Integer programing.ppt

7/28/2019 Integer programing.ppt

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Integer

Programming


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Introduction

Integer programmingis the extension ofLP that solves problems requiringinteger solutions.

Solution values must be whole numbers

in integer programming . There are three types of integer

programs:

pure integer programming;

mixed-integer programming;

01 integer programming.


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Integer Programming(continued)

1. The Pure Integer Programming

problems are cases in which all

variables are required to have integer

values.

2. The Mixed-Integer Programming

problems are cases in which some, but

not all, of the decision variables are

required to have integer values.3. The ZeroOne Integer Programming

problems are special cases in which all

the decision variables must have integer

solution values of 0 or 1.


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Integer ProgrammingExample:

Harrison Electric Company

The Company produces two products

popular with home renovators: old-

fashioned chandeliers and ceiling fans.

Both the chandeliers and fans require atwo-step production process involving

wiring and assembly.

It takes about 2 hours to wire each

chandelier and 3 hours to wire a ceilingfan. Final assembly of the chandeliers and

fans requires 6 and 5 hours, respectively.

The production capability is such that

only 12 hours of wiring time and 30 hoursof assembly time are available.


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Integer Programming:Example (continued)

If each chandelier produced nets the firm

$7 and each fan $6, Harrisons production

mix decision can be formulated using LP as

follows:maximize profit = $7X1 + $6X2

subject to: 2X1 + 3X2 12 (wiring hours)

6X1

+ 5X2 30 (assembly hours)

X1,X2 0 (nonnegative)

X1 = number of chandeliers produced

X2 = number of ceiling fans produced


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Integer Programming:Example (continued)

With only two variables and two constraints,

the graphical LP approach to generate the

optimal solution is given below:

6X1 + 5X2

30

+ = Possible Integer Solution

Optimal LP Solution

(X1 = 33/4,X2 = 1

1/2,

Profit = $35.25

2X1 + 3X2 12


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Integer Solution toHarrison Electric Co.

Optimal

solution

Solution if

rounding

off


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Integer Solution toHarrison Electric Co.

(continued)

Rounding off is one way to reachinteger solution values, but it often

does not yield the best solution.

An important concept tounderstand is that an integer

programming solution can never

be better than the solution to the

same LP problem. The integer problem is usually

worse in terms of higher cost or

lower profit.


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Branch and BoundMethod

Branch and Bound break the feasible

solution region into sub-problems

until an optimal solution is found.

There are Six Steps in Solving IntegerProgramming Maximization Problems

by Branch and Bound.

The steps are given over the next

several slides.


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Branch and BoundMethod: The Six Steps

1. Solve the original problem using

LP.

If the answer satisfies the integer

constraints, it is done. If not, this value provides an

initial upper bound.

2. Find any feasible solution that meets

the integer constraints for use as a

lower bound.

Usually, rounding down each

variable will accomplish this.


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Branch and BoundMethod Steps: (continued)

3. Branch on one variable from Step 1

that does not have an integer value.

Split the problem into two sub-

problems based on integer valuesthat are immediately above and

below the non-integer value.

For example, ifX2 = 3.75 was in the

final LP solution, introduce theconstraint X2 4 in the first sub-

problem and X2 3 in the second

sub-problem.

4. Create nodes at the top of these newbranches by solving the new

problems.


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5.

a) If a branch yields a solution to the LP

problem that is not feasible, terminate the

branch.

b) If a branch yields a solution to the LP

problem that is feasible, but not an

integer solution, go to step 6.

c) If the branch yields a feasible integer

solution, examine the value of the

objective function.

If this value equals the upper bound, an

optimal solution has been reached.

If it is not equal to the upper bound, but

exceeds the lower bound, set it as the

new lower bound and go to step 6.

Finally, if it is less than the lower bound,

terminate this branch.


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. Examine both branches again and

set the upper bound equal to the

maximum value of the objective

function at all final nodes. If the upper bound equals the lower

bound, stop.

If not, go back to step 3.

Minimization problems involve reversinghe roles of the upper and lower bounds.


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Harrison Electric Co:Revisited

Figure 11.1 shows graphically that theoptimal, non-integer solution is

X1 = 3.75 chandeliers

X2 = 1.5 ceiling fans

profit = $35.25

SinceX1 andX2 are not integers, this

solution is not valid.

The profit value of $35.25 will provide

as an initial upper bound.

Note that rounding down givesX1 = 3,

X2 = 1, profit = $27, which is feasible

and can be used as a lower bound.


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Integer Solution:Creating Sub-problems

The problem is now divided into two

sub-problems:A andB.

Consider branching on either variable

that does not have an integer solution;pickX1 this time.

Subproblem Amaximize profit = $7X1 + $6X2

Subject to: 2X1 + 3X2 126X1 + 5X2 30

X1 4

Subproblem B

maximize profit = $7X1 + $6X2Subject to: 2X1 + 3X2 12

6X1 + 5X2 30

X1 3


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Optimal Solution forSub-problems

Optimal solutions are:Sub-problem A: X1 = 4;X2 = 1.2,

profit=$35.20

Sub-problem B:X1=3,X2=2,

profit=$33.00see figure on next slide)

Stop searching on the Subproblem B

branch because it has an all-integerfeasible solution.

The $33 profit becomes the lower

bound.

Subproblem As branch is searched

further since it has a non-integer solution.

The second upper bound becomes

$35.20, replacing $35.25 from the first

node.


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To accompany Quantitative Analysis

for Management, 9e

by Render/Stair/Hanna11-17

2006 by Prentice Hall, Inc.

Upper Saddle River, NJ 07458

Optimal Solution forSub-problem


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Sub-problems C and D

Subproblem As branching yieldsSubproblems C and D.

Subproblem C

maximize profit = $7X1 + $6X2Subject to: 2X1 + 3X2 12

6X1 + 5X2 30

X1 4

X2 2

Subproblem Dmaximize profit = $7X1 + $6X2Subject to: 2X1 + 3X2 12

6X1 + 5X2 30

X1 4X2 1


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Sub-problems C and D(continued)

Subproblem Chas no feasible solution

at all because the first two constraints

are violated if the X1 4 and X2 2

constraints are observed. Terminate this branch and do not

consider its solution.

SubproblemDs optimal solution is

X1 = 4 ,X2 = 1, profit = $35.16. This non-integer solution yields a

new upper bound of $35.16,

replacing the original $35.20.

Subproblems C and D, as well as thefinal branches for the problem, are

shown in the figure on the next slide.


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Harrison Electrics FullBranch and Bound

Solution


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Subproblems E and F

Finally, create subproblemsEandFand

olve forX1 andX2 with the added

onstraintsX1 4 andX1 5. The

ubproblems and their solutions are:

Subproblem Emaximize profit = $7X1 + $6X2

Subject to: 2X1 + 3X2 126X1 + 5X2 30

X1 4

X1 4

X2 1Optimal solution for E:

X1 = 4, X2 = 1, profit = $34


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Subproblems E and F(continued)

Subproblem Fmaximize profit = $7X1 + $6X2Subject to: 2X1 + 3X2 12

6X1 + 5X2 30

X1 4X1 5

X2 1

Optimal solution for F:

X1 = 5, X2 = 0, profit = $35

Integer programing.ppt

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