(not in text). an lp with additional constraints requiring that all the variables be integers is...
TRANSCRIPT
- Slide 1
- (Not in text)
- Slide 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.
- Slide 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
- Slide 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
- Slide 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
- Slide 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.
- Slide 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
- Slide 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.
- Slide 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
- Slide 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
- Slide 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
- Slide 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 Excels Solver package.
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- 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 cant do one without the other) 4. Projects 18 and 19 are mutually exclusive
- Slide 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 cant do one without the other) X 15 -X 16 = 0 4. Projects 18 and 19 are mutually exclusive X 18 +X 19 1