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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…