CHAPTER 7: INTEGER PROGRAMMING
7.1-1.
(a)
" if the decision is to build a factory in city 4,
B4 œ œ
! otherwise
"
C4 œ œ
!
if the decision is to build a factory in city 4,
otherwise
for 4 œ LA, SF, SD.
maximize
NPV œ *BLA  &BSF  (BSD  'CLA  %CSF  &CSD
subject to
'BLA  $BSF  %BSD  &CLA  #CSF  $CSD Ÿ "!
CLA  CSF  CSD Ÿ "
BLA  CLA Ÿ !
BSF  CSF Ÿ !
BSD  CSD Ÿ !
BLA ß BSF ß BSD ß CLA ß CSF ß CSD binary
(b)
7.1-2.
(a)
"
Q4 œ œ
!
"
H4 œ œ
!
if 4 does marketing,
otherwise
if 4 does dishwashing,
otherwise
" if 4 does cooking,
G4 œ œ
! otherwise
"
P4 œ œ
!
if 4 does laundry,
otherwise
for 4 œ E ÐEveÑß S ÐStevenÑ.
min
T œ %Þ&QE  (Þ)GE  $Þ'HE  #Þ*PE  %Þ*QS  (Þ#GS  %Þ$HS  $Þ"PS
st
QE  GE  HE  PE œ #
QS  GS  HS  PS œ #
QE  QS œ "
GE  GS œ "
H E  HS œ "
PE  PS œ "
QE ß QS ß GE ß GS ß HE ß HS ß PE ß PS binary
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(b)
11.1-3.
(a)
" if the decision is to invest in project 4,
B4 œ œ
! otherwise
for 4 œ "ß #ß $ß %ß &.
maximize
NPV œ B"  "Þ)B#  "Þ'B$  !Þ)B%  "Þ%B&
subject to
'B"  "#B#  "!B$  %B%  )B& Ÿ #!
B" ß B# ß B$ ß B% ß B& binary
(b) - (c)
7.1.3
(a)
B4 œ œ
" if the decision is to invest in opportunity 4,
! otherwise
for 4 œ "ß #ß $ß %ß &ß '.
Let :4 denote the estimated profit of opportunity 4 and -4 the capital required for
opportunity 4 in millions of dollars.
! B4 :4
'
maximize
4œ"
! B4 -4 Ÿ "!!
'
subject to
4œ"
B"  B# Ÿ "
B$  B% Ÿ "
B$ Ÿ B"  B#
B% Ÿ B"  B#
B4 binary, for 4 œ "ß á ß '
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(b) Solution: Invest in opportunities 1, 3 and 5.
11.1-5.
Each swimmer can swim only one stroke and each stroke can be assigned to only one
swimmer.
11.1-6.
(a) Let X be the number of tow bars produced and W be the number of stabilizer bars
produced.
maximize
T œ "$!X  "&!W
subject to
$Þ#X  #Þ%W Ÿ "'
#X  $W Ÿ "&
X ß W ! integers
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7.3.4
(a) Let Q be a very large number, say "!! million.
max
(!B"  &!ß !!!C"  '!B#  %!ß !!!C#  *!B$  (!ß !!!C$  )!B%  '!ß !!!C%
st
C"  C#  C$  C% Ÿ #
C$ Ÿ C"  C#
C% Ÿ C"  C#
&B"  $B#  'B$  %B% Ÿ '!!!  Q C&
%B"  'B#  $B$  &B% Ÿ '!!!  Q Ð"  C& Ñ
! Ÿ B3 Ÿ Q C3 , for 3 œ "ß #ß $ß %
C3 binary, for 3 œ "ß #ß $ß %
(b)
7.3.5
B"  B# œ !C"  $C#  $C$  'C%  'C& , C3 − Ö!ß "×, for 3 œ "ß á ß &.
11.3-3.
1.
$B"  B#  B$  B% Ÿ "#  Q C"
B"  B#  B$  B% Ÿ "&  Q Ð"  C" Ñ
C" binary
2.
#B"  &B#  B$  B% Ÿ $!  Q C#
B"  $B#  &B$  B% Ÿ %!  Q C$
$B"  B#  $B$  B% Ÿ '!  Q C%
C#  C$  C% Ÿ "
C3 binary, for 3 œ #ß $ß %
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11.3-4.
(a) Let C" and C# be binary variables that indicate whether or not toys 1 and 2 are
produced. Let B" and B# be the number of toys 1 and 2 that are produced. Also, let D be !
if factory 1 is used and " if factory 2 is used.
maximize
"!B"  "&B#  &!ß !!!C"  )!ß !!!C#
subject to
B" Ÿ Q C"
B# Ÿ Q C#
"
&! B"

"
%! B#
Ÿ &!!  Q D
"
%! B"

"
#& B#
Ÿ (!!  Q Ð"  DÑ
B" ß B# ! integers
C" ß C# ß D binary
(b)
7.3.7
(a) Let P, Q , and W be the number of long-, medium-, and short-range jets to buy
respectively.
maximize
T œ %Þ#P  $Q  #Þ$W
subject to
'(P  &!Q  $&W Ÿ "&!!
P  Q  W Ÿ $!
&
$P
 %$ Q  W Ÿ %!
Pß Q ß W ! integers
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(b)
(c)
$! %!
P Ÿ minš "&!!
'( ß " ß &Î$ › œ #%
$! %!
Q Ÿ minš "&!!
&! ß " ß %Î$ › œ $!
$! %!
W Ÿ minš "&!!
$& ß " ß " › œ $!
P œ #! 6!  #" 6"  ## 6#  #$ 6$  #% 6%
Q œ #! 7!  #" 7"  ## 7#  #$ 7$  #% 7%
W œ #! =!  #" ="  ## =#  #$ =$  #% =%
T œ %Þ#! #3 63  $! #3 73  #Þ$! #3 =3
%
maximize
%
3œ!
subject to
%
3œ!
3œ!
'(! #3 63  &!! #3 73  $&! #3 =3 Ÿ "&!!
%
%
%
3œ!
3œ!
3œ!
! #3 63  ! #3 73  ! #3 =3 Ÿ $!
%
%
%
3œ!
3œ!
3œ!
%
&! 3
# 63
$
3œ!
 %$ ! #3 73  ! #3 =3 Ÿ %!
%
%
3œ!
3œ!
63 ß 73 ß =3 binary, for 3 œ !ß "ß #ß $ß %
(d) Solution: 6! œ 6% œ !ß 6" œ 6# œ 6$ œ "ß !%3œ! #3 63 œ "%
7! œ 7" œ 7# œ 7$ œ 7% œ !ß !%3œ! #3 73 œ !
=! œ =" œ =# œ =$ œ !ß =% œ "ß !%3œ! #3 =3 œ "'
T œ $*&Þ' (same as in (b))
11.3-6.
(a) B" œ C""  #C"# ß B# œ C#"  #C##
maximize
^ œ C""  #C"#  &C#"  "!C##
subject to
C""  #C"#  "!C#"  #!C## Ÿ #!
C""  #C"# Ÿ #
C34 binary, for 3ß 4 œ "ß #
(b) Solution: C"" œ C"# œ ! Ê B" œ !, C#" œ !, C## œ " Ê B# œ #, ^ œ "!
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7.3.8
(a) Let B3 be the number of units to produce of product 3 œ "ß #ß $.
C3 œ œ
"
!
if product 3 is produced,
otherwise
maximize
#B"  $B#  !Þ)B$  $C"  #C#
subject to
!Þ#B"  !Þ%B#  !Þ#B$ Ÿ "
B" Ÿ Q C"
B# Ÿ Q C#
! Ÿ B" Ÿ $ integer
! Ÿ B# Ÿ # integer
! Ÿ B$ Ÿ & integer
C" ß C# binary
(b)
11.4-1.
(a)
C34 œ œ
"
!
if B3 œ 4 (i.e., produce 4 units of 3),
otherwise
for 3 œ "ß #ß $ and 4 œ "ß #ß $ß %ß &.
max
C""  #C"#  %C"$  C#"  &C##  C$"  $C$#  &C$$  'C$%  (C$&
st
C""  C"#  C"$ Ÿ "
C#"  C## Ÿ "
C$"  C$#  C$$  C$%  C$& Ÿ "
C""  #C"#  $C"$  #C#"  %C##  C$"  #C$#  $C$$  %C$%  &C$& Ÿ &
C34 binary
(b) Solution: C34 œ ! except for Ð3ß 4Ñ œ Ð$ß &Ñ, C$& œ " Ê B$ œ &, ^ œ (
(c)
C34 œ œ
"
!
if B3 4,
otherwise
for 3 œ "ß #ß $ and 4 œ "ß #ß $ß %ß &.
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7.4.9
(a) Let C" and C# be binary variables that indicate whether or not toys 1 and 2 are
produced. Let B" and B# be the number of toys 1 and 2 that are produced. Also, let D be !
if factory 1 is used and " if factory 2 is used.
maximize
"!B"  "&B#  &!ß !!!C"  )!ß !!!C#
subject to
B" Ÿ Q C"
B# Ÿ Q C#
"
&! B"

"
%! B#
Ÿ &!!  Q D
"
%! B"

"
#& B#
Ÿ (!!  Q Ð"  DÑ
B" ß B# ! integers
C" ß C# ß D binary
(b)
7.3.7
(a) Let P, Q , and W be the number of long-, medium-, and short-range jets to buy
respectively.
maximize
T œ %Þ#P  $Q  #Þ$W
subject to
'(P  &!Q  $&W Ÿ "&!!
P  Q  W Ÿ $!
&
$P
 %$ Q  W Ÿ %!
Pß Q ß W ! integers
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7.3.8
(a) Let B3 be the number of units to produce of product 3 œ "ß #ß $.
C3 œ œ
"
!
if product 3 is produced,
otherwise
maximize
#B"  $B#  !Þ)B$  $C"  #C#
subject to
!Þ#B"  !Þ%B#  !Þ#B$ Ÿ "
B" Ÿ Q C"
B# Ÿ Q C#
! Ÿ B" Ÿ $ integer
! Ÿ B# Ÿ # integer
! Ÿ B$ Ÿ & integer
C" ß C# binary
(b)
7.4.10
(a)
C34 œ œ
"
!
if B3 œ 4 (i.e., produce 4 units of 3),
otherwise
for 3 œ "ß #ß $ and 4 œ "ß #ß $ß %ß &.
max
C""  #C"#  %C"$  C#"  &C##  C$"  $C$#  &C$$  'C$%  (C$&
st
C""  C"#  C"$ Ÿ "
C#"  C## Ÿ "
C$"  C$#  C$$  C$%  C$& Ÿ "
C""  #C"#  $C"$  #C#"  %C##  C$"  #C$#  $C$$  %C$%  &C$& Ÿ &
C34 binary
(b) Solution: C34 œ ! except for Ð3ß 4Ñ œ Ð$ß &Ñ, C$& œ " Ê B$ œ &, ^ œ (
(c)
C34 œ œ
"
!
if B3 4,
otherwise
for 3 œ "ß #ß $ and 4 œ "ß #ß $ß %ß &.
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max
C""  $C"#  #C"$  C#"  %C##  C$"  #C$#  #C$$  C$%  C$&
st
C"$ Ÿ C"# Ÿ C""
C## Ÿ C#"
C$& Ÿ C$% Ÿ C$$ Ÿ C$# Ÿ C$"
C""  C"#  C"$  #C#"  #C##  C$"  C$#  C$$  C$%  C$& Ÿ &
C34 binary
(d) Solution: C34 œ ! for 3 œ "ß #, C$4 œ " for 4 œ "ß á ß & Ê B$ œ &, ^ œ (
7.4.11
Introduce the binary variables C" and C# and add constraints B" Ÿ Q C" , B# Ÿ Q C# ,
C"  C# œ ".
11.4-3.
(a) Introduce the binary variables C" , C# , and C$ to represent positive (nonzero)
production levels.
maximize
^ œ &!B"  #!B#  #&B$
subject to
*B"  $B#  &B$ Ÿ &!!
&B"  %B#
Ÿ $&!
$B"
 #B$ Ÿ "&!
B$ Ÿ #!
B" Ÿ Q C" , B# Ÿ Q C# , B$ Ÿ Q C$
C"  C#  C$ Ÿ #
B" ß B# ß B$ !
C" ß C# ß C$ binary
(b)
7.4.12
(a)
"
C34 œ œ
!
if B3 œ 4,
otherwise
for 3 œ "ß # and 4 œ "ß #ß $.
Work out by hand the objective function contribution for B" ß B# œ !ß "ß #ß $.
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maximize
subject to
$C""  )C"#  *C"$  *C#"  #%C##  *C#$
C""  C"#  C"$ Ÿ "
C#"  C##  C#$ Ÿ "
C""  C#$ Ÿ "
C"$  C#$ Ÿ "
C"#  C#$ Ÿ "
C"#  C## Ÿ "
C"$  C## Ÿ "
C"$  C#" Ÿ "
C34 binary
(b) Solution: C34 œ ! except C"" œ C## œ " Ê B" œ ", B# œ #, ^ œ #(
C34 œ œ
(c)
"
!
if B3 4,
otherwise
for 3 œ "ß # and 4 œ "ß #ß $.
Work out by hand the objective function contribution for B" ß B# œ !ß "ß #ß $.
maximize
subject to
$C""  &C"#  C"$  *C#"  "&C##  "&C#$
C"$ Ÿ C"# Ÿ C""
C#$ Ÿ C## Ÿ C#"
C""  C#$ Ÿ "
C"#  C## Ÿ "
C"$  C#" Ÿ "
C34 binary
(d) Solution: C34 œ ! except C"" œ C#" œ C## œ " Ê B" œ ", B# œ #, ^ œ #(
7.4.13
B34 œ œ
(a)
min
st
"
!
if arc from node 3 to node 4 is in the shortest path,
otherwise
$B"#  'B"$  'B#%  &B#&  %B$%  $B$&  $B%'  #B&'
B"#  B"$ œ "
(1)
B#%  B#&  B$%  B$& œ "
(2)
B%'  B&' œ "
(3)
B#%  B#& Ÿ B"#
(4)
B$%  B$& Ÿ B"$
(5)
B%' Ÿ B#%  B$%
(6)
B&' Ÿ B#&  B$&
(7)
B34 binary
(1), (2), (3) ensure that exactly one arc is used at each stage and they represent mutually
exclusive alternatives. (4), (5), (6) ensure that node 3 is left only if it is entered and they
represent contingent decisions.
(b) Solution: B34 œ ! except B"# œ B#& œ B&' œ ", ^ œ "!
Shortest path: 1 Ä 2 Ä 5 Ä 6
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7.5.14
(a) The dots represent the feasible solutions in the graph below.
Optimal Solution: ÐB" ß B# Ñ œ Ð#ß $Ñß ^ œ &B"  B# œ "$
(b) The optimal solution of the LP relaxation is ÐB" ß B# Ñ œ Ð#Þ'ß "Þ'Ñß ^ œ "%Þ'. The
nearest integer point is ÐB" ß B# Ñ œ Ð$ß #Ñ, which is not feasible, since % † $  #  "#.
Rounded Solutions
Ð$ß #Ñ
Ð$ß "Ñ
Ð#ß #Ñ
Ð#ß "Ñ
Violated Constraints
3rd
2nd and 3rd
none
none
^


"#
""
Hence, none of the feasible rounded solutions is optimal for the IP problem.
11.5-3.
(a) The dots represent the feasible solutions in the graph below.
Optimal Solution: ÐB" ß B# Ñ œ Ð#ß $Ñß ^ œ ##!B"  )!B# œ ')!
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(b) The optimal solution of the LP relaxation is ÐB" ß B# Ñ œ Ð!ß !Þ*Ñ. The nearest integer
point is ÐB" ß B# Ñ œ Ð!ß "Ñ, which is not feasible. The other rounded solution is Ð!ß !Ñ,
which is feasible, but not optimal.
7.5.16
(a) TRUE, Sec. 11.5, 4th paragraph, p. 501.
(b) TRUE, Sec. 11.5, 9th paragraph, p. 502.
(c) FALSE, the result need not be feasible, see Fig. 11.2 for a counterexample, p. 503.
Sec. 11.5, 11th paragraph explains this pitfall.
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7.6.17
Optimal Solution: Ð!ß !ß "ß "ß "Ñ, ^ œ '
7.6.18
Optimal Solution: Ð"ß !ß "ß !ß !Ñ, ^ œ "#
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11.6-5.
7.6.19
(a) FALSE. The feasible region for the IP problem is a subset of the feasible region for
the LP relaxation. It is called a relaxation because it relaxes the feasible region.
(b) TRUE. If the optimal solution for the LP relaxation is integer, then it is feasible for
the IP problem and since the solution for the latter cannot be better than the solution for
the former, it has to be optimal.
(c) FALSE. Figure 11.2 is a counterexample for this statement.
7.6.20
(a) Initialization: Set ^ ‡ œ _. Apply the bounding and fathoming steps and the optimality test as described below for the whole problem. If the whole problem is not
fathomed, then it becomes the initial subproblem for the first iteration below.
Iteration:
1. Branching: Choose the most recently created unfathomed subproblem (in case of a tie,
select the one with the smallest bound). Among the assignees not yet assigned for the
current subproblem, choose the first one in the natural ordering to be the branching
variable. Subproblems correspond to each of the possible remaining assignments for the
branching assignee. Form a subproblem for each remaining assignment by deleting the
constraint that each of the unassigned assignees must perform exactly one assignment.
2. Bounding: For each new subproblem, obtain its bound by choosing the cheapest
assignee for each remaining assignment and totaling the costs.
3. Fathoming: For each new subproblem, apply the two fathoming tests:
Test 1. bound ^ ‡
Test 2. The optimal solution for its relaxation is a feasible assignment (If this
solution is better than the incumbent, it becomes the new incumbent and Test 1 is
reapplied to all unfathomed subproblems with the new smaller ^ ‡ ).
Optimality Test: Identical to the one given in the text.
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(b) Matchings are indicated with the notation (assignee, assignment).
Optimal matching: Ð"ß "Ñß Ð#ß $Ñß Ð$ß #Ñß Ð%ß %Ñß Ð&ß &Ñ, with total cost "&%.
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11.6-8.
(a)
Branch Step: Use the best bound rule.
Bound Step: Given a partial sequencing N" ß á ß N5 of the first 5 jobs, a lower
bound on the time for the setup of the remaining &  5 jobs is found by adding the
minimum elements of the columns corresponding to the remaining jobs, excluding those
elements in rows "None", N" ß N# ß á ß N5" .
Fathoming Step: see the summary of the Branch-and-Bound technique in Sec. 11.6.
(b) The optimal sequence is #  "  %  &  $, with a total setup time of $'.
11.6-9.
Optimal Solution: ÐB" ß B# ß B$ ß B% Ñ œ Ð!ß "ß "ß !Ñ, ^ ‡ œ $'
7.6.21
(a) The only constraints of the Lagrangian relaxation are nonnegativity and integrality.
Since x is feasible for an MIP problem, it already satisfies these constraints, so it is
feasible for the corresponding Lagrangian relaxation.
(b) x‡ is feasible for an MIP problem, so from (a), it has to be feasible for its Lagrangian
relaxation. Also, Ex‡ Ÿ , and - !, so ^V‡ - x‡  -ÐEx‡  ,Ñ - x‡ œ ^ .
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11.7-1.
Prior to this study, Waste Management, Inc. (WM) encountered several operational
inefficiencies concerning the routing of its trucks. The routes served by different trucks
had overlaps and route planners or drivers determined in what order they were going to
visit the stops. The result was inefficient sequences and communication gaps between
customers and customer-service personnel. The problem is formulated as a mixed integer
program, or more specifically as a vehicle routing problem with time windows. The goal
is to obtain routes with minimum number of vehicles and travel time, maximum visual
attractiveness and a balanced workload. First, a network with nodes that represent actual
stops, landfills, lunch break and the depot is constructed. The binary variables B345 refer
to whether arc Ð3ß 4Ñ is included in the route of vehicle 5 or not. The integer variables R5
denote the number of disposal trips and the continuous variables A35 correspond to the
beginning time of service for node 3 by vehicle 5 . The objective function to be minimized
is the total travel time. The constraints make sure that each stop is served by exactly one
truck, each truck starts at the depot, the amount of garbage at the stops does not exceed
the vehicle capacity and each route includes a lunch break. An iterative two-phase
algorithm enhanced with metaheuristics is employed to solve the problem.
Financial benefits of this study include savings of approximately $18 million in 2003 and
estimated savings of $44 million in 2004. WM expects to save more and to increase its
cash flow by $648 million over a five-year interval. The savings in operational costs over
five years is expected to be $498 million. By using mathematical modeling, WM now
generates more efficient routes with minimal overlaps, a reduced number of vehicles and
cost-effective sequences. All these contribute to the decrease in operational costs. At the
same time, centralized routing made communication in the organization and with the
customers easier. Customer-service personnel can now address customer problems more
quickly, since they know the routes of the vehicles. As a result, WM provides a more
reliable customer service. Operational efficiency also affected the environment and the
employees positively. Emissions and noise are reduced. Finally, the benefits from this
study led WM to exploit operations research techniques in other operational areas, too.
7.7.22
(a)
Corner Points
Ð$ß "Þ("%$Ñ
Ð!Þ'ß !Ñ
Ð$ß !Ñ
^
!Þ%#*
"Þ)
*
Ricerca operativa - Fondamenti 9/ed
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Copyright © 2010 - The McGraw-Hill Companies srl
Optimal solution for the LP relaxation: Ð$ß "Þ("%$Ñ with ^ ‡ œ !Þ%#*
Optimal integer solution: Ð#ß "Ñ with ^ ‡ œ "
(b) LP relaxation of the entire problem:
Optimal Solution: ÐB" ß B# Ñ œ Ð$ß "#Î(Ñ, ^ œ $Î(
Branch B# #:
This subproblem is infeasible, so the branch is fathomed.
Ricerca operativa - Fondamenti 9/ed
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Copyright © 2010 - The McGraw-Hill Companies srl
Branch B# Ÿ ":
Optimal Solution: ÐB" ß B# Ñ œ Ð#ß "Ñ, ^ œ ", feasible for the original problem
Hence, the optimal solution for the original problem is ÐB" ß B# Ñ œ Ð#ß "Ñ with ^ œ ".
(c) Let B" œ C""  #C"# and B# œ C#"  #C## .
maximize
subject to
^ œ $C""  'C"#  &C#"  "!C##
&C""  "!C"#  (C#"  "%C## $
C"" ß C"# ß C#" ß C## binary
(d) Optimal Solution: ÐC"" ß C"# ß C#" ß C## Ñ œ Ð!ß "ß "ß !Ñ, ^ œ ", so B" œ # and B# œ " as
in (a).
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(b) Optimal Solution: ÐB" ß B# Ñ œ Ð#ß $Ñ, ^ œ "$
(c) Solution: ÐB" ß B# Ñ œ Ð#ß $Ñ, ^ œ "$
7.7.24
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11.7-6.
Optimal Solution: x œ Ð"%ß !ß "'Ñ, ^ œ *&Þ'.
7.7.25
(a) Let B3 be the number of ¼ units of product 3 to be produced, for 3 œ "ß #.
maximize
%B"  #Þ&B#
subject to
$
% B"
 "# B# Ÿ )
"
# B"
 $% B# Ÿ (
B" ß B# ! integers
(b)
Optimal Solution: ÐB" ß B# Ñ œ Ð"!Þ''(ß !Ñ, ^ œ %#Þ''(
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(c)
Branch B" "": Infeasible
Branch B" Ÿ "!:
Optimal Solution: ÐB" ß B# Ñ œ Ð"!ß "Ñ, ^ œ %#Þ&, feasible for the original problem
Hence, the optimal solution for the original problem is ÐB" ß B# Ñ œ Ð"!ß "Ñ with ^ œ %#Þ&.
(d) Optimal Solution: ÐB" ß B# Ñ œ Ð"!ß "Ñ, ^ œ %#Þ&
(e) Solution: ÐB" ß B# Ñ œ Ð"!ß "Ñ, ^ œ %#Þ&
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11.7-10.
Optimal Solution: x œ Ð"ß !ß "ß !ß #Ñ and x œ Ð#ß #ß !ß !ß !Ñ, ^ œ "#
7.8.26
(a) B" œ !ß B$ œ !
(b) B" œ !
(c) B" œ "ß B$ œ "
11.8-2.
(a) B" œ !
(b) B" œ "ß B# œ !
(c) B" œ !ß B# œ "
7.8.27
From the first equation, ß B$ œ !. Then, this equation becomes redundant. From the third
equation, B& œ ! and B' œ ". Now, this equation is redundant, too. Since B' œ ", from
the second equation, B# œ B% œ ! and this equation becomes redundant. Finally, the
fourth equation reduces to B" œ !. Consequently, all equations become redundant. The
solution is then fixed to Ð!ß !ß !ß !ß !ß "ß B( Ñ.
Ricerca operativa - Fondamenti 9/ed
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Copyright © 2010 - The McGraw-Hill Companies srl
11.8-4.
(a) Redundant. Even if all the variables are set to their upper bounds, B3 œ ", #  "  # Ÿ &.
(b) Not redundant. For example, Ð"ß !ß "Ñ violates this constraint.
(c) Not redundant. For example Ð!Þ!Þ!Ñ violates this constraint.
(d) Redundant. The least value of $B"  B#  #B$ is attained by Ð!ß "ß "Ñ and it is $, so
the constraint is still satisfied.
7.8.28
%B"  $B#  B$  #B% Ÿ &
, œ &ß W œ (ß W  ,  l+" l Ê +" œ W  , œ #ß , œ W  +" œ $
Ê
#B"  $B#  B$  #B% Ÿ $
, œ $ß W œ &ß W  ,  l+# l Ê +# œ ,  W œ #
Ê
#B"  #B#  B$  #B% Ÿ $
, œ $ß W œ &ß W ,  l+4 l for 4 œ "ß #ß $ß %
7.8.29
&B"  "!B#  "&B$ Ÿ "&
, œ "&ß W œ #!ß W  ,  l+# l Ê +# œ ,  W œ &
Ê
&B"  &B#  "&B$ Ÿ "&
, œ "&ß W œ #!ß W  ,  l+$ l Ê +$ œ W  , œ &ß , œ W  +$ œ &
Ê
&B"  &B#  &B$ Ÿ &
, œ &ß W œ "!ß W ,  l+4 l for 4 œ "ß #ß $
11.8-7.
B"  B#  $B$  %B% "
Í
B"  B#  $B$  %B% Ÿ "
, œ "ß W œ "ß W  ,  l+$ l Ê +$ œ ,  W œ #
Ê
B"  B#  #B$  %B% Ÿ "
, œ "ß W œ "ß W  ,  l+% l Ê +% œ ,  W œ #
Ê
B"  B#  #B$  #B% Ÿ "
, œ "ß W œ "ß W ,  l+4 l for 4 œ "ß #ß $ß %
11.8-8.
(a)
B"  $B#  %B$ Ÿ #
, œ #ß W œ %ß W  ,  l+# l Ê +# œ W  , œ #ß , œ W  +# œ "
Ê
B"  #B#  %B$ Ÿ "
, œ "ß W œ $ß W  ,  l+$ l Ê +$ œ ,  W œ #
Ê
B"  #B#  #B$ Ÿ "
, œ "ß W œ $ß W ,  l+4 l for 4 œ "ß #ß $
(b)
$B"  B#  %B$ "
Í
$B"  B#  %B$ Ÿ "
, œ "ß W œ "ß W  ,  l+" l Ê +" œ ,  W œ #
Ê
#B"  B#  %B$ Ÿ "
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, œ "ß W œ "ß W  ,  l+$ l Ê +$ œ ,  W œ #
Ê
#B"  B#  #B$ Ÿ "
, œ "ß W œ "ß W ,  l+4 l for 4 œ "ß #ß $
7.8.30
The minimum cover for the constraint #B"  $B# Ÿ % is ÖB" ß B# ×, so the resulting cutting
plane is B"  B# Ÿ ", which is the same constraint obtained using the tightening
procedure.
7.8.31
ÖB# ß B% × Ä B#  B% Ÿ "
ÖB$ ß B% × Ä B$  B% Ÿ "
ÖB" ß B# ß B$ × Ä B"  B#  B$ Ÿ #
11.8-11.
ÖB" ß B# × Ä B"  B# Ÿ "
ÖB" ß B$ × Ä B"  B$ Ÿ "
ÖB# ß B$ ß B% × Ä B#  B$  B% Ÿ #
7.8.32
ÖB" ß B% × Ä B"  B% Ÿ "
ÖB# ß B% × Ä B#  B% Ÿ "
ÖB$ ß B% × Ä B$  B% Ÿ "
ÖB" ß B# ß B$ × Ä B"  B#  B$ Ÿ #
11.8-13.
ÖB" ß B$ × Ä B"  B$ Ÿ "
ÖB" ß B& × Ä B"  B& Ÿ "
ÖB# ß B$ × Ä B#  B$ Ÿ "
ÖB$ ß B% × Ä B$  B% Ÿ "
ÖB$ ß B& × Ä B$  B& Ÿ "
ÖB% ß B& × Ä B%  B& Ÿ "
ÖB" ß B# ß B% × Ä B"  B#  B% Ÿ #
11.8-14.
(1) $B#  B%  B& $ Ê B# œ "
(2) B"  B# Ÿ " and B# œ " Ê B" œ !
(3) B#  B%  B&  B' Ÿ " and B# œ " Ê B% œ !ß B& œ B' œ "
(4) B#  #B'  $B(  B)  #B* % and B# œ B' œ "
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Ê $B(  B)  #B* " Ê B(  B)  B* "
(5) B$  #B&  B'  #B(  #B)  B* Ÿ & and B& œ B' œ "
Ê B$  #B(  #B)  B* Ÿ # Ê B$  B(  B)  B* Ÿ "
Hence, the problem is reduced to finding binary variables B$ ß B( ß B) ß B* that
maximize
B$  #B(  B)  $B*
subject to
B(  B)  B* "
B$  B(  B)  B* Ÿ ".
The objective is maximized when all variables with positive coefficients are set to their
upper bounds, so when B$ œ B( œ B) œ B* œ ". This solution also satisfies the
constraints, so it is optimal.
Optimal Solution: x œ Ð!ß "ß "ß !ß "ß "ß "ß "ß "Ñ, ^ œ "&
11.9-1.
Since the variables B" ß B# ß B$ take values from the set Ö#ß $ß %× and all the variables must
have different values, B% œ ". There are two feasible solutions, Ð#ß %ß $ß "Ñ and Ð$ß #ß %ß "Ñ.
Their objective function values are ^ œ #*! and ^ œ #)! respectively, so Ð#ß %ß $ß "Ñ is
optimal.
7.9.34
B" œ "#: B% œ ', B# œ $, and B$ œ *, but "#  *  '  #&, so this is not feasible.
B" œ ': B# œ $, B% œ "#, and B$ œ *, '  *  "#  #&, so this is not feasible.
B" œ $: B# œ ', B% œ "#, and B$ œ *, $  *  "# Ÿ #&, so this is feasible. There are two
feasible solutions, Ð$ß 'ß *ß "#ß "&Ñ with ^ œ "$) and Ð$ß 'ß *ß "#ß ")Ñ with ^ œ **.
Hence, Ð$ß 'ß *ß "#ß "&Ñ is optimal.
11.9-3.
B" œ #&: B% œ #! and B$ œ $!, but #&  $!  &&, so this is not feasible.
B" œ $!: B$ − Ö#!ß #&×, but $!  #&  &&, so B$ œ #! and B% œ #&. There are two
feasible solutions, Ð$!ß $&ß #!ß #&Ñ with ^ œ "")#& and Ð$!ß %!ß #!ß #&Ñ with ^ œ ""*&!,
so Ð$!ß %!ß #!ß #&Ñ is optimal.
11.9-4.
Let C3 denote the task to which the assignee 3 is assigned.
minimize
subject to
D "  D #  D $  D%
elementÐC" ß Ò"$ß "'ß "#ß ""Óß D" Ñ
elementÐC# ß Ò"&ß Mß "$ß #!Óß D# Ñ
elementÐC$ ß Ò&ß (ß "!ß 'Óß D$ Ñ
elementÐC% ß Ò!ß !ß !ß !Óß D% Ñ
all-differentÐC" ß C# ß C$ ß C% Ñ
C3 − Ö"ß #ß $ß %×, for 3 œ "ß #ß $ß %
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