1. ENGM 435/535
Integer Programming

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. np hard
Consider just 3 variables x1, x2, x3 . Possible solution set is then
(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1)
Or 8 possible solutions. In general, there are 2n possible solutions if variables are binary.
We can check all possible solutions but it is darn tough as n gets larger. We need a more efficient algorithm.

10. Branch and Bound Technique
Partition,
Let x1 = 1, Solution subset: (1,0,0), (1,0,1), (1,1,0), (1,1,1)
Let x1 = 0, Solution subset: (0,0,0), (0,0,1), (0,1,0), (0,1,1)
Lower Bound ZL
Create a relaxation by dropping some restrictions, find a lower bound on a given subset
Upper Bound ZU
Upper bound is Z evaluated for any feasible solution of original problem.
Fathoming
If ZL > ZU for a subset, then eliminate the subset
If subset contains no feasible solutions, it can be eliminated

11. Example Max Z = 9x1 + 5x2 + 6x3 + 4x4 s.t. 6x1+3x2+5x3+5x4 < 10
xj = 0 or 1, j=1, 2, 3, 4

12. Example Branch Let x1 = 0 , then Subset 1: Max Z = 5x2 + 6x3 + 4x4
s.t.
3x2+5x3+5x4 < 10
x3+ x4 < 1
x < 1
-1x x4 < 0
xj = 0 or 1, j=2, 3, 4,

13. Example Branch Let x1 = 1 , then Subset 1: Max Z = 9 +5x2 + 6x3 + 4x4
s.t.
3x2+5x3+5x4 < 4
x3+ x4 < 1
x < 1
-1x x4 < 0
xj = 0 or 1, j=2, 3, 4,

14. Example Branching Tree
All
X1=1

15. Find a Bound to Original
Max Z = 9x1 + 5x2 + 6x3 + 4x4
s.t.
6x1+3x2+5x3+5x4 < 10
x3+ x4 < 1
-1x x < 0
-1x x4 < 0
0 < xj < 1 or 1, j=1, 2, 3, 4

16. Obtain a Bound for Each Subset
Find a bound by relaxing one or more constraints
Let 0 < xj < 1, j=1, 2, 3, 4

17. Obtain a Bound for Each Subset
Find a bound by relaxing one or more constraints
Let 0 < xj < 1, j=1, 2, 3, 4
All xj must be integer, yields Z must be integer, yields Z=16 is a bound for the whole problem.

18. Example Branching Tree
Z < 16
(.83,1,0,1)
X1=0
All
X1=1
Now let us obtain a bound for each subproblem.

19. Example Branch (x1=0) Let x1 = 0 , then Subset 1:
Max Z = 5x2 + 6x3 + 4x4
s.t.
3x2+5x3+5x4 < 10
x3+ x4 < 1
x < 1
-1x x4 < 0
0 < xj < 1, j=2, 3, 4,

20. Example Bound (x1=0)

21. Example Branching Tree
Z = 9
(0,1,0,1)
Z < 16
(.83,1,0,1)
X1=0
All
X1=1

22. Example Branch (x1=1) Let x1 = 1 , then Subset 1:
Max Z = 9 +5x2 + 6x3 + 4x4
s.t.
3x2+5x3+5x4 < 4
x3+ x4 < 1
x < 1
-1x x4 < 0
0 < xj < 1, j=2, 3, 4,

23. Example Branch (x1=1)

24. Example Branching Tree
Z = 9
(0,1,0,1)
Z < 16
(.83,1,0,1)
X1=0
All
X1=1
Z = 16
(1,.8,0,.8)

25. Fathoming
Z* = 9
(0,1,0,1)
For x1=0, Z has a bound less than 16. It also has integer values so (0,1,0,1) is the optimal solution for subset 1.
(0,1,0,1) = incumbent
Z < 16
(.83,1,0,1)
X1=0
All
X1=1
For x1=1, Z has a bound equal to 16, but solution is not feasible. We need to do another iteration.
Z = 16
(1,.8,0,.8)

26. Iteration 2, x1=1 Max Z = 9 +5x2 + 6x3 + 4x4 s.t. 3x2+5x3+5x4 < 4
xj = 0 or 1, j=2, 3, 4,

27. Iteration 2, x1=1, x2=0 Max Z = 9 + 6x3 + 4x4 s.t. 5x3+2x4 < 4
xj = 0 or 1, j= 3, 4

28. Iteration 2, x1=1, x2=1 Max Z = 14 + 6x3 + 4x4 s.t. 5x3+2x4 < 1
xj = 0 or 1, j= 3, 4

29. Iteration 2, x1=1, x2=0, Relax Max Z = 9 + 6x3 + 4x4 s.t.
0 < xj < 1, j= 3, 4

30. Iteration 2, x1=1, x2=0, Relax

31. Fathoming Z* = 9 (0,1,0,1) Z < 16 (.83,1,0,1) Z = 13.8 (1,0,.8,0)
X1=0
All
X2=0
X1=1
Z = 16
(1,.8,0,.8)
X1=1

32. Iteration 2, x1=1, x2=1, Relax Max Z = 14 + 6x3 + 4x4 s.t.
0 < xj < 1, j= 3, 4

33. Iteration 2, x1=1, x2=1, Relax

34. Fathoming Z* = 9 (0,1,0,1) Z < 16 (.83,1,0,1) Z = 13 (1,0,.8,0)
X1=0
All
X2=0
X1=1
Both Bounds > 9, and have feasible solutions so fathoming test fails.
Z = 16
(1,.8,0,.8)
X2=1
Z = 16
(1,1,0,.5)

35. Iteration 3, x1=1, x2=1, x3=0 Max Z = 14 + 4x4 s.t. 2x4 < 1
xj = 0 or 1, j= 4

36. Iteration 3, x1=1, x2=1, x3=0, Relax
Max Z = x4
s.t.
2x4 < 1
x4 < 1
x4 < 1
0 < xj < 1, j= 4

37. Iteration 3, x1=1, x2=1, x3=0, Relax

38. Fathoming Z* = 9 (0,1,0,1) Z < 16 (.83,1,0,1) Z = 13 (1,0,.8,0)
X1=0
All
X2=0
Z = 16
(1,1,0,.5)
X1=1
X3=0
Z = 16
(1,.8,0,.8)
X2=1
Z = 16
(1,1,0,.5)
X3=1

39. Iteration 3, x1=1, x2=1, x3=1 Max Z = 20 + 4x4 s.t. 2x4 < -4
xj = 0 or 1, j= 4

40. Iteration 3, x1=1, x2=1, x3=1, Relax
Max Z = x4
s.t.
2x4 < -4
x4 < 0
x4 < 1
0 < xj < 1, j= 4
Infeasible

41. Infeasible

42. Fathoming Z* = 9 (0,1,0,1) Z < 16 (.83,1,0,1) Z = 13 (1,0,.8,0)
X1=0
All
X2=0
Z = 16
(1,1,0,.5)
X1=1
X3=0
Z = 16
(1,.8,0,.8)
X2=1
Z = 16
(1,1,0,.5)
X3=1
Infeasible

43. Fathoming Z* = 9 (0,1,0,1) Z < 16 (.83,1,0,1) Z = 13 (1,0,.8,0)
X1=0
All
X2=0
Z = 16
(1,1,0,.5)
X4=0
X1=1
X3=0
Z = 16
(1,.8,0,.8)
X4=1
X2=1
Z = 16
(1,1,0,.5)
X3=1
Infeasible

44. Iteration 4, x1=1, x2=1, x3=0, x4=1 Max Z = 14 + 4x4 s.t. 2x4 < 1
xj = 0 or 1, j= 4

45. Iteration 4, x1=1, x2=1, x3=0, x4=0

46. Fathoming Z* = 9 (0,1,0,1) Z < 16 (.83,1,0,1) Z = 13 (1,0,.8,0)
X1=0
Z = 14
(1,1,0,0)
All
X2=0
Z = 16
(1,1,0,.5)
X4=0
X1=1
X3=0
Z = 16
(1,.8,0,.8)
X4=1
X2=1
Z = 16
(1,1,0,.5)
X3=1
Infeasible

47. Iteration 3, x1=1, x2=1, x3=0, x4=1 Max Z = 14 + 4x4 s.t. 2x4 < 1
0 < xj < 1, j= 4

48. Iteration 3, x1=1, x2=1, x3=0, x4=1
Infeasible

49. Fathoming Z = 9 (0,1,0,1) Z < 16 (.83,1,0,1) Z = 13 (1,0,.8,0) X1=0
(1,1,0,0)
All
X2=0
Z = 16
(1,1,0,.5)
X4=0
X1=1
X3=0
Z = 16
(1,.8,0,.8)
X4=1
X2=1
Infeasible
Z = 16
(1,1,0,.5)
X3=1
Infeasible