Lecture 7_IP Graphical and B&B

8/12/2019 Lecture 7_IP Graphical and B&B

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EMG181 Quarter 2 SY2013-14

Integer Programming:Graphical MethodBranch and Bound Method

Meeting 13

Lecture 7


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Definition

Integer Programming (IP) models arecharacterized by the fact that someorof the decision variables are required tobe integers.


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Difference between LP and IP Cost of Indivisibility Adding the indivisibility/integer require

results in additional constraints, making the optimal integersolution either inferior(the usual case) or, at best as goodoptimal noninteger solution.

Solution Method There is no simple solution to the IP proOne may use near-optimal rather than optimizing technique

Number of Solutions IP has only a finite number of possibsolutions

(as compared to an infinite number in LP) but thisnumber can still be very large.

Special Cases Transportation and Assignment Problems arspecial cases of IP.

LP Software can be used to solve some special cases of problems. However, the sensitivity analyses generated by thecodes do not usually have any meaning.

Optimal Solutions with LP, there is either one optimal soluor an infinite number of them. With IP, there is either one o

several (finite) optimal solutions.


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Three Types of IP Models

1. Total (Or Pure) Integer Modelall decision variableshave integer solution values.

2. 0-1 Integer Model (Or Binary Model)solution valuesthe decision variables are zero or one.

3. Mixed Integer Modelsome solution values for thedecision variables are integers and others can benonintegers (fractional).


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Methods of Solution

Complete Enumeration

Graphical Method

Rounding of LP Solution

Branch and Bound

Implicit Enumeration

More Complicated Methods

Computer software Dynamic Programming Heuristics Gomory Cutting Plane Method


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Example 1: Graphical MethodMaximize Z = 10x1+ 15x2

s.t. 8x1 + 4x2 40

15x1 + 30x2 200

x1, x2 0 and integer

LP-relaxation solution: Opt. (20/9, 50/9) Z = 950/9 ~105.56

x21110

9

8

6

4

2

1 2 3 4 5 6 7 8 9 10 11 12 13 x1(1)

(2)

Z

Opt. (20/9, 50/9) Z = 950/9


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Example 2: Graphical Method

Objective: Maximize Z = 5x1 + x2

Subject to: -x1 + 2x2 < 4

x1 - x2 < 14x1 + x2 < 12


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BRANCH AND BOUND METHOD

Uses a tree diagram of nodesand branch

to organize the solution partitioning.

This is an intelligent search procedure foreither an optimal or a close-to-optimal

solution to certain managerial problems,including all-integer and mixed-integerproblems.


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Branch and Bound Procedure

1. Solve the LP relaxation. If the optimal LP solution is integer,it is optimal for the IP.

2. Divide the problem into two (or more) subproblems (branchithat divides the feasible area into regions that removes thecurrent LP optimal solution from the new feasible region. Anupper bound (UB) and a lower bound (LB) on the value of thobjective function (Z) is set.

3. Start branching from the variable with the greatest fractionaThe variable is branched out to include only values > the inabove and < the integer value below the optimal LP solutionThe branches represent additional constraints to the originaproblem.


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4. The optimal solution for each branch is determined.Subproblems whose objective function is worse than thestablished feasible bounds are eliminated from furtheconsideration (inferior solutions).

5. The remaining subproblems are used to modify thebounds (LB or UB), then subdivided and investigated.

6. This process is repeated until no further subdivision is

possible, at which point the optimal (or near-optimal)solution has been reached.


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Method of division

x mple.If LP relaxation solution of a pure IP problem is x1= 5, x2= then subdivide the problem using x2as follows:

Case1 = x2 6

Case2 = x2 7 In the maximization problem, the initial UB is the Z value of theoptimal LP solution since the IP solution will be < than this value

The initial LB must always be feasible and is obtained by roundidown the initial optimal solution to the LP relaxation. All thesucceeding branching will have lower UB since the Z value decreawith every branching. For a maximization problem, subsequentbranching must alwaysbe from the node with the highest UB.

In the minimization problem, the initial LB is the Z value of theoptimal LP solution. The initial UB is obtained by rounding up thinitial optimal solution to the LP relaxation. All the succeedingbranching will have higher LB since the Z value increases with evebranching. For the minimization problem, subsequent branching malways be from the node with the lowest LB.


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Maximize Z = 10x1+ 15x2

s.t. 8x1 + 4x2 40

15x1 + 30x2 200


LP-relaxation solution: Opt. (20/9, 50/9) Z = 950/9 ~105.56

x21110

9

8

6

4

2

1 2 3 4 5 6 7 8 9 10 11 12 13 x1(1)

(2)

Z

Opt. (20/9, 50/9) Z = 950/9


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Partition the feasible region into two parts that excludeunwanted optimal solution. Get the integers that bordethe optimal LP solution.

x1= 2..22

x2= 5.56

Z = 105.56

From the Graphical Method, Optimal LP solution:

x1= 20/9 = 2.22x2= 50/9 = 5.56

Z = 950/9 = 105.56

Branch and Bound: x2 first

X2> 6

Upper bound = 105.56Lower bound = 95

X2< 5


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Branch first on higher fractional part: x26 and x25

x211

10

98

6

4

2

1 2 3 4 5 6 7 8 9 10 11 12 13 x1

(1)

(2)

Z

x26

x25

Opt. (4/3, 6) Z = 310/3 ~ 103.3

Opt (5/2, 5) Z = 100


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Optimal Solution from First Branch

x1= 2.22

x2= 5.56

Z = 105.56

x1= 1.33

x2= 6

Z = 103.3

x1= 2.5

x2= 5Z = 100

x2> 6

Upper bound = 105.56

Lower bound = 95

x2< 5

FIS


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Branch on higher Z at x26 branch on: x12 & x

x2

1110

9

8

6

1 2 3 4 5 6 7 8 x1

(1)(2)

Z

x26

O t. 1 37/6 Z = 205/2 ~ 102.5

x11

x12

Second Branching


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Optimal Solution from Second Branch

x1= 2.22

x2= 5.56

Z = 105.56

x1= 1.33

x2= 6

Z = 103.3

x1= 2.5

x2= 5

Z = 100

NFS

x1= 1

x2= 6.17

Z = 102.5

x1> 2

Upper bound = 103.3

Lower bound = 95

x2> 6

x1< 1


Lower bound = 95

x2< 5


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Third BranchingBranch on higher Zat x11 branch on: x27 & x

Feasible region of x26 is segment from (0, 6) to (1,

x2

1110

9

8

6

1 2 3 4 5 6 7 8 x1

Since the x25 branch cannot improve on this answ

this is the optimal integer solution.

(1)(2)

Z

x26

O t 1, 6 Z = 100

x11

x26

x27


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Optimal Solution from Third Branch

x1= 2.22

x2= 5.56

Z = 105.56

x1= 1.33

x2= 6

Z = 103.3

x1= 2.5

x2= 5

Z = 100

NFS

x1= 1

x2= 6.17

Z = 102.5

NFS

x1= 1x2= 6

Z = 100

x1> 2

Upper bound = 103.3

Lower bound = 95 x2> 7

x2> 6

x1< 1 Upper bound = 102.5Lower bound = 95


Lower bound = 95

x2< 6

x2< 5


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Example 2: MinimizationMinimize Z = 6x1+ 3x2

s.t. x1 + x2 4

8x1 + 2x2 16


LP-relaxation solution: Opt. (4/3, 8/3) Z = 16

x298

6

4

2

1 2 3 4 5 6 x1

(1)(2) Z

Opt. (4/3, 8/3) Z = 16


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Branch first on higher fractional part: x23 and x2

x298

6

43

2

1 2 3 4 5 6 x1

(1)(2) Z

Opt. (5/4, 3) Z = 33/2 ~ 16.5

Opt. (2, 2) Z = 18

x23

x22


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Branch on lower Zat x23 branch on: x12 and

x29

8

6

4

3

2

1 2 3 4 5 6 x1

Since all the branchings gave integer solutions, ther

need to branch any more.

(1)(2) Z

Opt. (1, 4) Z = 18

Opt. (2, 3) Z = 21

x23

x22

x12x11


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Optimal IP solution is the best of the integer solutionsfound.

(Minimize)

Multiple optimal solutions with Z = 18.

(2, 2) and (1,4)

(4/3, 8/3) Z = 16UB (2, 3) = 21, LB = 16

(2, 2) Z = 18UB = 21, LB = 18

(5/4, 3) Z = 33/2 16.UB = 21, LB = 33/2

x22 x23

(1, 4) Z = 18UB = 21, LB = 18

(2, 3) Z = 2UB = 21, LB

x1x11STOP branching. Integer

Optimal

STOP. Integer

Optimal

STOP. Inte

Inferior


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Example 3

Solve the following LP problem using the graphical methoand the corresponding pure IP using Branch and Bou

Obj. Max Z = 30 X1 + 40 X2

Subject to:

-2 X1 + X2 < 100

5 X1 + 2 X2 > 250

5 X1 + 10 X2 < 1500X1 X2 < 50

X1 + 6 X2 > 300


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Graphical Solution (LP Problem)


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Branch and Bound Solution (IP Problem)


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Branch and Bound for the Mixed Integer

Same basic steps as in the pure integer mod

with only a few differences: Only variables that are required to be integers arebranched, and only they are to be rounded down up) to get an initial LB (or UB).

Always branch (from among the variables that musinteger) the one with the greatest fractional part.


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Solving the 0-1 Model by Branch and Boun

Same basic steps as the pure-integer with only few changes.

The LP relaxation to be solved should have thefollowing constraints added: x1 1, x2 1, , x1 for all variables that are binary.

Find the variable with the greatest fractional par

xj, and branch as: xj = 0 and xj = 1.


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Solving the 0-1 Model by Implicit Enumerati

Used when the number of constraints is less ththe number of variables.

Pursues only feasible solution branches.

Each branch is for values of x=0 or 1.

Decision variables branched one at a time.


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Implicit Enumeration Example

Obj. Min Z = 20 X1 + 25 X2 + 15 X3 + 10 X4

Subject To:X1 + 4 X2 + 2 X3 - X4 > 6

X1 + X2 - X3 + X4 > 2

3 X1 + 2 X2 + 4 X3 + X4 > 5

X1 + 2 X2 - X4 > 2

Xi = 0 or 1

Lecture 7_IP Graphical and B&B

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