4. Robust Design
• Real-world designs are subject to variations
• Variation may arise from manufacturing processes, material
properties, changing operating conditions, etc.
• Variation lead to poor performance, product failure, and customer
dissatisfaction.
• In optimal design problems, variations invariably reduce the feasible
region. Since optimal designs are normally defined by binding
constraints, variation may cause optimal designs to become
infeasible.
• A robust design is a design that can withstand variation.
• Feasibility robustness means the constraints will remain feasible
when subjected to variation.
5. Tolerance Box Approach
• Consider the design optimization problem:
min
𝒙
𝑓 𝒙 , subject to 𝑔𝑖 𝒙 ≤ 𝑏𝑖, 𝑖 = 1, … , 𝑚
• Using worst case tolerances in design variables, we may define a
tolerance box, where a robust design will have the entire box in the
feasible region
6. Transmitted Variation Approach
• Consider the design optimization problem:
min
𝒙
𝑓 𝒙, 𝒑 , subject to 𝑔𝑖 𝒙, 𝒑 ≤ 𝑏𝑖, 𝑖 = 1, … , 𝑚
where 𝒙 is a vector of design variables, and 𝒑 is a parameter vector
• Let Δ𝒙, Δ𝒑, Δ𝒃 denote the variation in 𝒙, 𝒑, 𝑎𝑛𝑑 𝒃 about nominal
values; then variation in the constraint functions is computed as:
Δ𝑔𝑖 =
𝜕𝑔𝑖
𝜕𝑥𝑗
Δ𝑥𝑗
𝑛
𝑗=1 +
𝜕𝑔𝑖
𝜕𝑝𝑗
Δ𝑝𝑗
𝑚
𝑗=1
• The total variation is given as: Δ𝑖 = Δ𝑔𝑖 + Δ𝑏𝑖
• A robust design solves the following optimization problem:
min
𝒙
𝑓 𝒙, 𝒑 , subject to 𝑔𝑖 𝒙, 𝒑 ≤ 𝑏𝑖 − Δ𝑖, 𝑖 = 1, … , 𝑚
i.e., the feasible region is reduced by the total variation in constraints
7. Transmitted Variation Approach
• Assuming the robust optimum is close to the nominal optimum, it
may be solved as:
– Compute the nominal optimum point
– Reduce the rhs by the transmitted variation
– Re-compute the optimum
8. Design Example: Symmetric Two-Bar Truss
Problem: design a symmetrical two-bar truss of minimum mass to
support a fixed load 𝑃. The truss has height 𝐻, and span 𝐵.
Design variables: diameter 𝑑 , thickness (𝑡), height 𝐻
𝑙 = (𝐵/2)2+𝐻2, 𝐴 = 𝜋𝑑𝑡
Total weight: 𝑊 = 2𝜌𝑙𝐴,
Constraints:
Axial stress: 𝜎 =
𝑃𝑙
2𝜋𝑑𝑡𝐻
≤ 𝜎𝑎
Buckling stress: 𝜎𝑏 =
𝜋2 𝑑2+𝑡2 𝐸
8𝑙2 ≤ 𝜎𝑎
Deflection: 𝜀 =
𝑃𝑙3
2𝜋𝑑𝑡𝐻2𝐸
≤ 𝜀𝑚𝑎𝑥
9. Design Example: Symmetric Two-Bar Truss
The design optimization problem is defined as:
Objective: min
d,H
2𝜋𝑑𝑡𝜌 (𝐵/2)2+𝐻2
Subject to:
𝜎𝑏
𝜎
− 1 ≤ 0,
𝜎
𝜎𝑎
− 1 ≤ 0,
𝜀
𝜀𝑚𝑎𝑥
− 1 ≤ 0,
For a particular problem, let:
𝑃 = 66 𝑘𝑖𝑝𝑠; 𝐵 = 60 𝑖𝑛; 𝑡 = 0.15; 𝜌 = 0.3
𝑙𝑏
𝑖𝑛3
; 𝐸 = 30 × 106
𝑙𝑏
𝑖𝑛2
;
𝜎𝑎 = 1 × 105 𝑝𝑠𝑖; 𝜀𝑚𝑎𝑥 = 0.25 𝑖𝑛
The optimum design, obtained via SA, is given as:
𝐻 =29.9010 in, 𝑑 =1.8631 in, 𝑡 =0.0799 in; 𝑓 =11.8801 lbs
The stress and elongation at the optimum point are:
𝜎 = 99.96 𝑘𝑖𝑝𝑠, 𝜎𝑏 = 71.74 𝑘𝑖𝑝𝑠, 𝜀 = 0.2 𝑖𝑛
10. Design Example: Symmetric Two-bar Truss
• Consider the following variations in design parameters
Load: Δ𝑃 = 2 𝑘𝑖𝑝𝑠
Span: Δ𝐵 = 1.0 𝑖𝑛
• The transmitted variations to the constraints are computed as:
Δ𝜎 =
𝜕𝜎
𝜕𝑃
Δ𝑃 +
𝜕𝜎
𝜕𝑙
𝜕𝑙
𝜕𝐵
Δ𝐵
Δ𝜎𝑏
=
𝜕𝜎𝑏
𝜕𝑃
Δ𝑃 +
𝜕𝜎𝑏
𝜕𝑙
𝜕𝑙
𝜕𝐵
Δ𝐵
Δ𝜀 =
𝜕𝜀
𝜕𝑃
Δ𝑃 +
𝜕𝜀
𝜕𝑙
𝜕𝑙
𝜕𝐵
Δ𝐵
Then Δ𝜎 = 3.865 𝑘𝑖𝑝𝑠; Δ𝜎𝑏
= −1.196 𝑘𝑖𝑝𝑠; Δ𝜀 = 0.011 𝑖𝑛
11. Design Example: Symmetric Two-bar Truss
• The robust design optimization problem is defined as:
Objective: min
d,H
2𝜋𝑑𝑡𝜌 (𝐵/2)2+𝐻2
Subject to: 𝜎 + Δ𝜎 − 𝜎𝑎 ≤ 0,
𝜎𝑏 + Δ𝜎 + Δ𝜎𝑏 − 𝜎 ≤ 0,
𝜀 + Δ𝜀 − 𝜀𝑚𝑎𝑥 ≤ 0,
• The robust optimum design obtained via SA is given as:
𝐻 =30.537 in, 𝑑 =1.931 in, 𝑡 =0.0792 in; 𝑓 =12.34 lbs
Compare with nominal optimum design:
𝐻 =29.9010 in, 𝑑 =1.8631 in, 𝑡 =0.0799 in; 𝑓 =11.88 lbs
15. Statistical Variation
• Assume that design variables and parameters are independent
random variables, normally distributed about the mean, i.e.,
𝑋𝑖~𝑁(𝜇𝑖, 𝜎𝑖)
• Then the transmitted variation is computed as:
𝜎𝑔𝑖
2
=
𝜕𝑔𝑖
𝜕𝑥𝑗
𝜎𝑥𝑗
2
𝑛
𝑗=1 +
𝜕𝑔𝑖
𝜕𝑝𝑗
𝜎𝑝𝑗
𝑚
𝑗=1
2
• The total constraint variance (including rhs variability) is:
𝜎𝑖
2
= 𝜎𝑔𝑖
2
+ 𝜎𝑏𝑖
2
• For robust design, constraints are modified as:
𝑔𝑖 ≤ 𝑏𝑖 − 𝑘𝜎𝑖
2
where a value of 𝑘 = 3 results in 99.865% feasibility
16. Design Example: Symmetrical Three-bar Truss
• Consider the optimum design problem for three-bar truss
Objective: min
d,H
2𝜋𝑑𝑡𝜌 (𝐵/2)2+𝐻2
Subject to:
𝜎𝑏
𝜎
− 1 ≤ 0,
𝜎
𝜎𝑎
− 1 ≤ 0,
𝜀
𝜀𝑚𝑎𝑥
− 1 ≤ 0,
• The nominal design is given as:
𝐻 =29.9010 in, 𝑑 =1.8631 in, 𝑡 =0.0799 in; 𝑓 =11.88 lbs
• Assume the following distribution for load 𝑃 and span 𝐵
𝑃~𝑁 66,0.667 𝑘𝑖𝑝𝑠, 𝐵~𝑁 60,0.333 𝑖𝑛
19. Design Example: Symmetric Two-bar Truss
• A comparison of robust design methods
• Notes
– These results were obtained via simulated annealing
– Worst case design using transmitted variation is quite conservative
– The statistical variation design is less conservative, however
guarantees 99.865% feasibility
Parameter Nominal
design
Transmitted
variation
Statistical
variation
H (in) 29.9 30.54 29.68
D (in) 1.86 1.93 1.93
T (in) 0.08 0.079 0.08
W (lbs) 11.88 12.34 12.25
20. Multi-Objective Optimization
• Practical optimization problems possess multiple objectives:
– For example, we want to minimize production and maintenance
costs, maximize safety, aesthetic appeal, space usage, etc.
– Suppose we desire to minimize weight as well as deflection at a
particular location. These two objectives are competing.
21. Multi-Objective Optimization
• A multi-objective optimization problem is defined in terms of a
vector of objectives 𝐹(𝑥) that is the subject to inequality and/or
equality constraints and/or bounds on design variables.
Objectives: 𝐹 𝑥 = 𝐹1 𝑥 , 𝐹2 𝑥 , … , 𝐹𝑀 𝑥
Subject to: 𝐺𝑖 𝑥 ≤ 0, 𝑖 = 1, … , 𝑘; 𝐺𝑖 𝑥 = 0, 𝑖 = 𝑘 + 1, … , 𝑚;
𝑥𝑗
𝐿
≤ 𝑥𝑗 < 𝑥𝑗
𝑈
, 𝑗 = 1, … , 𝑛.
• As there is no unique solution, the concept of Pareto optimality is
used to characterize the objectives.
At the optimal point an improvement in one of the objectives is
accompanied by an equivalent degradation in another.
22. Multi-Objective Optimization
• A general goal in multi-objective optimization is construction of
Pareto optima, i.e., a series of non-dominated solutions.
23. Multi-Objective Optimization Methods
• Weighted sum and weighted metric approaches
• Multi-objective linear programming
• Multi-objective combinatorial optimization
• Goal programming approaches (minimize deviation from goal)
• Evolutionary algorithms (EA are naturally suited to MOOPs)
– Multi-objective genetic algorithm
– Multi-objective ant colony optimization
– Multi-objective particle swarm optimization
• Archived simulated annealing
– Non-dominated solutions found during SA are archived
• Game theoretic approaches (multi-player competing games)
24. Design Example: Symmetric Two-Bar Truss
Problem: design a symmetrical two-bar truss of minimum mass to
support a fixed load 𝑃. The truss has height 𝐻, and span 𝐵.
Design variables: diameter 𝑑 , height 𝐻
Total weight: 𝑊 = 2𝜌𝑙𝐴, where 𝑙 = (𝐵/2)2+𝐻2, 𝐴 = 𝜋𝑑𝑡
Dual Objective: Minimize weight and end deflection
Constraints:
Axial stress: 𝜎 =
𝑃𝑙
2𝜋𝑑𝑡𝐻
≤ 𝜎𝑎
Buckling stress: 𝜎𝑏 =
𝜋2 𝑑2+𝑡2 𝐸
8𝑙2 ≤ 𝜎𝑎
Deflection: 𝜀 =
𝑃𝑙3
2𝜋𝑑𝑡𝐻2𝐸
≤ 𝜀𝑚𝑎𝑥
28. MOOP Example: CNC Lathe Cutting Process
• Problem: Find the Pareto front for CNC lathed cutting, where the
objectives are: 1) minimize process time; 2) maximize tool life
Design variables: cutting speed 𝑣
𝑚
𝑚𝑖𝑛
, feed 𝑓
𝑚𝑚
𝑟𝑒𝑣
, depth 𝑎 [𝑚𝑚]
Process time is given as: 𝜏 = 𝜏𝑠 +
𝑉
𝑀
1 +
𝜏𝑇𝐶
𝑇
+ 𝜏0
where 𝜏𝑠 is setup time, 𝑉 is material removed, 𝑀 is removal rate, 𝑇 is
tool life, and 𝜏0 is inactive time
Material removal rate is: 𝑀 = 1000𝑣𝑓𝑎
Used tool life is: 𝜉 =
𝑉
𝑀𝑇
× 100%
• Empirical models for tool life and cutting force are obtained as:
𝑇 =
5.48×109
𝑣3.46𝑓0.696𝑎0.46 ; 𝐹𝑐 = 6.56 × 103 𝑓0.917𝑎1.1
𝑣0.286
29. MOOP Example: CNC Lathe Cutting Process
• The multi-objective optimization problem is formulated as:
Objective 𝐹1: min
𝑣,𝑓,𝑎
𝜏
Objective 𝐹2: max
𝑣,𝑓,𝑎
𝜉
Subject to: 𝑔1:
100𝑃
𝜂𝑃𝑚𝑜𝑡
− 1 ≤ 0
𝑔2:
𝐹𝑐
𝐹𝑐𝑚𝑎𝑥
− 1 ≤ 0
𝑔3:
𝑅
𝑅𝑚𝑎𝑥
− 1 ≤ 0
where 𝑅 =
125𝑓2
𝑟𝐸
, 𝑃 =
𝐹𝑐𝑣
60,000
, 𝑃𝑚𝑜𝑡 is motor power, 𝜂 is efficiency
Let 𝑃𝑚𝑜𝑡 = 10𝑘𝑊, 𝜂 = 0.75, 𝑟𝐸 = 0.8 𝑚𝑚, 𝑅𝑚𝑎𝑥 = 50𝜇𝑚, 𝐹𝑐𝑚𝑎𝑥
= 5𝑘𝑁
Let 200 ≤ 𝑣 ≤ 400, 0.15 ≤ 𝑓 ≤ 0.55, 1.5 ≤ 𝑎 ≤ 6
30. MOOP Example: CNC Lathe Cutting Process
%cnc lathe cutting example (Ref: Deb, Optimization for Engineering Design)
%objectives: 1. minimize operation time; 2. max tool life
%variables: cutting speed(v), feed(f), depth of cut(a)
xl=[200,.15,2]; xu=[400,.55,6]; %variable bounds
x0=(xl+xu)/2; %trial design [v,f,a]
Pmax=10e3; %[W]
Fcmax=5000; %[N] max force
Rmax=50; %[mu-m] roughness
rn=0.8; %[mm] tool radius
eta=.75; %efficiency
V=219912; %[mm3] material removed
ts=.15; %[min] setup time
tc=.20; %[min] tool change time
ti=.05; %[min] inactive time
T=@(x) 5.48e9/prod(x.^[3.46,.696,.46]); %tool life
Fc=@(x) 6560*prod(x.^[-.286,.917,1.1]); %cutting force
P=@(x) Fc(x)*x(1)/60e3; %power used
M=@(x) 1000*prod(x); %material removal rate
Tp=@(x) ts+ti+V*(1+tc/T(x))/M(x); %process time
TL=@(x) 100*V/(M(x)*T(x)); %used tool life
R=@(x) 125*x(2)^2/rn; %surface roughness
fmo=@(x) [Tp(x);TL(x)]; %objective
g=@(x) [100*P(x)/(eta*Pmax)-1; Fc(x)/Fcmax-1; R(x)/Rmax-1]; %constraints
31. MOOP Example: CNC Lathe Cutting Process
• Use MATLAB multi-objective GA solver
Opt.solver='gamultiobj';
Opt.nvars=3;
Opt.fitnessfcn=fmo;
Opt.options =
optimoptions('gamultiobj','PlotFcn',@gaplotpareto);
Opt.nonlcon=@(x) deal(g(x),[]);
Opt.x0=x0;
Opt.lb=xl;
Opt.ub=xu;
[x,fval]=gamultiobj(Opt)
34. MOOP Example: Minimum Thrust Design
• Problem: minimize engine thrust requirement for a business jet
• Background: engine thrust requirements are dictated by the
minimum thrust required during:
– Take off
– Climb
– Cruise
– Sustained turn
– Service ceiling
