EML4552 - Engineering Design Systems II (Senior Design Project)

EML 4550 - Spring’08

EML4552 - Engineering Design Systems II(Senior Design Project)

Optimization Theory andOptimum DesignDynamic Programming

Hyman: Chapter 10


Basic Concepts

Optimization in DesignFrom Concept Selection to Optimum

DesignOptimization Theory and Methods

Large number of design choices: Dynamic Programming

Optimization with continuous variablesLinear programmingNon-linear programming and search methodsLagrange multipliers


Why Optimum Design?

Find system with minimum ‘cost’-’weight’-’fuel usage’-…etc. that will fulfill the functional specification

Find system with maximum ‘capability’ within certain constraints (cost, weight, etc.)

Competitive pressure drives towards optimum design


Optimization

Minimize (Maximize) an Objective Function of certain Variables subject to Constraints

0),...,(

...

0),...,(

0),...,(

),...,(min

21

212

211

21

nm

n

n

n

xxxg

xxxg

xxxg

xxxf


Design Optimization

Concept GenerationConcept SelectionSystem ArchitectureDetailed DesignManufacturingOperational Experience

Design Optimization starts with System Architecture and becomes an integral part of the design process through the

lifetime of the product

OPTIMIZATION


Dynamic Programming

Optimization of systems that feature ‘stages’Large number of stages Large number of choices per stageApparently very large number of choices (yet

finite) can be efficiently explored and an optimum found with dynamic programming

Dynamic programming allows for a consistent search of the optimum in multi-stage problems

“Efficiency” of dynamic programming increases with the problem size


Dynamic Programming - Example:Optimum Routing of a Transmission Line

A

B1

B2

C1

C2

D1

D2

E

14

16

20 1815

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1318

10

12

Find least cost to build transmission between A and E and going through (B1 or B2), (C1 or C2), and (D1 or D2)


Dynamic Programming - Example

Route Costs TotalA-B1-C1-D1-E 16+17+20+10 63A-B1-C1-D2-E 16+17+10+15 58A-B1-C2-D1-E 16+18+12+10 56A-B1-C2-D2-E 16+18+18+15 67A-B2-C1-D1-E 14+13+20+10 57A-B2-C1-D2-E 14+13+10+15 52A-B2-C2-D1-E 14+20+12+10 56A-B2-C2-D2-E 14+20+18+15 67



In this case the combination set of paths is very small, optimum can be found by exhaustive search and inspection

We needed to compute the ‘objective function’ 8 times to determine the minimum

What happens if the number of choices is so large that it becomes impractical to conduct an exhaustive search?

We need a structured approach to find the optimum



Most D.P. problems can be solved by moving forward or backwards through the stages analyzing one stage at a time

Consider working backwards from point E There are only two paths leading to point E Tabulate costs for all the paths leading to the

last stage


Dynamic Programming - Identify “Stages”

A

B1

B2

C1

C2

D1

D2

E

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17 2010

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10

12

Stage 1Stage 2Stage 3Stage 4


Stage 1

Beginning End SegmentCost

Cost fromBeginning ofStage to E

D1 E 10 10

D2 E 15 15


Stage 2

There are four possible paths to consider in this stage, paths that begin in C1 or C2, and end on D1 or D2

Tabulate all the costs for the paths in this stage

Combine with costs from previous stage to compute total cost for Stage 1 + Stage 2

For each beginning point of Stage 2, pick an optimum to arrive at the end point and eliminate those paths that cannot be optimum (basic principle of D.P.)


Stage 2

Start End Cost CostStage 1

Cost - Beg.Stage 2 to E

Opt

C1 D1 20 10 30

C1 D2 10 15 25 X

C2 D1 12 10 22 X

C2 D2 18 15 33


Stage 3

Repeat previous approach and prepare a table with the four possible paths for this stage

Only consider the optimum possibilities for the paths from the end of Stage 3 (beginning of Stage 2) to the end point E

identify the optimum paths that go from the beginning of Stage 3 to the end point E (basic principle of D.P.)


Stage 3

Start End Cost CostStg. 1+2

Cost - Beg.Stage 3 to E

Opt

B1 C1 17 25 42

B1 C2 18 22 40 X

B2 C1 13 25 38 X

B2 C2 20 22 42


Stage 4

Repeat procedure for the last stage, now there are only 2 paths to consider in in this stage

Apply basic principle of D.P. to determine the optimum path that covers all four stages


Stage 4

Start End Cost Cost1+2 +3

Cost - Stage4 (A) to E

Opt

A B1 16 40 56

A B2 14 38 52 X


Reconstruct the “Optimum Path”

Stage Segment Cost

4 A-B2 14

3 B2-C1 13

2 C1-D2 10

1 D2-E 15

52


Dynamic Programming - Example:Optimum Routing of a Transmission Line

A

B1

B2

C1

C2

D1

D2

E

14

16

20 1815

17 2010

1318

10

12

In this example the optimum could be determined by inspection, but as system complexity increases, dynamic programming is needed


Dynamic Programming

n

iir1

min(max)

Stage n Stage n-1 Stage 1nS

1 nn SS12 SS 1S

nd 1nd 1d

nr 1nr 1r


Example: Gas Pipeline Operation

Minimize Fuel Consumption through Compressor Pressure Settings

EML4552 - Engineering Design Systems II (Senior Design Project)

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