MATH 3170 Assignment #5 Solution (20 marks)
Questions marked: 4.8.6, 4.11.4, 4.12.2 and 4.12.5
4.8.4
We wish to solve
max z = x1 - 0.25x2
st
x1 - 0.25x20 (dollar constraint)
-3x1 +
x20 (franc constraint)
x1,
x20
z
x1
x2
s1
s2
RHS
1
-1
0.25 0
0
0
-------------------------------0
-1
0.25 1
0
0
-------------------------------0
3
-1
0
1
0
-------------------------------x1 enters the basis
z
x1
x2
s1
s2
RHS
1
0 -1/12 0 1/3
0
-------------------------------0
0 -1/12 1
1/3
0
-------------------------------0
1 -1/3 0
1/3
0
-------------------------------The x2 column now indicates that the LP problem is unbounded.
4.8.6 (5 marks)
-Z
X1
X2
S1
S2
RHS
1
-1
-3
0
0
0
0
1
-2
1
0
4
0
-1
1
0
1
3
X2 enters the basis in ROW 2 yielding the following tableau:
-Z
X1
X2
S1
S2
RHS
1
-4
0
0
3
9
0
-1
0
1
2
10
0
-1
1
0
1
3
We would like to enter X1 into the basis but there is no row in
which X1 has a positive coefficient. Therefore, the LP is
unbounded.
4.11.2
-z
x1
x2
s1
s2
RHS
----------------------------1 -1
-1
0
0
0
----------------------------0
1
1
1
0
1* Enter x1 in row 1
----------------------------0 -1
1
0
1
0
-z
x1
x2
s1
s2
RHS
----------------------------1
0
0
1
0
1
----------------------------0
1
1
1
0
1
----------------------------0
0
2
1
1
1
----------------------------This is an optimal tableau with an optimal solution being x1 = 1,
x2 = 0.
4.11.4 (5 marks)
Here are the pivots:
Z
1
0
0
0
Z
1
0
0
0
Z
1
0
0
0
X1
X2
X3
X4
S1
3
-1
6
0
0
9
1
-9
-2
1
1
1/3
-2
-1/3 0
-9
-1
9
2
0
BV={S1,S2,S3}.
X2 now enters in Row 1 yielding
X1
X2
X3
X4
S1
S2
0
0
1
0
S3
0
0
0
1
the following tableau.
S2
S3
RHS
12
0
-3
-2
1
0
0
9
1
-9
-2
1
0
0
-2
0
1
1/3
-1/3 1
0
0
0
0
0
1
0
1
BV={X2,S2,S3}.
We now enter X3 into the basis in Row 2.
X1
X2
X3
X4
S1
RHS
0
0
0
1
S2
S3
6
0
0
-1
0
3
0
-9
1
0
1
-2
9
0
-2
0
1
1/3
-1/3 1
0
0
0
0
0
1
0
1
BV={X2,X3,S3}.
We now enter X4 into the basis in Row 1.
0
0
0
1
RHS
0
0
0
1
Z
1
0
0
0
Z
1
0
0
0
Z
1
0
0
0
X1
X2
X3
-3
1
0
-9
1
0
1
-1/3 1
0
0
0
BV={X4,X3,S3}.
X1 now enters basis
X1
X2
X3
X4
S1
S2
S3
RHS
0
1
0
0
-2
-2
1/3
1
12
9
-2
0
0
0
0
1
0
0
0
1
S1
S2
S3
RHS
-1
1
1/3
1
6
-9
-2
0
0
0
0
1
0
0
0
1
S2
S3
RHS
-3
-9
1
9
0
0
0
1
0
0
0
1
in Row 2.
X4
0
0
3
0
0
-2
9
1
1
-1/3 1
0
0
0
0
0
BV={X4,X1,S3}.
We now choose to enter S1
X1
X2
X3
X4
in Row 1.
S1
0
-2
12
1
0
0
-2
9
1
1
1
1/3
-2
-1/3 0
0
2
-9
-1
0
BV={S1,X1,S3}.
S2 would now enter basis in Row
the initial tableau, so cycling
2. This will bring us back to
has occurred.
4.12.2 (5 marks)
After adding slack, excess and artificial variables, we obtain the
following:
min z = 2x1 + 3x2 + Ma1
s.t.
2x1 + x2 - e1 + a1 = 4
-x1 + x2 + s2 = 1
with all variables nonnegative. After eliminating a1 from -z +2x1
+3x2 +Ma1 = 0 we obtain -z +(2 - 2M)x1 + (-M+3)x2 + Me1 = -4M.
Proceeding with the simplex we obtain
-z
x1
x2
e1
s2
a1
RHS
____________________________________________________
1
-2M+2
-M+3
M
0
0
-4M
____________________________________________________
0
2
1
-1
0
1
4
____________________________________________________
0
-1
1
0
1
0
1
____________________________________________________
x1 enters and a1 leaves.
____________________________________________________
1
0
2
1
0
M-1
-4
____________________________________________________
0
1
1/2
-1/2
0
1/2
2
____________________________________________________
0
0
3/2
-1/2
1
1/2
3
____________________________________________________
Note that we could have omitted the column of a1 since it turns
nonbasic in the above iteration.
This is an optimal tableau with optimal solution x1 = 2, s2 = 3,
e1 = x2 = 0.
4.12.5 (5 marks)
Initial tableau is
-Z
X1
1
-3M+1
0
2
0
1
We now enter X3 in
X2
-2M+1
1
1
the basis
X3
A1
-3M
0
1
1
2
0
in row 2:
A2
0
0
1
RHS
-6M
4
2
-Z
X1
X2
X3
A1
A2
RHS
1
-3M/2+1
-M/2+1
0
0
3M/2
-3M
0
3/2
1/2
0
1
-1/2
3
0
1/2
1/2
1
0
1/2
1
The column of A2 could have been omitted since it is a non-basic
artificial variable. We now enter X1 in the basis in row 2.
This is done according to the smallest subscript rule when there
is a tie in choosing the leaving variable. Both X3 and A1 are
candidates for leaving variable; X3 is chosen since it precedes
A1 in our ordering of the variables.
-Z
X1
X2
X3
A1
A2
RHS
1
0
M
-2+3M
0
-1+3M
-2
0
0
-1
-3
1
-2
0
0
1
1
2
0
1
2
This optimal tableau yields the optimal solution X1=2, X2=X3=0
with Z=2 for the original problem, since all the artificial
variables are now zero (A2 is non-basic, A1 is basic but
degenerate).
Note that if we leave A1 in the second iteration, we would obtain
another optimal tableau for the basis {X1,X3}. Both bases
{X1,X3} and {A1,X1} correspond to the same optimal BFS (of the
Big M problem) since the problem is degenerate. Both cases
correspond to the same optimal solution for the original problem:
(2,0,0).