L3 Formulation
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Homework
Review
Column example
Constraint satisfaction
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Standard Design
Optimzation Model
Find x * such that
Design Variables
MINIMIZE : f (x )
Subject
Objective function
To :
h j (x ) = 0
j = 1 p
gi (x )  0
i = 1 m
(L )
(U )
xi  xi  xi
Constraints
i = 1 n
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Design of a can - Summary
Min f(D,H) = πD2 /2 + πDH (cm2)
Subject to:
(πD2/4)H ≥ 400 (cm3)
3.5 ≤ D
D≤ 8
8≤H
H ≤ 18
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Formulating Process (Eggert)
Step 1. Describe problem
Separate into one-line phrases
Step 2. Collect info
Diagram, handbooks, notes
Step 3. Define DVs
“Form” which influences behavior
Name, symbol, units
Step 4. Determine objective function
performance criterion, f(DVs)
Step 5. Formulate constraints
Laws of nature, man, economics
Failure modes
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Min Weight Column
Step 1. Describe problem
Separate into one-line phrases
Min mass tubular column
Length l
Supports load P
No buckling
No overstressing
“fixed” end-conditon
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Min Weight Column
Step 2. Collect info
Diagram, handbooks, notes
mass  density( volume)
 density( area )length
area  circum(thick )
area  2Rt
mass  ( 2Rt )l
  a
P  Pcr
P
P

 a
A 2Rt
3 ER 3t
P
4l 2
Crush
Buckle
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Min Weight Column
Step 3. Define DVs
“Form” which influences behavior
Name, symbol, units
x  [ x1.x2 ]  [ R, t ]
Name
Radius
Thickness
Symbol
R
t
Units
inches
Inches
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Min Weight Column
Step 4. Determine objective function
performance criterion, f(DVs)
MIN
f (x )  (2Rt )l
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Min Weight Column
Step 5. Formulate constraints
Laws of nature, man, economics
Failure modes
P
 a
2Rt
3 ER 3t
P
4l 2
Rmin  R  Rmax
tmin  t  tmax
PDPs for this problem?
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Min Weight Column - Summary
MIN
f (x )  (2Rt )l
Subject to:
P
 a
2Rt
3 ER 3t
P
4l 2
Rmin  R  Rmax
tmin  t  tmax
How many equality or inequality eqns?
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Maximizing versus Minimizing
Maximizing F(x)
is “equivalent” to
Minimizing f(x)= - F(x)
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Feasible Designs
feasible design - A design candidate that meets design
specifications and/or satisfies design constraints.
feasible design - a point in the design space that satisfies
all constraints.
feasible design region (i.e. feasible design space) – the set of points
in the design space that satisfies all constraints.
feasible region = feasible design space
When a point violates any constraint it is said to be INFEASIBLE.
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Satisfy an equality constraint?
An equality constraint is satisfied iff h(x )  0
For example, let x (1)  1, 1
T
h(x )  x1  x2  0
h( x )  1  1  0
h( x )  0
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Satisfy an inequality constraint
An inequality constraint is satisfied iff g ( x )  0 ,
i.e. when either g (x )  0 or g ( x )  0 .
Case A: g (x )  0
For example, let x (1)  1, 1 , then for
g (x )  x1  x2  5  0
T
g (x )  1  1  5  3
g (x )  3  0
When g(x) < 0 the constraint is said to be
INACTIVE or NONBINDING
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Active inequality constraint
Case B: g ( x )  0
For example let x ( 2 )  2.5, 2.5 , then for
g (x )  x1  x2  5  0
T
g (x )  2.5  2.5  5  0
g (x )  0
When g(x) = 0 the constraint is said to be
ACTIVE or BINDING
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Violated inequality constraint
Case C: g (x )  0
For example let x ( 3)  3, 3 , then for
g (x )  x1  x2  5  0
g (x )  3  3  5  1
g (x )  1, therefore
g (x )  0
When g(x) > 0, the constraint is VIOLATED
T
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Constraint Activity/Condition
Constraint Type Satisfied
Violated
Equality
h(x )  0
h(x )  0
Inequality
g (x )  0 inactive
g (x )  0 active
g (x )  0
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Summary
• Formulating: 5 step process to determine
DVs, obj. function, and constraints.
• Constraints derive from
nature, man, economics
• Constraints can be satisfied or violated.
• Design candidates (i.e. points in x) that
satisfy all constraints are “feasible .”
• Satisfied inequality constraints are either
active or inactive
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