EMGT 501
HW #2
4.4-6(b) (c)
6.1-4,
6.1-5
Due Day: Sep. 21
4.4-6 Consider the following problem.
Maximize
subject to
Z  2 x1  4 x2  3x3 ,
3x1  4 x2  2 x3  60
2 x1  x2  2 x3  40
x1  3x2  2 x3  80
and
x1  0, x2  0, x3  0.
(b) Work through the simplex method step by step in tabular form.
(c) Use a software package based on the simplex method to solve
the problem.
4.4-6 (b)
Basis
Z
X4
X5
X6
Z
X2
X5
X6
Z
X2
X3
X6
(c)
Z
1
0
0
0
1
0
0
0
1
0
0
0
X1
-2
3
2
1
1
3/4
5/4
-5/4
11/6
1/3
5/6
-5/3
X2
-4
4
1
3
0
1
0
0
0
1
0
0
X3
-3
2
2
2
-1
1/2
3/2
1/2
0
0
1
0
X4
0
1
0
0
1
1/4
-1/4
-3/4
5/6
1/3
-1/6
-2/3
X5
0
0
1
0
0
0
1
0
2/3
-1/3
2/3
-1/3
X6
0
0
0
1
0
0
0
1
0
0
0
1
Optimal solution ( x1 *, x2 *, x3 *, x4 *, x5 *, x6 *)  (0, 20 / 3, 50 / 3, 0, 0, 80 / 3)
with Z  230 / 3.
RHS
0
60
40
80
60
15
25
35
230/3
20/3
50/3
80/3
6.1-4
For each of the following linear programming models, give your recommendation on
which is the more efficient way (probably) to obtain an optimal solution: by applying the
simplex method directly to this primal problem or by applying the simplex method directly
to the dual problem instead. Explain.
(a) Maximize
Z  10 x1  4 x2  7 x3 ,
subject to
3 x1  x2  2 x3  25
x1  2 x2  3x3  25
5 x1  x2  2 x3  40
x1  x2  x3  90
2 x1  x2  x3  20
and
x1  0, x2  0, x3  0.
(b) Maximize
Z  2 x1  5x2  3x3  4 x4  x5 ,
subject to
x1  3x2  2 x3  3x4  x5  6
4 x1  6 x2  5 x3  7 x4  x5  15
and
x j  0,
for j  1, 2, 3, 4, 5.
6.1-4 (a) Dual formulation becomes
Min
s.t.
25 y1  25 y2  40 y3  90 y4  20 y5
3 y1  1y2  5 y3  1y4  2 y5  10
 1y1  2 y2  1y3  1y4  1y5  4
2 y1  3 y2  2 y3  1y4  1y5  7
y1 , y2 , y3 , y4 , y5  0.
# of constraints of Dual = 3
# of constraints of Primal = 5
So, Dual is better than Primal because the size
of B-1 in Dual is smaller than that of Primal.
(b) Dual formulation becomes
Min
6 y1  15 y2
s.t.
y  4y  2
1
2
3 y1  6 y2  5
2 y1  5 y2  3
3 y1  7 y2  4
y1  y2  1
# of constraints of Dual = 5
# of constraints of Primal = 3
So, Primal is better than Dual because the size
of B-1 in Primal is smaller than that of Dual.
6.1-5
Consider the following problem.
Maximize
Z   x1  2 x2  x3 ,
subject to
x1  x2  2 x3  12
x1  x2  x3  1
and
x1  0, x2  0, x3  0.
(a) Construct the dual problem.
(b) Use duality theory to show that the optimal solution
for the primal problem has
Z  0.
6.1-5
(a)
Min 12 y1  1y2
s.t.
y1  y2  1
y1  y2  2
2 y1  y2  1
y1  0, y2  0
(b)
It is clear that Z*=0, y1*=0, y2*=0.