1. (Not in text)

2.  An LP with additional constraints requiring that all the variables be integers is called an all-integer linear program (IP).  The LP that results from dropping the integer requirements is called the LP Relaxation of the IP.  If only a subset of the variables are restricted to be integers, the problem is called a mixed-integer linear program (MIP).  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 IP.

3.  Consider the following IP Max 3 x 1 + 2 x 2 s.t. 3 x 1 + x 2 < 9 x 1 + 3 x 2 < 7 - x 1 + x 2 < 1 x 1, x 2 > 0 and integer

4.  LP Relaxation Solving the LP relaxation (i.e. ignoring the integer constraints), gives fractional values for both x 1 and x 2. From the graph on the next slide, we see that the optimal solution to the LP relaxation is: x 1 = 2.5, x 2 = 1.5, z = 10.5 But we need the values for x 1 & x 2 to be integer…

5.  LP Relaxation LP Optimal (2.5, 1.5) Max 3 x 1 + 2 x 2 Max 3 x 1 + 2 x 2 - x 1 + x 2 < 1 x2x2x2x2 x1x1x1x1 3 x 1 + x 2 < 9 1 3 2 5 4 1 2 3 4 5 6 7 x 1 + 3 x 2 < 7 x 1 + 3 x 2 < 7

6.  Rounding Up If we round up the fractional solution ( x 1 = 2.5, x 2 = 1.5) to the LP relaxation problem, we get x 1 = 3 and x 2 = 2. From the graph on the next slide, we see that this point lies outside the feasible region, making this solution infeasible.

7.  Rounded Up Solution LP Optimal (2.5, 1.5) Max 3 x 1 + 2 x 2 Max 3 x 1 + 2 x 2 x2x2x2x2 x1x1x1x1 IP Infeasible (3, 2) IP Infeasible (3, 2) 1 2 3 4 5 6 7 1 3 2 5 4 - x 1 + x 2 < 1 3 x 1 + x 2 < 9 x 1 + 3 x 2 < 7 x 1 + 3 x 2 < 7

8.  Rounding Down By rounding the optimal solution down to x 1 = 2, x 2 = 1, we see that this solution indeed is an integer solution within the feasible region, and substituting in the objective function, it gives z = 8. We have found a feasible all-integer solution, but have we found the OPTIMAL all-integer solution? --------------------- The answer is NO! The optimal solution is x 1 = 3 and x 2 = 0 giving z = 9, as evidenced in the next two slides.

9.  Complete Enumeration of Feasible IP Solutions There are eight feasible integer solutions to this problem, where z = 3 x 1 + 2 x 2 : x 1 x 2 z 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

10. IP Optimal (3, 0) Max 3 x 1 + 2 x 2 Max 3 x 1 + 2 x 2 x2x2x2x2 x1x1x1x1 1 3 2 5 4 1 2 3 4 5 6 7 - x 1 + x 2 < 1 3 x 1 + x 2 < 9 x 1 + 3 x 2 < 7 x 1 + 3 x 2 < 7

11. IP Optimal (3, 0) Max 3 x 1 + 2 x 2 Max 3 x 1 + 2 x 2 x2x2x2x2 x1x1x1x1 1 3 2 5 4 1 2 3 4 5 6 7 - x 1 + x 2 < 1 3 x 1 + x 2 < 10 x 1 + 3 x 2 < 7 x 1 + 3 x 2 < 7 (This was 9) LP extreme point

12.  While complete enumeration is one possible solution method, it is not efficient on larger problems  Specialized solution techniques have been developed to solve IPs that have been incorporated into commercially-available software packages.  Premium Solver, included in your textbook Student CD, can solve IPs. It is an enhanced version of Excel’s Solver package.

14.  Suppose selecting some of the projects is conditional on whether others are selected. 1. Project1 can only be selected if Project 5 is selected 2. At most 2 of these projects can be selected:  Project6, Project7, Project8, Project9, Project10 3. Projects 15 and 16 must both be selected or not. (i.e. we can’t do one without the other) 4. Projects 18 and 19 are mutually exclusive

15. 1. Project1 can only be selected if Project 5 is selected X 1 - X 5 ≤ 0 2. At most 2 of these projects can be selected:  Project6, Project7, Project8, Project9, Project10 X 6 + X 7 + X 8 + X 9 + X 10 ≤ 2 3. Projects 15 and 16 must both be selected or not. (i.e. we can’t do one without the other) X 15 -X 16 = 0 4. Projects 18 and 19 are mutually exclusive X 18 +X 19 ≤ 1