Integer Programming (IP)

Prof. Yongwon Seo

(seoyw@cau.ac.kr)

College of Business Administration, CAU

Types of Integer Programming(IP) Problems

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Total Integer Model: All decision variables required to have

integer solution values.

0-1 Integer Model: All decision variables required to have

integer values of zero or one.

Mixed Integer Model: Some of the decision variables (but not all) required to have integer values.

Total Integer Model

• Machine shop obtaining new presses and lathes.

• Marginal profitability: each press $100/day; each lathe $150/day.

• Resource constraints: $40,000 budget, 200 sq. ft. floor space.

• Machine purchase prices and space requirements:

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Machine

Required Floor Space (ft.2)

Purchase Price

Press Lathe

15

30

$8,000

4,000

Total Integer Model

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Maximize Z = $100x1 + $150x2

subject to:

$8,000x1 + 4,000x2 $40,000

15x1 + 30x2 200 ft2

x1, x2 0 and integer

0-1 Integer Model

• Recreation facilities selection to maximize daily usage by residents.

• Resource constraints: $120,000 budget; 12 acres of land.

• Selection constraint: either swimming pool or tennis center (not both).

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Recreation

Facility

Expected Usage (people/day)

Cost ($)

Land Requirement (acres)

Swimming pool Tennis Center Athletic field Gymnasium

300 90 400 150

35,000 10,000 25,000 90,000

4 2 7 3

0-1 Integer Model

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x1 = construction of a swimming pool

x2 = construction of a tennis center

x3 = construction of an athletic field

x4 = construction of a gymnasium

Maximize Z = 300x1 + 90x2 + 400x3 + 150x4

subject to:

$35,000x1 + 10,000x2 + 25,000x3 + 90,000x4 $120,000

4x1 + 2x2 + 7x3 + 3x4 12 acres

x1 + x2 1 facility

x1, x2, x3, x4 = 0 or 1

Mixed Integer Model

• $250,000 available for investments providing greatest return after one year.

• Data: – Condominium cost $50,000/unit; $9,000 profit if sold after one

year.

– Land cost $12,000/ acre; $1,500 profit if sold after one year.

– Municipal bond cost $8,000/bond; $1,000 profit if sold after one year.

– Only 4 condominiums, 15 acres of land, and 20 municipal bonds available.

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Mixed Integer Model

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x1 = condominiums purchased

x2 = acres of land purchased

x3 = bonds purchased

Maximize Z = $9,000x1 + 1,500x2 + 1,000x3

subject to:

50,000x1 + 12,000x2 + 8,000x3 $250,000

x1 4 condominiums

x2 15 acres

x3 20 bonds

x2 0

x1, x3 0 and integer

Why Integer Programming Complex

• Rounding non-integer solution values up to the nearest integer value can result in an infeasible solution.

• A feasible solution by rounding down non-integer solution values may result in a less than optimal (sub-optimal) solution.

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Example of IP solution

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Maximize Z = $100x1 + $150x2subject to:

8,000x1 + 4,000x2 $40,000

15x1 + 30x2 200 ft2

x1, x2 0 and integer

LP Optimal Solution:

Z = $1,055.56

x1 = 2.22 presses

x2 = 5.55 lathes

How to solve?

• Branch-and-Bound Method– Traditional approach to solving integer programming problems.

– Feasible solutions can be partitioned into smaller subsets

– Smaller subsets evaluated until best solution is found.

– Method is a tedious and complex mathematical process

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0-1 Model with Excel

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Recreational Facilities Example:

Maximize Z = 300x1 + 90x2 + 400x3 + 150x4

subject to:

$35,000x1 + 10,000x2 + 25,000x3 + 90,000x4 $120,000

4x1 + 2x2 + 7x3 + 3x4 12 acres

x1 + x2 1 facility

x1, x2, x3, x4 = 0 or 1

0-1 Model with Excel

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0-1 Model with Excel

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Restricts variables, C12:C15, to integer

and 0-1 values

0-1 Model with Excel

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Click on “bin” for 0-1.

0-1 Model with Excel

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Deactivate

Return to solver

window

Total Integer Model with Excel

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Integer Programming Model of Machine Shop:

Maximize Z = $100x1 + $150x2

subject to:

8,000x1 + 4,000x2 $40,000

15x1 + 30x2 200 ft2

x1, x2 0 and integer

Total Integer Model with Excel

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Objective function

Slack, =G6-E6

=C6*B10+D6*B11Decision variables—B10:B11

Total Integer Model with Excel

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Integer variables

Total Integer Model with Excel

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Click on “int”

Mixed Integer Model with Excel: Try!

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Integer Programming Model for Investments Problem:

Maximize Z = $9,000x1 + 1,500x2 + 1,000x3

subject to:

50,000x1 + 12,000x2 + 8,000x3 $250,000

x1 4 condominiums

x2 15 acres

x3 20 bonds

x2 0

x1, x3 0 and integer

Integer Programming Problems and Sensitivity Analysis

• Integer programming problems do not readily lend themselves to sensitivity analysis.– Small changes in the problem result in completely different

solutions.

– Because only a relatively few of the infinite solution possibilities in a feasible solution space will meet integer requirements.

• Completely solving each modified problem is required.

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FORMULATION TECHNIQUES

Integer Programming

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Formulating 0-1 IP

• Either-Or Alternatives– two alternatives x1, x2 (for example, backup machine)

• x1+x2 = 1, x1, x2=0 or 1

– if it is allowed to use neither machine

• x1+x2 1, x1, x2=0 or 1

• k-Out-of-n Alternatives– for example, choosing 2 machines out of 5

• x1+x2+x3+x4+x5 = 2, all xj=0 or 1

– at least 2 machines are required out of 5

• x1+x2+x3+x4+x5 2, all xj=0 or 1

– no more than 2 machines are required out of 5

• x1+x2+x3+x4+x5 2, all xj=0 or 1

– 2~4 machines?

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Formulating 0-1 IP (cont’d)

• If-Then Alternatives– If you buy x2, you should also buy x1.

• x1 x2 , x1, x2=0 or 1

– If the purchase of either machine requires the purchase of the other,

• x1 = x2 , x1, x2=0 or 1

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Formulating 0-1 IP (cont’d)

• Either-Or Constraints (Activating or De-activating certain constraints)– If you use machine x8, 5x4+3x5 100 should be hold. (Turn on

the constraint 5x4+3x5 100)

• 5x4+3x5 100x8 , x8=0 or 1

– If you use machine x8, 5x4+3x5 >=100. If not, 5x4+3x5 50

• 5x4+3x5 100x8• 5x4+3x5 50(1-x8) , x8=0 or 1

– ( case, different) If you execute project y1, 2x1+x2 16. If not, 4x1+8x2 40.

• 2x1+x2 16+M(1-y1)

• 4x1+8x2 40+My1 , y1=0 or 1, M is a very large number

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Formulating 0-1 IP (cont’d)

• Variables That Have Minimum Level Requirements– If you decide to buy x1, the minimum quantity is 200 units.

• x1 200y1• My1 x1 , y1=0 or 1

• Note that,

– x1=0 y1=0

– x1>0 y1=1 x1>=200

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Capital Budgeting

• University bookstore expansion project.

• Not enough space available for both a computer department and a clothing department.

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Project NPV Return

($1,000s) Project Costs per Year ($1000)

1 2 3

1. Web site 2. Warehouse 3. Clothing department 4. Computer department 5. ATMs Available funds per year

$120 85

105 140

75

$55 45 60 50 30

150

$40 35 25 35 30

110

$25 20 --

30 --

60

Capital Budgeting

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x1 = selection of web site project

x2 = selection of warehouse project

x3 = selection clothing department project

x4 = selection of computer department project

x5 = selection of ATM project

xi = 1 if project “i” is selected, 0 if project “i” is not selected

Maximize Z = $120x1 + $85x2 + $105x3 + $140x4 + $70x5subject to:

55x1 + 45x2 + 60x3 + 50x4 + 30x5 150

40x1 + 35x2 + 25x3 + 35x4 + 30x5 110

25x1 + 20x2 + 30x4 60

x3 + x4 1

xi = 0 or 1

Fixed Charge and Facility

• Which of six farms should be purchased that will meet current production capacity at minimum total cost, including annual fixed costs and shipping costs?

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Plant

Available Capacity

(tons,1000s)

A B C

12 10 14

Farms Annual Fixed Costs

($1000)

Projected Annual Harvest (tons, 1000s)

1 2 3 4 5 6

405 390 450 368 520 465

11.2 10.5 12.8 9.3 10.8 9.6

Farm

Plant ($/ton shipped)

A B C

1 2 3 4 5 6

18 15 12 13 10 17 16 14 18 19 15 16 17 19 12 14 16 12

Fixed Charge and Facility

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yi = 0 if farm i is not selected, and 1 if farm i is selected; i = 1,2,3,4,5,6

xij = potatoes (1000 tons) shipped from farm I to plant j; j = A,B,C.

Minimize Z = 18x1A+ 15x1B+ 12x1C+ 13x2A+ 10x2B+ 17x2C+ 16x3+ 14x3B

+18x3C+ 19x4A+ 15x4b+ 16x4C+ 17x5A+ 19x5B+12x5C+ 14x6A

+ 16x6B+ 12x6C+ 405y1+ 390y2+ 450y3+ 368y4+ 520y5+ 465y6subject to:

x1A + x1B + x1B - 11.2y1 ≤ 0 x2A + x2B + x2C -10.5y2 ≤ 0

x3A + x3A + x3C - 12.8y3 ≤ 0 x4A + x4b + x4C - 9.3y4 ≤ 0

x5A + x5B + x5B - 10.8y5 ≤ 0 x6A + x6B + X6C - 9.6y6 ≤ 0

x1A + x2A + x3A + x4A + x5A + x6A = 12

x1B + x2B + x3B + x4B + x5B + x6B = 10

x1C + x2C + x3C + x4C + x5C + x6C = 14

xij ≥ 0 yi = 0 or 1

Set Covering

• Deciding on which nodes provide coverage so that all areas are served.

• A telecommunication company is considering 7 location nodes to reach all 10 neighborhoods. – The cost of opening 7 nodes are : 125, 85, 70, 60, 90, 100, 110

– The coverage of 7 nodes are

• node 1: neighborhoods 1,3,4,6,9,10

• node 2: neighborhoods 2,4,6,8

• node 3: neighborhoods 1,2,5

• node 4: neighborhoods 3,6,7,10

• node 5: neighborhoods 2,3,7,9

• node 6: neighborhoods 4,5,8,10

• node 7: neighborhoods 1,5,7,8,9

• Determine which nodes should be opened to provide coverage to all neighborhoods at a minimum cost.

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Set Covering

Min 125X1 + 85X2 + 70X3 + 60X4 + 90X5 + 100X6 + 110X7

ST

X1+X3+X7>=1

X2+X3+X5>=1

X1+X4+X5>=1

X1+X2+X6>=1

X3+X6+X7>=1

X1+X2+X4>=1

X4+X5+X7>=1

X2+X6+X7>=1

X1+X5+X7>=1

X1+X4+X6>=1

Xj = 0 or 1

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