University of Southern California
Daniel J. Epstein Department of Industrial and Systems Engineering
ISE 330: Introduction to Operations Research
Homework 2 Solution: prepared by Xumei Tan
4.6-2 a), b)
Eq
BV
No
Z
0
1
-10M
-4
-4M
-2
-5M
-3
-7M
-5
0
Right
side
-600
M
0
_
1
0
2
3
4
2
1
0
2
0
8
1
1
5
0
1
x5
_
x6
Z
x1
x2
x3
x4
_
_
x5
x6
Initial Artificial BF solution is (0, 0, 0, 0, 300, 300)
Eq Z
BV
x1
x2
x3
x4
No
_
_
x6
0
1
0
-2.75M
-1.5
-3.75M
-2.5
-0.75M
-2.5
0
_
x5
1
0
0
2.75
3.75
0.75
1
x1
2
0
1
0.125
0.125
0.625
0
Eq Z
No
x1
x2
x3
x4
_
x6
0
1
0
0.3333
0
-2
x3
1
0
0
0.7333
1
0.2
x1
2
0
1
0.0333
0
0.6 -0.033
Eq Z
No
x1
x2
x3
x4
Z
0
1
3.333
3
0.4444
0
0
x3
1
0 -0.333
0.7222
1
0
0.125
_
1M
+0.66
7
0.266
7
37.5
Right
side
300
1M
+0.333
60
-0.067
30
0.1333
_
_
x5
x6
1M
+0.55
6
0.277
8
Right
side
1.25M
-225M
+0.5
+150
225
-0.25
x5
Z
BV
300
x5
Z
BV
300
1M
+0.778
-0.111
Right
side
400
50
x4
2
0
1.666
7
0.0556
0
1 -0.056
Optimal solution: (x1, x2, x3, x4) = (0, 0, 50, 50)
C)-f) phase 1
Eq
Z
BV
No
x1
Z
0
1
_
1
0
2
0
x5
_
x6
x2
x3
x4
Z*=400
_
_
x5
x6
-4
-5
-7
0
0
2
3
4
2
1
0
8
1
1
5
0
1
-10
50
0.2222
Right
side
-600
300
300
Initial Artificial BF Solution: (0, 0, 0, 0, 300, 300)
BV
Z
_
Eq Z
No
0
x1
1
x2
x3
x4
0
-2.75
-3.75
-0.75
_
_
x5
x6
0
Right
side
1.25
x5
1
0
0
2.75
3.75
0.75
1
-0.25
x1
2
0
1
0.125
0.125
0.625
0
0.125
BV
Eq Z
No
x1
x2
x3
x4
_
_
x5
x6
0
1
0
0
0
0
1
1
x3
1
0
0
0.7333
1
0.2
0.266
7
-0.067
x1
2
0
1
0.0333
0
0.6 -0.033
0.1333
BV
Eq Z
No
x1
x2
x3
Right
side
x4
Z
0
1
0
0.3333
0
-2
x3
1
0
0
0.7333
1
0.2
300
60
225
37.5
Right
side
Z
Phase 2:
-225
0
60
30
x1
BV
2
0
Eq Z
No
1
x1
0.0333
x2
0
x3
30
0.6
Right
side
x4
Z
0
1
3.333
3
0.4444
0
0
x3
1
0 -0.333
0.7222
1
0
x4
2
0
1.666
7
0.0556
0
1
400
50
50
Optimal Solution: (x1, x2, x3, x4) = (0, 0, 50, 50), Z*=400
(g)
The basic solutions of the two methods coincide. They are artificial basic feasible
_
_
solutions for the revised problem until both artificial variables x 5 and x 6 are
driven out of the basis, which in the two phase method is the end of phase 1.
4.6-5
Once all artificial variables are driven from the basis in a maximization (or
minimization) problem, choosing an artificial variable to reenter the basis can
only lower (raise) the objective value by an arbitrarily large amount depending
on M.
4.6-15 (a)
x2
-3x1+x2=6
(-1.14, 2.57)
Optimal
Z=11.43
x1+2x2=4
x1
x2=-3
(x1, x2)= (-1.14, 2.57) is optimal with Z=11.43
(b) Add new variable x3 . Let x2(old) +3 = x3 ,
And let x1(old) = x1  x2 (new), we get
Maximize  x1  x2  4 x3  12
 3x1  3x 2  x3  9
Subject to x1  x 2  2 x3  10
x1  0, x 2  0, x3  0
C)
BV
Eq
No
Z
x1
x2
x3
x4
Right
side
x5
Z
0
1
1
-1
-4
0
0
x4
1
0
-3
3
1
1
0
x5
2
0
1
-1
2
0
1
Z
0
1
3
-3
0
0
2
x4
1
0
-3.5
3.5
0
1
-0.5
x3
2
0
0.5
-0.5
1
0
0.5
Z
0
1
0
0
0
x2
1
0
-1
1
0
x3
2
0
0
0
1
0
9
10
20
4
5
0.851 1.571 23.4286
7
4
0.285
1.14286
-0.143
7
0.142 0.428 5.57143
9
6
Optimal solution for original problem is
(x1, x2) = (-1.14, 2.57), Z*= 11.43
For the revised problem the optimal solution is (0, 1.14, 5.57) with Z*=23.43
5.1-4
a) (x1, x2, x3) = (10, 0, 0)
b)x2=0
x3=0
x1-x2+2x3=10
5.1-13
a) True. If there are no optimal solutions, then the problem must have no feasible
solution or the objective value can be increased indefinitely (Chap. 3). The former
is not the case (assumed in the problem) and the latter cannot be true since the
feasible region is bounded. Thus, there must be at least one optimal solution.
b) False. If a solution is optimal, it need not be a BF solution. A convex combination
of BF solutions can give an optimal but not basic (i.e. CP) solution. However, it is
true that if optimal solutions exist, then at least one of them must be a BF solution.
This follows straight from Property 1 (since BF solutions
CPF solutions).
c) True. Since BF solutions correspond to CPF solutions, this follows directly from
Property 2.