ba 452 lesson b.3 integer programming 11readingsreadings chapter 7 integer linear programming

BA 452 Lesson B.3 Integer Programming 11

Readings

Readings

Chapter 7Integer Linear Programming


Overview

Overview


Overview

Rounding Off solutions in continuous variables to the nearest integer (like 2.67 rounded off to 3) is an unreliable way to solve a linear programming problem when decision variables should be integers.

Sensitivity Analysis with Integer Variables is more important than with continuous variables because a small change in a constraint coefficient can cause a relatively large change in the optimal solution.

Assignment Problems with Valuable Time minimize the total time of assigning workers to jobs. Minimizing total time is appropriate when each worker has the same value of time.

Assignment Problems with Supply and Demand are Transportation Problems of suppliers to demanders except that each demand is assigned to exactly one supplier.


Tool Summary Do not make integer restrictions, and maybe the variables at an

optimum will be integers.• First Example: Pi = (integer) number of producers in month i.

Use compound variables:• First Example: Pi = number of producers in month i

Use dynamic or recursive constraints:• First Example: Define the constraint that the number of

apprentices in a month must not exceed the number of recruits in the previous month: A2 - R1 < 0; A3 - R2 < 0

Constrain one variable to be a proportional to another variable:• First Example: Define the constraint that each trainer can train

two recruits: 2T1 - R1 > 0; 2T2 - R2 > 0 Use inventory variables:

• Second Example: P2 + I1–I2 = 150 (production-net inventory = demand)

Overview


Tool Summary Use binary variables to model fixed cost constraints.

• For example, consider a linear programming problem with a constraint• M1 < 15

• Introduce binary variable U1 and constraints:• M1 > 5 U1 • 15U1 > M1

• Then, U1 = 1 if M1 > 0, and either M1 = 0 or M1 > 5. Do not try to solve an integer linear program by rounding off (up or

down) a solution to the problem without integer constraints. Rather, graph all feasible integer solutions and use iso-value linear, or use the Management Scientist module.

Tool Summary


Rounding Off

Rounding Off


Overview

Rounding Off solutions in continuous variables to the nearest integer (like 2.67 rounded off to 3) is an unreliable way to solve a linear programming problem when decision variables should be integers.

Rounding Off


A LP in which all the variables are restricted to be integers is called an all-integer linear program (ILP).

The LP that results from dropping the integer requirements is called the LP Relaxation of the ILP.

If only a subset of the variables are restricted to be integers, the problem is called a mixed-integer linear program (MILP).

Binary variables are variables whose values are restricted to be 0 or 1. If all variables are restricted to be 0 or 1, the problem is called a 0-1 or binary integer linear program.

Rounding Off


Solve the following all-integer linear program, and compare that solution to the problem where the decision variables do not have to be integers:

Max 3x1 + 2x2

s.t. 3x1 + x2 < 9

x1 + 3x2 < 7

-x1 + x2 < 1

x1, x2 > 0 and integer

Rounding Off


LP Relaxation. If we drop the integer constraints, we can graph the optimal solution to the linear program. And the optimal solution has fractional values: x1 = 2.5, x2 = 1.5, Max 3x1 + 2x2 = 10.5

Max 3x1 + 2x2

s.t. 3x1 + x2 < 9

x1 + 3x2 < 7

-x1 + x2 < 1


LP Optimal (2.5, 1.5)

Max 3x1 + 2x2

- x1 + x2 < 1

x2

x1

3x1 + x2 < 9

x1 + 3x2 < 7

1 2 3 4 5 6 7

1

3

2

5

4

Rounding Off


Rounding Up. If we round up the fractional solution (x1 = 2.5, x2 = 1.5) to the previous relaxed LP problem, we get x1 = 3 and x2 = 2. But the graph shows those values are infeasible.

Max 3x1 + 2x2

s.t. 3x1 + x2 < 9

x1 + 3x2 < 7

-x1 + x2 < 1


ILP Infeasible (3, 2)

LP Optimal (2.5, 1.5)

Max 3x1 + 2x2

- x1 + x2 < 1

x2

x1

3x1 + x2 < 9

x1 + 3x2 < 7

1 2 3 4 5 6 7

1

3

2

5

4

Rounding Off


Rounding Down. By rounding the non-integer solution down to x1 = 2, x2 = 1, we have a feasible solution, with objective function 3x1 + 2x2 = 8. But that solution is not optimal for the integer program.

Max 3x1 + 2x2

s.t. 3x1 + x2 < 9

x1 + 3x2 < 7

-x1 + x2 < 1


ILP Optimal (3, 0)

Max 3x1 + 2x2

- x1 + x2 < 1

x2

x1

3x1 + x2 < 9

x1 + 3x2 < 7

1 2 3 4 5 6 7

1

3

2

5

4

Optimal ILP solution. The optimal ILP solution (3,0) is not the closest feasible point to the non-integer solution (2.5,1.5).

Rounding Off


x1 x2 3x1 + 2x2

1. 0 0 0 2. 1 0 3 3. 2 0 6 4. 3 0 9 optimal

solution 5. 0 1 2 6. 1 1 5 7. 2 1 8

8. 1 2 7

Max 3x1 + 2x2

s.t. 3x1 + x2 < 9

x1 + 3x2 < 7

-x1 + x2 < 1


Exhaustive Search. One way to solve the integer linear program is to evaluate the objective function at each feasible solution. There are 8 alternative feasible solutions in Example 1.

Rounding Off


Max 3x1 + 2x2

s.t. 3x1 + x2 < 9

x1 + 3x2 < 7

-x1 + x2 < 1


Rounding Off


Max 3x1 + 2x2

s.t. 3x1 + x2 < 9

x1 + 3x2 < 7

-x1 + x2 < 1


ILP Optimal (3, 0)

Rounding Off


Sensitivity Analysis with Integer Variables



Overview

Sensitivity Analysis with Integer Variables is more important than Sensitivity Analysis with Continuous Variables because a small change in a constraint coefficient can cause a relatively large change in the optimal solution, and in the objective-function value of the optimal solution.



Sensitivity analysis often is more crucial for ILP problems than for LP problems.

A small change in a constraint coefficient can cause a relatively large change in the optimal solution, and the objective-function value of the optimal solution.

Recommendation: Resolve the ILP problem several times with slight variations in the coefficients before choosing the “best” solution for implementation.



Sensitivity Analysis

Max 3x1 + 2x2 = 9 at old optimum (3,0)

x2

x11 2 3 4 5 6 7

1

3

2

5

4

Max 3x1 + 2x2 = 10 at new optimum (2,2)

A small change in the constraints (adding the red feasible area) can cause a jump in the objective function at the optimum, from 9 to 10.


Assignment with Valuable Time



Overview

Assignment Problems with Valuable Time minimize the total time of assigning workers to jobs. Minimizing total time is appropriate when each worker has the same value of time.



Question: The NPD Group is a market research firm with three clients that each request the firm conduct a sample survey. Four statisticians can be assigned to these three projects; however, all four are busy, and therefore can handle only one client. The following are the number of hours required for each statistician to complete each job. The differences in time are based on experience and ability of the statisticians.


Client A Client B Client C

Statistician 1 150 210 270





What is the most natural objective function for the firm to optimize?

Formulate the firm’s optimization problem.

Are there any implicit assumptions in your formulation?



Answer: Linear programming formulation (supply inequality, demand equality). Variables: Xij = 1 if Statistician i is assigned to Client j, else 0 Objective (minimize time, assuming all time values equal):

Min 150X11 + 210X12 + 270X13 + 170X21 + 230X22 + 220X23

+ 180X31 + 230X32 + 225X33 + 160X41 + 240X42 + 230X43 Supply Constraints (Statisticians to at most 1 Client):

X11 + X12 + X13 < 1, X21 + X22 + X23 < 1,

X31 + X32 + X33 < 1, X41 + X42 + X43 < 1 Demand Constraints (each Client served):

X11 + X21 + X31 + X41 = 1

X12 + X22 + X32 + X42 = 1

X13 + X23 + X33 + X43 = 1



Assignment with Supply and Demand



Overview

Assignment Problems with Supply and Demand are Transportation Problems of suppliers to demanders except that each demand is assigned to exactly one supplier.



Question: Dow Chemical uses the chemical Rbase in production operations at five divisions. Only six suppliers of Rbase meet Dow’s quality standards. The quantity of Rbase needed by each Dow division and the price per gallon charged by each supplier are as follows:


Price per Gallon ($)

Sup 1 12.60

Sup 2 14.00

Sup 3 10.20

Sup 4 14.20

Sup 5 12.00

Sup 6 13.00

Demand (1000s of gallons)

Div 1 40

Div 2 45

Div 3 50

Div 4 35

Div 5 45


The cost per gallon ($) for shipping from each supplier to each division are as follows:


Cij Div 1 Div 2 Div 3 Div 4 Div 5

Sup 1 2.75 0.80 4.70 2.60 3.40

Sup 2 2.50 0.20 2.60 1.80 0.40

Sup 3 3.15 5.40 5.30 4.40 5.00

Sup 4 2.80 1.20 2.80 2.40 1.20

Sup 5 2.75 3.40 6.00 5.00 2.60

Sup 6 2.75 1.00 5.60 2.80 3.60


Dow wants to diversify by spreading its business so that each division’s demand is assigned to exactly one supplier.

Formulate the optimal assignment of suppliers to divisions as a linear-programming problem.



Answer: Linear programming formulation (supply inequality, demand equality). Variables: Xij = 1 if Supplier i is assigned to Division j,

else 0 Assignment Costs

The total cost is the sum of the purchase cost and the transportation cost.

Supplier 1 assigned to Division 1 (cost in $1000s): Purchase cost: (40 x $12.60) = $504 Transportation Cost: (40 x $2.75) = $110 Total Cost: $614



Assignment Costs: Cij = Cost of assigning Supplier i to Division j

Cij Div 1 Div 2 Div 3 Div 4 Div 5

Sup 1 614 603 865 532 720

Sup 2 660 639 830 553 648

Sup 3 534 702 775 511 684

Sup 4 680 693 850 581 693

Sup 5 590 693 900 595 657

Sup 6 630 630 930 553 747



Linear programming formulation (supply inequality, demand equality). Objective (minimize cost): Min S Cij Xij Demand Constraints (since each division’s demand is

assigned to exactly one supplier):X11 + X21 + X31 + X41 + X51 + X61 = 1

X12 + X22 + X32 + X42 + X52 + X62 = 1

X13 + X23 + X33 + X43 + X53 + X63 = 1

X14 + X24 + X34 + X44 + X54 + X64 = 1

X15 + X25 + X35 + X45 + X55 + X65 = 1



Optional: There is no mention of supply constraints, which are common in assignment problems. Here is what those common constraints would be in this problem. Supply Constraints (Each supplier can supply at most 1

Division): X11 + X12 + X13 + X14 + X15 < 1

X21 + X22 + X23 + X24 + X25 < 1

X31 + X32 + X33 + X34 + X35 < 1

X41 + X42 + X43 + X44 + X45 < 1

X51 + X52 + X53 + X54 + X55 < 1

X61 + X62 + X63 + X64 + X65 < 1



BA 452 Quantitative Analysis

End of Lesson B.3

ba 452 lesson b.3 integer programming 11readingsreadings chapter 7 integer linear programming

Documents

integer programming

integer variables

integer constraints

integer number of producers

nearest integer

integer restrictions

linear programming problem

feasible integer solutions