Probabilistic Optimization Exercise
The Snack Can
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The Assignment
• Review the different approaches to solving the problem
–
–
–
–
GRC - Section 2B (an example of setting up the problem incorrectly)
Dave Rumpf - Section 2C
Don Beeson - Section 2D
Filtered Monte Carlo (homework folder)
• Work the problem using the Section 2D approach in Excel
– Use the “Sensitivity add-in” to do the partial derivatives (homework
folder)
– Next, work the problem using Crystal BallTM / OptQuestTM (homework
folder)
• The goal is to achieve 4.5 sigma long term capability for the
Customer CTQ (volume), while minimizing cost
• Compare your answers with Don’s answers from PEZ
– Hint - they will be different
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Section 2B
Probabilistic Optimization
(GRC Approach)
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Optimal Design Problem Formulation – A Simple Example
Designing a Snack Food Can


The Customer CTQ – Can volume must be at
least 500 cm3
Aesthetic, fabrication, and shipping
considerations impose the following conditions:
 The diameter must be no larger than 10
cm and no smaller than 5 cm
 The can should be no higher than 12 cm
and no shorter than 6 cm


Finance – Make cans for < $0.045 each
Manufacturing – design is consistent with
process capability

Height sigma long term = 0.065

Diameter sigma long term = 0.030
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Which design best meets
all these requirements?
What does “best” mean in
this problem?
4
The Challenge in Optimal Design
We have a design that depends upon a number of Xs and which is also
subject to a number of operational constraints
What is our Optimal Design?
Optimization Theory:
 Identify the design variables (Xs)
 Develop a transfer function(s) which models design
performance (mean and variance) in terms of the design
variables
 Use the transfer function(s) to construct an objective
function
 Identify design constraints and convert them into
mathematical expressions
 Use a numerical optimization method to find the optimal
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Design Variables and Transfer Functions
Design Variables
 H = Height
 D = Diameter
Transfer Functions
Volume  V 

4
D2H
Surface Area  A  DH 
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
2
D2
6
Defining the Objective Function
The criterion (Objective) by which we judge the suitability of a design must be
a scalar quantity whose value is a function of the design variables (X’s).
 Do we want to maximize or minimize the mean performance?
Maximize Y = f1(xn)
 Do we want to minimize variation?
Minimize s = f2(xn)
 Do we want to maximize Z?
Maximize Z = f1(xn) / f2(xn)
 Do we want to minimize cost?
Minimize Area * Cost/Area
 Some other criteria?
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The Objective Function -- Challenges
 Complex Designs – The formulation of a meaningful objective function
may not be easy
 What is the best optimization function to use for a complex system?
 Multi-Objective Problems – There may be several objective functions
that we want to optimize simultaneously
 There is no general method for solving this class of problems
 There are approaches – for example:
•
Develop an objective function that is a composite weighted sum
of all the objective functions
•
Select the most important objective for optimization and treat the
others as constraints
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Objective Function for the Snack Food Can Problem
GRC Approach
Designing a Snack Food Can
 Marketing – Can must contain at least
500 cm3
 Aesthetic, fabrication, and shipping
considerations impose the
following conditions.
 The diameter must be no larger
than 10 cm and no smaller than
5 cm.
 The can should be no higher
than 12 cm and no shorter than
6 cm.
Objective Function
Cost  SheetMetal SurfaceArea
Sheet metal cost = $0.0001 cm-2
USL = 450 cm2 of sheet metal
Area  DH 

2
D2
Manufacturing says maximize ZA :
MaximizeZ A 
450 A
sA
Z will be our Objective Function
 Finance – Make cans for < $0.045 each.
We’ll need a transfer function for sA
 Manufacturing – Also want best possible
process capability
 Use propagation of errors.
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Standard Deviation of Surface Area
SurfaceArea
500
Transfer Function for Surface Area
Area  DH 

2
437.5-500
437.5
375-437.5
375
312.5-375
312.5
250-312.5
250
187.5-250
125-187.5
187.5
D2
62.5-125
125
6
0-62.5
62.5
H
0
10
D
5
Transfer Function for Standard Deviation of Surface Area
Using Propagation of Errors
dArea
  H  D
dD
Area-Sigma
30
26.25
dArea
 D
dH
26.25-30
22.5
22.5-26.25
18.75
18.75-22.5
15
15-18.75
11.25
11.25-15
7.5-11.25
7.5
3.75
3.75-7.5
0-3.75
0
10
6
D

s Area    H  D s D   Ds H 
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2

1
2 2
H
12
5
10
The Complete Objective Function
MaximizeZ A 
where
450 A
sA
A  DH 

2
D2
s A  ( ( D  H )s D ) 2  (Ds H ) 2
Optimization Objective: Maximize Z Subject to the Design Constraints
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Design Constraints
 All real systems must be designed and fabricated within given resource
and materials limitations as well as performance requirements.
 Examples
• Structures must not fail under load
• Natural vibration frequency of a structure must differ from the
operating frequency of the machine.
 All constraints must be expressed in terms of at least one of
the design variables
 Most design constraints are expressed as inequalities, < or >, and in
some cases as equalities, =.
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Design Constraints for the Snack Food Can Problem
Designing a Snack Food Can
 Marketing – Can must contain at least
500 cm3
 Aesthetic, fabrication, and shipping
considerations impose the
following conditions.
Design Constraints
 The diameter must be no larger
than 10 cm and no smaller than
5 cm.
• 6 cm  H  12 cm
 The can should be no higher
than 12 cm and no shorter than
6 cm.
• Volume  V 
• 5 cm  D cm

4
D 2 H  500 cm 3
 Finance – Make cans for < $0.045 each.
 Manufacturing – Also want best possible
process capability
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The Probabilistic Optimization Problem
Maximize
where
ZA 
450 A
sA
A  DH 

2
D2
s A  ( ( D  H )s D ) 2  (Ds H ) 2
Subject to the Constraints
5  D  10
6  H  12
V
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D 2 H
4
 500
14
Enter Equations in Excel
SNACK FOOD CAN OPTIMIZATION
This is where you
enter your starting
values of H and D
Cell D4
Height in cm
Diameter in cm
6.0
5.0
Surface Area in cm2
133.5
Cell D5
Enter Transfer Function for Area
s - Surface Area in cm
2
14.6
Enter Transfer Function for sArea
Z-Value
2
USL = 450 cm
21.8
Volume in cm3
117.81
Enter Transfer Function for Z
Enter Transfer Function for Volume
Constraints
Volume 500 cm3
You will enter these constraints into the
Excel Solver. For Volume you will reference
the transfer function for Volume.
6 H 12
5 D 10
Manufacturing Capability
Cell D27
Height - s in cm
0.065
Diameter - s in cm
0.030
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Cell D29
Manufacturing capability
data is from a process
capability database.
15
How to Enter the Transfer Functions in Excel
Surface Area
=(PI()*D5*D4) + (PI()*D5^2)/2
Standard Deviation of Area
= SQRT(((PI()*(D4+D5)*D29)^2) + (PI()*D5*D27)^2)
ZA - Value
= (450 - D7)/D9
Volume
= (PI()*D5^2*D4)/4
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Set Up the Excel Solver
SNACK FOOD CAN OPTIMIZATION
Height in cm
Diameter in cm
6.0
5.0
2
133.5
s - Surface Area in cm2
14.6
Z-Value
USL = 450 cm2
21.8
Volume in cm3
117.81
Surface Area in cm
Select Tools->Solver in the
Excel menu to invoke the
Excel Solver dialog box.
Constraints
Volume 500 cm3
6 H 12
5 D 10
Manufacturing Capability
Height - s in cm
0.065
Diameter - s in cm
0.030
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Optimal Solution
The “Best” Can:
V = 500 cm3 
A = 349.1 cm2 : DPMO = 0 
Cost = 3.49¢ < 4.5¢ 
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Optimal Solution Design Review
Really The “Best”
Can???
V = 500 cm3 
A = 349.1 cm2 : DPMO = 0 
Cost = 3.49¢ < 4.5¢ 
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Optimal Solution Design Review
The volume constraint V  500
is deterministic. The optimal
solution resulted in V = 500.
However, V is a random
variable – it has a mean of 500
and some standard deviation
due to variance in D and H.
Therefore, we will fail to meet
the volume constraint V  500
50% of the time!
P(d) = 50%
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The Enhanced Probabilistic Optimization Problem
Minimize
A
Subject to the Constraints
where
5  D  10
A  DH 
6  H  12
D 2 H
V
V  500
ZA 
ZA 
450 A
sA
V  500
sV

2
D2
4
s A  ( ( D  H )s D ) 2  (Ds H ) 2
6
2
 DHs D   D s H

sV  


2   4

2
6



2
Update the Excel worksheet to include the process capability for V, and find
the optimal solution to this problem.
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Optimal Solution
The “Best” Can:
A = 363.5 cm2
Cost = 3.64¢ < 4.5¢ 
V = 532 cm3  500 
ZA = 35.6  6 
ZV = 6  6 
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How Robust is the Design?
ZV vs D at H = 8.99
10
10
5
5
ZV
ZA
ZA vs D at H = 8.99
0
5
6
7
8
9
0
10
5
-5
-5
-10
-10
6
7
D
5
5
ZV
ZA
10
0
8
9
-5
10
ZV vs H at D = 8.68
10
7
9
D
ZA vs H at D = 8.68
6
8
10
11
0
6
12
7
8
9
10
11
12
-5
-10
-10
H
H
When it comes to the surface area response, the design is very robust – small variations
in D about the optimal value D=8.68 will not result in any decrease in process capability
for A or V. However, the design is not robust for the volume response. Small reductions in
either D or H will significantly increase the defect rate on the V>500 spec limit.
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A Robust Design
ZV vs D at H = 8.99
10
10
5
5
ZV
ZA
ZA vs D at H = 8.99
0
5
6
7
8
9
0
10
5
-5
6
7
-10
Optimal
Design
10
D
ZA vs H at D = 8.68
ZV vs H at D = 8.68
10
10
5
5
ZV
ZA
9
-10
D
Robust
Design
8
-5
0
6
7
8
9
-5
10
11
0
6
12
7
8
9
10
11
12
-5
-10
-10
H
H
A robust design can be attained by increasing the optimal values of D and H (i.e. the values
from the optimization analysis) by 6 times their standard deviations. This pushes the design
point (D, H) into a region of the design space where changes in the design variables over
even 6 standard deviations (unlikely) will still yield 6s (or higher) process capability. The final
design values are D = 8.86 and H = 9.34. Note that the cost of the robust design is only 6%
more than the cost for the optimal design.
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Section 2C
Solution to the Can Example
by Dave Rumpf
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A Non-linear Optimization Example
We know the constraints
• 5<diameter<10
• 6<height<12
• Volume > 500 cc
• Surface area < 450 cm2 (from the sheet metal cost per cm2 relationship)
Plot constraints for better understanding. One approach:
• Create multiple combinations of d, h in Excel (see optimizing snack food can.xls)
• Calculate A and V, the non-linear constraints.in Excel
• Transfer to Minitab. Go to graph/contour plot and choose Z=volume, Y=diameter
and X=height to get plot on left. Repeat with Z = area to get plot on right.
Contour Plot of Volume
Contour Plot of SurfArea
10
min volume
curve
9
Diameter
100
200
300
400
450
500
600
700
800
900
8
7
6
5
11
100
200
300
400
450
500
600
700
800
900
10
max cost
curve
9
Diameter
11
8
7
6
5
5
6
7
8
9
10
11
Height
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5
6
7
8
9
10
11
12
Height
26
Constrained Region
Contour Plot of SurfArea
11
100
200
300
400
450
500
600
700
800
900
10
max cost
curve
Diameter
9
8
7
6
5
5
6
7
8
9
10
11
12
Height
Cross hatched region satisfies
all constraints
Now that we know the region of possibilities. Move on to optimizing.
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What Is Our Objective
and the Objective Function
Goal: Best possible process capability. Which outputs do we care about?
• Surface area since it directly translates to cost, SA has upper spec only
• Volume since it is a design requirement, V has lower spec only
Functional representation of goal:
• Z cost = average cost/std dev cost
• Z volume = average volume/std dev volume
• Use partial derivative approach to find std dev equations for SA and V
• Use known std dev for h and d as needed
So which is the objective function and which is the constraint
• Choose maximize Z cost as the objective
• Set a constraint on Z volume, Z volume > 6 is a logical choice
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Using Excel Solver
Suggested Process:
• Decide on a matrix of starting points which are systematically spaced
over the feasible region.
• A modification of a DOE approach can work for this step
• Run solver for the various starting points
• Evaluate solutions to minimize chance of stopping at a local optima
First Step for the snack can example. Start with 5 starting points making a
tipped box with centerpoint
Starting points
1
2
3
4
5
Height
7
9
12
12
10
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Diameter
10
10
8.6
7.5
9
comments
1st two points
max d intersections
next two points
max h intersections
last the centerpoint
29
Set up and Run Solver for the 5 Starting Points
SNACK FOOD CAN OPTIMIZATION
Height in cm
9.47
Diameter in cm
8.45
Surface Area in cm2
363.7
s - Surface Area in cm
2
2.4
Add constraint for Volume
Z volume =
std dev V=
6.000
5.248
Z-Value
USL = 450 cm2
35.7
Volume in cm3 531.488
• All 5 starting points yield same optimal solution
• Optimal solution has nearly identical Z for cost and Z for Volume but
different solution values versus GRC solution
• This is an indication of likely ridge of optimality in the solution space
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Solution Using Dave’s Spreadsheet
SNACK FOOD CAN OPTIMIZATION
Height in cm
8.99
Diameter in cm
8.68
Surface Area in cm2
363.5
s - Surface Area in cm
2
2.4
Add constraint for Volume
Z volume =
std dev V=
6.008
5.321
Z-Value
USL = 450 cm2
35.6
Volume in cm3 531.972
Different H and D value but almost identical Z, SA and V and std dev of SA and V
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Location of Optimal Ridge of Solutions
Contour Plot of SurfArea
11
100
200
300
400
450
500
600
700
800
900
10
max cost
curve
Diameter
9
8
7
6
5
5
6
7
8
9
10
11
12
Height
Cross hatched region satisfies
all constraints
= The two optimal solutions
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Section 2D
Optimal Can Design Problem
Don Beeson
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Heuristics
We always try to emphasize basic problem solving “heuristics”
One of them is …
“Given a complex problem, first solve a related
simpler problem. - apply what you learn from the
simpler problem solution to solve the complex
problem”
This optimal can design problem provides a nice
opportunity to use this concept
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The Simple Problem
The related simpler problem would be, ignoring all
variation, what can design will give a volume of V0 = 500
and have a minimum surface area
This problem can easily be solved by basic calculus or
any optimization software (Excel Solver, PEZ, etc.)
When the problem is solved, you should
discover that, for all values of V0, the minimum
area is always achieved when D = H
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In fact, from calculus it can easily be shown that
.
1
D
 4 V0 


  
3
.
1
4 V0 
H  

  
3
gives the minimum surface area
Regardless of the method, for V0 = 500, we find that,
if there is no variation, the optimal design is
D = H = 8.60254
A = 348.734205
What do we now know from solving this simpler problem?
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36
(1) We know the entitlement of the manufacturing process
If all variation is removed from the manufacturing
process, the lowest cost per can is 3.49 cents
(can’t do any better than this)
(2) For any can with no aesthetic, fabrication or shipping
restrictions on D and H, the minimum area will happen
when D = H (regardless of any variation present)
Best Practice: Make D = H
(3) Since manufacturing variations are small and
3.49 cents/can << 4.50 cents/can, the effects of
variation on the condition V > 500 will be the main
concern (the Customer CTQ)
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37
Now we can investigate how manufacturing variation
will modify the solution to the simpler problem
Known
D
Known
Noise
sD
sH
H
D
Control
H
???
Cylinder
Equations
V
A
???
Note: We may want to bring the standard deviation
parameters around and make them control
parameters also. (i.e., reduce cost by improving
the manufacturing processes)
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To estimate the variation in volume, V, and the area, A,
we can use approximation formulas
These formulas come from a Taylor’s series
approximation and are usually referred to as “the partial
derivative method”
Also the fact that correlation can be included in the
formulas should be pointed out (in some problems it
may make a big difference)
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39
For this problem, the formulas, including the correlation
term (with rho = 0) and D = H = 8.60254 give
.
2
2


sV    D H   sD 
2

2
   D2   sH2 


4

   D H      D2     sD sH



2
4

sV  7 .75 44 91 02 8 03 74 1
sA 
  D   H 2 sD2    D 2 sH2    D   H    D    sD sH
sA  3 .60 56 75 02 3 13 89
Since the variation in D and H are both small compared
to the non-linearity of V and A, we can expect the
distributions of both V and A to be approximately normal
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40
Also it should be pointed out that these approximate
formulas can be used to resolve the total output variance
into the key contributing factors
For the volume, V (with no correlation between D and H)
.
2
   D H   sD2


2


p ct_ D 
 1 00
p ct_ D  9 4.9 43 82 0 22 47 19 1
2
sV
2
   D2   sH2


4


p ct_ H 
 1 00
2
p ct_ H  5 .05 61 79 7 75 28 09
sV
   D H      D2     sD sH



2
4




p ct_ corr 
 1 00
2
p ct_ corr  0
sV
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41
We see that almost all of the volume variation comes
from the variation in D (if we were interested in improving
the manufacturing process, we should focus our attention
on D)
Also, if possible, the accuracy of the approximate formulas
should be checked by running a quick Monte Carlo analysis
For this problem, a 100,000 Monte Carlo gives the results
on the next page …
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42
Monte Carlo
(N = 100,000)
Formulas
mean ( D)  8 .60 25 68 81 3 93 06 5
D  8 .60 25 40 13 8 28 09 9
st dev( D)  0 .06 49 93 19 0 06 33 34
sD  0 .06 5
mean ( H)  8 .60 24 17 17 6 02 81 1
H  8 .60 25 40 13 8 28 09 9
st dev( H)  0 .02 99 98 88 2 05 70 43
sH  0 .03
mean ( V)  5 00 .02 49 01 2 07 83 8
V0  5 00
st dev( V)  7 .76 47 30 12 3 67 10 3
sV  7 .75 44 91 02 8 03 74 1
mean ( A )  3 48 .73 91 08 1 74 69 2
st dev( A )  3 .61 04 72 30 1 98 39 9

2
2
 D   D H  3 48 .73 42 05 4 52 88 8
sA  3 .60 56 75 02 3 13 89
Use of the approximate formulas appears to be ok
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Now if we look at the results graphically, we see that
the volume needs to be moved to the right until the
the number of defects (V < 500) is at an acceptable
level …
.
Vo lu me Di strib u ti on
0.06
???
0.04
0.02
0
460
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480
500
520
Volum e
540
560
580
44
.
Suppose a defect rate of
p 
1
1 00 00 0
is acceptable
The corresponding Z value is -4.2648907
Therefore, the required shift in V is
.
. V0_shift  V0  Z  sV
V0_shift  533.072
Finding the D and H that will give a minimum area
for this volume yields
D = H = 8.788
and also a slightly different value for the volume standard
deviation (because the design space is not linear)
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Repeating the process, with the new volume standard
deviation gives the following final result …
For a volume defect of 1 in 100,000, manufacture the
cans with mean values for D and H of
D = H = 8.796 cm
This will result in an average cost of 3.65 cents per can
The average volume of cans will be 534.6 cubic cm
Further investigation will show that any correlation of
D and H has little effect on these results
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