1. BINARY/MIXED-INTEGER PROGRAMMING ( A SPECIAL TYPE OF INTEGER PROGRAMMING)

2. Defining Variables as Integers limits the values given to such variables to Integer values. The formulation of constraints given this condition do not require any other formulation technique except the definition of such variables to be integers.

3. Binary Programming is usually used when we want to define variables to have only either the value of ‘0’ or ‘1’. We assign a value of ‘0’ if a certain condition is not met and ‘1’ if the condition is met.

4. By defining variables to be binaries, we can increase the versatility of our spreadsheet modeling techniques. The following formulation models show how we can use this 0-1 variable to define certain conditions.

5. Limiting the Number of Alternatives Selected: 1 Suppose you want to select no more than two of three projects: X1 + X2 + X3 <= 2 2. Suppose you want to select exactly two of three projects: X1 + X2 + X3 = 2

6. Dependent Selections: 1.The selection of one project (X1) depends in the selection of another project (X2) : X1 <= X2 or X1 – X2 <= 0 2.If we wish to either both be selected or both not be selected: X1 = X2or X1 – X2 = 0

7. MiXED – INTEGER PROGRAMMING We use different types of variables to be able to formulate certain conditions. For example, suppose a company is planning to build at least one more plant to have a capacity of at least 38,000 units and three cities are being considered. The cost associated with the possible locations are given below: SITE ANNUAL FIXED COST VARIABLE COST PER UNIT ANNUAL PLANT CAPACITY CITY 1Php 340,000Php 3221,000 units CITY 2Php 270,000Php 3320,000 units CITY 3Php 290,000Php 3019,000 units

8. We can define the Decision Variables as: X1 = Binary where X1 = 1 if factory is built in City 1; X1 = 0 if otherwise X2 = Binary where X2 = 1 if factory is built in City 2; X2 = 0 if otherwise X3 = Binary where X3 = 1 if factory is built in City 3; X3 = 0 if otherwise X4 = number of units produced at City 1 plant X5 = number of units produced at City 2 plant X6 = number of units produced at City 3 plant

9. Minimize Cost = 340000X1 + 270000X2 + 29000X3 + 32X4 + 33X5 + 30X6 Subject to: Total capacity requirement: X4 + X5 + X6 <= 38,000 Capacity at City 1 plant: X4 <= 21,000 X1 Capacity at City 2 plant: X5 <= 20,000 X2 Capacity at City 3 plant: X6 <= 19,000 X3 Where: X1, X2, X3 = 0 or 1; X4, X5, X6 >= 0 and integer

10. Batch Size Problem Another use of binary variables in situations where a company can only purchase a minimum quantity of products or in batches. For example, a firm may have to buy a minimum quantity of at least 50 units of a certain product or none at all. Let X be the number of units bought. Then either X = 0 or X >= 50. The situation can be formulated by introducing a binary variable Y and adding the following constraints: X <= MY (M is a very large number) X >= 50Y Note that if Y = 0, then, by the first constraint, X must be 0. If Y = 1, then by the second constraint, X must be at least 50.

11. Either – Or Constraints “Sometimes, in a decision situation, either one or another constraint must hold, but not both. For example: Either:5X1 + 2X2 <= 10 Or :3X1 + 4X2 <= 24 But not both We can handle this … by using both of the following… 5X1 + 2X2 <= 10 + MY 3X1 + 4X2 <= 24 + M(1-Y) Note that when Y=0, the first constraint is binding, the right hand side of the second constraint becomes very large and hence not binding. And conversely when Y =1, the second constant binds but the first does not because MY is very large.”

12. END INTEGER / BINARY PROGRAMMING