EML4552 - Engineering Design Systems II (Senior Design Project)
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EML 4550 - Spring’08
EML4552 - Engineering Design Systems II(Senior Design Project)
Optimization Theory andOptimum DesignDynamic Programming
Hyman: Chapter 10
EML 4550 - Spring’08
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
EML 4550 - Spring’08
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
EML 4550 - Spring’08
Optimization
Minimize (Maximize) an Objective Function of certain Variables subject to Constraints
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...
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0),...,(
),...,(min
21
212
211
21
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xxxf
EML 4550 - Spring’08
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
EML 4550 - Spring’08
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
EML 4550 - Spring’08
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
Find least cost to build transmission between A and E and going through (B1 or B2), (C1 or C2), and (D1 or D2)
EML 4550 - Spring’08
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
EML 4550 - Spring’08
Dynamic Programming - Example
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
EML 4550 - Spring’08
Dynamic Programming - Example
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
EML 4550 - Spring’08
Dynamic Programming - Identify “Stages”
A
B1
B2
C1
C2
D1
D2
E
14
16
20 1815
17 2010
1318
10
12
Stage 1Stage 2Stage 3Stage 4
EML 4550 - Spring’08
Stage 1
Beginning End SegmentCost
Cost fromBeginning ofStage to E
D1 E 10 10
D2 E 15 15
EML 4550 - Spring’08
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.)
EML 4550 - Spring’08
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
EML 4550 - Spring’08
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.)
EML 4550 - Spring’08
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
EML 4550 - Spring’08
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
EML 4550 - Spring’08
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
EML 4550 - Spring’08
Reconstruct the “Optimum Path”
Stage Segment Cost
4 A-B2 14
3 B2-C1 13
2 C1-D2 10
1 D2-E 15
52
EML 4550 - Spring’08
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
EML 4550 - Spring’08
Dynamic Programming
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iir1
min(max)
Stage n Stage n-1 Stage 1nS
1 nn SS12 SS 1S
nd 1nd 1d
nr 1nr 1r
EML 4550 - Spring’08
Example: Gas Pipeline Operation
Minimize Fuel Consumption through Compressor Pressure Settings
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