L 32 Constrained Optimization Optimizing Gear Ratio Distribution

L 32. Constrained Optimization Optimizing Gear Ratio Distribution

An Optimal Design Problem

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

It’s a Matter of Teeth E. g. , 13 teeth E. g. , 48 teeth E. g. , Gear ratio = 48/13 = 3. 692

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

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.

Things to Do 1. 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

The Quality of a Gear Ratio Distribution Ideal: 1 Good: Poor: 4

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

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

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

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

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

How Constraints Arise Purchasing says that pedal sprockets only come in six sizes: C 1: p(i) is one of 52 48 42 39 32 28.

How Constraints Arise Marketing says the best bike must have a maximum gear ratio exactly equal to 4: C 2: p(1)/w(7) = 4 This means that p(1) must be a multiple of 4.

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

How Constraints Arise Purchasing says that wheel sprockets are available in 31 sizes… C 4: w(i) is one of 12, 13, …, 42.

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

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

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);

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;

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)