, 0, 1/2)
The values of W at the points are
w = 34y1 + 320 y3 +320 y3 = 34 × 0 + 320 ×
9
64
+ 320 ×
9
2
41
64
= 0 + 45 + 205 = 250
1
2
w = 34y1 + 320 y3 +320 y3 = 34 × + 320 × 0 + 320 × = 153 + 0 + 160 = 313
The optimal solution is y1 = 9/2, y2 = 0, y3 = ½, W = 313
Now consider
Minimize z = 24x1 + 9x2
Subject to
X1 + x2 ≥ 34
9x1+23x2 ≥ 320
23 x1 + 9x2 ≥ 320
The dual is
Maximize w = 34y1 + 320y2 + 320y3
Subject to
Y1 + 9y2 + 23y3 = 24
Y1 + 23y2 + 9y3 = 9
Y1, y2, y3 ≥0
Y1 =0
24
𝑦2 = 9
9
|
23
|
9
23
𝑦3 =
9
|
23
|
Y2=0
24-23y3=9-9y3
23
|
9 =− 9
23
448
|
9
24
|
9 = −471 = 471
23
| −448 448
9
15 = 14y3
Y3=15/14
Y1 = 9 – 135/14 = -9/14
Y3 = 0
24 – 9y2 = 9 -23 y2
14 y2 = -15
Y2 = -15/14
Y1 = 24 + 135/14 = 471/14
Since points of intersection of the line with coordinate planes are not feasible.
There is no feasible solution.
Therefore, the primal problem has an unbounded solution.
In the graph, if we move along the line BA beyond A, the value of the objective function goes on
decreasing without limit and tends to -∞.