277
16.
cj Basic 300 400 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 18 3 2 1 0 0
0 s2 20 2 4 0 1 0
0 s3 4 0 1 0 0 1
zj 0 0 0 0 0 0
cj β zj 300 400 0 0 0
cj Basic 300 400 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 10 3 0 1 0 β2
0 s2 4 2 0 0 1 β4
400 x2 4 0 1 0 0 1
zj 1,600 0 400 0 0 400
cj β zj 300 0 0 0 β400
cj Basic 300 400 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 4 0 0 1 β3/2 4
300 x1 2 1 0 0 1/2 β2
400 x2 4 0 1 0 0 1
zj 2,200 300 400 0 150 β200
cj β zj 0 0 0 β150 200
cj Basic 300 400 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s3 1 0 0 1/4 β3/8 1
300 x1 4 1 0 1/2 β1/4 0
400 x2 3 0 1 β1/4 3/8 0
zj 2,400 300 400 50 75 0
cj β zj 0 0 β50 β75 0
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 277
278
17.
cj Basic 5 4 0 0
Variables Quantity x1 x2 s1 s2
0 s1 150 3/10 1/2 1 0
0 s2 2,000 10 4 0 1
zj 0 0 0 0 0
cj β zj 5 4 0 0
cj Basic 5 4 0 0
Variables Quantity x1 x2 s1 s2
0 s1 90 0 19/50 1 β3/100
5 x1 200 1 2/5 0 1/10
zj 1,000 5 2 0 1/2
cj β zj 0 2 0 β1/2
cj Basic 5 4 0 0
Variables Quantity x1 x2 s1 s2
4 x2 4,500/19 0 1 50/19 β3/38
5 x1 2,000/19 1 0 β20/19 5/38
zj 28,000/19 5 4 100/19 13/28
cj β zj 0 0 β100/19 β13/38
Optimal
18.
cj Basic 100 150 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 160 10 4 1 0 0
0 s2 20 1 1 0 1 0
0 s3 300 10 20 0 0 1
zj 0 0 0 0 0 0
cj β zj 100 150 0 0 0
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 278
279
cj Basic 100 150 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 20 0 0 1 β16 3/5
100 x1 10 1 0 0 2 β1/10
150 x2 10 0 1 0 β1 1/10
zj 2,500 100 150 0 50 5
cj β zj 0 0 0 β50 β5
Optimal
19.
cj Basic 100 20 60 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s1 60 3 5 0 1 0 0
0 s2 100 2 2 2 0 1 0
0 s3 40 0 0 1 0 0 1
zj 0 0 0 0 0 0 0
cj β zj 100 20 60 0 0 0
cj Basic 100 20 60 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
100 x1 20 1 5/3 0 1/3 0 0
0 s2 60 0 β4/3 2 β2/3 1 0
0 s3 40 0 0 1 0 0 1
zj 2,000 100 500/3 0 100/3 0 0
cj β zj 0 β440/3 60 β100/3 0 0
(continued)
cj Basic 100 150 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 100 8 0 1 0 β1/5
0 s2 5 1/2 0 0 1 β1/20
150 x2 15 1/2 1 0 0 1/20
zj 2,250 75 150 0 0 15/2
cj β zj 25 0 0 0 β15/2
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 279
21.
cj Basic 120 40 240 0 0 M M
Variables Quantity x1 x2 x3 s1 s2 A1 A2
M A1 27 4 1 3 β1 0 1 0
M A2 30 2 6 3 0 β1 0 1
zj 54M 6M 7M 6M βM βM M M
zj β cj 6M β 120 7M β 40 6M β 240 βM βM 0 0
cj Basic 120 40 240 0 0 M
Variables Quantity x1 x2 x3 s1 s2 A1
M A1 22 11/3 0 5/2 β1 1/6 1
40 x2 5 1/3 1 1/2 0 β1/6 0
zj 19M + 200 11M/3 + 40/3 40 5M/2 + 20 βM M/6 β 20/3 M
zj β cj 11M/3 β 320/3 0 5M/2 β 220 βM M/6 β 20/3 0
cj Basic 120 40 240 0 0
Variables Quantity x1 x2 x3 s1 s2
120 x1 6 1 0 15/22 β3/11 1/22
40 x2 3 0 1 6/22 1/11 β2/11
zj 840 120 40 1,020/11 β320/11 β20/11
cj β zj 0 0 β1,620/11 β320/11 β20/11
Optimal
20. a. maximization, because cj β zjb. x2 = 10, s2 = 20, x1 = 10, Z = 30c. maximize Z = x1 + 2x2 β x3d. 3e. No, because there are three constraints and
three βslackβ variablesf. s1 = 0g. yes, because cj β zj = 0 for s1h. x2 = 3 1/3, s2 = 26 2/3, x1 = 23 1/3, Z = 30
280
cj Basic 100 20 60 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
100 x1 20 1 5/3 0 1/3 0 0
60 x3 30 0 β2/3 1 β1/3 1/2 0
0 s3 10 0 2/3 0 1/3 β1/2 1
zj 3,800 100 380/3 60 40/3 30 0
cj β zj 0 β320/3 0 β40/3 β30 0
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 280
281
22.
cj Basic .05 .10 0 0 M M
Variables Quantity x1 x2 s1 s2 A1 A2
M A1 36 6 2 β1 0 1 0
M A2 50 5 5 0 β1 0 1
zj 86M 11M 7M βM βM M M
zj β cj 11M β .05 7M β .10 βM βM 0 0
cj Basic .05 .10 0 0 M
Variables Quantity x1 x2 s1 s2 A2
.05 x1 6 1 1/3 β1/6 0 0
M A2 20 0 10/3 5/6 β1 1
zj 20M + .3 .05 10M/3 + .02 5M/6 β .01 βM M
zj β cj 0 10M/3 β .08 5M/6 β .01 βM 0
cj Basic .05 .10 0 0
Variables Quantity x1 x2 s1 s2
.05 x1 4 1 0 β1/4 1/10
.10 x2 6 0 1 1/4 β3/10
zj .80 .05 .10 .0125 β.025
zj β cj 0 0 .0125 β.025
cj Basic .05 .10 0 0
Variables Quantity x1 x2 s1 s2
.05 x1 10 1 1 0 β1/5
0 s1 24 0 4 1 β6/5
zj .50 .05 .05 0 β.01
zj β cj 0 β.05 0 β.01
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 281
282
23.
0 4 8 12 16 20 24 28 32x
x
1
4
8
12
16
20
24
2
Firsttableau
Thirdtableau
Second tableauFourth tableau
24.
cj Basic 10 12 7 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s1 300 20 15 10 1 0 0
0 s2 120 10 5 0 0 1 0
0 s3 40 1 0 2 0 0 1
zj 0 0 0 0 0 0 0
cj β zj 10 12 7 0 0 0
cj Basic 10 12 7 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
12 x2 20 4/3 1 2/3 1/15 0 0
0 s2 20 10/3 0 β10/3 β1/3 1 0
0 s3 40 1 0 2 0 0 1
zj 240 16 12 8 4/5 0 0
cj β zj β6 0 β1 β4/5 0 0
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 282
283
25.
cj Basic 6 2 12 0 0
Variables Quantity x1 x2 x3 s1 s2
0 s1 24 4 1 3 1 0
0 s2 30 2 6 3 0 1
zj 0 0 0 0 0 0
cj β zj 6 2 12 0 0
cj Basic 6 2 12 0 0
Variables Quantity x1 x2 x3 s1 s2
12 x3 8 4/3 1/3 1 1/3 0
0 s2 6 β2 5 0 β1 1
zj 96 16 4 12 4 0
cj β zj β10 β2 0 β4 0
Optimal
26.
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
0 s1 40 3 2 0 0 1 0 0 0
0 s2 25 0 0 4 1 0 1 0 0
0 s3 2,000 200 0 250 0 0 0 1 0
0 s4 2,200 100 0 0 200 0 0 0 1
zj 0 0 0 0 0 0 0 0 0
cj β zj 100 75 90 95 0 0 0 0
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
0 s1 10 0 2 β3.75 0 1 0 β.015 0
0 s2 25 0 0 4 1 0 1 0 0
100 x1 10 1 0 1.25 0 0 0 .005 0
0 s4 1,200 0 0 β1.25 200 0 0 β.50 1
zj 1,000 100 0 1.25 0 0 0 .50 0
cj β zj 0 75 β35 95 0 0 β.50 0
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 283
284
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
0 s1 10 0 2 β3.75 0 1 0 β.015 0
0 s2 19 0 0 4.6 0 0 1 .002 β.005
100 x1 10 1 0 1.25 0 0 0 .005 0
95 x4 6 0 0 β.625 1 0 0 β.002 .005
zj 1,570 100 0 65.6 95 0 0 .26 .475
cj β zj 0 75 24.4 0 0 0 β.26 β.475
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
75 x2 5 0 1 β1.875 0 .50 0 β.007 0
0 s2 19 0 0 4.6 0 0 1 .002 β.005
100 x1 10 1 0 1.25 0 0 0 .005 0
95 x4 6 0 0 β.625 1 0 0 β.002 .005
zj 1,945 100 75 β75 95 37.5 0 β.30 .475
cj β zj 0 0 165 0 β37.5 0 .30 β.475
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
75 x2 12.7 0 1 0 0 .5 .405 β.006 β.002
90 x3 4.1 0 0 1 0 0 .216 .001 β.001
100 x1 4.9 1 0 0 0 0 β.270 .004 .001
95 x4 8.6 0 0 0 1 0 .135 β.002 .004
zj 2,623 100 75 90 95 37.5 35.7 β.211 .297
cj β zj 0 0 0 0 β37.5 β35.7 .211 β.297
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
75 x2 20 1.5 1 0 0 .50 0 0 0
90 x3 3.5 β.125 0 1 0 0 .25 0 β.001
0 s3 1,125 231.25 0 0 0 0 β62.5 1 .313
95 x4 11 .50 0 0 1 0 0 0 .005
zj 2,860 148.75 75 90 95 37.5 22.5 0 .36
cj β zj β48.75 0 0 0 β37.5 β22.5 0 β.36
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 284
285
27.
cj Basic 20 16 0 0 0 M M M
Variables Quantity x1 x2 s1 s2 s3 A1 A2 A3
M A1 6 3 1 β1 0 0 1 0 0
M A2 4 1 1 0 β1 0 0 1 0
M A3 12 2 6 0 0 β1 0 0 1
zj 22M 6M 8M βM βM βM M M M
zj β cj 6M β 20 8M β 16 βM βM βM 0 0 0
cj Basic 20 16 0 0 0 M M
Variables Quantity x1 x2 s1 s2 s3 A1 A2
M A1 4 8/3 0 β1 0 1/6 1 0
M A2 2 2/3 0 0 β1 1/6 0 1
16 x2 2 1/3 1 0 0 β1/6 0 0
zj 32 + 6M 10M/3 + 16/3 16 βM βM M/3 β 8/3 M M
zj β cj 10M/3 β 44/3 0 βM βM M/3 β 8/3 0 0
cj Basic 20 16 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 A2
20 x1 3/2 1 0 β3/8 0 1/16 0
M A2 1 0 0 1/4 β1 1/8 1
16 x2 3/2 0 1 1/8 0 β3/16 0
zj M + 27 20 16 M/4 β 53/6 βM + 16/3 M/8 β 25/12 M
zj β cj 0 0 M/4 β 53/6 βM + 16/3 M/8 β 25/12 0
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 285
28.
0 2 4 6 8 10 12 14 16x
x
1
2
4
6
8
10
12
2
First tableau
Second tableau
Fourth tableau
Fifth tableau
Third tableau
286
cj Basic 20 16 0 0 0
Variables Quantity x1 x2 s1 s2 s3
20 x1 3 1 0 0 β3/2 1/4
0 s1 4 0 0 1 β4 1/2
16 x2 1 0 1 0 1/2 β1/4
zj 76 20 16 0 β22 1
zj β cj 0 0 0 β22 1
cj Basic 20 16 0 0 0
Variables Quantity x1 x2 s1 s2 s3
20 x1 1 1 0 β1/2 1/2 0
0 s3 8 0 0 2 β8 1
16 x2 3 0 1 1/2 β3/2 0
zj 68 20 16 β2 β14 0
zj β cj 0 0 β2 β14 0
Optimal
29. Minimize Z = 8x1 + 2x2 + 7x3 + 0s1 + 0s2 + 0s3β MA1 β MA2 β MA3
subject to
2x1 + 6x2 + x3 + A1 = 303x2 + 4x3 β s1 + A2 = 604x1 + x2 + 2x3 + s2 = 50x1 + 2x2 β s3 + A3 = 20
x1, x2, x3 β₯ 0
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 286
287
30. Minimize Z = 40x1 + 55x2 + 30x3 + 0s1 + 0s2 + 0s3 + MA1 + MA2 + MA3
subject to
x1 + 2x2 + 3x3 + s1 = 602x1 + x2 + x3 + A1 = 40
x1 + 3x2 + x3 β s2 + A2 = 505x2 β3x3 β s3 + A3 = 100
x1, x2, x3 β₯ 0
31.
cj Basic 40 60 0 0 0 0 βM βM
Variables Quantity x1 x2 s1 s2 s3 s4 A1 A2
0 s1 30 1 2 1 0 0 0 0 0
0 s2 72 4 4 0 1 0 0 0 0
βM A1 5 1 0 0 0 β1 0 1 0
βM A2 12 0 1 0 0 0 β1 0 1
zj β17M βM βM 0 0 M M βM βM
cj β zj M + 40 M + 60 0 0 βM βM 0 0
cj Basic 40 60 0 0 0 0 βM
Variables Quantity x1 x2 s1 s2 s3 s4 A1
0 s1 6 1 0 1 0 0 2 0
0 s2 24 4 0 0 1 0 4 0
βM A1 5 1 0 0 0 β1 0 1
60 x2 12 0 1 0 0 0 β1 0
zj β5M + 720 βM 60 0 0 M β60 βM
cj β zj M + 40 0 0 0 βM 60 0
cj Basic 40 60 0 0 0 0
Variables Quantity x1 x2 s1 s2 s3 s4
0 s1 1 0 0 1 0 1 2
0 s2 4 0 0 0 1 4 4
40 x1 5 1 0 0 0 β1 0
60 x2 12 0 1 0 0 0 β1
zj 920 40 60 0 0 β40 β60
cj β zj 0 0 0 0 40 60
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 287
cj Basic 40 60 0 0 0 0
Variables Quantity x1 x2 s1 s2 s3 s4
0 s4 0 0 0 1 β1/4 0 1
0 s3 1 0 0 β4 1/2 1 0
40 x1 6 1 0 β1 1/2 0 0
60 x2 12 0 1 1 β1/4 0 0
zj 960 40 60 20 5 0 0
cj β zj 0 0 β20 β5 0 0
Optimal
32.
cj Basic 1 5 0 0 0 βM
Variables Quantity x1 x2 s1 s2 s3 A1
βM A1 25 5 5 β1 0 0 1
0 s2 16 2 4 0 1 0 0
0 s3 5 1 0 0 0 1 0
zj β25M β5M β5M M 0 0 βM
cj β zj 5M + 1 5M + 5 βM 0 0 0
cj Basic 1 5 0 0 0 βM
Variables Quantity x1 x2 s1 s2 s3 A1
βM A1 5 5/2 0 β1 β5/4 0 1
5 x2 4 1/2 1 0 1/4 0 0
0 s3 5 1 0 0 0 1 0
zj β5M + 20 β5M/2 + 5/2 5 M β5M/4 + 5/4 0 βM
cj β zj 5M/2 β 3/2 0 βM 5M/4 β 5/4 0 0
(continued)
288
cj Basic 40 60 0 0 0 0
Variables Quantity x1 x2 s1 s2 s3 s4
0 s4 1/2 0 0 1/2 0 1/2 1Tie
0 s2 2 0 0 β2 1 2 0
40 x1 5 1 0 0 0 β1 0
60 x2 25/2 0 1 1/2 0 1/2 0
zj 950 40 60 30 0 β10 0
cj β zj 0 0 β30 0 10 0
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 288
289
cj Basic 1 5 0 0 0
Variables Quantity x1 x2 s1 s2 s3
1 x1 2 1 0 β2/5 β1/2 0
5 x2 3 0 1 1/5 1/2 0
0 s3 3 0 0 2/5 1/2 1
zj 17 1 5 3/5 2 0
cj β zj 0 0 β3/5 β2 0
Optimal
33.
cj Basic 3 6 0 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 s4 A1
0 s1 18 3 2 1 0 0 0 0
M A1 5 1 1 0 β1 0 0 1
0 s3 4 1 0 0 0 1 0 0
0 s4 7 0 1 0 0 0 1 0
zj 5M M M 0 βM 0 0 M
zj β cj M β 3 M β 6 0 βM 0 0 0
cj Basic 3 6 0 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 s4 A1
0 s1 6 0 2 1 0 β3 0 0
M A1 1 0 1 0 β1 β1 0 1
3 x1 4 1 0 0 0 1 0 0
0 s4 7 0 1 0 0 0 1 0
zj M + 12 3 M 0 βM βM + 3 0 M
zj β cj 0 M β 6 0 βM βM + 3 0 0
cj Basic 3 6 0 0 0 0
Variables Quantity x1 x2 s1 s2 s3 s4
0 s1 4 0 0 1 2 β1 0
6 x2 1 0 1 0 β1 β1 0
3 x1 4 1 0 0 0 1 0
0 s4 6 0 0 0 1 1 1
zj 18 3 6 0 β6 β3 0
zj β cj 0 0 0 β6 β3 0
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 289
290
34.
cj Basic 10 5 0 0 βM βM
Variables Quantity x1 x2 s1 s2 A1 A2
βM A1 10 2 1 β1 0 1 0
βM A2 4 0 1 0 0 0 1
0 s2 20 1 4 0 1 0 0
zj β14M β2M β2M M 0 βM βM
cj β zj 2M + 10 2M + 5 βM 0 0 0
cj Basic 10 5 0 0 βM
Variables Quantity x1 x2 s1 s2 A2
10 x1 5 1 1/2 β1/2 0 0
βM A2 4 0 1 0 0 1
0 s2 15 0 7/2 1/2 1 0
zj β4M + 50 10 βM + 5 β5 0 βM
cj β zj 0 M 5 0 0
cj Basic 10 5 0 0
Variables Quantity x1 x2 s1 s2
10 x1 3 1 0 β1/2 0
5 x2 4 0 1 0 0
0 s2 1 0 0 1/2 1
zj 50 10 5 β5 0
cj β zj 0 0 5 0
cj Basic 10 5 0 0
Variables Quantity x1 x2 s1 s2
10 x1 4 1 0 0 1
5 x2 4 0 1 0 0
0 s1 2 0 0 1 2
zj 60 10 5 0 10
cj β zj 0 0 0 β10
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 290
291
35.
cj Basic 1 2 β1 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s1 40 0 4 1 1 0 0
0 s2 20 1 β1 0 0 1 0
0 s3 60 2 4 3 0 0 1
zj 0 0 0 0 0 0 0
cj β zj 1 2 β1 0 0 0
cj Basic 1 2 β1 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
2 x2 10 0 1 1/4 1/4 0 0
0 s2 30 1 0 1/4 1/4 1 0
0 s3 20 2 0 2 β1 0 1
zj 20 0 2 1/2 1/2 0 0
cj β zj 1 0 β3/2 β1/2 0 0
cj Basic 1 2 β1 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
2 x2 10 0 1 1/4 1/4 0 0
0 s2 30 0 0 β3/4 3/4 1 β1/2
1 x1 10 1 0 1 β1/2 0 1/2
zj 30 1 2 3/2 0 0 1/2
cj β zj 0 0 β5/2 0 0 β1/2
Multiple optimum solution
Alternate solution:
cj Basic 1 2 β1 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
2 x2 10/3 0 1 1/2 0 β1/3 1/6
0 s1 80/3 0 0 β1 1 4/3 β2/3
1 x1 70/3 1 0 1/2 0 2/3 1/6
zj 30 1 2 3/2 0 0 1/2
cj β zj 0 0 β5/2 0 0 β1/2
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 291
292
36.
cj Basic 1 2 2 0 0 βM βM
Variables Quantity x1 x2 x3 s1 s2 A1 A2
0 s1 12 1 1 2 1 0 0 0
βM A1 20 2 1 5 0 0 1 0
βM A2 8 1 1 β1 0 β1 0 1
zj β28M β3M β2M β4M 0 M βM βM
cj β zj 1 + 3M 2 + 2M 2 + 4M 0 βM 0 0
cj Basic 1 2 2 0 0 βM
Variables Quantity x1 x2 x3 s1 s2 A2
0 s1 4 1/5 3/5 0 1 0 0
2 x3 4 2/5 1/5 1 0 0 0
βM A2 12 7/5 6/5 0 0 β1 1
zj β12M + 8 β4/5 β 7M/5 2/5 β 6M/5 2 0 M βM
cj β zj 1/5 + 7M/5 8/5 + 6M/5 0 0 βM 0
cj Basic 1 2 2 0 0
Variables Quantity x1 x2 x3 s1 s2
0 s1 16/7 0 3/7 0 1 1/7
2 x3 4/7 0 β1/7 1 0 2/7
1 x1 60/7 1 6/7 0 0 β5/7
zj 68/7 1 4/7 2 0 β1/7
cj β zj 0 10/7 0 0 1/7
cj Basic 1 2 2 0 0
Variables Quantity x1 x2 x3 s1 s2
2 x2 16/3 0 1 0 7/3 1/3
2 x3 4/3 0 0 1 1/3 1/3
1 x1 4 1 0 0 β2 β1
zj 52/3 1 2 2 10/3 1/3
cj β zj 0 0 0 β10/3 β1/3
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 292
293
cj Basic 400 350 450 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 120 2 3 2 1 0 0 0
0 s2 160 4 3 1 0 1 0 0
0 s3 100 3 2 4 0 0 1 0
0 s4 40 1 1 1 0 0 0 1
zj 0 0 0 0 0 0 0 0
cj β zj 400 350 450 0 0 0 0
37.
cj Basic 400 350 450 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 70 1/2 2 0 1 0 β1/2 0
0 s2 135 13/4 5/2 0 0 1 β1/4 0
450 x3 25 3/4 1/2 1 0 0 1/4 0
0 s4 15 β1/4 1/2 0 0 0 β1/4 1
zj 11,250 1,350/4 450/2 450 0 0 450/4 0
cj β zj 250/4 250/2 0 0 0 β450/4 0
cj Basic 400 350 450 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 10 β1/2 0 0 1 0 1/2 β4
0 s2 60 2 0 0 0 1 1 β5
450 x3 10 1/2 0 1 0 0 1/2 β1
350 x2 30 1/2 1 0 0 0 β1/2 2
zj 15,000 400 350 450 0 0 50 250
cj β zj 0 0 0 0 0 β50 β250
Multiple optimum solution at x1
Alternate solution:
cj Basic 400 350 450 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 20 0 0 1 1 0 1 β5
0 s2 20 0 0 β4 0 1 β1 β1
400 x1 20 1 0 2 0 0 1 β2
350 x2 20 0 1 β1 0 0 β1 3
zj 15,000 400 350 450 0 0 50 250
cj β zj 0 0 0 0 0 β50 β250
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 293
39. (a)
0 1 2 3 4 5 6x
x
1
Unbounded solution
Z1
2
3
4
5
2
-1-2-3-4
294
38. (a)
0 1 2 3 4 5 6 7x
x
1
Infeasible-no common solution space
1
2
3
4
5
2
(b)
cj Basic 3 2 0 0 βM
Variables Quantity x1 x2 s1 s2 A1
0 s1 1 1 1 1 0 0
βM A1 2 1 1 0 β1 1
zj β2M βM βM 0 M βM
cj β zj M + 3 M + 2 0 βM 0
cj Basic 3 2 0 0 βM
Variables Quantity x1 x2 s1 s2 A1
3 x1 1 1 1 1 0 0
βM A1 1 0 0 β1 β1 1
zj 3 β M 3 3 M M βM
cj β zj 0 β1 βM βM 0
Infeasible solution
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 294
295
(b)
cj Basic 1 1 0 0
Variables Quantity x1 x2 s1 s2
0 s1 1 β1 1 1 0
0 s2 4 β1 2 0 1
zj 0 0 0 0 0
cj β zj 1 1 0 0
Tie for entering variable; if x1 is chosen, the solution is unbounded. Select x2 arbitrarily.
cj Basic 1 1 0 0
Variables Quantity x1 x2 s1 s2
1 x2 1 β1 1 1 0
0 s2 3 1 0 β1 1
zj 1 β1 1 1 0
cj β zj 2 0 β1 0
cj Basic 1 1 0 0
Variables Quantity x1 x2 s1 s2
1 x2 4 0 1 0 1
1 x1 3 1 0 β1 1
zj 7 1 1 β1 2
cj β zj 0 0 1 β2
Unbounded; no pivot row available
40.
cj Basic 7 5 5 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 25 1 1 1 1 0 0 0
0 s2 40 2 1 1 0 1 0 0
0 s3 25 1 1 0 0 0 1 0
0 s4 6 0 0 1 0 0 0 1
zj 0 0 0 0 0 0 0 0
cj β zj 7 5 5 0 0 0 0
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 295
Alternate Solution:
cj Basic 7 5 5 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
5 x2 4 0 1 0 2 β1 0 β1
7 x1 15 1 0 0 β1 1 0 0
0 s3 6 0 0 0 β1 0 1 1
5 x3 6 0 0 1 0 0 0 1
zj 155 7 5 5 3 2 0 0
cj β zj 0 0 0 β3 β2 0 0
296
cj Basic 7 5 5 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 5 0 1/2 1/2 1 β1/2 0 0
7 x1 20 1 1/2 1/2 0 1/2 0 0
0 s3 5 0 1/2 β1/2 0 β1/2 1 0
0 s4 6 0 0 0 0 0 0 1
zj 140 7 7/2 7/2 0 7/2 0 0
cj β zj 0 3/2 3/2 0 β7/2 0 0
Tie
cj Basic 7 5 5 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
5 x2 10 0 1 1 2 β1 0 0
7 x1 15 1 0 0 β1 1 0 0
0 s3 0 0 0 β1 β1 0 1 0
0 s4 6 0 0 1 0 0 0 1
zj 155 7 5 5 3 2 0 0
cj β zj 0 0 0 β3 β2 0 0
Multiple optimum (continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 296
297
41.
cj Basic 15 25 0 0 0 M M
Variables Quantity x1 x2 s1 s2 s3 A1 A2
M A1 12 3 4 β1 0 0 1 0
M A2 6 2 1 0 β1 0 0 1
0 s3 9 3 2 0 0 1 0 0
zj 18M 5M 5M βM βM 0 M M
zj β cj 5M β 15 5M β 25 βM βM 0 0 0
cj Basic 15 25 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 A1
M A1 3 0 5/2 β1 1 0 1
15 x1 3 1 1/2 0 0 0 0
0 s3 0 0 1/2 0 0 1 0
zj 3M + 45 15 5M/2 + 15/2 βM 3M/2 β 15/2 0 M
zj β cj 0 5M/2 β 35/2 βM 3M/2 β 15/2 0 0
cj Basic 15 25 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 A1
M A1 3 0 0 β1 β6 β5 1
15 x1 3 1 0 0 β2 β1 0
25 x2 0 0 1 0 3 2 0
zj 3M + 45 15 15 βM β6M +45 β5M +45 M
zj β cj 0 0 βM β6M +45 β5M +45 0
Infeasible solution
42. a).minimize Zd = 90y1 + 60y2subject to
y1 + 2y2 β₯ 64y1 + 2y2 β₯ 10
y1, y2 β₯ 0
b) y1 = the marginal value of one additional lb of brass = $1.33
y2 = the marginal value of one additional hrof labor = $2.33
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 297
298
cj Basic 6 + β 10 0 0
Variables Quantity x1 x2 s1 s2
10 x2 20 0 1 1/3 β1/6
6 + β x1 10 1 0 β1/3 2/3
zj 260 + 10β 6 + β 10 4/3 β β/3 7/3 + 2β/3
cj β zj 0 0 β4/3 + β/3 β7/3 β 2β/3
Solving for the cj β zj inequalities:
β4/3 + β/3 β€ 0β/3 β€ 4/3
β β€ 4
Since c1 = 6 + β; β = c1 β 6. Thus
c1 β 6 β€ 4c1 β€ 10
β7/3 β 2β/3 β€ 0β2β/3 β€ 7/3
β2β β€ 7β β₯ β7/2
Since c1 = 6 + β; β = c1 β 6. Thus
c1 β 6 β₯ β7/2c1 β₯ 5/2
Summarizing, 5/2 β€ c1 β€ 10.
c2, basic:
cj Basic 6 10 + β 0 0
Variables Quantity x1 x2 s1 s2
10 + β x2 20 0 1 1/3 β1/6
6 x1 10 1 0 β1/3 2/3
zj 280 + 20β 6 10 + β 4/3 + β/3 7/3 β β/6
cj β zj 0 0 β4/3 β β/3 β7/3 + β/6
c) c1, basic:
Solving for the cj β zj inequalities:
β4/3 β β/3 β€ 0ββ/3 β€ 4/3
β β₯ β4
Since c2 = 10 + β; β = c2 β 10. Thus
c2 β 10 β₯ β4c2 β₯ 6
β7/3 + β/6 β€ 0β/6 β€ 7/3
β β€ 14
Since c2 = 10 + β; β = c2 β 10. Thus
c2 β 10 β€ 14c2 β€ 24
Summarizing, 6 β€ c2 β€ 24.
d) q1:
x2: 20 + β/3 β₯ 0 x1: 10 β β/3 β₯ 0β/3 β₯ β20 ββ/3 β₯ β10
β β₯ β60 β β€ 30
Therefore, β60 β€ β β€ 30. Since
q1 = 90 + ββ = q1 β 90
β60 β€ q1 β 90 β€ 3030 β€ q1 β€ 120
q2:
x2: 20 β β/6 β₯ 0 x1: 10 + 2β/3 β₯ 0β β/6 β₯ β20 2β/3 β₯ β10
β β€ 120 β β₯ β15
Therefore, β15 β€ β β€ 120. Since
q2 = 60 + ββ = q2 β 60
β15 β€ q2 β 60 β€ 12045 β€ q2 β€ 180
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 298
299
e) The marginal value of 1 hr of labor is $2.33.From part d, the sensitivity range for q2, labor,is 45 β€ q2 β€ 180. Thus, the company wouldpurchase up to 180 hr at the marginal valueprice.
43. a) Minimize Zd = 500y1 + 800y2subject to
10y1 + 34y2 β₯ 20050y1 + 20y2 β₯ 300
y1, y2 β₯ 0
y1 = $4.13 = the marginal value of oneadditional lb of chili beans; y2 = $4.67 = the marginal value of one additional lb of ground beef.
b)
Point C must become the optimal solution for x2 = 0; therefore the slope of the objectivefunction must be greater than the slope of theconstraint for ground beef, β34/20. Solving thefollowing for the profit, p, of Razorback chiliyields
βp/300 = β34/20p = $510
Thus, if the profit of Razorback chili is greater than $510, no Longhorn chili will be produced.The new optimal solution will be x1 = 23.5 andx2 = 0.
c)
The constraint line for chili beans rotates,creating a new, smaller solution space, and theoptimal solution shifts from point B to point BΒ΄where x1 = 21.43 and x2 = 3.57.
0 10 20 30 40 50x
x
1C
BA
10
20
30
40
50
60
2
60
0 10 20 30 40 50x
x
1C
BBΒ΄
A10
20
30
40
50
60
2
60
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 299
300
cj Basic 200 + β 300 0 0
Variables Quantity x1 x2 s1 s2
300 x2 6 0 1 17/750 β1/150
200 + β x1 20 1 0 β1/75 1/30
zj 5,800 + 20β 200 + β 300 310/75 β β/75 70/15 + β/30
cj β zj 0 0 β310/75 + β/75 β70/15 β β/30
d) c1, basic:
Solving for the cj β zj inequalities:
β310/75 + β/75 β€ 0β/75 β€ 3310/75
β β€ 310
Since c1 = 200 + β; β = c1 β 200. Thus
c1 β 200 β€ 310c1 β€ 510
β70/15 β β/30 β€ 0ββ/30 β€ 70/15
β β₯ β140
Since c1 = 200 + β; β = c1 β 200. Thus
c1 β 200 β₯ β140c1 β₯ 60
Summarizing, 60 β€ c1 β€ 510.
c2, basic:
Since c2 = 300 + β; β = c2 β 300. Thus
c2 β 300 β€ 700c2 β€ 1,000
Summarizing, 117.65 β€ c2 β€ 1,000.
e) q1:
x2: 6 + 17β/750 β₯ 0 x1: 20 β β/75 β₯ 017β/750 β₯ β6 ββ/75 β₯ β20
β β₯ β264.7 β β€ 1,500
Therefore, β264.7 β€ β β€1,500. Since
q1 = 500 + ββ = q1 β 500
β264.7 β€ q1 β 500 β€ 1,500235.3 β€ q1 β€ 2,000
cj Basic 200 300 + β 0 0
Variables Quantity x1 x2 s1 s2
300 + β x2 6 0 1 17/750 β1/150
200 x1 20 1 0 β1/75 1/30
zj 5,800 + 6β 200 300 + β 310/75 + 17β/750 70/15 β β/150
cj β zj 0 0 β310/75 β 17β/750 β70/15 + β/150
Solving for the cj β zj inequalities:
β310/75 β 17β/750 β€ 0β17β/750 β€ 310/75
β β₯ β182.35
Since c2 = 300 + β; β = c2 β 300. Thus
c2 β 300 β₯ β182.35c2 β₯ 117.65
β70/15 + β/150 β€ 0β/150 β€ 70/15
β β€ 700
q2:
x2: 6 β β/150 β₯ 0 x1: 20 + β/30 β₯ 0β β/150 β₯ β6 β/30 β₯ β20
β β€ 900 β β₯ β600
Therefore, β600 β€ β β€ 900. Since
q2 = 800 + ββ = q2 β 800
β600 β€ q2 β 800 β€ 900200 β€ q2 β€ 1,700
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 300
301
f) The marginal value of 1 lb of chili beans is$4.13. The sensitivity range for q1, chili beans,is 235.3 β€ q1q β€ 2,000. Thus, the companywould purchase up to 2,000 lb at the marginalvalue price.
g) Groundbeef
h) No effect
44. a) Minimize Zd = 60y1 + 40y2subject to
12y1 + 4y2 β₯ 94y1 + 8y2 β₯ 7
y1, y2 β₯ 0
b) y1 = the marginal value of one additional hr ofprocess 1; y2 = the marginal value of oneadditional hr of process 2
For the s1 column, the cj β zj value of $.55 is the marginal value of 1 hr of process 1production time. For the s2 column, the cj β zjvalue of $0.60 is the marginal value of 1 hr ofprocess 2 production time.
cj Basic 9 + β 7 0 0
Variables Quantity x1 x2 s1 s2
9 + β x1 4 1 0 1/10 β1/20
7 x2 3 0 1 β1/20 3/20
zj 57 + 4β 9 + β 7 11/20 β β/10 12/20 + β/20
cj β zj 0 0 β11/20 β β/10 β12/20 + β/20
c) cj, basic:
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 301
302
Solving for the cj β zj inequalities:
β11/20 β β/10 β€ 0ββ/10 β€ 11/20
ββ β€ 11/2β β₯ β11/2
Since c1 = 9 + β , β = c1 β 9. Thus
c1 β 9 β₯ 11/2c1 β₯ 7/2
β12/20 + β/20 β€ 0β/20 β€ 12/20
β β€ 12
Since c1 = 9 + β , β = c1 β 9. Thus
c1 β 9 β€ 12c1 β€ 12
Summarizing, 7/2 β€ c1 β€ 12.
c2, basic:
Since c2 = 7 + β; β = c2 β 7. Thus
c2 β 7 β₯ β4c2 β₯ 3
Summarizing, 3 β€ c2 β€ 18.
d) q1:
x1: 4 + β/10 β₯ 0 x2: 3 β β/20 β₯ 0β/10 β₯ β4 ββ/20 β₯ β3
ββ β₯ β80 ββ β₯ β60β β₯ β40 β β€ 60
Therefore, β40 β€ β β€60. Since
q1 = 60 + ββ = q1 β 60
β40 β€ q1 β 60 β€ 6020 β€ q1 β€ 120
cj Basic 9 7 + β 0 0
Variables Quantity x1 x2 s1 s2
9 x1 4 1 0 1/10 β1/20
7 + β x2 3 0 1 β1/20 3/20
zj 57 + 3β 9 7 + β 11/20 β β/20 12/20 + 3β/20
cj β zj 0 0 β11/20 + β/20 β12/20 β 3β/20
Solving for the cj β zj inequalities:
β11/20 + β/20 β€ 0β/20 β€ 11/20
β β€ 11
Since c2 = 7 + β; β = c2 β 7. Thus
c2 β 7 β€ 11c2 β€ 18
β12/20 β 3β/20 β€ 0β3β/20 β€ 12/20
β β₯ β4
q2:
x1: 4 β β/20 β₯ 0 x2: 3 +3 β/20 β₯ 0ββ/20 β₯ β4 3β/20 β₯ β3
ββ β₯ β80 β β₯ β20β β€ 80
Therefore, β20 β€ β β€ 80. Since
q2 = 40 + ββ = q2 β 40
β20 β€ q2 β 40 β€ 8020 β€ q2 β€ 120
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 302
303
e) The marginal value of 1 hr of process 1production time is $.55. The sensitivity rangefor q1, production hours, is 20 β€ q1 β€ 120.Thus, the company would purchase up to 120hr at the marginal value price.
45. a) Minimize Zd = 180y1 + 135y2subject to
2y1 + 3y2 β₯ 2005y1 + 3y2 β₯ 300
y1, y2 β₯ 0
b) y1 = $33.33 = the marginal value of an additional hr of labor; y2 = $44.44 = themarginal value of an additional bd. ft. of wood
c)
Point A must become the optimal solution forx1 = 0; therefore the slope of the objectivefunction must be less than the slope of theconstraint for labor, β2/5. Solving the followingfor the profit, p, of coffee tables gives
β200/p = β2/5p = $500
Thus, if the profit for coffee tables is greaterthan $500, no end tables will be produced. Thenew optimal solution will be x1 = 0 and x2 = 36.
d)
The constraint line for wood moves outward,creating a new solution space, and the optimalsolution point shifts from point B to point BΒ΄where x1 = 31.67 and x2 = 23.33.0 10 20 30 40 50 60 70
x
x
1C
BA
10
20
30
40
50
60
2
80 90 100
0 10 20 30 40 50 60 70x
x
1C CΒ΄
B BΒ΄
A
10
20
30
40
50
60
2
80 90 100
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 303
304
cj Basic 200 + β 300 0 0
Variables Quantity x1 x2 s1 s2
300 x2 30 0 1 1/3 β2/9
200 + β x1 15 1 0 β1/3 5/9
zj 12,000 + 15β 200 + β 300 100/3 β β/3 400/9 + 5β/9
cj β zj 0 0 β100/3 + β/3 β400/9 β 5β/9
e) c1, basic:
Solving for the cj β zj inequalities:
β100/3 + β/3 β€ 0β/3 β€ 100/3
β β€ 100
Since c1 = 200 + β; β = c1 β 200. Thus
c1 β 200 β€ 100c1 β€ 300
β400/9 β 5β/9 β€ 0β5β/9 β€ 400/9
ββ β€ 80β β₯ β80
Since c1 = 200 + β; β = c1 β 200. Thus
c1 β 200 β₯ β80c1 β₯ 120
Summarizing, 120 β€ c1 β€ 300.
c2, basic:
Since c2 = 300 + β; β = c2 β 300. Thus
c2 β 300 β€ 200c2 β€ 500
Summarizing, 200 β€ c2 β€ 500.
f) q1:
x2: 30 + β/3 β₯ 0 x1: 15 β β/3 β₯ 0β/3 β₯ β30 ββ/3 β₯ β15
β β₯ β90 ββ β€ β45β β€ 45
Therefore, β90 β€ β β€ 45. Since
q1 = 180 + ββ = q1 β 180
β90 β€ q1 β 180 β€ 4590 β€ q1 β€ 225
cj Basic 200 300 + β 0 0
Variables Quantity x1 x2 s1 s2
300 + β x2 30 0 1 1/3 β2/9
200 x1 15 1 0 β1/3 5/9
zj 12,000 + 300β 200 300 + β 100/3 + β/3 400/9 β 2β /9
cj β zj 0 0 β100/3 β β/3 β400/9 + 2β /9
Solving for the cj β zj inequalities:
β100/3 β β/3 β€ 0ββ/3 β€ 100/3
ββ β€ 100β β₯ β100
Since c2 = 300 + β; β = c2 β 300. Thus
c2 β 300 β₯ β100c2 β₯ 200
β400/9 + 2β/9 β€ 02β/9 β€ 400/9
β β€ 200
q2:
x2: 30 β 2β/9 β₯ 0 x1: 15 + 5β/9 β₯ 0β2β/9 β₯ β30 5β/9 β₯ β15
ββ β₯ β135 β β₯ β27β β€ 135
Therefore, β27 β€ β β€ 135. Since
q2 = 135 + ββ = q2 β 135
β27 β€ q2 β 135 β€ 135108 β€ q2 β€ 270
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 304
305
g) The marginal value of 1 lb of wood is $44.44.From part f, the sensitivity range for q2, wood,is 108 β€ q2 β€ 270. Thus, the company wouldpurchase up to 270 bd. ft. of wood at themarginal value price.
g) The marginal value of labor is $33.33 and themarginal value of wood is $44.44; thus, woodshould be purchased.
46. a) Minimize Zd = 19y1 + 14y2 + 20y3subject to
2y1 + y2 + y3 β₯ 70y1 + y2 + 2y3 β₯ 80
y1, y2, y3 β₯ 0
b) y1 =$20 = the marginal value of an additionalhr of production time; y2 =$0 = the marginalvalue of an additional lb of steel; y3 = $30 = themarginal value of an additional ft of wire
cj Basic 70 + β 80 0 0 0
Variables Quantity x1 x2 s1 s2 s3
70 + β x1 6 1 0 2/3 0 β1/3
0 s2 1 0 0 β1/3 1 β1/3
80 x2 7 0 1 β1/3 0 2/3
zj 980 + 6β 70 + β 80 20 + 2β/3 0 30 β β/3
cj β zj 0 0 β20 β 2β/3 0 β30 + β/3
c) c1, basic:
Solving for the cj β zj inequalities:
β20 β 2β/3 β€ 0β2β/3 β€ 20
ββ β€ 30β β₯ β30
Since c1 = 70 + β; β = c1 β 70. Thus
c1 β 70 β₯ β30c1 β₯ 40
β30 + β/3 β€ 0β/3 β€ 30
β β€ 90
Since c1 = 70 + β; β = c1 β 70. Thus
c1 β 70 β€ 90c1 β€ 160
Summarizing, 40 β€ c1 β€ 160.
c2, basic:
Solving for the cj β zj inequalities:
β20 + β/3 β€ 0β/3 β€ 20
β β₯ 60
Since c2 = 80 + β; β = c2 β 80. Thus
c2 β 80 β€ 60c2 β€ 140
β30 β 2β/3 β€ 0β2β/3 β€ 30
ββ β€ 45β β₯ β45
Since c2 = 80 + β; β = c2 β 80. Thus
c2 β 80 β₯ β45c2 β₯ 35
Summarizing, 35 β€ c2 β€ 140.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 305
306
cj Basic 70 80 + β 0 0 0
Variables Quantity x1 x2 s1 s2 s3
70 x1 6 1 0 2/3 0 β1/3
0 s2 1 0 0 β1/3 1 β1/3
80 + β x2 7 0 1 β1/3 0 2/3
zj 980 + 7β 70 80 + β 20 β β/3 0 30 + 2β/3
cj β zj 0 0 β20 + β/3 0 β30 β 2β/3
d) q1:
x1: 6 + 2β/3 β₯ 0 x2: 7 β β/3 β₯ 0 s2: 1 β β/3 β₯ 02β/3 β₯ β6 ββ/3 β₯ β7 ββ/3 β₯ β1
β β₯ β9 ββ β₯ β21 ββ β₯ β3β β€ 21 β β€ 3
Therefore, β9 β€ β β€ 3. Since
q1 = 19 + ββ = q1 β 19
β9 β€ q1 β 19 β€ 310 β€ q1 β€ 22
q2:
x1: 6 + 0β β₯ 0 x2: 7 + 0β β₯ 0 s2: 1 + β β₯ 0β β₯ β1
Therefore, β β₯ β1. Since
q2 = 14 + ββ = q2 β 14
q2 β 14 β₯ β1q2 β₯ 13
q3:
x1: 6 β β/3 β₯ 0 x2: 7 + 2β/3 β₯ 0 s2: 1 β β/3 β₯ 0ββ/3 β₯ β6 2β/3 β₯ β7 ββ/3 β₯ β1
ββ β₯ β18 β β₯ β21/2 ββ β₯ β3β β€ 18 β β€ 3
Therefore, β21/2 β€ β β€ 3. Since
q3 = 20 + β β21/2 β€ q3 β 20 β€ 3β = q3 β 20 19/2 β€ q3 β€ 23
e) The sensitivity range for production hours is 10 β€ q1 β€ 22. Since 25 hr exceeds the upper limit of the range, it would change the optimal solution.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 306
307
47. a) Minimize Zd = 64y1 + 50y2 + 120y3 + 7y4 + 7y5subject to
4y1 + 5y2 + 15y3 + y4 β₯ 98y1 + 5y2 + 8y3 + y5 β₯ 12
y1, y2, y3, y4, y5 β₯ 0
b) y1 = $.75 = the marginal value of one additionalhr of labor for process 1; y2 = $1.20 = the marginal value of one additional hr of labor for process 2; y3, y4, y5 = $0; these resources have no value since there were units available which were not used.
c) c1, basic:
Solving for the cj β zj inequalities:
β3/4 + β/4 β€ 0 β6/5 β 2β/5 β€ 0β/4 β€ 3/4 β2β/5 β€ 6/5
β β€ 3 β β₯ β3
Since c1 = 9 + β; β = c1 β 9. Thus
β3 β€ β β€ 3β3 β€ c1 β 9 β€ 36 β€ c1 β€ 12
c2, basic:
Solving for the cj β zj inequalities:
β3/4 + β/4 β€ 0 β6/5 + β/5 β€ 0ββ/4 β€ 3/4 β/5 β€ 6/5
β β₯ β3 β β€ 6
Since c2 = 12 + β; β = c1 β 12. Thus
β3 β€ c2 β 12 β€ 69 β€ c2 β€ 18
d) q1:
x1: 4 β β/4 β₯ 0 s5: 1 β β/4 β₯ 0 s3: 12 + 7β/4 β₯ 0ββ/4 β₯ β4 ββ/4 β₯ β1 7β/4 β₯ β12
β β€ 16 β β€ 4 β β₯ β6.86
s4: 3 + β/4 β₯ 0 x2: 6 + β/4 β₯ 0β/4 β₯ β3 β/4 β₯ β6
β β₯ β12 β β₯ β24
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 307
308
Summarizing,
β24 < β12 < β6.86 β€ β β€ 4 β€ 16
and, therefore,
β6.86 β€ β β€ 4
Since q1 = 64 + β; β = q1 β 64. Therefore,
β6.86 β€ β q1 β 64 β€ 457.14 β€ q1 β€ 68
e) q3:
x1: 4 + 0β β₯ 0 s5: 1 + 0β β₯ 0 s3: 12 + β β₯ 00β β₯ β4 β β€ β β β₯ β12β β€ β
s4: 3 + 0β β₯ 0 x2: 6 + 0β β₯ 0β β€ β β β€ β
β12 β€ β β€ β
Since q3 = 120 + β; β = q3 β 120. Therefore,
β12 β€ β β€ ββ12 β€ q3 β 120 β€ β108 β€ q3 β€ β
Since 100 pounds is less than the lower limitof the range, the optimal solution mix will change. s2 enters the solution and s3 leaves. The new solution is,
x1 = 3.27s2 = 1.82s5 = 0.64s4 = 3.73x2 = 6.36Z = 105.82
48. a) Minimize Zd = 120y1 + 160y2 + 100y3 + 40y4subject to
2y1 + 4y2 + 3y3 + y4 β₯ 403y1 + 3y2 + 2y3 + y4 β₯ 352y1 + y2 + 4y3 + y4 β₯ 45
y1, y2, y3, y4 β₯ 0
b) y1, y2 = 0; y3 = $5 = the marginal value of 1 hrof operation 3 time; y4 = $25 = the marginal value of 1 ft2 of storage space
c) It does not have an effect. In the alternatesolution the dual values remain the same, i.e.,y3 = $5 and y4 = $25.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 308
309
cj Basic 40 35 + β 45 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 10 β1/2 0 0 1 0 1/2 β4
0 s2 60 2 0 0 0 1 1 β5
45 x3 10 1/2 0 1 0 0 1/2 β1
35 + β x2 30 1/2 1 0 0 0 β1/2 2
zj 1,500 + 35β 40 + β/2 35 + β 45 0 0 5 β β/2 25 + 2β
cj β zj ββ/2 0 0 0 0 β5 + β/2 β25 β 2β
d) c2, basic:
Solving for the cj β zj inequalities:
ββ/2 β€ 0β β₯ 0
Since c2 = 35 + β; β = c2 β 35. Thus
c2 β 35 β₯ 0c2 β₯ 35
β5 + β/2 β€ 0β/2 β€ 5
β β€ 10
Since c2 = 35 + β; β = c2 β 35. Thus
c2 β 35 β€ 10c2 β€ 45
β25 β 2β β€ 0β2β β€ 25
β β₯ β12.5
Since c2 = 35 + β; β = c2 β 35. Thus
c2 β 35 β₯ β12.5c2 β₯ 22.5
Summarizing, 35 β€ c2 β€ 45.
e) q4:
s1: 10 β 4β β₯ 0 s2: 60 β 5β β₯ 0β4β β₯ β10 β5β β₯ β60
β β€ 5/2 β β€ 12
x3: 10 β β β₯ 10 x2: 30 + 2β β₯ 0ββ β₯ β10 2β β₯ β30β β€ 10 β β₯ β15
Therefore, β15 β€ β β€ 5/2. Since
q4 = 40 + ββ = q4 β 40
β15 β€ q4 β 40 β€ 5/225 β€ q4 β€ 42.5
f) The marginal value of 1 ft2 of storage is $25.From part e, the sensitivity range for q4 is 25 β€ q4 β€ 42.5. Thus, the company wouldpurchase up to 42.5 ft2 of storage space at themarginal value price.
49. a) Maximize Zd = 20y1 + 30y2 + 12y3subject to
4y1 + 12y2 + 3y3 β€ .035y1 + 3y2 + 2y3 β€ .02
y1, y2, y3 β₯ 0
b) y1 = $0 = marginal value of 1 mg of protein;y2 = $0 = marginal value of 1 mg of iron; y3 = $.01 = marginal value of 1 mg ofcarbohydrate (i.e., if one less mg ofcarbohydrate was required, it would be worth$.01 to the dietitian)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 309
310
cj Basic .03 + β .02 0 0 0
Variables Quantity x1 x2 s1 s2 s3
.02 x2 3.6 0 1 0 .20 β.80
.03 + β x1 1.6 1 0 0 β.13 .20
0 s1 4.4 0 0 1 .47 β3.2
zj .12 + 1.6β .03 + β .02 0 0 β .133β β.01 + .2β
zj β cj 0 0 0 0 β .133β β.01 + .2β
c) c1, basic:
Solving for the zj β cj inequalities:
0 β .133β β€ 0β.133β β€ 0
ββ β€ 0β β₯ 0
Since c1 = .03 + β; β = c1 β .03. Thus
c1 β .03 β₯ 0c1 β₯ .03
β.01 + .2β β€ 0.2β β€ .01
β β€ .05
Since c1 = .03 + β; β = c1 β .03. Thus
c1 β .03 β€ .05c1 β€ .08
Summarizing, .03 β€ c1 β€ .08.
c2, basic:
Since c2 = .02 + β; β = c1 β .02. Thus
c2 β .02 β€ 0c2 β€ .02
β.01 + .8β β€ 0.8β β€ .01
β β€ .0125
Since c2 = .02 + β; β = c1 β .02. Thus
c2 β .02 β€ .0125c2 β₯ .0075
Summarizing, c2 β€ .0375.
d) When determining sensitivity ranges for qivalues in a minimization problem, sinceartificial variables are eliminated, the surplusvariable column coefficients must be used. Thiscorresponds to a qi β β change.
cj Basic .03 .02 + β 0 0 0
Variables Quantity x1 x2 s1 s2 s3
.02 + β x2 3.6 0 1 0 .20 β.80
.03 x1 1.6 1 0 0 β.13 .20
0 s1 4.4 0 0 1 .47 β3.2
zj .12 + 3.6β .03 .02 + β 0 0 + .2β β.01 + .8β
zj β cj 0 0 0 0 + .2β β.01 + .8β
Solving for the zj β cj inequalities:
0 + .2β β€ 0.2β β€ 0
β β€ 0
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 310
311
q1:
x2: 3.6 + 0β β₯ 0 x1: 1.6 + 0β β₯ 0 s1: 4.4 + β β₯ 0β β₯ β4.4
Therefore, β β₯ β4.4. Since
q1 = 20 β ββ = 20 β q1
20 β q1 β₯ β4.4q1 β€ 24.4
q2:
x2: 3.6 + .2β β₯ 0 x1: 1.6 β .133β β₯ 0.2β β₯ β3.6 β.133β β₯ β1.6
β β₯ β18 ββ β₯ β12β β€ 12
s1: 4.4 + .47β β₯ 0.47β β₯ β4.4
β β₯ β9.36
Therefore, β9.36 β€ β β€ 12. Since
q2 = 30 β ββ = 30 β q2
β9.36 β€ 30 β q2 β€ 1218 β€ q2 β€ 39.36
q3:
x2: 3.6 β .8β β₯ 0 x1: 1.6 + .2β β₯ 0β.8β β₯ β3.6 .2β β₯ β1.6
ββ β₯ β4.5 β β₯ β8β β€ 4.5
s1: 4.4 β 3.2β β₯ 0β3.2β β₯ β4.4
ββ β₯ β1.375β β€ β1.375
Therefore, β8 β€ β β€ 1.375. Since
q3 = 12 β ββ = 12 β q3
β8 β€ 12 β q3 β€ 1.37510.625 β€ q3 β€ 20
e) The marginal value of 1 mg of carbohydrates is $.01. From part d, the sensitivity range for q3, carbohydrates, is 10.625 β€ q3 β€ 20. Thus, the dietitian could lower the requirements for carbohydrates to10.625 at the marginal value without the solution becoming infeasible.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 311
312
50. a) Minimize Zd = 1,200y1 + 500y3subject to
.50y1 + y2 β₯ 1.251.2y1 β y2 + y3 β₯ 2.00
.80y1 + y2 β₯ 1.75y1, y2, y3 β₯ 0
y1 = the marginal value of an additional hour ofproduction time
= $0y2 = the marginal value of increasing the
combined demand for cheese sandwichesby one sandwich
= $1.75y3 = the marginal value of producing an
additional ham salad sandwich= $3.75
b) c1, non-basic:
β5 + β β€ 0β β€ .50
Since c1 = 1.25 + β; β = c1 β 1.25. Therefore,
c1 β 1.25 β€ .50c1 β€ 1.75
c2, basic:
β3.75 β β β€ 0ββ β€ 3.75β β₯ β3.75
Since c2 = 2 + β; β = c2 β 2. Therefore,
c2 β2 β₯ β3.75c2 β₯ β1.75
c3, basic:
β.5 β β β€ 0 β3.75 β β β€ 0ββ β€ .5 ββ β€ 3.75β β₯ β.5 β β₯ β3.75
Summarizing,
β3.75 β€ β.5 β€ β
and, therefore,
β.5 β€ β
Since c3 = 1.75 + β; β = c3 β 1.75. Therefore,
β.5 β€ ββ.5 β€ c3 β 1.75
1.25 β€ c3
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 312
313
c) q3:
s1: 200 β 2β β₯ 0 x3: 500 + β β₯ 0 x2: 500 + β β₯ 0β2β β₯ β200 β β₯ β500 β β₯ β500
β β€ 100
Summarizing,
β500 β€ β β€ 100
Since q3 = 500 + β, β = q3 β 500. Therefore,
β500 β€ q3 β 500 β€ 1000 β€ q3 β€ 600
d) The marginal value for demand for cheese sandwiches is $1.75. The range for q2 iscomputed as follows.
s1: 200 β .8β β₯ 0 x3: 500 + β β₯ 0 x2: 500 + 0β β₯ 0β.8β β₯ β200 β β₯ β500 0β β₯ β500
β β€ 250 β β€ β
Summarizing,
β500 β€ β β€ 250
Since q2 = 0 + β, β = q2. Therefore,
β500 β€ q2 β€ 250
Thus, the demand for cheese sandwiches canbe increased up to a maximum of 250sandwiches.
The additional profit for 200 more cheesesandwiches would be,
($1.75) (200) = $350
Since the cost of advertising is $100, a $250profit would result, therefore the company should advertise.
51. a) c3, nonbasic:
β13 β β β€ 0ββ β€ 13
β β₯ β13
Since c3 = 2 + β, β = c3 β 2. And
c3 β 2 β₯ β13c3 β₯ β11
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 313
314
c1, basic:
cj Basic 3 + β 5 2 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s2 15 0 0 β4 β3/2 1 β1/2
3 + β x1 5 1 0 β2 β1/2 0 1/2
5 x2 30 0 1 β1 β1/2 0 β1/2
zj 165 + 5β 3 + β 5 β11 β 2β β4 β β/2 0 β1 + β/2
zj β cj 0 0 β13 β 2β β4 β β/2 0 β1 + β/2
Solving for the zj β cj inequalities:
β13 β 2β β€ 0β2β β€ 13
β β₯ β13/2
Since c1 = 3 + β, β = c1 β 3. Thus
c1 β 3 β₯ β13/2c1 β₯ β7/2
β4 β 2β β€ 0β2β β€ 4
β β₯ β2
Since c1 = 3 + β, β = c1 β 3. Thus
c1 β 3 β₯ β2c1 β₯ 1
β1 + β/2 β€ 0β/2 β€ 1
β β€ 2
Since c1 = 3 + β, β = c1 β 3. Thus
c1 β 3 β€ 2c1 β€ 5
Summarizing, 1 β€ c1 β€ 5.
c2: basic:
Solving for the zj β cj inequalities:
β13 β 2β β€ 0β2β β€ 13
β β₯ β13/2
Since c2 = 5 + β, β = c2 β 5. Thus
c2 β 5 β₯ β13/2c2 β₯ β3/2
β4 β β/2 β€ 0ββ/2 β€ 4
β β₯ β8
Since c2 = 5 + β, β = c2 β 5. Thus
c2 β 3 β₯ β8c2 β₯ β3
β1 β β/2 β€ 0ββ/2 β€ 1
β β₯ β2
Since c2 = 5 + β, β = c2 β 5. Thus
c2 β 5 β₯ β2c2 β₯ 3
Summarizing, c2 > 3.
cj Basic 3 5 + β 2 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s2 15 0 0 β4 β3/2 1 β1/2
3 x1 5 1 0 β2 β1/2 0 1/2
5 + β x2 30 0 1 β1 β1/2 0 β1/2
zj 165 + 30β 3 5 + β β11 β 2β β4 β β/2 0 β1 β β/2
zj β cj 0 0 β13 β 2β β4 β β/2 0 β1 β β/2
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 314
315
q1:
s2: 15 β 3β/2 β₯ 0 x1: 5 β β/2 β₯ 0 x2: 30 β β/2 β₯ 0β3β/2 β₯ β15 ββ/2 β₯ β5 ββ/2 β₯ β30
β β€ 10 β β€ 10 β β€ 60
Therefore, β β€ 10. Since
q1 = 35 β ββ = 35 β q1
35 β q1 β€ 10q1 β₯ 25
q2:
s2: 15 + β β₯ 0 x1: 5 β 0β β₯ 0 x2: 30 β 0β β₯ 0β β₯ β15
Therefore, β β€ β15. Since
q2 = 50 β ββ = 50 β q2
50 β q2 β€ β15q2 β€ 65
q3:
s2: 15 β β/2 β₯ 0 x1: 5 + β/2 β₯ 0 x2: 30 β β/2 β₯ 0ββ/2 β₯ β15 β/2 β₯ β5 ββ/2 β₯ β30
β β€ 30 β β₯ β10 β β€ 60
Therefore, β10 β€ β β€ 30. Since q3 = 25 β β and β = 25 β q3
β10 β€ 25 β q3 β€ 30β5 β€ q3 β€ 35
b) When determining sensitivity ranges for qivalues, since artificial values are eliminated,the surplus variable column coefficients mustbe used. This corresponds to a qi β β change.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 315
316
52. a) y1 = 7/15 = $.467
Range for q1:
Since q1 = 320 + β,
β = q1 β 320β120 β₯ q1 β 320 β€ 80200 β€ q1 β€ 400
As many as 400 pears can be purchased.
b) y2 = 1/15 = $.067
Range for q2:
x1: 8 + β/30 β₯ 0 x2: 16 + β/15 β₯ 0β/30 β₯ β8 β/15 β₯ β16
β β₯ β240 β β₯ β240
x3: 3 β β/10 β₯ 0 s4: 36 β β/30 β₯ 0ββ/10 β₯ β3 ββ/30 β₯ β36
β β€ 30 β β€ 1,080
β240 β€ β β€ 30
Since q2 = 400 + β,
β = q2 β 400β240 β€ q2 β 400 β€ 30160 β€ q2 β€ 430
Range over which the value of peaches is valid
c) Range for q3:
s3: 3 + β β₯ 0β β₯ β3
Since q3 = 43 + β,
β = q3 β 43q3 β 43 β₯ β3
q3 β₯ 40
No, increasing q3 from 43 to 60 will not affectthe optimal solution.
x1: 8 + β/15 β₯ 0 x2: 16 β β/30 β₯ 0β/15 β₯ β8 ββ/30 β₯ β16
β β₯ β120 β β€ 480
x3: 3 β β/40 β₯ 0β β₯ 120
s4: 36 β β/30 β₯ 0 s5: 8 β β/10 β₯ 0ββ/30 β₯ β 36 ββ/10 β₯ β8
β β€ 1,080 β β€ 80β120 β€ β β€ 80
d) Range for q5:
s5: 8 + β β₯ 0β β₯ β8
Since q5 = 40 + β,
β = q5 β 40q5 β 40 β₯ β8
q5 β₯ 32
No, increasing q5 from 40 to 50 will have noaffect on the optimal solution.
e) Since y1 = 7/15 = $.467, pears should be secured.
53. a) y2 = $0, spruce has no marginal value
q2:
s2: 70 + β β₯ 0β β₯ β70
Since q2 = 160 + β,
β = q2 β 160q2 β 160 β₯ β70
q2 β₯ 90
b) y3 = $2, marginal value of cutting hours
q3:
s1: 80 β 4β/3 β₯ 0 s2: 70 + β/3 β₯ 0β4β/3 β₯ β80 β/3 β₯ β70
β β€ 60 β β₯ β210
x3: 20 + 2β/3 β₯ 0 x2: 10 β β/3 β₯ 02β/3 β₯ β20 ββ/3 β₯ β10
β β₯ β30 β β€ 30
Therefore, β30 β€ β β€ 30. Since q3 = 50 + β,
β = q3 β 5020 β€ q3 β€ 80
c) y3 = $2, cutting hours; y4 = $2, pressing hours.Since they both have the same marginal value,management could choose either.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 316
317
d) From part a, q2 β₯ 90; thus, a decrease from 160to 100 lb of spruce will not affect the solution.
e) Compute the range for c1, a nonbasic cj value.
c1 = 4 + ββ2 + β β€ 0
β β€ 2c1 β 4 β€ 2
c1 β€ 6
The unit profit from Western paneling would have to be $6 or more before it would beproduced.
f) Compute the range for c3.
c3 = 8 + β
β2 β β/3 β€ 0 β2 β 2β/3 β€ 0 β2 +β/6 β€ 0ββ/3 β€ 2 β2β/3 β€ 2 β/6 β€ 2
β β₯ β6 β β₯ β3 β β€ 12
Since β = c3 β 8,
c3 β 8 β₯ β6 c3 β 8 β₯ β3 c3 β 8 β€ 12c3 β₯ 2 c3 β₯ 5 c3 β€ 20
Summarizing, 5 β€ c3 β€ 20. If the unit profit ofColonial paneling is increased to $13, thepercent solution would not be affected.
54. y1 = $1.33
q1:
x4: 80 + 2β/3 β₯ 0 x2: 40 β β/3 β₯ 02β/3 β₯ β80 ββ/3 β₯ β40
β β₯ β120 β β€ 120
β120 β€ β β€ 120
Since q1 = 200 + β,
β = q1 β 200β120 β€ q1 β 200 β€ 120
80 β€ q1 β€ 320
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 317