Integer Programming II ENGM 435/535 Integer Programming II

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.

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.

Prototype Formulation Let Max Z=2yA1+4yB1+6yC1+5yA2+7yB2+11yC2+ 5yA3+--+12yC3+ 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.

Prototype Formulation Let Max Z=2yA1+4yB1+6yC1+5yA2+7yB2+11yC2+ 5yA3+--+12yC3+ 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 yA3+ yC3 < 1 yA4+yB4+yC4 < 1

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+ 5yA3+--+12yC3+ 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 yA3+ yC3 < 1 yA4+yB4+yC4 = 1 yA1+yA2+yA3 < 3yA4 yA1+yA2+yA3 - 3yA4 < 0

Prototype Formulation B1, B2, and C4 cannot all be built because of competitive bids Max Z=2yA1+4yB1+6yC1+5yA2+7yB2+11yC2+ 5yA3+--+12yC3+ 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 yA3+ yC3 < 1 yA4+yB4+yC4 = 1 yA1+yA2+yA3 - 3yA4 < 0 yB1+yB2+yC4 < 2

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

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

Relaxed Feasible Region 7X1+8X2 = 56 9X1+6X2 = 54 X2 = 5 b c X1 = 5 d e f a 0 1 2 3 4 5 6

Real Feasible Region 7X1+8X2 = 56 9X1+6X2 = 54 X2 = 5 X1 = 5 0 1 2 3 4 5 6

Solution Techniques Optimization Complete Enumeration

Real Feasible Region 7X1+8X2 = 56 9X1+6X2 = 54 X2 = 5 15 12 9 6 3 22 22 16 13 10 7 4 26 20 17 14 11 8 24 21 18 15 12 25 22 19 16 23 20 X1 = 5 0 1 2 3 4 5 6

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

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) 3.000000014 Total Profit $25.00 X1=4 Integer Solution

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

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

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

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

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

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

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

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

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

Traveling Salesman Problem Let

Traveling Salesman Problem

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

Traveling Example

Traveling Example

Traveling Example

Traveling Example

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

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

Plant Location

Capital Budgeting

Traveling Salesman Problem

Line Balancing