Solutions Manual and Workbook
73
Chapter 7 Integer Linear Programming
7.1
Z = 210
7.2
Z = 28
7.3
Z = 116
7.4
Z = 115.33
7.5
Let X
1
= number of QM110 students
X
2
= number of QM210 students
MAX Z: 6X
1
+ 9X
2
(Profit, $)
Subject to: X
1
+ 2X
2
< 10 (Max. hours per week)
X
1
< 3 (Max. QM110 students)
X
2
< 5 (Max. QM210 students)
X
1
, X
2
= integer (Integer Restriction)
X
1
, X
2
> 0 (Nonnegativity)
7.6 Let X
1
= Pinecrest site
X
2
= Woodlands site
X
3
= Arbor Oaks site
X
4
= Regency site
X
5
= Hillsboro site
X
6
= Hillwood site
MAX Z: 10000X
1
+12000X
2
+20000X
3
+24000X
4
+16000X
5
+6000X
6
(Profit, $)
Subject to: 120000X
1
+ 100000X
2
+ 164000X
3
+ 206000X
4
+ 100000X
5
+ 82000X
6
< 600000 (Total Invest.)
X
1
, X
2
, X
3
, X
4
, X
5
, X
6
< 1 (Binary Restriction)
X
1
, X
2
, X
3
, X
4
, X
5
, X
6
> 0 (Nonnegativity)
74 Chapter 7 Integer Linear Programming
7.7 Let X
1
= investment in Stock A
X
2
= investment in Stock B
X
3
= investment in Stock C
X
4
= investment in Income Bond A
X
5
= investment in Income Bond B
X
6
= investment in Income Bond C
MAX Z: 100X
1
+ 200X
2
+ 60X
3
+ 120X
4
+ 200X
5
+ 80X
6
(Return, $)
S.T: 1000X
1
+2000X
2
+700X
3
+1400X
4
+1600X
5
+800X
6
< 5000 (Total Investment)
X
1
+ X
2
+ X
3
< 2 (Max. # of stocks)
X
4
+ X
5
+ X
6
> 2 (Min. # of bonds)
X
1
+ X
2
+ X
3
+ X
4
+ X
5
+ X
6
> 4 (Min. # of invest.)
X
1
, X
2
, X
3
, X
4
, X
5
, X
6
< 1 (Binary Restriction)
X
1
, X
2
, X
3
, X
4
, X
5
, X
6
> 0 (Nonnegativity)
7.8 Let X
1
= number of salesmen allocated to East Region
X
2
= number of salesmen allocated to South Region
X
3
= number of salesmen allocated to Southwest Region
MAX Z: 14000X
1
+ 15000X
2
+ 15500X
3
S.T.: 2750X
1
+3000X
2
+2500X
3
< 150000 (Max. Sales Expense)
X
1
< 40 (Max. Salesmen, E)
X
2
< 40 (Max. Salesmen, S)
X
3
< 40 (Max. Salesmen, SW)
X
1
> 15 (Min. Salesmen, E)
X
2
> 15 (Min. Salesmen, S)
X
3
> 15 (Min. Salesmen, SW)
X
1
+ X
2
+ X
3
< 100 (Total Salesmen)
X
1
, X
2
, X
3
= integer (Integer Restriction)
X
1
, X
2
, X
3
> 0 (Nonnegativity)
7.9 Let X
1
= number of sets of men's clubs
X
2
= number of sets of women's clubs
X
3
= number of sets of junior clubs
MAX Z: 170X
1
+ 165X
2
+ 145X
3
(Profit, $)
S.T.: 1.25X
1
+1.30X
2
+0.75X
3
< 100 (Total Hours per Month)
X
1
> 5 (Men’s backlog)
X
2
> 3 (Women’s backlog)
X
3
> 5 (Junior’s backlog)
X
1
, X
2
, X
3
= integer
(Integer Restriction)
X
1
, X
2
, X
3
> 0 (Nonnegativity)
Solutions Manual and Workbook
75
7.10 Let X
1
= number of lighters
X
2
= number of mirrors
X
3
= number of knives
MAX Z: 10X
1
+ 8X
2
+ 7X
3
(Profit, $)
S.T.: 2X
1
+ 8X
2
+ 10X
3
< 32 (Total weight, ozs.)
X
1
> 1 (Min. # of lighters)
X
2
> 1 (Min. # of mirrors)
X
3
> 1 (Min. # of knives)
X
1
, X
2
, X
3
= integer (Integer restriction)
X
1
, X
2
, X
3
> 0 (Nonnegativity)
7.11
X
1
= 4, X
2
= 4, Z
MAX
= 32
7.12
UB = 27.818 (X
1
= 0, X
2
= 3.909, X
3
= 1.182)
LB = 22 (X
1
= 0, X
2
= 3, X
3
= 1)
1
Z = 27.82
2
X
1
=0
X
2
=3
X
3
=1.57
3
X
1
=0
X
2
=4
X
3
=1
UB=27.0
UB=26.0
Stop
*Inferior*
Stop
*Optimal*
X
2
≤
3
X
2
≥
4
7.13
X
1
= 8; X
2
= 1; Z
MIN
= 60
7.14
X
1
= 8; X
2
= 13; Z
MIN
= 139
76 Chapter 7 Integer Linear Programming
7.15
UB = 380.5 (X
1
= 22.167, X
2
= 0.167)
LB = 374 (X
1
= 22, X
2
= 0)
1
Z=380.5
X
1
<
22 X
1
>
23
2
X
1
= 22
X
2
= 0.29
3
Infeasible
Solution
UB = 380.5
LB = 380.29
5
X
1
= 21
X
2
= 1
Stop
4
X
1
= 22
X
2
= 0
Stop
*Inferior*
X
2
<
0X
2
>
1
UB = 380.5
LB = 379
UB=380.5
LB = 374
Stop
*Optimal*
7.16
X
1
= 2; X
2
= 3; X
3
= 2; Z
MAX
= 1260.
7.17
X
1
= 2; X
2
= 4; Z
MAX
= 48
7.18
X
1
= 0; X
2
= 1; X
3
= 1; X
4
= 1; X
5
= 1; X
6
= 0; Z
MAX
= 72000.
7.19
X
1
= 1; X
2
= 0; X
3
= 0; X
4
= 1; X
5
= 1; X
6
= 1: Z
MAX
= 500.
7.20
X
1
= 15; X
2
= 15; X
3
= 25: Z
MAX
= 822500.
7.21
X
1
= 6; X
2
= 3; X
3
= 118; Z
MAX
= 18625.
7.22
X
1
= 7; X
2
= 1; X
3
= 1; Z
MAX
= 85.
7.23
X
1
= 6; X
2
= 0.667; Z
MAX
= 32.
7.24
X
1
= 6; X
2
= 1.6; Z
MIN
= 55.2.
7.25
Z = 360000.
7.26
Z = 314.
7.27
Z = 318.
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