l32. constrained optimization

L32. Constrained Optimization

Optimizing Gear Ratio Distribution

An Optimal Design Problem

Gear Ratio DistributionAssume 7 wheelsprockets

Assume 3pedal

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 wheelsprockets so that the 21 gear ratios areas 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 RatioDistribution

Ideal:

1 4

Good:

Poor:

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 = 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;

There Are Other ReasonableObjective 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 thatobjF(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

21 gear ratios.

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.

How Constraints Arise

Marketing says the best bike must havea maximum gear ratio exactly equal to4: C2: 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:

C3: p(3)/w(1) = 1

How Constraints Arise

Purchasing says that wheel sprockets are available in 31 sizes…

C4: w(i) is one of 12, 13,…,42.

Choosing Pedal Sprockets

Possible values…

Front = [52 48 42 39 32 28];

Constraint C1 says that p(1) must bedivisible by 4.

Also: p(3) < p(2) < p(1).

The Possibilities..

52 48 42 52 39 32 48 39 2852 48 39 52 39 28 48 32 2852 48 32 52 32 28 42 39 3252 48 28 48 42 39 42 39 2852 42 39 48 42 32 42 32 2852 42 32 48 42 28 52 42 28 48 39 32

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);

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) = p(3); w(7) = p(1)/4;

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)

Select w(2:6)

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

Reduce the Size of TheSearch Space

Build an environment that supportssomething better than brute forcesearch…