Operations Research Ch02

8/9/2019 Operations Research Ch02

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Shiqian Ma, SEEM 3440, Dept. of SEEM, CUHK 4

Example2.1 (Assignment Problem).

s1+ s2+ s3+ s4= 4 d1+ d2+ d3+ d4= 4

Machine 1

Machine 2

Machine 3

Job 1

Job 2

Job 3

Job 4

s1 = 1

s2 = 1

s3 = 1

d4= 1

d3= 1

d2= 1

d1= 1

c11= 1

c12= 4

c13= 6c14= 3

c21= 9

c22= 7

c23= 10

c24

= 9

c31= 4

c32= 5

c33= 11

c34= 7

Assignee Task

Machine 4s4 = 1

c41= 8c42= 7

c43= 8

c44= 5


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Letz= total cost.Fori= 1, 2, 3, 4 andj = 1, 2, 3, 4, let

xij =

1, if machineiis assigned to do jobj,

0, otherwise.


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The IP model is:Minimize z= x11+ 4x12+ 6x13+ 3x14

+ 9x21+ 7x22+ 10x23+ 9x24

+ 4x31+ 5x32+ 11x33+ 7x34+ 8x41+ 7x42+ 8x43+ 5x44 (Cost)

subject to x11+x12+ x13+ x14= 1 (Machine 1)

x21+x22+ x23+ x24= 1 (Machine 2)

x31+x32+ x33+ x34= 1 (Machine 3)

x41+x42+ x43+ x44= 1 (Machine 4)

x11+x21+ x31+ x41= 1 (Job 1)

x12+x22+ x32+ x42= 1 (Job 2)x13+x23+ x33+ x43= 1 (Job 3)

x14+x24+ x34+ x44= 1 (Job 4)

xij = 0 or 1, for alliandj.


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2.1.2. Pure Integer Programming (PIP) Problem All decision variables are required to have integer values.


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Example2.2 (Transportation Problem).

Factory 1

Factory 2

Factory 3

Shop 1

Shop 2

Shop 3

Shop 4

s1 = 15

s2 = 25

s3 = 10

d4= 15

d3= 15

d2= 15

d1= 5c11= 10

c12= 2

c13= 20c14= 11

c21= 12c22= 7

c23= 9

c24= 20

c31= 4

c32= 14

c33= 16

c34= 18

Source Destination

s1+ s2+ s3= 50 d1+ d2+ d3+ d4= 50


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Letz= total cost.Letxij = number of units transport from sourceito destinationj,

fori= 1, 2, 3 andj = 1, 2, 3, 4.

The IP model is:

Minimize z= 10x11+ 2x12+ 20x13+ 11x14+ 12x21+ 7x22+ 9x23+ 20x24+ 4x31+ 14x32+ 16x33+ 18x34 (Cost)

subject to x11+ x12+ x13+ x14= 15 (Factory 1)x21+ x22+ x23+ x24= 25 (Factory 2)

x31+ x32+ x33+ x34= 10 (Factory 3)

x11+ x21+ x31= 5 (Shop 1)x12+ x22+ x32= 15 (Shop 2)

x13+ x23+ x33= 15 (Shop 3)

x14+ x24+ x34= 15 (Shop 4)

xij 0, integer, for alliandj


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2.1.3. Mixed Integer Programming (MIP)Problem

Only some of the decision variables are required to have integervalues (so the divisibility assumption holds for the rest).

Example2.3.

Maximize z= 4x1 2x2 + 7x3 x4subject to x1 + 5x3 10

x1 + x2 x3 1

6x1 5x2 0

x1 + 2x3 2x4 3

xj 0, forj = 1, 2, 3, 4

x1, x2, x3 integers.


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2.2. Some Perspectives on Solving IPProblems

2.2.1. IP Problems Are Hard to SolveIP problems are, in general, much more difficult to solve than LP

problems, because

We are able to characterize the optimal solutions to LP problems:

Primal feasible

Dual feasible Complementary slackness

With IP, no such characterization exists.


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By ignoring integrality restrictions, we have the followingLP Relaxation Problem:

Maximize z= 5x1 + 4x2

subject to x1 + x2 510x1 + 6x2 45

x10, x20.

The optimal solution to the LP relaxation problem is

(x1, x2) = (3.75, 1.25) withz

= 23.75.


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By rounding up or down of the above solution, we have the followingclose-to-optimal solutions:

x1 x2 Feasible? z

3 1 Yes 193 2 Yes 23 *

4 1 No

4 2 NoQuestion: Is the solution (x1, x2) = (3, 2) with z= 23 optimal or

only close-to-optimal?


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2. An optimal LP solution is not necessarily feasible after it isrounded.

Example2.6.IP Problem:

Maximize z= x2subject to x1 + x2 0.5

x1 + x2 3.5

x10, x20x1, x2 integers.

LP Relaxation Problem:

Maximize z= x2subject to x1 + x2 0.5

x1 + x2 3.5

x10, x20.


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The optimal solution to the LP relaxation problem is(x1, x

2) = (1.5, 2).

However, both (1, 2) and (2, 2) are infeasible to the original

IP problem.


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3. There is no guarantee that the rounded solution will be the op-timal integer solution. In fact, it may even be far from optimal.

Example2.7.IP Problem:

Maximize z= 3x1 + 10x2subject to 2x1 + 7x2 14

x1 6


LP Relaxation Problem:


x1 6

x10, x20.


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The optimal solution to the LP relaxation problem is(x1, x

2) = (6, 2/7).

By rounding up or down of the above solution, we have the

following close-to-optimal solutions:

x1 x2 Feasible? z

6 0 Yes 18 *

6 1 No

Notice that (x1, x2) = (6, 0) withz= 18 is far from the optimal

solution to the IP problem (x1, x2) = (0, 2) withz

= 20.


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2.2.3. Exhaustive EnumerationThe Method

Checking each solution for feasibility and, if it is feasible, calculating

the value of the objective function.

Example2.8.Consider the following IP Problem again:


x1 6



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2

x2

2 x14 6 80


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(x1, x2) Feasible? z(x1, x2) Feasible? z(x1, x2) Feasible? z(0,0) Yes 0 (0,1) Yes 10 (0,2) Yes 20

(1,0) Yes 3 (1,1) Yes 13 (1,2) No

(2,0) Yes 6 (2,1) Yes 16 (2,2) No(3,0) Yes 9 (3,1) Yes 19 (3,2) No

(4,0) Yes 12 (4,1) No (4,2) No

(5,0) Yes 15 (5,1) No (5,2) No

(6,0) Yes 18 (6,1) No (6,2) No

Optimal Solution: (x1, x2) = (0, 2) withz

= 20.


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Exponential growth of difficulty Consider the simple case of BIP problems.

With n variables, there are 2n solutions to be considered (where

some of these solutions can subsequently be discarded because theyviolate the functional constraints).

Thus, each time n is increased by 1, the number of solutions is

doubled.

Even the fastest computer can solve IP problems with only a few

dozens of variables by exhaustive enumeration.


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Example2.9. Assume that there is a 1000-core 10 GHz computer system.

Assume that each core of this system can calculate and compare

the objective function values of 1010 integer points in 1 second.

Then, this computer solve a BIP problem with only 80 variables in

about 3830 years by exhaustive enumeration.

n Number of solutions Time required

40 240 0.11 seconds

50 250 1.88 minutes

60 260 1.33 days

70 270 3.74 years

80 280 3830.85 years


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2.3. Modeling Techniques andIllustrative Applications

2.3.1. Binary ChoicesForj = 1, 2, . . . , N , let

yj =

1, if decisionj is yes,

0, if decisionj is no.


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Example2.10 (Capital Budgeting Problem).Five projects are being evaluated over a 3-year planning horizon.

The following table gives the expected returns for each project and the

associated yearly expenditures. (M$ = million dollars)

Expenditures (M$)/year Expected

Year 1 Year 2 Year 3 Returns (M$)

Project 1 5 1 8 20Project 2 4 7 10 40

Project 3 3 9 2 20

Project 4 7 4 1 15

Project 5 8 6 10 30

Available funds (M$) 25 25 25

Which projects should be selected over the 3-year horizon?


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SolutionLetz= total expected returns (million dollars).

Forj = 1, 2, 3, 4, 5, let

yj =

1, if projectj is selected,0, if projectj is not selected.

The IP model is:

Max z= 20y1 + 40y2 + 20y3 + 15y4 + 30y5s.t. 5y1 + 4y2 + 3y3 + 7y4 + 8y5 25

y1 + 7y2 + 9y3 + 4y4 + 6y5 25

8y1 + 10y2 + 2y3 + y4 + 10y5 25y1, y2, y3, y4, y5 binary.


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2.3.2. Relations Between Variables At LeastKDecisions in the Group Must be YES

N

j=1 yj K.

Example2.11 (Set Covering Problem).

To promote on-campus safety, the Security Department of a Uni-versity is in the process of installing emergency telephones at selected

locations. The department wants to install the minimum number of

telephones provided that each of the campus main streets is served by

at least one telephone. The following figure shows that the layout ofthe streets:


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Street A Street B

Street C

Street DStreet E

Street

F

Street

G

Street

I

Street

HStreet

J

Street

K

1 2 3

4 5

6 7 8


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SolutionLetz= number of telephones to be installed.

Forj = 1, 2, . . . , 8, let

yj =

1, if a telephone is installed at locationj,0, otherwise.


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The IP model is:Min z= y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8s.t. y1 + y2 1

y2 + y3 1y4 + y5 1

y7 + y8 1

y6 + y7 1

y2 + y6 1y1 + y6 1

y4 + y7 1

y2 + y4 1y5 + y8 1

y3 + y5 1

yj binary, forj = 1, 2, . . . , 8.


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ExactlyKDecisions in the Group Must be YESN

j=1yj =K.

At MostKDecisions in the Group Can be YES

N

j=1yj K.

y2 is Allowed to be YES only ify1 is YES

y2y1.


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Example2.12 (Capital Budgeting Problem).The California Manufacturing Company is considering expansion by

building a new factory in either Los Angeles (LA) or San Francisco (SF),

or perhaps even in both cities. It also is considering building at most

one new warehouse, but the choice of location is restricted to a citywhere a new factory is being built. The objective is to find the feasible

combination of alternatives that maximizes the total net present value

(NPV).


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Decision Yes-or-No Decision CapitalNumber Question Variable NPV Required

1 Build factory in LA? y1 $9M $6M

2 Build factory in SF? y2 $5M $3M

3 Build warehouse in LA? y3 $6M $5M

4 Build warehouse in SF? y4 $4M $2M

Capital available: $10M

(M = million)


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SolutionLetz= total net present value ($).

Forj = 1, 2, 3, 4, let

yj =

1, if decisionj is yes,0, if decisionj is no.

The IP model is:

Max z= 9y1 + 5y2 + 6y3 + 4y4s.t. 6y1 + 3y2 + 5y3 + 2y4 10

y3 + y4 1

y1 + y3 0 y2 + y4 0

yj binary, forj = 1, 2, 3, 4.


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2.3.3. Function withNPossible Values

g(x1, x2, . . . , xn) =b1 or b2 or . . . or bN

is equivalent to

g(x1, x2, . . . , xn) =N

i=1biyi

N

i=1yi = 1

yi binary, fori= 1, 2, . . . , N .


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Sh M SEEM 44 D f SEEM CUHK


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18 hours of production time per week in Plant 3 currently is unusedand available for the two new products or for certain future products

that will be ready for production soon. In order to leave any remain-

ing capacity in usable blocks for these future products, management

now wants to impose the restriction that the production time used byProducts 1 and 2 be 6 or 12 or 18 hours per week.

Determine what the production rates should be for the two products

in order to maximize the total profit subject to the foregoing restrictions.

Shi i M SEEM 3440 D t f SEEM CUHK 39


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SolutionLetz= total profit (thousand dollars)

Letxj = number of productj produced per week (j = 1, 2)

Let

y1 =

1, if 6 hours/week of production time in Plant 3 being used,

0, otherwise,

y2 =


0, otherwise,

y3 =


0, otherwise,


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Shiqian Ma SEEM 3440 Dept of SEEM CUHK 41


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2.3.4. Fixed-Charge ProblemRequirement

Minimize the cost function:

f(x) =

k+ cx, ifx >0,

0, ifx= 0,

wherek >0,c >0 andx0.



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IP FormulationLetMbe a positive number such thatMx, for all feasiblex, and

let

y=

1, ifx >0,

0, ifx= 0.

Then, the problem can be reformulated as follows:

f(x, y) = ky+ cx

x M y

x 0

y binary.



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Note:In practice, some care is taken to choose a value ofMthat definitely

is large enough to avoid eliminating any feasible solution, but as small

as possible otherwise in order to avoid unduly enlarging the feasibleregion for the LP-relaxation and to avoid numerical instability.



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Example2.14 (Fixed-Charge Problem)

.

John has been approached by three telephone companies to subscribe

to their long distance service in the United States.

Company Flat Fee ($ Per Month) Charge ($ Per Minute)1 16 0.25

2 25 0.21

3 18 0.22

He usually makes an average of 200 minutes of long-distance calls a

month. Assuming that he does not pay the flat monthly fee unless

he makes calls and that he can apportion his calls among all threecompanies as he pleases, how would he use the three companies to

minimize his monthly telephone bill?



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SolutionLetz= total cost ($ per month).

Forj = 1, 2, 3, let

xj = companyj long-distance minutes per month

yj =

1, ifxj >0,

0, ifxj = 0.

LetMj 200 be a constant (j = 1, 2, 3).


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IP FormulationLetMbe a positive number such that

g1(x1, x2, . . . , xn) b1+ M

g2(x1, x2, . . . , xn) b2+ M

for all feasible solutions (x1, x2, . . . , xn).

Then, the problem can be reformulated as follows:

g1(x1, x2, . . . , xn) b1+ M y

g2(x1, x2, . . . , xn) b2+ M(1 y)

y binary.



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Example2.15 (Resource Allocation Problem).

Product 1 Product 2 Product 3 Available

Steel (tons) 1.5 3 5 6000

Labor (hours) 30 25 40 60000Profit ($) 2000 3000 4000

For production of a type of product to be economically feasible, at

least 1000 units of that type must be produced.


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The IP model is:

Max z= 2x1 + 3x2 + 4x3s.t. 1.5x1 + 3x2 + 5x3 6000

30x1 + 25x2 + 40x3 60000

x1 M1y1x1 1000 + M1(1 y1)

x2 M2y2

x2 1000 + M2(1 y2)x3 M3y3

x3 1000 + M3(1 y3)

x10, x20, x30x1,x2,x3 integer

y1,y2,y3 binary.



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2.3.6.Kout ofNConstraintsRequirement

Kout of the followingNconstraints must hold:

g1(x1, x2, . . . , xn) b1g2(x1, x2, . . . , xn) b2

...

gN(x1, x2, . . . , xn) bN



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IP FormulationLetMbe a positive number such that

g1(x1, x2, . . . , xn) b1+ M

g2(x1, x2, . . . , xn) b2+ M...

gN(x1, x2, . . . , xn) bN+ M

for all feasible solution (x1, x2, . . . , xn).



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Then, the problem can be reformulate as follows:

g1(x1, x2, . . . , xn) b1+ M(1 y1)

g2(x1, x2, . . . , xn) b2+ M(1 y2)...

gN(x1, x2, . . . , xn) bN+ M(1 yN)N

i=1yi = K

yi binary, fori= 1, 2, . . . , N .



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2.3.7. If-Then ConstraintsRequirement

h(x1, x2, . . . , xn)>0 = g(x1, x2, . . . , xn)0.IP Formulation

LetMbe a positive number such that

g(x1, x2, . . . , xn) M

h(x1, x2, . . . , xn) M

for all feasible solution (x1, x2, . . . , xn).

Then, the problem can be reformulate as follows:

g(x1, x2, . . . , xn) M yh(x1, x2, . . . , xn) M(1 y)

y binary.



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Example2.16 (Capital Budgeting Problem).

In Example 2.10, if project 5 is selected, then no other project can

be selected.

SolutionLetM4 be a constant.

IP formulation:

y1+ y2+ y3+ y4 M yy5 M(1 y)

y binary



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2.3.8. Binary Representation of General IntegerVariables

0xu, integer

is equivalent to

x=N

i=02i yi

yi binary, fori= 0, 1, 2, . . . , N

whereNis the integer such that 2

N

u


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Example2.17.

Use the binary representation for integer variables to reformulate the

following PIP problem as BIP problem.

Max z= 3x1 + 4x2s.t. x1 5

2x1 + 3x2 30

x10, x20

x1,x2 are integers.


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2.4. The Branch-and-Bound (B&B)Method for IP

2.4.1. Introduction The first B&B algorithm was developed in 1960 by A. Land andG. Doig.

The basic concept underlying the B&B technique is to divide andconquer.

Since the original large problem is too difficult to be solved di-

rectly, it is divided into smaller and smaller subproblems until thesesubproblems can be conquered.

The B&B method has three basic steps: branching, bounding and

fathoming.



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Branching Consider the following problems:

(P) Maximize z=cTx

subject to xX=XA XB(PA) Maximize z=c

Tx

subject to xXA

(PB) Maximize z=c

T

xsubject to xXB

Theorem 1. If zA and zB are the optimal values of problems

(PA) and (PB), respectively, thenz

= max{zA, z

B}is the optimal

value of problem (P).



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Branching involves selecting one remaining subproblem (P) and

dividing it into smaller subproblems (PA) and (PB).

By comparing the optimal values, zA and zB, of the subproblems,

and selecting the maximum value between them, the original prob-

lem can be solved.

Some popular rules for branching are:

Select the subproblem with the best bound. Select the most recently created subproblem.



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Bounding Consider the following problems:

(P) Maximize z=cTx

subject to xX= Z(P) Maximize z=cTx

subject to x

Theorem 2. Ifz and z are the optimal values of problems (P)and (P), respectively, thenz z.

Theorem 3. If

x is an optimal solution to problem (P) and

x Z,

then x

is an optimal solution to problem (P) andz

= z

.


B di ll i d b l i l i


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Bounding usually is done by solving a relaxation.

Some popular relaxations for bounding are:

LP relaxation

Lagrangian relaxation



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Fathoming Fathoming of a subproblem, and thereby dismissing from furtherconsideration, generally is done by an analysis of its relaxation in

the three ways described below.

Criterion 1: An upper bound of the optimal objective function

value of the subproblem the objective function value of the in-

cumbent, or Criterion 2: The subproblem has no feasible solutions, or

Criterion 3: An optimal solution of the subproblem has been

found.



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2.4.2. The Branch-and-Bound AlgorithmInitialization

Setz =(incumbent optimal value).

Apply the bounding step, fathoming step, and optimality test de-

scribed below to the whole problem.

If not fathomed, classify this problem as the one remaining sub-problem for performing the first full iteration below.



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IterationStep 1: Branching

Among the remaining (unfathomed) subproblems, select the one

with the best bound or the one that was created most recently. Among the integer-restricted variables that have a non-integer value

in the optimal solution for the LP relaxation of the subproblem,

choose the first one in the natural ordering of the variable to be thebranching variable.

Let xj be this variable and xj its value in this solution. Branch

from the node for the subproblem to create two new subproblemsby adding the respective constraintsxj xjandxj xj + 1.

(Note: xj= greatest integer xj.)


St 2 B di


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Step 2: Bounding

For each new subproblem, obtain its bound by applying the sim-

plex method (or the dual simplex method when reoptimizing) to its LP

relaxation and using the value ofz for the resulting optimal solution.


St 3 F th i


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Step 3: Fathoming

For each new subproblem, apply the three fathoming tests given be-

low, and discard those subproblems that are fathomed by any of the

tests.

Test 1: Its upper bound z, where z is the value ofz for the

incumbent.

Test 2: Its LP relaxation has no feasible solution.

Test 3: The optimal solution for its LP relaxation has integer

values for the integer-restricted variables. (If this solution is better

than the incumbent, it becomes the new incumbent and test 1 isreapplied to all unfathomed subproblems with the new larger z.)



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Optimality TestStop when there are no remaining subproblems; the incumbent isoptimal. (If there is no incumbent, the conclusion is that the problem

has no feasible solutions.) Otherwise, perform another iteration.



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2.4.3. A 2-D Pure IP ExampleMaximize z= 5x1 + 4x2subject to x1 + x2 5

10x1 + 6x2 45x10, x20

x1, x2 integers.


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Performing Fathoming Tests for the Original Problem


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Test 1: Bound = 23.75> z ==failed.

Test 2: (3.75, 1.25) is a feasible solution to the LP relaxation of the

original problem =failed. Test 3: (3.75, 1.25) violates the integer restrictions for x1 and x2

=failed.

The original problem is not fathomed.



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Iteration 1Creating Branches for the Original Problem

The first integer-restricted variable that has a non-integer value is

x1= 3.75, sox1 becomes the branching variable. Since 3x14, we then create the following two subproblems:

Subproblem 1: Original problem plus additional constraint:

x13.


x14.


Obtaining a Bound for the Subproblem 1


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Solving the LP relaxation of the Subproblem 1 by the graphical

method or the dual simplex method, we obtain

(x1, x2) = (3, 2) with z= 23. Set the bound for the Subproblem 1 = 23.

Performing Fathoming Tests for the Subproblem 1

Since the optimal solution to the LP relaxation of the Subproblem 1,(x1, x2) = ( 3, 2), has integer values for all its integer-restricted

variables, the Subproblem 1 is fathomed by test 3.

Since the optimal value of the LP relaxation of the Subproblem 1,z= 23, satisfiesz > z =, we then update

(x1, x2) = (3, 2) and z

= 23.




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(x1, x2) = (4, 5/6) with z= 70/323.33.

Set the bound for the Subproblem 2 = 70/3.

Performing Fathoming Tests for the Subproblem 2 Test 1: Bound23.33> z = 23 =failed.

Test 2: (4, 5/6) is a feasible solution to the LP relaxation of the

Subproblem 2 =failed.

Test 3: (4, 5/6) violates the integer restriction forx2 =failed.

The Subproblem 2 is not fathomed.


Iteration 2


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Iteration 2

Select a Subproblem for Branching

With only one remaining subproblem, we select Subproblem 2 for

branching.Create Branches for the Subproblem 2


x2= 5/6, sox2 becomes the branching variable.

Since 0x21, we then create the following two subproblems:

Subproblem 3: Subproblem 2 plus additional constraint:

x20.


x21.




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O g S p 3



(x1, x2) = (4.5, 0) with z= 22.5.

Set the bound for the Subproblem 3 = 22.5.

Performing Fathoming Tests for the Subproblem 3 The Subproblem 3 is fathomed by test 1 because its bound = 22.5

z = 23.




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g p


method or the dual simplex method, we find that it has no feasible

solutions.


The Subproblem 4 is fathomed by test 2.

Result

There are no remaining subproblems, so the current solution is opti-

mal. Thus (x1, x2) = (3, 2) withz

= 23.


Branch-and-Bound Tree/Solution Tree/Enumeration Tree


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/S /

LP0

1, x2 = 3.75, 1.25 , z = 23.75

LP1

x1, x2 = 3, 2 , z= 23

LP2

x x = 4 0.833 z 23.3 > 23

LP3

x x = 4.5 0 z = 22.5 23

LP4

Infeasible

1 3 1 4

2 0 2 1

Fathomed

1*, x2* = 3, 2 , z* = 23

z* =

(x1*, x2*) = (3, 2), z* = 23

Fathomed

x1*, 2* = 3, 2 , z* = 23

Fathomed

x1*, x2* = 3, 2 , z* = 23



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2.4.4. Another 2-D Pure IP ExampleMaximize z= 8x1 + 5x2subject to x1 + x2 6

9x1 + 5x2 45x10, x20

x1, x2 integers.


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Test 2: (3.75, 2.25) is a feasible solution to the LP relaxation of the

original problem =failed. Test 3: (3.75, 2.25) violates the integer restrictions for x1 and x2

=failed.



Iteration 1


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Creating Branches for the Original Problem




x13.


x14.




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(x1, x2) = (3, 3) with z= 39.

Set the bound for the Subproblem 1 = 39.




Since the optimal value of the LP relaxation of the Subproblem 1,z= 39, satisfiesz > z =, we then update

(x1, x2) = (3, 3) and z

= 39.




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(x1, x2) = (4, 1.8) with z= 41.


Performing Fathoming Tests for the Subproblem 2 Test 1: Bound = 41 > z = 39 =failed.

Test 2: (4, 1.8) is a feasible solution to the LP relaxation of the


Test 3: (4, 1.8) violates the integer restriction forx2 =failed.



Iteration 2


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x21.


x22.




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(x1, x2) = (40/9, 1)(4.44, 1) with z= 365/940.56.


Performing Fathoming Tests for the Subproblem 3 Test 1: Bound40.56> z = 39 =failed.

Test 2: (40/9, 1) is a feasible solution to the LP relaxation of the


Test 3: (40/9, 1) violates the integer restriction forx1 =failed.





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method or the dual simplex method, we find that it has no feasible

solutions.




Iteration 3


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x14.44, sox1 becomes the branching variable. Since 4x15, we then create the following two subproblems:


x14.


x15.




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(x1, x2) = (4, 1) with z= 37.


Performing Fathoming Tests for the Subproblem 5 The Subproblem 5 is fathomed by test 1 because its bound = 37

z = 39.



S l h LP l f h S b bl b h h l


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(x1, x2) = (5, 0) with z= 40.





Since the optimal value of the LP relaxation of the Subproblem 6,z= 40, satisfiesz > z = 39, we then update

(x1, x2) = (5, 0) and z

= 40.


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Branch-and-Bound Tree


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LP0

x1, x2 = 3.75, 2.25 , z = 41.25

LP1

x1, x2 = 3, 3 , z= 39

LP2

x1, x2 = 4, 1.8 , z = 41 > 39

LP3

x = 4.44 1 z 40.6 > 39

LP4

Infeasible

1 3 1 4

2 1 2 2

LP5

x = 4 1 z= 37 39

LP6

x1, x2 = 5, 0 , z = 40 > 39

1 4 1 5

z* =

Fathomed

x1*,

x2* = 3, 3 ,

z

* = 39

(x1*, x2*) = (3, 3), z* = 39

(x1*, x2*) = (3, 3), z* = 39 Fathomed

1*, x2* = 3, 3 , z* = 39

Fathomed

x1*, x2* = 3, 3 , z* = 39

Fathomed

x1*, x2* = 5, 0 , z* = 40


2 4 5 An MIP E ample


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2.4.5. An MIP ExampleMaximize z= 4x1 2x2 + 7x3 x4subject to x1 + 5x3 10

x1 + x2 x3 16x1 5x2 0

x1 + 2x3 2x4 3

xj 0, forj = 1, 2, 3, 4x1, x2, x3 integer.


Initialization

I i i li i h C O i l V l


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Initializing the Current Optimal Value

Set the current optimal valuez =.

Obtaining a Bound for the Original Problem

Deleting the integer constraints and solving the resulting LP relax-

ation of the original problem by the simplex method, we obtain

(x1, x2, x3, x4) = (1.25, 1.5, 1.75, 0) with z= 14.25.

Set the bound for the original problem = 14.25.



Test 1: Bound = 14 25 > z = = failed


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Test 2: (1.25, 1.5, 1.75, 0) is a feasible solution to the LP relaxation

of the original problem =failed. Test 3: (1.25, 1.5, 1.75, 0) violates the integer restrictions forx1,x2

andx3 =failed.



Iteration 1

C ti B h f th O i i l P bl


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x1= 1.25, sox1 becomes the branching variable.

Since 1x12, we then create the following two subproblems:


x11.


x12.



Solving the LP relaxation of the Subproblem 1 by the dual simplex


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method, we obtain

(x1, x2, x3, x4) = (1, 1.2, 1.8, 0) with z= 14.2.




Test 2: (1, 1.2, 1.8, 0) is a feasible solution to the LP relaxation of

the Subproblem 1 =failed.

Test 3: (1, 1.2, 1.8, 0) violates the integer restriction for x2 and x3=failed.



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Iteration 2



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branching.

Create Branches for the Subproblem 1




x21.Subproblem 4: Subproblem 1 plus additional constraint:

x22.





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method, we obtain

(x1, x2, x3, x4) = (5/6, 1, 11/6, 0) with z= 85/614.17.



Test 1: Bound14.17> z ==failed.

Test 2: (5/6, 1, 11/6, 0) is a feasible solution to the LP relaxation

of the Subproblem 3 =failed.

Test 3: (5/6, 1, 11/6, 0) violates the integer restriction for x1 and

x3 =failed.






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So v g t e e a at o o t e Subp ob e by t e dua s p e

method, we obtain

(x1, x2, x3, x4) = (5/6, 2, 11/6, 0) with z= 73/612.17.



Test 1: Bound12.17> z ==failed.

Test 2: (5/6, 2, 11/6, 0) is a feasible solution to the LP relaxation

of the Subproblem 4 =failed.

Test 3: (5/6, 2, 11/6, 0) violates the integer restriction for x1 and

x3 =failed.



Iteration 3



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With more than one subproblem, the one with the largest bound

(Subproblem 3) is selected for branching.



x1= 5/6, sox1 becomes the branching variable. Since 0x11, we then create the following two subproblems:


x10.Subproblem 6: Subproblem 3 plus additional constraint:

x11.


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g p y p

method, we find that it has no feasible solutions.

Performing Fathoming Tests for the Subproblem 6 The Subproblem 6 is fathomed by test 2.


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Branch-and-Bound Tree

LP0


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x= 1.25, 1.5, 1.75, 0 , z = 14.25

z* =

LP1

x= 1 1.2 1.8 0 z= 14.2

z* =

LP2

InfeasibleFathomed

z* =

LP3

x= 5/6 1 11/6 0 z= 14.17

z* =

LP4

x = 5/6 2 11/6 0 z= 12.17

z* =

LP5

= 0, 0, 2, 0.5 , z= 13.5Fathomed

x* = 0, 0, 2, 0.5 , z* = 13.5

LP6

InfeasibleFathomed

x* = 0, 0, 2, 0.5 , z* = 13.5

1 1 1 2

2 12 2

1

0

1 1 z= 12.17 13.5 Fathomed


2.4.6. A BIP Example


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Maximize z= 9x1 + 5x2 + 6x3 + 4x4subject to 6x1 + 3x2 + 5x3 + 2x4 10

x3 + x4 1x1 + x3 0

x2 + x4 0

xj binary, forj = 1, 2, 3, 4.


Initialization

Initializing the Current Optimal Value


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g

Set the current optimal valuez =.

Obtaining a Bound for the Original Problem Note that the binary restriction on each variablexj (j = 1, 2, 3, 4)

is equivalent to the following set of constraints:

0xj 1xj integer.


Hence, the original problem can be rewritten as:

Maximize z= 9x1 + 5x2 + 6x3 + 4x4


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subject to 6x1 + 3x2 + 5x3 + 2x4 10

x3 + x4 1

x1 + x3 0 x2 + x4 0

x1 1

x2 1x3 1

x4 1

xj 0, forj = 1, 2, 3, 4

xj integer, forj = 1, 2, 3, 4.


Deleting the integer constraints and solving the resulting LP relax-

ation of the original problem by the simplex method, we obtain


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(x1, x2, x3, x4) = (5/6, 1, 0, 1) with z= 16.5.

Set the bound for the original problem = 16.5.



Test 2: (5/6, 1, 0, 1) is a feasible solution to the LP relaxation of

the original problem =failed.

Test 3: (5/6, 1, 0, 1) violates the integer restrictions for x1 =failed.



Iteration 1



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x1= 5/6, sox1 becomes the branching variable.

Sincex1is a binary variable, we then create the following two sub-

problems:

Subproblem 1: Fixx1= 0 in the original problem.

Maximize z= 5x2 + 6x3 + 4x4subject to 3x2 + 5x3 + 2x4 10

x3

+ x4

1

x3 0

x2 + x4 0

xj binary, forj = 2, 3, 4.


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Solving the LP relaxation of the Subproblem 2 by the dual simplexth d bt i


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method, we obtain

(x1, x2, x3, x4) = (1, 0.8, 0, 0.8) with z= 16.2.



Test 1: Bound = 16.2> z = 9 =failed.

Test 2: (1, 0.8, 0, 0.8) is a feasible solution to the LP relaxation of

the Subproblem 2 =failed.

Test 3: (1, 0.8, 0, 0.8) violates the integer restriction for x2 and x4=failed.



Iteration 2



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branching.



x2= 0.8, sox2 becomes the branching variable. Since x2 is binary variable, we then create the following two sub-

problems:


Subproblem 3: Fixx2= 0 in the Subproblem 2.

Maximize z= 6x3 + 4x4 + 9subject to 5x3 + 2x4 4


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subject to 5x3 + 2x4 4

x3 + x4 1

x3 1x4 0

xj binary, forj = 3, 4.

Subproblem 4: Fixx2= 1 in the Subproblem 2.

Maximize z= 6x3 + 4x4 + 14

subject to 5x3 + 2x4 1

x3 + x4 1

x3 1x4 1

xj binary, forj = 3, 4.



Solving the LP relaxation of the Subproblem 3 by the dual simplexmethod or the graphical method we obtain


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method or the graphical method, we obtain

(x1, x2, x3, x4) = (1, 0, 0.8, 0) with z= 13.8.



Test 1: Bound = 13.8> z = 9 =failed.

Test 2: (1, 0, 0.8, 0) is a feasible solution to the LP relaxation of the

Subproblem 3 =failed. Test 3: (1, 0, 0.8, 0) violates the integer restriction forx3= failed.




Solving the LP relaxation of the Subproblem 4 by the dual simplexmethod or the graphical method we obtain


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method or the graphical method, we obtain

(x1, x2, x3, x4) = (1, 1, 0, 0.5) with z= 16.



Test 1: Bound = 16 > z = 9 =failed.

Test 2: (1, 1, 0, 0.5) is a feasible solution to the LP relaxation of the

Subproblem 4 =failed. Test 3: (1, 1, 0, 0.5) violates the integer restriction forx4= failed.



Iteration 3


W h h b bl h h h l b d


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x4= 0.5, sox4 becomes the branching variable. Sincex4is a binary variable, we then create the following two sub-

problems:


Subproblem 5: Fixx4= 0 in Subproblem 4.

Maximize z= 6x3 + 14

subject to 5x3 1


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subject to 5x3 1

x3 1

x3 1x3 binary.

Subproblem 6: Fixx4= 1 in Subproblem 4.

Maximize z= 6x3 + 18

subject to 5x3 1

x3 0

x3 1x3 binary.


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Solving the LP relaxation of the Subproblem 6 by the dual simplexmethod or the graphical method, we find that it has no feasible


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g p ,

solutions.




Iteration 4


With more than one subproblem the one with the largest bound


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x3= 0.2, sox3 becomes the branching variable. Since the resulting branching variable x3 is the last variable, fixing

its value at either 0 or 1 actually creates a single solution rather

than subproblems requiring further investigation.


Fixx3= 0 in Subproblem 5

(x1, x2, x3, x4) = (1, 1, 0, 0) withz= 14 is a feasible solution to theoriginal problem.


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g p

Since (x1, x2, x3, x4) = ( 1, 1, 0, 0) has integer values for all its

integer-restricted variables, it is fathomed by test 3.

Sincez= 14, satisfiesz > z = 9, we then update

(x1, x2, x3, x4) = (1, 1, 0, 0) with z = 14.

Fixx3= 1 in Subproblem 5

(x1, x2, x3, x4) = (1, 1, 1, 0) is an infeasible solution to the originalproblem.

It is fathomed by test 2.


Perform Fathoming Test 1 for all the Remaining Subproblems

Subproblem 3: Bound = 13.8z = 14 =fathomed.

R lt


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Result

There are no remaining subproblems, so the current solution is opti-mal. Thus (x1, x2, x

3, x

4) = (1, 1, 0, 0) withz

= 14.


Branch-and-Bound TreeLP0

= 5/6, 1, 0, 1 , z = 16.5

z* =


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LP1

x= (0, 1, 0, 1), z = 9Fathomed

* = 0, 1, 0, 1 , z* = 9

LP2

x = (1, 0.8, 0, 0.8), z = 16.2x* = (0, 1, 0, 1), z* = 9

LP4

x = (1, 1, 0, 0.5), z = 16

x* = (0, 1, 0, 1), z* = 9

LP5

= 1, 1, 0.2, 0 , z = 15.2

x* = (0, 1, 0, 1), z* = 9

LP6

Infeasible

Fathomed

x* = (0, 1, 0, 1), z* = 9

LP7

x= (1, 1, 0, 0), z= 14 > 9

Fathomed

x* = 1, 1, 0, 0 , z* = 14

LP8

Infeasible

Fathomed

x* = 1, 1, 0, 0 , z* = 14

LP3

= (1, 0, 0.8, 0), z= 13.8

x* = (0, 1, 0, 1), z* = 9

z= 13.8 14 Fathomed

1= 0 1= 1

2= 02= 1

4= 0 4= 1

3= 0 3= 1

Operations Research Ch02

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Transcript of Operations Research Ch02