1. Integer Programming II
ENGM 435/535
Integer Programming II

2. Prototype Problem
Consider an investment problem in a firm is planning construction of new buildings at 4 sites in the city. These sites are designated 1, 2, 3, and 4. At each site there are three possible designs designated A, B, C. The OR Analyst is to model the problem to help management decide on which sites to build and which design to use at the selected sites.

3. Prototype Problem
Consider an investment problem in a firm is planning construction of new buildings at 4 sites in the city. These sites are designated 1, 2, 3, and 4. At each site there are three possible designs designated A, B, C. The OR Analyst is to model the problem to help management decide on which sites to build and which design to use at the selected sites. The table below shows alternatives, investment required and annual income.
Option A1 B1 C1 A2 B2 C2 A3 B3 C3 A4 B4 C4
Investment 12 20 24 30 39 45 10 -- 28 42 50 55
Return 2 4 6 5 7 11 16 19 22
The company has a total investment budget of 80. Our objective is to maximize yearly income.

4. Prototype Formulation
Let
Max Z=2yA1+4yB1+6yC1+5yA2+7yB2+11yC2+ 5yA yC3+ 16yA4+19yB4+22yC4
Subject to
12yA1+20yB1+24yC1+30yA2+39yB2+45yC2+10yA3+28yC3+42yA4+50yB4+55yC4 < 80
In addition to the budget constraint, we have several logical constraints. For example we will not put more than one design on a site.

5. Prototype Formulation
Let
Max Z=2yA1+4yB1+6yC1+5yA2+7yB2+11yC2+ 5yA yC3+ 16yA4+19yB4+22yC4
Subject to
12yA1+20yB1+24yC1+30yA2+39yB2+45yC2+10yA3+28yC3+42yA4+50yB4+55yC4 < 80
yA1+yB1+yC1 < 1
yA2+yB2+yC2 < 1
yA yC3 < 1
yA4+yB4+yC4 < 1

6. Prototype Formulation
Contingent: Plan A will only be available for sites 1, 2, 3 if it is used at site 4
Max Z=2yA1+4yB1+6yC1+5yA2+7yB2+11yC2+ 5yA yC3+ 16yA4+19yB4+22yC4
Subject to
12yA1+20yB1+24yC1+30yA2+39yB2+45yC2+10yA3+28yC3+42yA4+50yB4+55yC4 < 80
yA1+yB1+yC1 < 1
yA2+yB2+yC2 < 1
yA yC3 < 1
yA4+yB4+yC4 = 1
yA1+yA2+yA3 < 3yA yA1+yA2+yA3 - 3yA4 < 0

7. Prototype Formulation
B1, B2, and C4 cannot all be built because of competitive bids
Max Z=2yA1+4yB1+6yC1+5yA2+7yB2+11yC2+ 5yA yC3+ 16yA4+19yB4+22yC4
Subject to
12yA1+20yB1+24yC1+30yA2+39yB2+45yC2+10yA3+28yC3+42yA4+50yB4+55yC4 < 80
yA1+yB1+yC1 < 1
yA2+yB2+yC2 < 1
yA yC3 < 1
yA4+yB4+yC4 = 1
yA1+yA2+yA3 - 3yA4 < 0
yB1+yB2+yC4 < 2

8. Standard Form This may require rearranging variables
Max in standard form equally valid

9. Non-binary Problems Max Z = 4x1 + 3x2 s.t. 9x1+ 6x2 < 54
x1 , x2 integer valued

10. Relaxed Feasible Region
7X1+8X2 = 56
9X1+6X2 = 54
X2 = 5 b c
X1 = 5 d e f a

11. Real Feasible Region 7X1+8X2 = 56 9X1+6X2 = 54 X2 = 5 X1 = 5

12. Solution Techniques
Optimization
Complete Enumeration

13. Real Feasible Region 7X1+8X2 = 56 9X1+6X2 = 54 X2 = 5 15 12 9 6 3 2222 16 13 10 7 4 26 20 17 14 11 8 24 21 18 15 12 25 22 19 16 23 20
X1 = 5

14. Solution Techniques Optimization Complete Enumeration Branch and Bound
Not very practical
Branch and Bound

15. Example Branching Tree
Z=25.12
X1=3
Z=25.4
All
Z=25 X1 X2
Answer
Profit Coefficients $4 $3
Constraint Coefficients
Used
Right Side C1 9 6 54
< C2 7 8 52 56 C3 1 4 5
C3b
> C4 3
Variables (Produced)
Total Profit
$25.00
X1=4
Integer
Solution

16. Example Branching Tree
Z=24.14
X1<3
X2=5
Z=25.12
X1<3 NI
Z=25.4
Z=24
All
X2=4
Z=25
X1=4
Integer
Solution
Integer
Solution

17. Example Branching Tree
Z=23
X1<2
X2=5
Z=24.14
X1<3
X2=5
Z=25.12
X1<3 NI
Z=25.4
Z=24
All
X2=4
Z=25
X1=4
Integer
Solution
Integer
Solution

18. Example Branching Tree
Z=23
X1<2
X2=5
Z=24.14
X1<3
X2=5
Z=25.12
X1<3 NI
X1=3
X2=5
No Feasible
Z=25.4
Z=24
All
X2=4
Z=25
X1=4
Integer
Solution
Integer
Solution

19. Example Branching Tree
Z=23
X1<2
X2=5
Z=24.14
X1<3
X2=5
Z=25.12
X1<3 NI
X1=3
X2=5
No Feasible
Z=25.4
Z=24
All
X2=4
Z=25
X1=4
Integer
Solution
Optimal

20. Solution Techniques Optimization Heuristics Complete Enumeration
Not very practical
Branch and Bound
Can be complicated, particularly for non-binary
Heuristics
Rounding

21. Non-binary Problems Max Z = 4x1 + 3x2 s.t. 9x1+ 6x2 < 54
x1 , x2 integer valued

22. Non-binary Problems Max Z = 4x1 + 3x2 s.t. 9x1+ 6x2 < 54

23. Non-binary Problems Max Z = 4x1 + 3x2 s.t. 9x1+ 6x2 < 54
X1=4, X2=3 Optimal but would we
round to here?

24. Solution Techniques Optimization Heuristics Complete Enumeration
Not very practical
Branch and Bound
Can be complicated, particularly for non-binary
Heuristics
Rounding
Nearest Neighbor

25. Traveling Salesman Problem
Let

26. Traveling Salesman Problem

27. Traveling Example Nearest Neighbor Start with any city
From all cities not yet on the path, find the closest city
Connect these two; find next closes
Complete loop by connecting last to first

28. Traveling Example

29. Traveling Example

30. Traveling Example

31. Traveling Example

32. Solution Techniques Optimization
Complete Enumeration
Not very practical
Branch and Bound
Can be complicated, particularly for non-binary
Heuristics (Optimality is not guaranteed)
Rounding
Nearest Neighbor

33. Classic IP Problems Plant Location Capital Budgeting
Traveling Salesman
Assembly Line Balancing

34. Plant Location

35. Capital Budgeting

36. Traveling Salesman Problem

37. Line Balancing