EML4552 - Engineering Design Systems II (Senior Design Project)

1. EML4552 - Engineering Design Systems II(Senior Design Project) Optimization Theory and Optimum Design Linear Programming Hyman: Chapter 10

2. Optimization • Minimize (Maximize) an Objective Function of certain Variables subject to Constraints

3. Linear Programming • Maximize a LinearObjective Function of ipositivevariables subject to j Linear Constraints

4. Linear Programming • Almost all linear optimization problems can be reduced to the canonical (standard) form • Minimization, change sign on objective function • Variables that can be negative, define: • Constraint reversal, ‘slack’ and ‘surplus’ variables

5. L.P.: Geometric Interpretation

6. (1) x2 (4) (2) (3) x1 L.P.: Geometric Interpretation 0 6.5 U= 2 4

7. L.P.: The “Simplex” Algorithm • Standard method to solve linear programming optimization problems • Geometric interpretation of linear programming problems (simple example) • Based on principle that optimum will be ON a constraint (typically a “corner”)

8. Simplex Algorithm: Example

9. Simplex Algorithm: Example Add x4 to make the Inequality an equality

10. Simplex Algorithm: Example • The Simplex “Algorithm” consists of a number of steps carried out in ‘cycles’ until an optimum is reached • “Geometrically”, the Simplex method explores the objective function by ‘walking’ along the constraints in the ‘steepest’ direction until it reaches the optimum • Based on: • Unique solution • Optimum on a constraint

11. Simplex Algorithm: Example • Cycle 1 • Select a starting point (a ‘corner’) • In this example a corner is set by 3 variables (the other 3 can be solved from the constraints) • Step 1: Select the 3 ‘basis’ variables such that they appear in only one constraint and have coefficient of unity (x4, x5, x6 in this case). Assign a value of 0 to the ‘non-basis’ variables. • Step 2:Determine coordinates of initial ‘corner’ (x1=0, x2=0, x3=0, x4=430, x5=460, x6=420)

12. Simplex Algorithm: Example • Step 3: Calculate objective function at the corner point (x1=0, x2=0, x3=0, x4=430, x5=460, x6=420), U=0 • Step 4: Move from corner to an ‘adjacent’ corner along the ‘edges’ represented by the constraints. Vary the non-basis variables to move along the edges. Move along the edge which will increase “U” the most. To do this, express the objective function in terms of the non-basis (U=3x1+2x2+5x3, already, if not, solve the constraints to do so)

13. Simplex Algorithm: Example • Step 5: From the present ‘corner’ (x1=0, x2=0, x3=0) move along the variable that will produce the largest increase in U (highest coefficient), in this case x3. We re-formulate the basis by introducing x3 and deleting from the basis one of the variables. • Step 6: We want to increase x3 by as much as possible without violating any constraint (i.e., move to the next ‘corner’). To do this, reformulate the constraints for the ‘incoming’ variable to the basis

14. Simplex Algorithm: Example • Step 6 (cont.): In this case for x1=x2=0 and the incoming x3, the constraints are: • Step 7: Determine how much x3 can increase before driving any variable negative. Limits are: (a) 430, (b) 230, (c)  therefore, x3 can be increased up to 230 and x5 leaves the basis

15. Simplex Algorithm: Example • Step 8: x3, x4, and x6 are the new basis, set the non-basis variables to zero, x1=0, x2=0, x5=0 • Prepare to start the next cycle (repeat procedure)

16. Simplex Algorithm: Example • Cycle 2 • Step 1: The new basis was chosen as the last step of the previous cycle (x3, x4, x6). • Step 2:Determine coordinates of new ‘corner’ by re-writing constraints: (x1=0, x2=0, x3=230, x4=200, x5=0, x6=420)

17. Simplex Algorithm: Example • Step 3: Calculate objective function at the corner point (x1=0, x2=0, x3=230, x4=200, x5=0, x6=420), U=3(0)+2(0)+5(230)=1150 • Step 4: Prepare to choose new basis. Re-write U in terms of non-basis variables (using the constraints to eliminate the basis variables):

18. Simplex Algorithm: Example • Step 5: Move away from present corner along the edge that will increase U the most. From the objective function, only an increase in x2 will yield improvement (all other coefficients are negative). Introduce x2 to the basis. • Step 6: Re-write constraints in terms of new basis variable:

19. Simplex Algorithm: Example • Step 7: Determine how much x2 can increase before driving any variable negative. Limits are: (a) 100 (for x3=230), (b) , (c) 105 therefore, x2 can be increased up to 100 and x4 leaves the basis • Step 8: x2, x3, and x6 are the new basis, set the non-basis variables to zero, x1=0, x4=0, x5=0 • Prepare to start the next cycle (repeat procedure)

20. Simplex Algorithm: Example • Cycle 3 • Repeat the same steps now with x2, x3, x6 as the basis. After steps 2 and 3 the solution is x1=0, x2=100, x3=230, x4=0, x5=0, x6=20, U=1350 (further improvement). At step 4, however, all coefficients on the objective function, U, are negative, so no further improvements are possible, we are done! We have found the optimum (x1=0, x2=100, x3=230, U=1350).

22. Simplex Algorithm • Found the optimum with exploration of just 3 ‘corners’ • Efficiency increases tremendously with the complexity of the problem (number of variables and constraints) • “New” L.P. algorithms may be even more efficient in extremely large problems