1. Linear Programming
Integer Linear Models

2. When Variables Have To Be Integers
Example – one time production decisions
Fractional values make no sense
But if ongoing process, fractional values could represent work in progress
Example -- building houses or planes, or scheduling crews
Binary variables
Restricted to be 0 or 1
Example – Is a plant built?

3. Types of Integer Programs (ILP)
All Integer Linear Programs (AILP)
All the decision variables are required to be integers
Mixed Integer Linear Programs (MILP)
Only some of the variables are required to be integers
Binary Integer Linear Programs (BILP)
Variables are restricted to be 0 or 1

4. Example
Boxcar Burger will build restaurants in the suburbs and downtown
Suburbs
Profit $12000/day
$2,000,000 investment
Requires 3 managers
Downtown
Profit $20000/day
$6,000,000 investment
Requires 1 manager
Constraints
$27,000,000 budget
At least 2 downtown restaurants
19 managers available

5. Decision Variables/Objective
X1 = Number of restaurants built in suburbs
X2 = Number of restaurants built downtown
MAX Expected Daily Profit
MAX 12X1 + 20X2 (in $1000’s)
MAX Expected Daily Profit

6. # downtown restaurants
Constraints
Cannot invest more than $27,000,000
At least 2 downtown restaurants
Number of managers used cannot exceed 19
Total Amount Invested
Cannot
Exceed 27
In $1,000,000’s
2X1 + 6X2 27 ≤
# downtown restaurants
Must be
At least 2 X2 2 ≥
# Managers used
Cannot
Exceed 19 19
3X1 + 1X2 ≤

7. The Complete Model MAX 12000X1 + 20000X2 s.t. 2X1 + 6X2  27 (Budget)
X2  2 (Downtown)
3X X2  19 (Managers)
Both X’s  0
Both X’s INTEGER!

8. The Linear Programming Feasible RegionX2 6 5 4 3 2 1
Max 12X1 + 20X2
12X1 + 20X2
LP Feasible Region
3X1 + 1X2 ≤ 19
2X1 + 6X2 ≤ 27
X2 ≥ 2
Rounded off
(5,3)
Rounded up
(6,3)
X1, X2 ≥ 0
LP Optimum (5 7/16, 2 11/16) Obj. Value = 119
Rounded down
FEASIBLE Objective Value = 100
(5,2) 1 2 3 4 5 6 X1

9. The Integer Programming Feasible RegionX2 6 5 4 3 2 1
Max 12X1 + 20X2
12X1 + 20X2
ILP Optimum (4,3) OBJ. VALUE = 108
2X1 + 6X2 ≤ 27
3X1 + 1X2 ≤ 19
X2 ≥ 2
3X1 + 1X2 ≤ 19
X1, X2 integer
X1, X2 ≥ 0 1 2 3 4 5 6 X1

10. Why Not Round To Get the Optimal Integer Solution?
Rounding may yield the optimal integer solution
None did in this example
But it may yield an infeasible solution
Both (5,3) and (6,3) are infeasible solutions
Or a feasible solution that is not optimal
(5,2) is feasible but not optimal
Many times a feasible rounded point gives a “good” solution (giving close to the optimal value of the objective function) -- BUT NOT ALWAYS

11. General Facts About Integer Models
The solution time to solve integer models is longer than that of linear programs
Because many linear programs are solved en route to obtaining an optimal integer solution
For maximization models, the optimal value of the objective function will be less (or at least not greater than) the value for the equivalent linear model
Because constraints have been added – the integer constraints
There is no sensitivity analysis
Because the feasible region is not continuous

12. Solving ILP’s Using SOLVER
The only change in SOLVER is to add the integer constraints
In the Add Constraints dialogue box, highlight the cells required to be integer and choose “int” from the pull down menu for the sign

14. Build 4 Suburban Restaurants Build 3 Downtown Restaurants
Optimal
Build 4 Suburban Restaurants
Build 3 Downtown Restaurants
Average Daily Profit $108,000

15. Review When to use integer models Why rounding will not always work
Solution time
No sensitivity analysis
Objective function value cannot improve
SOLVER solution approach