Solutions_ISE 230_HW3_Fall23
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School
San Jose State University *
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Course
230
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
Pages
7
Uploaded by NickTian on coursehero.com
ISE 230 HW 3
Due October 7, 2023 at 3pm
Question 1.
Consider the following LP and its optimal tableau:
Max ࠵? = 1.5 ࠵?
+
+ 4࠵?
.
− ࠵?
0
s.t.
2࠵?
+
+ 5࠵?
.
+ ࠵?
0
≤ 6
4࠵?
+
+ 2࠵?
.
− ࠵?
0
≤ 10
࠵?
+
, ࠵?
.
, ࠵?
0
≥ 0
Optimal Tableau:
Z
X1
X2
X3
S1
S2
RHS
1
1/10
0
9/5
4/5
0
24/5
0
2/5
1
1/5
1/5
0
6/5
0
16/5
0
-7/5
-2/5
1
38/5
From this table you can identify the optimal basis ( basic variables set).Using this information:
a)
Find the range of values for the right hand side for constraint 2 for which the current basis
remains optimal (be careful, I am not asking amount of change Delta, but asking the range of b2)
Find the new optimal solution if the right hand side of constraint 2 changes from 10 to 5.
b)
Find the range of values for the cost coefficient of
࠵?
0
such that the current basis remains
optimal.
What is the optimal objective function value if the cost coefficient of
࠵?
0
changes from
-1 to 0.5? Explain why z changes or does not change.
Solution:
a)
From the optimal tableau we know
࠵? = 8
5
0
2
1
9
:.; <=>?@A
BCCCCCCCD ࠵?
E+
= F
1/5
0
−2/5
1
H
In order to keep the current basis optimal
࠵?
E+
I
J
K.
L ≥ 0 → N
+
O
0
−
.
O
1
P I
J
K.
L ≥ 0 → ࠵?
.
≥ 12/5
So the range for the rhs of the second constraint is [0, 2.4]. As long as the right hand side of the
second constraint is bigger than 2.4, the current basis (intersection of the same constraints will
be optimal)
When rhs of the second constraint becomes 5, the optimal solution is
8
R
S
T
S
9 = F
1/5
0
−2/5
1
H I
J
O
L = 8
J/O
+0/O
9
,
࠵?
+
, ࠵?
0
, ࠵?
+
= 0
Calculate the new z value by multiplying c_BV*x_BV=24/5=4.8
b)
In order to keep the current basis optimal, all nonbasic variables’ row 0 values should stay >=0.
࠵?
W
࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? (࠵?࠵?࠵? 0 ࠵?࠵?࠵?࠵?࠵?࠵?) = ࠵?
fg
࠵?
E+
[࠵?
+
, ࠵?
0
, ࠵?
j+
] − ࠵?
l.f.g
≥ 0
( 4
0) m
1
5
0
−
2
5
1
n 8
2
1
1
4
−1
0
9 − (1.5, ࠵?3 ,0) ≥ 0
4
5
− ࠵?3 ≥ 0.
࠵?
0
≤ 0.8
As long as x3’s coefficient in the objective function stays <=0.8 same solution will be optimal.
The optimal objective function value will not change because we change c3 from -1 to 0.5,and
since x3 is a nonbasic variable an the optimal solution will not change, it will stay as nonbasic.
Thus x3 will be 0, and z will not change.
If we made a change on the objective function coefficient of a “basic variable”, and if the change
is within the allowable range, same solution will be optimal, but z value will change due to the
coefficient change on objective function.
Question 2)
Duality & Complementary slackness
The optimal solution to the following primal LP is x1=1, x2=x3=0,x4=2,
z*=13. Find the dual
solution using
complementary slackness:
Max
࠵?
= 3
࠵?
+
+ 4
࠵?
.
+
࠵?
0
+ 5
࠵?
q
s.t.
࠵?
+
+ 2
࠵?
.
+
࠵?
0
+
2
࠵?
q
≤ 5
2
࠵?
+
+ 3
࠵?
.
+
࠵?
0
+
3
࠵?
q
≤ 8
࠵?
+
,
࠵?
.
,
࠵?
0
,
࠵?
q
≥ 0
Solution
The optimal solution to the primal LP is x1=1, x2=x3=0,x4=2,
z*=13
The dual problem is as follows:
Min 5y1+8y2
s.t.
y1+2y2>=3
2y1+3y2>=4
y1+y2>=1
2y1+3y2>=5
(add excess variables ei to
dual problem)
From complementary slackness theorem we know that
S1*y1=0
(si:slack variables of primal LP)
S2*y2=0
and
e1*x1=0
e3*x3 =0
(ei=excess variables of dual LP)
e2*x2=0
e4*x4=0
Substituting optimal xi values we can find that s1=0 and s2=0 (also can be seen from LINGO
output).
Also, we know that x1>0 and x4>0 at the optimal solution. So we can decide that e1=0 and e4=0,
which means that dual constraints 1 and 4 won’t have excess at the dual optimum:
Y1+2y2=3
2y1+3y2=5
Solving these two equations we find that y1=1, y2=1, optimal dual objective=13.
Question 3)
Mr. Johnson has a plant that produces buns and hotdogs. The factory grinds its own
flour to produce buns at a maximum rate of 200 pounds per week. The factory has 800lb weekly
supply of meat for hotdogs. Each bun requires 0.1 pounds of flour and each hotdog requires
¼
lb
of meat. The factory has 5 workers who can work 40 hours a week. Producing a bun requires 2
minutes of labor and producing a hotdog requires a 3 minutes of labor.
Each hotdog yields a profit of $0.88, and each bun yields a profit of $0.33.
a)
Formulate a linear program for this problem to find the weekly production of the factory
b)
Solve it with a solver (Include solver report- you can upload it to CANVAS if you cannot
print)
Looking at the Sensitivity Analysis report from your software answer the following
questions:
c)
Which of the three amounts of available resources (meat, flour, labor) are completely
consumed at the
optimal solution.
d)
If the maximum available amount of flour is increased from 200lb to 201lb, what would
be the new objective function value?
e)
If the weekly supply of meat is increased from 800 lb to 801 lb what would be the new
objective function value?
Answer:
a) Let x1 be the number of hotdogs to produce, x2 be number of buns to produce
max
= 0.88*x1+0.33*x2;
0.1*x2<=200;
0.25*x1<=800;
3*x1+2*x2<=12000;
end
b)
Global optimal solution found.
Objective value:
3212.000
Infeasibilities:
0.000000
Total solver iterations:
1
Elapsed runtime seconds:
0.05
Model Class:
LP
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