1 OPTIMIZATION Sayeed N Ghani PhD (Univ London), DIC (Imperial College), CEng (UK), MIEE (UK) Quality Six Sigma Green Belt Certified(USA) Copyright2007.

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OPTIMIZATIONSayeed N Ghani

PhD (Univ London), DIC (Imperial College), CEng (UK), MIEE (UK)

Quality Six Sigma Green Belt Certified(USA)

Copyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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OPTIMIZATION 1Index 2

Systems Analysis vs.. Design 3Traditional Systems Design 4Optimum Systems Design 6Design Examples

Example: Design of A Grain Silo 13First Design --- By Inexperienced Engineer 15

Second Design --- By Experienced Engineer 17Third Design --- Optimum Design 20

Rural Area --- Cost of Land Low 20 First Pass --- Local Minimum 25 Second Pass --- Global Minimum 26 Urban Area --- Cost of Land High 27

First Pass --- Local Minimum 28Second Pass --- Global Minimum 29

Example: Design of an Electric Power Supply Pi-Section LC Filter 32 First Design --- By Inexperienced Engineer 34Second Design --- By Experienced Engineer 35Third Design --- Optimum Design 37

Minimum Cost Filter Design When Explicit Inequality 51 Constraints Were Not Accounted ForComparison of Above Three Designs 53

Maximization 55Copyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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Systems Analysis vs.. Design


System Design Details Provided A Priori

System Performance

Analyze/

Calculate 1:1

Many Systems Can Be Synthesized

Synthesize/

Designn:1

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Traditional Systems Design

• In traditional systems design the objective is merely to meet the specifications. There is no formal attempt to reach the best design in the strict mathematical sense of minimizing cost or weight or volume or maximizing profit.


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Traditional Systems Design (cont..)

• Further in traditional systems design a highly skilled engineer is in the design loop making sound engineering decisions at every stage of the design process.

•The process undergoes many manual iterations before the design can be


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Optimum Systems Design

finalized making it a slow and very costly process.

•The science of optimization is a formalism that allows not only all specifications (design constraints) to be met, but would also yield design which is the best in terms of some figure(s) of merit.


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Systems Analysis vs. Design


System Design Details Provided A Priori

System Performance

Analyze/

Calculate 1:1

Best System Synthesized

Synthesize/

Optimum Design

1:1

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Optimum Systems Design (cont.)


•It is a completely automated process that allows lesser skilled and experienced engineers to create optimum design.

•Optimization is applicable to all numerate disciplines including fuzzy systems.

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•Fuzzy logic, fuzzy sets, fuzzy relations and fuzzy reasoning allows synthesis of high performance control systems that would beat, hands down, any linear counterpart (if properly designed).


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•Fuzzy concept is ideally suited to model poorly defined processes that could only be described in qualitative terms via linguistic variables. “If temperature is extremely low then set fuel injection high.”


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•Objective functions involving fuzzy systems are known to possess multiple minima and difficult to optimize.

•Optimization is one of the core concepts used in DFSS (Design for Six Sigma) for product to service and for manufacturing to transaction.


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•Optimization is the only way to ensure that an enterprise not only meets all its requirements but is also the best in its widest possible meaning.


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Example: Design of a Grain Silo

dh

d + 4

All dimensions are in meters


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Volume V of the grain silo is: V = Pi/4 x d2 x h m3

Specification V = 200 m3

Substituting we obtain 200 = Pi/4 x d2 x h

or d2 x h = 254.655 m3


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First design --- By Inexperienced EngineerArbitrarily choose d = 1 mTherefore, h = 254.655 m

The design will of course work, but the problem is that it will be too expensive. What we have actually done really is to apply rule of thumb saying that we shall arbitrarily choose diameter d = 1 m.


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This applies an artificial constraint on our design variable ‘d’ to yield a unique solution.

Now in practice rules of thumb used by competent engineers are never so unrealistic. But nevertheless they are only guess work. If we study a number of grain silo designs we will repeatedly come across designs more or less of square shape.


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Second design --- By Experienced Engineer

The engineer has been making similar grain silos for many years. From past experience he bases his design on roughly a square shape.

With d2 x h = 254.655 m3, and d = h

d3 = 254.655 or d = h = 6.34 m Copyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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But he has not attempted, in any way, to obtain the best design in terms of some figure of merit which is cost in this case.

In this example we want most economical design !

So we see traditional design approach (use of rules of thumb) has no capability to satisfy our quest for the best.


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Since we are seeking the most cost effective design we shall have to bring, somehow, the concept of total system cost in our design formulae.

(In general terms this is how we do it. From the manufacturer’s catalogues we derive the cost of a component to its size. We next obtain a best curve fit to obtain cost formulae.)


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Optimum Design

We perform a cost analysis and deduce the following econometric models.

Cost of base including land $5000 + $4000/m2

Cost of roof $1000 + $1500/m2

Cost of silo walls $3000 + $4000/m2 of wall area + $1000/m of height


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Total cost = Cost of base + Cost of roof + Cost of silo wall.

Cost of base including land $5000 + $400 x Pi/4 x (d + 4)2

Cost of roof $1000 + $20 x Pi/4 x (d + 4)2

Cost of silo wall $3000 + $40 x Pi x d x h + $1000h

Total cost of the silo = $9000 + ${420 x Pi/4 x (d + 4)2 }+ $(40 x Pi x d x h) + $(1000 x h)

Copyright 2007 by Sayeed Nurul Ghani. All rights reserved.

C

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The optimization model (cost function for constrained optimization) for the grain silo is then

Minimize cost function F(d, h) = $9000 + ${420 x Pi/4 x (d + 4)2 }+ $(40 x Pi x d x h) + $(1000 x h)

Subject to explicit constraints 0 =< d <= infinity 0 =< h <= infinity


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and implicit constraint

254.655 =< d2h <= infinity

Defining design variables x1 = d and x2 = h the optimization model for the grain silo then becomes in generalized terms

Minimize cost function F(x1, x2) = $9000 + ${420 x Pi/4 x (x1 + 4)2 }+ $(40 x Pi x1 x2) + $(1000 x2)


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Subject to explicit constraints 0 =< x1 <= 99999 0 =< x2 <= 99999

and implicit constraint

254.655 =< x12 x2 <= 99999

Next a constrained optimizer ‘EVOP’ developed by this author was used to minimize the above objective function.


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And here are the results of optimization from subroutine EVOP.

First Pass and A Local Minimum:

Diameter d = 6.34 m Height h = 6.34 m

Implicit constraint XX = d2h = 254.655 m3 Volume = Pi/4 x d2h = 200 m3

Cost = $55,643Identical to Rule of Thumb Design by Experienced Engineer.Copyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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Second Pass and Global Minimum:

Diameter d = 4.85 m Height h = 10.84 m

Implicit constraint XX = d2h = 254.655 m3 Volume = Pi/4 x d2h = 200 m3

Cost = $52,259Copyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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Saving over rule of thumb design =($ 55,643 - $52,259)/ $52,2596.5 % only.

Design for Urban Area With Ten Fold Increase in Cost of Land:

Base = $50,000 + ${4000 x Pi/4 * (d +4)2} . Cost of roof and wall remains unchanged (Page 21).


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Design for Urban Area (cont.):

First Pass and A Local Minimum:

Diameter d = 6.33 m Height h = 6.33 mImplicit constraint XX = d2h = 254.655 m3 Volume = Pi/4 x d2h = 200 m3

Cost = $402,841


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Design for Urban Area (cont.):Identical to Rule of Thumb Design by Experienced Engineer.

Second Pass and Global Minimum:Diameter d = 2.44 m Height h = 42.92 mImplicit constraint XX = d2h = 254.655 m3


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Design for Urban Area (cont.):Volume = Pi/4 x d2h = 200 m3

Cost = $240,828

Saving over rule of thumb design = ($402,841 - $240,828)/$240,828= 67.3 %

For a silo of 500 m3 volume a saving of staggering 87 % has been achieved by EVOP.


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FINISHED FOR NOW

FOLKS


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2nd Example

Design of An Electric Power

Supply PI-Section L-C Filter


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Example: Design a Power Supply L-C Filter

R = 1 K

L

C1C2


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Ripple factor r = 4.31 x 108/(f3 C1 C2 L R)

Substituting r = 0.01, f = 50 and R = 1000 we obtain L = 346/(C1 C2)

First design --- By Inexperienced Engineer

To obtain a design we may choose C1 = C2 = 1 uF and obtain L = 346 H


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Second design --- By Experienced Engineer

Now in practice rules of thumb used by competent engineers are never so unrealistic. But nevertheless they are only guess work. If we study a number of power supply designs we will repeatedly come across figures like 32 uF.


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This will yield L = 338 mH – a more realistic value to use for the inductor. But we have not attempted, in any way, to obtain the best design in terms of some figure of merit which is cost in this case.

We want most economical design !


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So we see traditional design approach (use of rules of thumb) has no capability to satisfy our quest for the best.

Third Design --- Optimum Design

Since we are seeking the most cost effective design we shall


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have to bring, somehow, the concept of total system cost in our design formulae. This is how we do it.

From the manufacturer’s catalogues we derive the cost of a component to its size. We next obtain a best curve fit to obtain a mathematical expression relating cost to size.


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This we can do either on a computer, or manually on a graph paper as shown below. From the two figures below the functional relationship between cost and

Cents

uFHenry

X

X

XX

X

X

Cents


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component size were determined to be

Cost of a capacitor = 5 + 1/uF cents

Cost of an inductor = 50 + 5/H + 1/H2 cents

Total system cost was then

System Cost = Cost of C1 + Cost of C2 + Cost of L


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System Cost = (5 + C1) + (5 + C2) + (50 + 5L + L2) cents

= 60 + (C1 + C2) + L(5 + L) cents

Having obtained the total system cost, we have to next introduce the constraints on the design variables imposed by the specification on the ripple factor in our optimization model.


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The constraint due to specification on ripple factor, as derived earlier, is

L = 346/(C1 C2)

So, the mathematical model for optimization becomes

Minimize cost F(C1, C2, L) = 60 + (C1 + C2) + L(5 + L)

Subject to the following constraints on the design variables


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0 < L < infinity 0 < C1 < infinity 0 < C2 < infinity

And implicit equality constraint

L = 346/(C1 C2)

This equality constraint is next introduced in the cost function F(C1, C2, L) to yield the cost function of reduced dimensionalityCopyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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F(C1, C2) = 60 + (C1 + C2) + {346/(C1* C2)} * {5 + 346/(C1 + C2)}

We next account for non-negativity of our design variables C1 and C2 by introducing the following two transformations.

x12 = C1 and x2

2 = C2

The cost function for unconstrained optimization now becomes

Cost = F(x1, x2) = 60 + (x12 + x2

2)Copyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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+ {346/(x12 * x2

2)} * {5 + 346/(x12 * x2

2)} cents

x1 and x2 can now take negative values but capacitor values C1 and C2 will always remain positive.

Next a versatile optimizer EVOP due to Ghani was used to minimize the above cost function.


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The problem presented to EVOP was Minimize F(x1, x2) = F(x1, x2) = 60 + (x1

2 + x22) + + {346/(x1

2 * x22)} * {5 +

346/(x12 + x2

2)} cents

Subject to explicit constraints: -99999 <= x1 <= 99999 -99999 <= x2 <= 99999

And implicit constraint: -99999 <= (x1 + x2) <= 99999


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EVOP calculated the following values of the objective function at the minimum:

X1* = 3.778; X2

* = 3.777 and F* = 99.92

Hence, C1* = 3.7782 = 14.273 uF

and C2* = 3.7772 =

14.273 uF and L* = 346/(C1C2) = 1.698 HCopyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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Using the relation Cost = 60 + (C1 + C2) + L(5 + L) cents we hand calculate the optimalCost F* = 60 + (14.268 + 14.268) + 1.698(5 + 1.698) = 99.91 cents.

Fletcher Powell’s famous gradient based variable metric quadratic convergent algorithm for unconstrained objective function yields the following optimum values.


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C1o = 14.4 uF, C2

o = 14.4 uF, Cost Fo = 99.93 cents.

The value of L was rounded up to 1.7 H.

Minimum Cost Filter Design

C1 = 14.4 uF C2 = 14.4 uF

L = 1.7 H

Cost = 100.19 centsCopyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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If explicit inequality constraints on the design variables were not accounted for, the computer being nothing more than a fast number crunching idiot will churn out an impossible, unrealistic design as follows:


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Minimum Cost Filter Design When Explicit Inequality Constraints Were Not Accounted For

C1 = -1.007 x 104 F C2 = -8.392 x 104 F L = 0.4112 uH

Cost = -18460 Mega DollarsCopyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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Only if I could make a negative capacitor capable of carrying

currents at power level, wouldn’t I have been a double billionaire by

just making (not even selling) only 1 number of such a power

supply !!

AND wouldn’t all power electronics engineers like

myself !!


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Cost Comparison

Finally, let us make a cost comparison of all three designs.

Designed by popular vote by students.

C1 = C2 = 1 uf L = 346 H

Cost = $ 1215.08


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Designed by experienced engineer using rule of thumb.

C1 = C2 = 32 uf L = 0.338 H

Cost = $ 1.26

Optimized design is $1 only. (26 % cheaper and no skilled designer overhead)

Without design experience optimization is indispensable tool.Copyright 2007 by Sayeed Nurul Ghani. All rights reserved.C

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Maximization

•Finally, if maximization is needed (instead of minimization), then negate the objective function and then minimize as usual.

Minimize -Fo (x1, x2, …, xn) will yield:

Maximize Fo (x1, x2, …, xn)

Constraints should be kept unchanged.


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FINISHED FOR NOW

FOLKS


1 OPTIMIZATION Sayeed N Ghani PhD (Univ London), DIC (Imperial College), CEng (UK), MIEE (UK) Quality Six Sigma Green Belt Certified(USA) Copyright2007.

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