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Integer Programming (IP)
Prof. Yongwon Seo
College of Business Administration, CAU
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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0-1 Model with Excel
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Click on “bin” for 0-1.
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0-1 Model with Excel
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Deactivate
Return to solver
window
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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
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Total Integer Model with Excel
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Objective function
Slack, =G6-E6
=C6*B10+D6*B11Decision variables—B10:B11
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Total Integer Model with Excel
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Integer variables
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Total Integer Model with Excel
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Click on “int”
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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
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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
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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
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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
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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
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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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