Table Of ContentChapter 9
Corrections and Exercise Solutions
Integrated Methods for Optimization, 2nd ed.
J. N. Hooker
Corrections
• Page 241. P should be defined P = {(x,y) ≥ (0,0) | Ax+By ≥ b} in
Exercises 6.10 and 6.11.
• Page 259. Theorem 6.11 should state “...if and only if b ≤k and
0
a + a <a ”
j j 0
jX<b0 Xj>k
• Page 269. Replace v∗ on line 8 with v∗+f(x¯). Replace v∗ with v∗+f(x)
i i i i
in formulas (6.50) and (6.51). The last line of (6.51) should be
ui(b−Dx)≤B +Mδ , i∈L
i i 1
• Page 270. Replace 12 and 13 with 12+3x +4x and 13+3x +4x ,
1 2 1 2
respectively, in the Benders cut and its linearized form. The solution of
the second master problem is z = 12 with (x¯ ,x¯ ) = (0,0), not z = 101
1 2 4
with(x¯ ,x¯ )=(0,1).Becausetheprevioussubproblemhasoptimalvalue
1 2
12,thealgorithmterminateswithoptimalsolutionx=(0,0),y=(1,0,1).
• Page 280. Replace the second min in the formulas for U′ and U′ with
z x
max.
• Page 281. Replace “(6.70)–(6.72)” with “(6.73)–(6.75)” in Exercise 6.46.
• Page 287. Replace “element” in line 1 of (6.77) with “alldiff.”
• Page 308. The entry for b(3,b) in Table 6.3 should be 3, not 2.
• Page 310. Replace r with r¯ in line 8 of Fig. 6.17.
front(Q)j front(Q)j
• Page 347. Replace ≫ with ≪ in Exercise 6.69.
• Page 354. Replace e∗ =∞ with e∗ =0 in line 3 of Fig. 6.30.
J J
• Page388.ImmediatelyafterTheorem7.3,replace“C definesanoddhole”
with “C defines an odd hole and contains all vertices of G.”
• Page 390. Replace >1 with <1 in line 1 (Exercise 7.7).
575
576 9 Corrections and Exercise Solutions
• Page 419. Replace d −d ≥−1 just above the figure with x −x ≥−1.
1 2 1 2
• Page 448. Replace “Let x be” with “Let x ≥0 be” in Exercise 7.50.
j j
• Page 469.Replacey≤x1 ≤4y with3y≤x1 ≤5y,andreplace3(1−y)≤
2 2
x2 ≤ 5(1−y) with 1−y ≤ x2 ≤ 4(1−y), immediately below formula
1 1
(7.149).
• Page 491. Replace “has no integrality gap” with “has an integrality gap”
in Exercise 7.91.
• Page 502. Add domain x ∈{6} to Exercise 7.102.
5
Exercise Solutions
Chapter 2 Examples
2.1 The first round of domain propagation reduces the domains to
D ={1,2}, D ={1,2}, D ={0,1,2}
1 2 3
The second round results in
D ={2}, D ={2}, D ={0,1}
1 2 3
and the third round yields
D ={2}, D ={2}, D ={0}
1 2 3
atwhichpointnofurtherreductionispossible.Theinequalitieshaveonly
one feasible solution, x=(2,2,0).
2.2 The packings are on the left below, and the corresponding knapsack
cuts are on the right:
{1} x +x +x ≥5
2 3 4
{1,2} x +x ≥2
3 4
{1,3} x +x ≥2
2 4
{1,4} x +x ≥3
2 3
{2} x +x +x ≥5
1 3 4
{2,3} x +x ≥3
1 4
{2,3,4} x ≥1
1
{2,4} x +x ≥3
1 3
{3} x +x +x ≥6
1 2 4
{3,4} x +x ≥3
1 2
{4} x +x +x ≥6
1 2 3
9 Corrections and Exercise Solutions 577
The cuts corresponding to {2}, {3} and {4} were strengthened using
bounds.
2.3 (a) First solve by branch and relax without bounds propagation. At
the root node we get (x ,x )=(1/2,3). Branching on x , the left branch
1 2 1
adds x ≤ 0 to the relaxation and is infeasible. The right branch adds
1
x ≥ 1 and yields the integer solution x = (2,2). The search tree has 3
1
nodes.
(b) Solving by bounds propagation alone, we obtain x ∈ {1,2,3} and
1
x ∈{2,3} from the constraint 2x +3x ≥10. Branching on x because
2 1 2 2
it has the smaller domain, we first set x = 2, and bounds propagation
2
yieldsx ∈{2,3}.Branchingonx ,wefirstsetx =2andgetthesolution
1 1 1
x=(2,2)withvalue14.Toseekabettersolutionweboundtheobjective
function by 14−1: 3x +4x ≤ 13. This yields x ≤ 1, when in turn
1 2 1
yields x ≥4 when propagated through the constraint. Thus x ’s domain
2 2
becomesempty,andx=(2,2)mustbeoptimal.Thesearchtreeagainhas
3 nodes.
(c)Augmentingbranchandrelaxwithpropagation,weaddtheconstraints
x ≥ 1 and x ≥ 2 to the relaxation, which now has the solution x =
1 2
(1,7/3).Branchingfirstonx ≤2,therelaxationhastheintergalsolution
2
x = (2,2) with value 14. Propagation of the objective function bound 13
reveals that there can be no better solution. The search tree has only 2
nodes.
2.4 Consider the inequalities 2x ≤ 1, 2x +2x ≥ 1 with domains x ∈
1 1 2 j
{0,1}.Thefirstinequalityallowsreductionofx ’sdomainto{0}.Thisin
1
turnsallowsthesecondinequalitytoreducex ’sdomainto{1}.However,
2
theminimumofx subjecttothecontinuousrelaxationoftheinequalities
2
is 0, which results in no domain reduction for x .
2
2.5 Considertheinequalitiesx +x ≥1,x −x ≥0,againwithdomains
1 2 1 2
x ∈ {0,1}. Bounds propagation has no effect on the domains, but the
j
minimum of x subject to the continuous relaxation of the inequalities is
1
1/2, which allows the domain of x to be reduced to {1}.
1
2.6 Letx bethetimbershippedfromforestitosawmillj.Aconditional
ij
formulation is
578 9 Corrections and Exercise Solutions
linear: min z+ c x
ij ij
ij
X
z =f +f +f
1A 1B 2B
δ ⇒ x =d
a 1A A
x +x =d
conditional: δb ⇒xzx111=ABBf+=1Axd2B2+ABf=1BdA+B f2A
z =f +f +f
logic:δa∨δb∨δδcc ⇒xx12AB+=1Axd2BA =2AdA 2B
domains:x ≥0, all i,j, δ ,δ ,δ ∈{true, false}
ij c b c
A disjunctive formulation is
Linear: min z+ c x
ij ij
ij
X
linear disjunction:
z =f +f +f z =f +f +f
1A 1B 2B 1A 1B 2A
x =d ∨ x +x =d
1A A 1A 2A A
x +x =d x =d
1B 2B B 1B B
z =f +f +f
1A 2A 2B
∨ x +x =d
1A 2A A
x =d
2B B
Domains:x ≥0, all i,j
ij
2.7 The model is
linear: min cx
logic:¬δ ∨δ
1 3
δ δ
1 ∨ 2
A1x≥b1 A2x≥b2
linear disjunction: (cid:18) (cid:19) (cid:18) (cid:19)
A2xδ2≥b2 ∨ A3xδ3≥b3
(cid:18) (cid:19) (cid:18) (cid:19)
2.8 Letw betheguardatstationsonnightd,andlety bethestation
sd id
assigned to guard i on night d (0 or −1 if none). Because two guards are
unassigned on a given night, we need two dummy stations to make the
channeling constraints work. The model is
9 Corrections and Exercise Solutions 579
(w )
alldiff: ·d , all d
(y )
·d
(cid:26) (cid:27)
cardinality: (w |(1,2,3,4,5,6),(4,4,4,4,4,4),(5,5,5,5,5,5))
··
nvalues:(w |1,3), s=1,2,3,4
s·
nvalues:(y )|3,5), all i
i·
stretchCycle: (y |(1,2,3,4),(1,1,1,1),(1,1,1,1),P)
i·
w =i, all i
linear: yidd , all d
y =s, all s
(cid:26) wsdd (cid:27)
w ∈{1,...,6}, s=−1,0,1,2,3,4
sd
domains: ,all d
y ∈{−1,0,2,3,4}, i=1,...,6
(cid:26) id (cid:27)
where P ={(i,j)|i,j ∈{−1,0,1,2,3,4}}\{(1,2),(2,1),(3,4),(4,3)}.
2.9 The patterns consistent with the stretch-cycle constraint are:
y y y y y y y
i1 i2 i3 i4 i5 i6 i7
0 2 2 0 1 0 0
0 2 2 0 1 0 1
0 2 2 0 1 1 0
0 2 2 0 1 1 1
2 2 2 0 1 0 0
2 2 2 0 1 0 2
2 2 2 0 1 1 0
3 3 3 0 1 0 0
3 3 3 0 1 1 0
3 3 3 0 1 0 3
Byscanningthecolumns,weseethatthereduceddomainsareasin(2.13).
2.10 Let J be the set of products. The model is
n n
min (h s +g x )+ c
i it i it yt−1yt
i t=1 t=2
XX X
linear: xssiit0===s0iq,,t−a,l1la+ill≥xti1t≥−1dit, all i,t≥1
ytt yt
stretch:(yXi≥,i0,xℓii,tu=i)q,yta,llaill≥t1≥1
domains:x ,s ≥0, all i,t, y ∈J∪{0}, all t
it it t
whereP ={(i,j)|i,j ∈J∪{0}}.Settingy =0indicatesthatnoproduct
t
is manufactured in period t. We can suppose that q =0. The constraint
0
x =q ensuresthatx =0whenproductiisnotmanufacturedin
i≥0 it yt it
P
580 9 Corrections and Exercise Solutions
periodt.Ifconditionalconstraintsareused,thisconstraintcanbereplaced
by
(i6=y )→(x =0), all i,t≥1
t it
The expression q is replaced by z and
ytt t
element(y ,z |q )
t t ·t
The expression c is replaced by w and
yt−1yt t
element((y ,y ),w |C)
t−1 t t
where C is the array of elements c and the element constraint picks out
ij
element (y ,y ) of C.
t−1 t
2.11 Lety =−iindicatethatnoproductismanufacturedinperiodt,and
t
thelastproductmanufacturedisi.Thenc isthesetupcostforproduct
|i|j
j if y = i in the previous period. We can suppose q = 0 for i < 1. The
t i
model becomes
n n
min (h s +g x )+ c
i it i it |yt−1|yt
i t=1 t=2
linear: xssiit0===Xs0iq,,tX−a,l1la+ill≥xti1t≥−1dit, all i,tX≥1
ytt yt
stretch:(yX,ii,xℓii,tu=i)q,yta,llaill≥t1≥1
domains:x ,s ≥0, all i,t, y ∈J∪{−i|i∈J}, all t
it it t
where
P ={(i,j)|i,j ∈J}∪{(i,−i),(−i,i),(−i,−i)|i∈J}
2.12 Let x be the quantity of electricity produced by plant i in period t.
it
One possible model is
min q x +c
i it iyi,t−1yit
i,t
X
linear: Xi xit ≥dt, all t
y =1 y =0
it it
∨ , all i,t
stretchCy(cid:18)clxe:it(≤yi·qi|(cid:19)(0,1(cid:18)),x(iℓt′i=,ℓi0)(cid:19),(n,n),P), all i
domains:x ≥0, y ∈{0,1}, all i,t
it it
9 Corrections and Exercise Solutions 581
where P = {(0,1),(1,0)} and y is identified with y . The expression
i0 in
c isinterpretedbyreplacingitwithz andaddingtheconstraints
iyi,t−1yit it
Element: ((y ,y ),C ,z ), all i,t
i,t−1 it i it
where Ci is the array of elements citt′.
2.13 Let x∈[L,U]. The relaxation for y=x2 is
y≥2Lx−L2, y≥2Ux−U2
y≤(L+U)x−LU
Therelaxationdescribestheregionabovethetangentsat(x,y)=(L,L2),(U,U2)
and below the secant connecting these two points. A similar approach to
y=ax yields the relaxation
y≥aL+L(lna)(ax−aL), y≥aU +U(lna)(ax−aU)
UaL−LaU aU −aL
y≤ + x
U −L U −L
The factored relaxation for
max ax1 +bx2
(x +2x )2 ≤1, L ≤x ≤U , j =1,2
1 2 j j j
uses auxiliary variables y1 = ax1, y1 = bx2, y3 = x1+2x2, y4 = y32. The
definition of y is already linear and needs no relaxation. The lower and
3
upperboundsony areL +2L andU +2U ,respectively.Sotherelaxed
3 1 2 1 2
problem is
max y +y
1 2
y ≤1
4
y ≥2(L +2L )y −(L +2L )2, y ≥2(U +2U )y −(U +2U )2
4 1 2 3 1 2 4 1 2 3 1 2
y ≤(L +U +2L +2U )y −(L +2L )(U +2U )
4 1 1 2 2 3 1 2 1 2
y =x +2x
3 1 2
y ≥aL1 +L (lna) ax1 −aL1 , y ≥aU1 +U (lna) ax1 −aU1
1 1 1 1
y ≤ aL1U1−aU1L1 + aU1 −a(cid:1)L1x (cid:1)
1 U −L U −L 1
1 1 1 1
y ≥bL2 +L (lnb) bx2 −bL2 , y ≥bU2 +U (lnb) bx2 −bU2
2 2 2 2
y ≤ bL2U2−bU2L2 + bU2 −a(cid:1)L2x (cid:1)
2 U −L U −L 2
2 2 2 2
L ≤x ≤U , j =1,2
j j j
2.14 To apply the method in the text, write the problem as
582 9 Corrections and Exercise Solutions
max −x2−x2
1 2
−3x −x ≤−15
1 2
x ,x ≥0and integral
1 2
This yields the inequality
v−L
−3x −x ≥−15−
1 2 λ
which must be satisfied by any optimal solution. Here, lower bound L is
the value of the best known feasible solution x = (4,3), upper bound v
the value of the solution x = (4.5,1.5) of the continuous relaxation, and
Lagrangemultiplierλ=3.SoL=−25andv=−22.5.Wethereforehave
−3x −x ≥−155
1 2 6
Because x ,x are integral, we can round up to obtain the inequality
1 2
−3x −x ≥−15, or 3x +x ≤15. We conclude that 3x +x =15.
1 2 1 2 1 2
2.15 Let D be the domain of y . If we assume each x ≥0, we have the
yi i i
valid knapsack cut
max {A }x + A x ≥b
ik i i i
i∈I k∈Dyi i∈I′
X X
2.16 Let y be the type of fertilizer applied to plot i∈{1,...,m}, and let
i
x the amount of fertilizer. The problem can be written
i
max a x
iyi i
i
X
c x ≤U
iyi i
i
X
nvalues: ({y ,...,y }|1,K)
1 m
domains:y ∈{types of fertilizer}, x ≥0, all i
i i
After unpacking in terms of indexed linear constraints, the problem is
max z
i
indexedLinear:
i
X
elementCoeff(yi,xi,zi |ai1,...,aiK), all i
w ≤U
i
indexedLinear:
i
X
elementCoeff(yi,xi,wi |ci1,...,ciK), all i
nvalues: ({y ,...,y }|1,K)
1 m
domains:y ∈{types of fertilizer}, x ≥0, all i
i i
9 Corrections and Exercise Solutions 583
Table 9.1 Solution of Exercise 2.17.
(t ,t ) J(t ,t ) Inequality
1 2 1 2
(0,7) {3,5} 3x +2x ≤7
A3 A5
(0,10) {1,2,3,4,5} x +3x +3x +4x +2x ≤10
A1 A2 A3 A4 A5
(2,7) {3,5} 3x +2x ≤5
A3 A5
(2,10) {3,4,5} 3x +4x +2x ≤8
A3 A4 A5
(4,7) {5} 2x ≤3
A5
(4,10) {5} 2x ≤6
A5
2.17 The inequalities appear in Table 9.1. All but the second and fourth
are redundant because x ∈{0,1}.
Aj
2.18 AminimummakespansolutiononmachineAis(s ,s ,s )=(0,2,5)
1 3 4
withmakespan9.Themakespanisthesameifjob1isremoved.Aminimal
Benders cut is therefore
M ≥9(x +x −1)
A3 A4
2.19 The time horizon can be divided into intervals [t ,t ] for i =
i i+1
1,...,m. Let x be the interval in which job j is scheduled. The prob-
j
lem becomes
(x ,...,x )
1 m
s +p ≤d , all j
subproblem:linear: j j j
(txj ≤sj ≤txj+1−pj,
noOverlap((s |x =i) |(p |x =i)), all i
j j j j
domains:sj ∈[rj,∞], all j
To solve the problem by logic-based Benders, let binary variable x = 1
ij
when x =i. The master problem is
j
Benders cuts
x ∈{0,1}, all i,j
ij
The subproblem on interval i is
s +p ≤d , all j
j j j
linear: t ≤s ≤t −p , all j with x¯ =1
i j i+1 j ij
noOverlap((sj|x¯ij =1) |(pj|x¯ij =1))
domains:s ∈[r ,∞], all j
j j
2.20 Let x = 1 when vehicle i is assigned to customer j. The master
ij
problem is
584 9 Corrections and Exercise Solutions
linear: min x
ij
ij
X
Benders cuts
domains:x ∈{0,1}, all i,j
ij
In subproblem i, let y be the customer that follows customer j in the
ij
routing of vehicle i, and let s be the time that the vehicle arrives at
j
customer j for delivery. Then the vehicle arrives at the next customer at
times .Ifx¯isthesolutionofthecurrentmasterproblem,thesubproblem
yij
for vehicle i is
e ≤s ≤d , all j with x¯ =1
j j j ij
linear:
s ≥s +p , all j with x¯ =1
(cid:26) yij j ijyij ij
circuit(y ,...,y )
i1 ini
domains:y ∈{1,...,n }, all j with x¯ =1
ij i ij
where n = x¯ .
i j ij
P
Chapter 3 Optimization Basics
3.1 Apply Gauss-Jordan elimination to
1110 2
x=
3101 3
(cid:20) (cid:21) (cid:20) (cid:21)
to obtain
1 0 −1/2 1/2 1/2
x=
0 1 3/2 −1/2 3/2
(cid:20) (cid:21) (cid:20) (cid:21)
So B−1 = −1/2 1/2 , u=c B−1 =[−7/2 −1/2], and r=c −uN =
3/2−1/2 B N
[7/2 1/2].hSince r≥0i, the solution is optimal.
3.2 The basic solutions in each iteration are
Iteration 1.(x ,x ,x )=(1,1,3),(r ,r )=(−3,2)
3 4 5 1 2
Iteration 2.(x ,x ,x )=(1,0,1),(r ,r )=(−4,3)
1 4 5 2 3
Iteration 3.(x ,x ,x )=(1,0,1),(r ,r )=(−1,4)
1 2 5 3 4
Iteration 4.(x ,x ,x )=(2,1,1),(r ,r )=(1,1)
1 2 3 4 5
The basic solutions in iterations 2 and 3 are degenerate.
3.3 The initial Phase I problem is
