1. Integer Linear Programming
Chapter 6
Integer Linear Programming
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2. Integer Linear Programming
All-Integer Linear Program
All variables must be integers
Mixed-Integer Linear Program
Some, but not all variables must be integers
0-1 Integer Linear Program
Integer variables must be 0 or 1, also known as binary variables
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3. Integer Programming – All Integers
Northern Airlines is a small regional airline. Management is now considering expanding the company by buying additional aircraft. One of the main decisions is whether to buy large or small aircraft to use in the expansion. The table below gives data on the large and small aircraft that may be purchased.
As noted in the table, management does not want to buy more than 2 small aircraft, while the number of large aircraft to be purchased is not limited.
How many aircraft of each type should be purchased in order to maximize annual profit?
Small
Large
Capital Available
Annual profit
$1 million
$5 million
Purchase cost
$5 million
$50 million
$100 million
Maximum purchase quantity 2
No maximum
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4. Define Variables - Northern Airlines
Let:
S = # of Small Aircraft
L = # of Large Aircraft
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5. General Form - Northern Airlines
Max
1S + 5L
s.t.
5S + 50L <= 100
S <= 2
S, L >= 0 & Integer
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6. Northern Airlines – Graph Solution
LP Relaxation
(2, 1.8)
Budget
Small AC
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7. Northern Airlines – Graph Solution
Budget
Small AC
Rounded
Solution
(2, 1)
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8. Northern Airlines – Graph Solution
Optimal
Solution
(0, 2)
Budget
Small AC
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13. Integer Linear Programming
All-Integer Linear Program
All variables must be integers
Mixed-Integer Linear Program
Some, but not all variables must be integers
0-1 Integer Linear Program
Integer variables must be 0 or 1, also known as binary variables
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14. Integer Programming – Mixed Integer
Hart Manufacturing, a mixed integer production problem:
Hart Manufacturing makes three products. Each product goes through three manufacturing departments, A, B, and C. The required production data are given in the table below. (All data are for a monthly production schedule.)
Production Department
Product 1
Product 2
Product 3
Hours available
A (hours/unit)
1.5 3 2
450
B (hours/unit) 1
2.5
350
C (hours/unit)
0.25 50
Profit Contributions per Unit
$25
$28
$30
Setup Costs per production run
$400
$550
$600
Max Production per production run (Units)
175
150
140
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15. General Form – Hart Manu.
Let:
X1= units of product 1
X2= units of product 2
X3= units of product 3
Y1= 1 if production run, else = 0
Y2= 1 if production run, else = 0
Y3= 1 if production run, else = 0
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16. General Form – Hart Manu.
Max
25X1 + 28X2 + 30X3 – 400Y1 – 550Y2 – 600Y3
s.t.
1.5X1 + 3X2 + 2X3 <= Dept. A
2X1 + X X3 <= Dept. B
.25X + .25X + .25X <= Dept. C
X1 <= 175Y1
X2 <= 150Y2
X3 <= 140Y3
Xi >= 0
Yi = integer 0,1
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22. Integer Linear Programming
All-Integer Linear Program
All variables must be integers
Mixed-Integer Linear Program
Some, but not all variables must be integers
0-1 Integer Linear Program
Integer variables must be 0 or 1, also known as binary variables
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23. 0-1 Integer Linear Program (Binary Integer Programming)
Assists in selection process
1 corresponding to undertaking
0 corresponding to not undertaking
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24. 0-1 Integer Linear Program (Binary Integer Programming)
Allows for modeling flexibility through:
Multiple choice constraints
k out of n alternatives constraint
Mutually exclusive constraints
Conditional & co-requisite constraint
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25. Integer Programming - Binary
CAPEX Inc. is a high technology company that faces some important capital budgeting decisions over the next four years. The company must decide among four opportunities:
1. Funding of a major R&D project.
2. Acquisition of an existing company, R&D Inc.
3. Building a new plant, and
4. Launching a new product.
CAPEX does not have enough capital to fund all of these projects. The table below gives the net present value of each item together with the schedule of outlays for each over the next four years. All values are in millions of dollars.
R&D Project
Acquisition of R&D Inc.
New Plant
Launch New Product
Capital Available
Net Present Value (NPV)
100 50 30
Year 1 10 5 40
Year 2 15 60
Year 3 80
Year 4 20 70
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26. General Form – CAPEX Inc.
Let:
X1= 1 if R&D Project funded, else = 0
X2= 1 if acquire company, else = 0
X3= 1 if build new plant, else = 0
X4= 1 if launch new project, else = 0
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27. General Form – CAPEX Inc
Max
100X1 + 50X2 + 30X3 + 50X4
s.t.
10X1 + 30X2 + 5X3 + 10X4 <= Yr 1
15X1 + 0X2 + 5X3 + 10X4 <= Yr 2
15X1 + 0X2 + 5X3 + 10X4 <= Yr 3
20X1 + 0X2 + 5X3 + 10X4 <= Yr 4
Xi = 0,1
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32. Review Problems Electrical Utility Distribution Co. Alpha Airlines
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33. Integer Programming - Review
Electrical Utility, a mixed integer set-up problem:
A problem faced by an electrical utility each day is that of deciding which generators to start up in order to minimize total cost. The utility in question has three generators with the characteristics shown in the table below. There are two periods in a day, and the number of megawatts needed in the first period is The second period requires 3900 megawatts. A generator started in the first period may be used in the second period without incurring an additional startup cost. All major generators (e.g. A, B, and C) are turned off at the end of the day. (Assume all startups occur in time period 1.)
Generator
Fixed Startup Cost
Cost Per Period Per Megawatt Used
Maximum Capacity In Each Period (MW) A
$3,000 $5
2,100 B
$2,000 $4
1,800 C
$1,000 $7
3,000
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34. General Form – Electrical Utility
Let:
XA1 = Power from Gen A in Period 1
XB1 = Power from Gen B in Period 1
XC1 = Power from Gen C in Period 1
XA2 = Power from Gen A in Period 2
XB2 = Power from Gen B in Period 2
XC2 = Power from Gen C in Period 2
YA = 1 if Generator A started; else = 0
YB = 1 if Generator A started; else = 0
YC = 1 if Generator A started; else = 0
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35. General Form – Electrical Utility
Min
5(XA1+XA2) + 4(XB1+XB2) + 7(XC1+XC2) YA YB YC
s.t.
XA1 + XB1 + XC1 >= 2900
XA2 + XB2 + XC2 >= 3900
XA1 <= 2100YA
XA2 <= 2100YA
XB1 <= 1800YB
XB2 <= 1800YB
XC1 <= 3000YC
XC2 <= 3000YC
Xij >= 0
Yi = 0, 1
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41. Integer Programming - Review
Distribution Company, a integer transportation problem:
A distribution company wants to minimize the cost of transporting goods from its warehouses A, B, and C to the retail outlets 1, 2, and 3. The costs (in $’s) for transporting one unit from warehouse to retailer are given in the following table.
The fixed cost of operating a warehouse is $500 for A, $750 for B, and $600 for C, and at least two of them have to be open. The warehouses can be assumed to have adequate storage capacity to store all units demanded, ie., assume each warehouse can store 525 units.
Retailer
Warehouse 1 2 3 A
$15
$32
$21 B $9 $7 $6 C
$11
$18 $5
Demand
200
150
175
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42. General Form – Distribution Co.
Let:
Xij = units shipped from i to j
YA = 1 if warehouse A opens, else = 0
YB = 1 if warehouse B opens, else = 0
YC = 1 if warehouse C opens, else = 0
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43. General Form – Distribution Co.
Min
500YA + 750YB + 600YC + 15XA1 + 32XA2 + 21XA3 + 9XB1 + 7XB2 + 6XB3 + 11XC1 + 18XC2 + 5XC3
s.t.
XA1 + XB1 + XC1 = 200
XA2 + XB2 + XC2 = 150
XA3 + XB3 + XC3 = 175
XA1 + XB1 + XC1 <= 525YA
XA2 + XB2 + XC2 <= 525YB
XA3 + XB3 + XC3 <= 525YC
YA + YB + YC >= 2
Xij >= 0
Yi = 0, 1
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48. Integer Programming - Review
Alpha Airlines, a integer scheduling problem:
Alpha Airlines wishes to schedule no more than one flight out of Chicago to each of the following cities: Columbus, Denver, Los Angeles, and New York. The available departure slots are 8 A.M., 10 A.M., and 12 noon. Alpha leases the airplanes at the cost of $5000 before and including 10 A.M. and $3000 after 10 A.M., and is able to lease at most two per departure slot. Also, if a flight leaves for New York in a time slot, there must be a flight leaving for Los Angeles in the same time slot. The expected profit contribution before rental costs per flight is shown below (in K$)
Time Slot
Cities
8:00 AM
10:00 AM
12:00 Noon
Columbus 10 6
Denver 9
Los Angeles 14 11
New York 18 15
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49. General Form – Alpha Airlines
Let:
Xij= 1 if flight to i occurs in time slot j, else = 0
Yj = number of planes leased for time slot j
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50. General Form – Alpha Airlines
Max
10XC1 + 6XC2 + 6XC3 + 9XD1 + 10XD2 + 9XD3 + 14XL1+ 11XL2 + 10XL3 + 18XN1 + 15XN2 + 10XN3 – 5Y1 – 5Y2 – 3Y3
s.t.
XC1 + XC2 + XC3 <= 1
XD1 + XD2 + XD3 <= 1
XL1 + XL2 + XL3 <= 1
XN1 + XN2 + XN3 <= 1
XC1 + XD1 + XL1 + XN1 = Y1
XC2 + XD2 + XL2 + XN2 = Y2
XC3 + XD3 + XL3 + XN3 = Y3
Y1 <= 2
Y2 <= 2
Y3 <= 2
XN1 <= XL1
XN2 <= XL2
XN3 <= XL3
Xij= 0,1
Yj = INTEGER
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