L32. Constrained Optimization

1. L32. Constrained Optimization Optimizing Gear Ratio Distribution

2. An Optimal Design Problem

3. Gear Ratio Distribution Assume 7 wheel sprockets Assume 3 pedal sprockets 21 = 7 x 3 possible gear ratios

4. It’s a Matter of Teeth E.g., 13 teeth E.g., 48 teeth E.g., Gear ratio = 48/13 = 3.692

5. Goal Choose 3 pedal sprockets and 7 wheel sprockets so that the 21 gear ratios are as evenly distributed across the interval [1,4].

6. Notation p(i) = #teeth on the i-th pedal sprocket, for i=1:3. w(i) = #teeth on the i-th wheel sprocket, for i=1:7. This is a 10—parameter design problem.

7. Things to Do • Define an Objective Function We need to measure the quality of a particular gear ratio distribution 2. Identify constraints. Sprockets are only available in certain sizes etc. Typical activity in Engineering Design

8. The Quality of a Gear RatioDistribution Ideal: 4 1 Good: Poor:

9. Average Discrepancy Sort the gear ratios: g(1) < g(2) <… < g(21) Compare g(i) with x(i) where x = linspace(1,4,21).

10. function tau = ObjF(p,w); g = []; for i=1:3 for j=1:7 g = [g p(i)/w(j)]; end end g = sort(g); dif = abs(g – linspace(1,4,21)); tau = sum(dif)/21;

11. There Are Other ReasonableObjective Functions g = sort(g); dif = abs(g –linspace(1,4,21)); tau = sum(dif)/21; Replace “sum” with “max”

12. Goal Choose p(1:3) and w(1:7) so that objF(p,w) is minimized. This defines the “best bike.” Our plan is to check all possible bikes. A 10-fold nested loop problem…

13. A Simplification We may assume that p(3) < p(2) < p(1) and w(7)<w(6)<w(5)<w(4)<w(3<w(2)<w(1) Relabeling the sprockets doesn’t change the 21 gear ratios.

14. How Constraints Arise Purchasing says that pedal sprockets only come in six sizes: C1: p(i) is one of 52 48 42 39 32 28.

15. How Constraints Arise Marketing says the best bike must have a maximum gear ratio exactly equal to 4: C2: p(1)/w(7) = 4 This means that p(1) must be a multiple of 4.

16. How Constraints Arise Marketing says the best bike must have a minimum gear ratio exactly equal to 1: C3: p(3)/w(1) = 1

17. How Constraints Arise Purchasing says that wheel sprockets are available in 31 sizes… C4: w(i) is one of 12, 13,…,42.

18. Choosing Pedal Sprockets Possible values… Front = [52 48 42 39 32 28]; Constraint C1 says that p(1) must be divisible by 4. Also: p(3) < p(2) < p(1).

19. The Possibilities.. 52 48 42 52 39 32 48 39 28 52 48 39 52 39 28 48 32 28 52 48 32 52 32 28 42 39 32 52 48 28 48 42 39 42 39 28 52 42 39 48 42 32 42 32 28 52 42 32 48 42 28 52 42 28 48 39 32

20. The Loops.. Front = [52 48 42 39 32 28]; for i = 1:3 for j=i+1:6 for k=j+1:6 p(1) = Front(i); p(2) = Front(j); p(3) = Front(k);

21. w(1) and w(7) “for free”.. Front = [52 48 42 39 32 28]; for i = 1:3 for j=i+1:6 for k=j+1:6 p(1) = Front(i); p(2) = Front(j); p(3) = Front(k); w(1) = p(3); w(7) = p(1)/4;

22. What About w(2:6) Front = [52 48 42 39 32 28]; for i = 1:3 for j=i+1:6 for k=j+1:6 p(1) = Front(i); p(2) = Front(j); p(3) = Front(k); w(1) = p(3); w(7) = p(1)/4; Select w(2:6)

23. All Possibilities? for a=12:w(1) for b = 12:a-1 for c = 12:b-1 for d = 12:c-1 for e = 12:d-1 w(2) = a; w(3) = b; etc

24. Reduce the Size of TheSearch Space Build an environment that supports something better than brute force search…