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Chapter 19 - Linear Programming
19-89
e. Determine values for new x2 row:
0
1
–0.3
0.1
0.3
–0.1
1.8
f. Determine new values for row Z:
Row
Cost
x1
x2
s1
s2
a1
a2
Quantity
x2
10
10(0)
10(1)
10(–0.3)
10(0.1)
10(0.3)
10(–0.1)
10(1.8)
x1
12
12(1)
12(0)
12(0.2)
12(–0.4)
12(–0.2)
12(0.4)
12(0.8)
Z
12
10
–0.6
–3.8
0.6
3.8
27.6
g. Determine values for the C – Z row:
x1
x2
s1
s2
a1
a2
C
12
10
0
0
999
999
Z
12
10
–0.6
–3.8
0.6
3.8
C – Z
0
0
0.6
3.8
998.4
995.2
h. Set up the next tableau. Since no C – Z values are negative, the solution is optimal. Hence,
x1 = 0.8, x2 = 1.8, and minimum cost is 27.60.
C
12
10
0
0
999
999
Variables
in solution
x1
x2
s1
s2
a1
a2
Quantity
10
x2
0
1
–0.3
0.1
0.3
–0.1
1.8
12
x1
1
0
0.2
–0.4
–0.2
0.4
0.8
Z
12
10
–0.6
–3.8
0.6
3.8
27.6
C – Z
0
0
0.6
3.8
998.4
995.2
Chapter 19 - Linear Programming
19-90
Problems for the enrichment module (simplex)
1. Given this information:
Maximize
Z = 10.50x + 11.75y + 10.80z
Subject to
Cutting
5x + 12y + 8z 1,400 minutes
Stapling
7x + 9y + 9z 1,250 minutes
Wrapping
4x + 3y + 6z 720 minutes
x, y, z 0
2. Use the simplex method to solve these problems:
a.
Minimize
Z = 21x1 + 18x2
Subject to
(1) 5x1 + 10x2 100
(2) 2x1 + 1x2 10
x1, x2 0
b.
Minimize
Z = 2x + 5y + 3z
Subject to
(1) 16x + 10y + 18z 340
(2) 11x + 12y + 13z 300
(3) 2x + 6y + 5z 120
x, y, z 0
3. Use the simplex method to solve the following problem.
Minimize Z = 3x1 + 4x2 + 8x3
4. Use the simplex method to solve the following problem.
Maximize Z = 8x1 + 2x2
Subject to 4x1 + 5x2 20
Chapter 19 - Linear Programming
19-91
Solutions-Enrichment Module (SIMPLEX)
1.
C
10.5
11.75
10.80
0
0
0
Var
x
y
z
S1
S2
S3
bi
ratio
0
S1
5
12
8
1
0
0
1,400
116.67
0
S2
7
9
9
0
1
0
1,250
138.89
4
S3
4
3
6
0
0
1
720
240
Z
0
0
0
0
0
0
0
C–Z
10.5
11.75
10.80
0
0
0
C
Var
x
y
z
S1
S2
S3
bi
ratio
11.45
y
5/12
1
2/3
1/12
0
0
1,400/12
280
0
S2
13/4
0
3
–3/4
1
0
200
61.54
0
S3
11/4
0
4
–1/4
0
1
370
134.54
Z
4.896
11.75
7.833
0.979
0
0
1,370.83
C–Z
5.604
0
2.967
–0.979
0
0
C
Var
x
y
z
S1
S2
S3
bi
ratio
11.75
y
0
1
11/39
7/39
–5/39
0
91.026
507.1
10.5
x
1
0
12/13
–3/13
4/13
0
61.54
0
S3
0
0
19/13
5/13
–11/13
1
2610/13
522
Z
10.5
11.75
13.01
–0.314
1.724
0
1,715.73
C–Z
0
0
–2.206
0.314
–1.724
0
C
Var
x
y
z
S1
S2
S3
bi
0
S1
0
39/7
11/7
1
–5/7
0
507.14
10.5
x
1
9/7
117/91
0
–1/7
0
178.57
0
S3
0
–15/7
78/91
0
–4/7
1
5.72
Z
10.5
13.5
13.5
0
1.5
0
1,874.99
C–Z
0
–1.75
–2.7
0
–1.5
0
Optimal solution is x = 178.57, y = 0, z = 0, and optimal solution = 1,874.99
Chapter 19 - Linear Programming
19-92
Solutions (continued)
2.
a.
Minimize Z =
21x1
+ 18x2
s.t.
5x1
+ 10x2
+ A1 – S1
= 100
2x1
+ 1x2
+ A2 – S2
= 10
C
21
18
M
0
M
0
I.
C
Var
x1
x2
A1
S1
A2
S2
bi
ratio
M
A1
5
10
1
–1
0
0
100
10
M
A2
2
1
0
0
1
–1
10
10
Z
7M
11M
M
–M
M
–M
110M
M
C–Z
[21–7M]
[18–11M]
–0
M
–0
M
C
21
18
M
0
M
0
II.
C
Var
x1
x2
A1
S1
A2
S2
bi
ratio
18
x2
0.5
1
0.1
–0.1
0
0
10
20
M
A2
1.5
0
–0.1
0.1
1
–1
0
0
Z
[1.5M+9]
18
[1.8–0.1M]
[0.1M–1.8]
M
–M
180
C–Z
[12–1.5M]
0
[1.1M–1.8]
[1.8–.1M]
–0
M
C
21
18
0
0
III.
C
Var
x1
x2
S1
S2
bi
18
x2
0
1
–0.1333
0.333
10
21
x1
1
0
+0.0667
–0.667
0
Z
21
18
–0.99999
–8.000
180
C–Z
0
0
0.99999
8.000
The optimal solution: x1 = 0; x2 = 10; Z = 180
Chapter 19 - Linear Programming
19-93
Solutions (continued)
2.
b.
I.
C
2
5
3
M
0
M
0
M
0
Var
x
y
z
A1
S1
A2
S2
A3
S3
bi
M
A1
16
10
18
1
–1
0
0
0
0
340
M
A2
11
12
13
0
0
1
–1
0
0
300
M
A3
2
6
5
0
0
0
0
+1
–1
120
Z
29M
28M
36M
M
–M
M
–M
M
–M
760M
C–Z
[–29M+2]
[–28M+5]
[–36M+3]
0
M
0
M
0
M
II.
C
Var
x
y
z
A1
S1
A2
S2
A3
S3
bi
3
Z
.8889
.5556
1
.0556
–.0556
0
0
0
0
18.89
M
A2
–.5556
4.778
0
–.722
.722
1
–1
0
0
54.44
M
A3
–2.444
3.222*
0
–.2778
.2778
0
0
1
–1
25.56
Z
[–3M+2.7]
[8M+1.7]
3
[–M+.17]
[M–.17]
M
–M
M
–M
80M+56.7
C–Z
[+3M+.7]
[–8M+3.3]
0
[2M–47]
[–M+.17]
0
M
0
M
III.
C
Var
x
y
z
S1
A2
S2
S3
bi
3
z
1.31
0
1
–.1034
0
0
.1724
14.48
M
A2
3.069*
0
0
.3103
1
–1
1.483
16.55
5
Y
–0.7586
1
0
0.08621
0
0
–0.3103
7.931
Z
[3M+.138]
5
3
[.3M+.121]
–M
[1.5M–1.03]
[16.55M+83.1]
C–Z
[–3.1M+1.86]
0
0
[–.3M–.121]
0
M
[–1.5M+1.03]
IV.
C
Var
x
y
z
S1
S2
S3
bi
3
z
0
0
1
–0.236
.427
–0.4607
7.416
2
x
1
0
0
.1011
–.3258
.4831
5.393
5
y
0
1
0
.1629
–.2472
.05618
12.02
Z
2
5
3
0.309
–0.6067
–0.1348
93.15
C–Z
0
0
0
–0.309
0.6067
0.1348
V.
C
Var
x
y
z
S1
S2
S3
bi
3
z
2.333
0
1
0
–0.3333
.6667
20
0
S1
9.889
0
0
1
–3.222
4.778
53.33
5
y
–1.611
1
0
0
.2778*
–0.7222
3.333
Z
–1.06
5
3
0
+0.3889
–1.6111
76.67
C–Z
3.056
0
0
0
–0.3889
1.6111
Chapter 19 - Linear Programming
19-94
Solutions (continued)
VI.
C
Var
x
y
z
S1
S2
S3
bi
3
z
.4
1.2
1
0
0
–0.200
24
0
S1
–8.8
11.6
0
1
0
–3.6
92
0
S2
–5.8
3.6
0
0
1
–2.6
23
Z
1.2
3.6
3
0
0
–0.6
72
C–Z
.8
1.4
0
0
0
.6
Optimal solution is: x = 0; y = 0; z = 24 and Z = 72.0
3.
C
3
4
8
0
0
M
M
Var
x1
x2
x3
S1
S2
A1
A2
bi
bi/aij
M
A1
2
1
0
–1
0
1
0
6
6/2 = 3
M
A2
0
1
2
0
–1
0
1
4
–
Zj
2M
2M
2M
–M
–M
M
M
10M
Cj–Zj
3–2M
4–2M
8–2M
M
M
0
0
C
3
4
8
0
0
Var
x1
x2
x3
S1
S2
bi
bi/aij
3
x1
1
½
0
–½
0
3
3 ½ = 6
M
A2
0
1
2
0
–1
4
4 1 = 4
Zj
3
M–3/2
2M
–3/2
–M
4M+9
Cj–Zj
0
5/2 –M
8–2M
3/2
M
C
3
4
8
0
0
Var
x1
x2
x3
S1
S2
bi
bi/aij
3
x1
1
½
0
–½
0
3
3 ½ = 6
8
x3
0
½
1
0
–½
2
2 ½ = 4
Zj
3
11/2
8
–3/2
–4
25
Cj–Zj
0
–3/2
0
3/2
4
C
3
4
8
0
0
Var
x1
x2
x3
S1
S2
bi
3
x1
1
0
–1
–½
½
1
4
x2
0
1
2
0
–1
4
Zj
3
4
5
–3/2
–5/2
19
Cj–Zj
0
0
3
3/2
5/2
The optimal solution is: x1 = 1; x2 = 4; x3 = 0 and Z = 19.
Chapter 19 - Linear Programming
19-95
Solutions (continued)
4.
Cj
8
2
0
0
Var
x1
x2
S1
S2
bi
0
S1
4
5
1
0
20
0
S2
2
6
0
1
18
Zj
0
0
0
0
0
Cj–Zj
8
2
0
0
Cj
Var
x1
x2
S1
S2
bi
8
x1
1
5/4
¼
0
5
0
S2
0
7/2
–1/2
1
8
Zj
8
10
2
0
40
Cj–Zj
0
–8
–2
0
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