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Optimum Design Problem Formulation Introduction
Optimum Design Problem Formulation
Cheng-Liang Chen
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Formulation of an optimum design problem involves involves transcribing a verbal description of a problem into a well-defined mathematical statement (Mathematical Model)
➢
A Mathematical Mathematical Model of a system is a set of mathematical (equal ualiti ities es,, in inequ equali aliti ties es,, log logica icall co condi nditi tions ons)) wh which ich relations (eq rep re pre rese sent nt an ab abst stra ract ctio ion n of th thee re real al worl rld d sys yste tem m un unde derr consideration
➢
PSE
Fundamental approaches (eg, Newton’s Law) ☞ Empirical methods (eg, empirical or black-box models) ☞ Methods based on analogy ☞
LABORATORY
Department of Chemical Engineering National TAIWAN University
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Mathematical Models can be developed using
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The correct formulation of a problem takes roughly 50% of the total effort needed to solve it
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Optimum Design Problem Formulation
Optimum Design Problem Formulation
Problem Formulation Process
Key Elements
Step 1: Project/Problem Statement Are the project goals clear ?
Step 2: Data and Information Collection
➢
Variables (continuous, integers)
➢
Parameters
➢
Constraints
Is all the information available to solve the problem ?
Step 3: Identification/Definition of Design of Design Variables What are these variables for describing the system ? How do I identify them ?
Step 4: Identifi Identificatio cation n of a Criterion (Objective Function Function)) to Be Optimized How do I know that my design is the best ?
☞
Equalities:
☞
Inequalities:
mass/energy balances, equilibrium relations, physical property calculations · · · allo al low wab able le ope opera rati ting ng re regi gime mes, s,
speci spe cifica ficati tion on of qu qual alit itie ies, s,
requirements, bounds on availabilities and demands, · · ·
(to judge whether or not a given design is better than others)
Step 5: Identification of of Constraints What restrictions do I have on my design (the system) ?
Mathematical relations (algebraic/differential, (algebraic/different ial, linear/nonlinea linear/nonlinear, r, · · · )
perfo per form rman ance ce
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Optimum Design Problem Formulation
A Two-Bar Structure Design
Model Structure
Problem Description
min f (x, y)
Objective, Performance measure
x,y
s.t.
h(x, y) =
0
Equalities constraints
g (x, y ) ≤
0
Inequalities constraints
x ∈ X ⊆
Rn Continuous variables
y ∈ Y
Problem: design a two-member bracket to support a force W without structural failure (given material) Objective: minimize its mass while also satisfying certain fabrication and space limitations Given: bracket material; σa; W, θ;
Integer or Binary variables
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A Two-Bar Structure Design
A Two-Bar Structure Design
Design Variables
Notes ➢
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First step in proper formulation: to identify Design Variables (parameters chosen to describe design of a system)
DVs for two-bar structures: cross-sectional shapes; h,s,do1 , di1 , do2 , di2 ?
DVs should be independent of each other as far as possible case (1): do, di or do, di, t and let t = 0.5(do − di) instead of d o, di, t case (2): do, r ≡ ddoi
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There is a minimum number of DVs required to formulate a design problem properly
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It is good to designate as many independent parameters as possible as DVs at the initial design problem formulation phase. Later on, some of the DVs can be always given a fixed value
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Vector of Design Variables :
x =
h s do1 di1 do2 d
=
x1 x2 x3 x4 x5
=
x
height h of the truss span s of the truss outer diameter of member 1 inner diameter of member 1 outer diameter of member 2 inner diameter of member 2
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A Two-Bar Structure Design
Design Constraints (Restrictions)
Cost (Objective) Function ➢
Cost (Objective) Function: a criterion to compare various designs (to be minimized )
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Cost (Objective) Function should be dependent on Design Variables
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Ex: Mass of a circular tube two-bar structure x1 = height h of the truss
x2 = span s of the truss
x3 = outer dia. of member 1
x4 = inner dia. of member 1
x5 = outer dia. of member 2
x6 = inner dia. of member 2
πρ 8
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Feasible (Acceptable) Design: a design that meets all requirements
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Infeasible (Unacceptable) Design
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Implicit Constraints
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Linear and Nonlinear Constraints
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Equality and Inequality Constraints
x3 2 2
f (x) = ρπ =
9
2 2 x4 2 + x25 − x26 2 x22 x23 + x25 − x24 − x26
x21 +
−
4x21 +
1/2 x2 2 2
(total mass)
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A Two-Bar Structure Design
A Two-Bar Structure Design
Design Constraints
Optimum Design Problem
−F 1 sin α + F 2 sin α = W cos θ −F 1 cos α − F 2 cos α = W sin θ =
h2 + (0.5s)2
F 1 = F 2 =
sin θ x1
sin θ x1
+ 2 coss θ − 2 coss θ
−F 1
A1
=
2W π(x23−x24 )
π 2 4 (x3 π 2 4 (x5
<
σ2 ≡
−F 2
A2
=
2W π(x25−x26 )
− x24) − x26)
min f (x) x∈Ω T x = [x1 x2 . . . x6] Ω = {x | all • are satisfied}
sin θ x1
θ + 2 cos x2
≤ σa
•
sin θ x1
θ − 2 cos x2
θ → F 2 is a tensile force 2 cos x2 F 2 −2W sin θ θ − 2 cos = π(x 2−x2) A2 x1 x2 5 6
xi ≤ xi ≤ xiu
i = 1
Find values of design variables x 1, x2, x3, x4, x5 and x6 to minimize the cost function f (x)
θ ≥ 2 cos → F 2 is a compressive force x2
σ2 ≡
IF
− W2 sinh θ − W2 sinh θ
stress σ1 ≡
IF
A1 = A2 =
⇓ F < 0 for compression
≤ σa
•
≤ σa
•
6
•
subject to all of the above-mentioned constraints
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Major Steps for Problem Formulation
Optimum Design Problem Formulation
Summary
Design of A Beer Can ➢
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Identify and define design variables x
Fabrication, handling, asthenic, shipping considerations impose following restrictions on size:
☞ ➢
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Identify cost function and develop an expression for it in terms of design variables, f (x) Identify constraints and develop expressions for them in terms of design variables ⇒ searching space Ω(x)
Diameter should be no more than 8
cm and
not be less than 3.5
cm ☞
Height should be no more than 18
cm and
no less than 8
cm
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The can is required to hold at least 400
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Design objective: to minimize total surface area of sheet metal
Formulate design problem
DVs
:
D, H
f (D, H ) = πDH + 2 π4 D 2
min f (x)
x∈Ω
π 2 D H 4
m
⇒
≥ 400
3.5 ≤ D ≤ 8
min f (x) x∈Ω x = {All Var.s} Ω = {x|All Constraints}
8 ≤ H ≤ 18 Chen CL
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Optimum Design Problem Formulation
Optimum Design Problem Formulation
Minimum Cost Cylindrical Tank Design
Saw Mill Operation
Problem: Design a minimum cost (surface area) cylindrical tank closed at both ends to contain a fixed volume of fluid V . DV: R, H
➢
Problem: A company owns two saw mills and two forests. Each forest can yield up to 200 logs/day. The cost to transport the logs is 15 cents/km/log. At least 300 logs are needed each day. Formulate the problem to minimize transportation cost each day.
A = 2πR 2 + 2πRH
Distance (km)
⇒ f (R, H ) = c(2πR 2 + 2πRH ) πR 2H = V Rmin ≤ R ≤ Rmax H min ≤ H ≤ H max
⇒
min f (x) x∈Ω x = {All Var.s} Ω = {x|All Constraints}
Mill capacity
Mill Forest 1 Forest 2 (log/day) A
24.0
20.5
240
B
17.2
18.0
300
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Optimum Design Problem Formulation
Optimum Design Problem Formulation
Saw Mill Operation
Saw Mill Operation
DVs:
➢
Constraints: x1 + x2 ≤ 240 (Mill A handling limitation)
x1 : number of logs shipped from Forest 1 to Mill A
x3 + x4 ≤ 300 (Mill B handling limitation)
x2 : number of logs shipped from Forest 2 to Mill A
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x3 : number of logs shipped from Forest 1 to Mill B
x1 + x3 ≤ 200 (Forest 1 supply limitation)
x4 : number of logs shipped from Forest 2 to Mill B
x2 + x4 ≤ 200 (Forest 2 supply limitation) x1 + x2 + x3 + x4 ≥ 300 (Market minimum demand)
Cost Function:
xi ≥ 0
f (x) = 24(0.15)x1 + 20.5(0.15)x2 + 17.2(0.15)x3 + 18(0.15)x4
i = 1, 2, 3, 4
⇒ min f (x)
x∈Ω x = {All Var.s}
= 3.6x1 + 3.075x2 + 2.58x3 + 2.7x4
Ω = {x|All Constraints} Chen CL
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Optimum Design Problem Formulation Oil Refinery Costs
Sales Prices gasoline 36/bbl crude oil #1 24/bbl ⇒ Refinery ⇒ kerosene 24/bbl crude oil #2 15/bbl fuel oil 21/bbl residual 10/bbl
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Let x1, x2 denote (bbl/day) of crude #1 and #2, respectively profit
income
Max: f (X ) = 36x3 + 24x4 + 21x5 + 10x6 − (24x1 + 15x2)− (0.5x1 + x2) processing cost
raw material cost
s.t. 0.8x1 + 0.44x2 = x 3 0.05x1 + 0.1x2 = x 4 0.1x1 + 0.36x2 = x 5 0.05x1 + 0.1x2 = x 6
gasoline (x3) kerosene (x4) fuel oil (x5) residual (x6) processing cost
vol % yield max allowable #1 #2 prod (bbl/day) 80 44 24, 000 5 10 2, 000 10 36 6, 000 5 10 − 0.5 1.0
x3 ≤ 24000 x4 ≤ 2000 x5 ≤ 6000 or Max: f (x1, x2) = 8.1x1 + 10.8x2 s.t. 0.8x1 + 0.44x2 ≤ 24000 0.05x1 + 0.1x2 ≤ 2000 0.1x1 + 0.36x2 ≤ 6000
x1 ≥ 0
x2 ≥ 0
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Optimum Design Problem Formulation Design of A Cabinet
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Formulation 1: x1 = number of C 1 to be bolted x3 = number of C 2 to be bolted x5 = number of C 3 to be bolted
Problem: A cabinet is assembled from components C 1, C 2, and C 3. Each cabinet requires eight C 1, five C 2, and fifteen C 3 components. Assembly of C 1 needs either five bolts or five rivets; C 2 six bolts or six rivets; and C 3 three bolts or three rivets. The cost of putting a bolt, including the cost of the bolt, is 0.70 for C 1, 1.00 for C 2 and 0.60 for C 3. Similarly, riveting costs are 0.60 for C 1, 0.80 for C 2 and 1.00 for C 3. A total of 100 cabinets must be assembled daily. Bolting and riveting capacities per day are 6000 and 8000, respectively. We wish to determine the number of components to be bolted and riveted to minimize the cost.
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➢ Formulation 2:
x2 = number of C 1 to be riveted x4 = number of C 2 to be riveted x6 = number of C 3 to be riveted
f (x) = 0.7(5)x1 + 0.60(5)x2 + 1.00(6)x3+ 0.80(6)x4 + 0.60(3)x5 + 1.00(3)x6 = 3.5x1 + 3.0x2 + 6.0x3 + 4.8x4 + 1.8x5 + 3.0x6 x1 + x2 = 8 × 100 (required # of C 1’s) x3 + x4 = 5 × 100 (required # of C 2’s) x5 + x6 = 15 × 100 (required # of C 3’s) 5x1 + 6x3 + 3x5 ≤ 6000
(bolting capacity)
5x2 + 6x4 + 3x6 ≤ 8000
(riveting capacity)
xi ≥ 0
i = 1, . . . , 6
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➢ Formulation 3:
x1 = total # of bolts required for all C 1;
x2 = total # of bolts required for all C 2 ;
x3 = total # of bolts required for all C 3; x4 = total # of rivets required for all C 1; x5 = total # of rivets required for all C 2; x6 = total # of rivets required for all C 3;
f (x) = 0.7x1 + 1.0x2 + 0.6x3 + 0.6x4 + 0.8x5 + 1.0x6
x1 = # of C 1’s to be bolted on 1 cabinet; x2 = # of C 1’s to be riveted on 1 cabinet; x3 = # of C 2’s to be bolted on 1 cabinet; x4 = # of C 2’s to be riveted on 1 cabinet; x5 = # of C 3’s to be bolted on 1 cabinet; x6 = # of C 3’s to be riveted on 1 cabinet;
f (x) = 100[5(0.7)x1 + 5(0.6)x2 + 6(1.0)x3+ 6(0.8)x4 + 3(0.6)x5 + 3(1.0)x6] = 350x1 + 300x2 + 600x3+ 480x4 + 180x5 + 300x6
x1 + x4 =
8 × 5 × 100 = 4000 (for C 1)
x1 + x2 = 8 (for C 1)
x2 + x5 =
5 × 6 × 100 = 3000 (for C 2)
x3 + x4 = 5 (for C 2)
x3 + x6 = 15 × 3 × 100 = 4500 (for C 3)
x5 + x6 = 15 (for C 3)
x1 + x2 + x3 ≤ 6000
(bolting capacity)
100(5x1 + 6x3 + 3x5) ≤ 6000
(bolting capacity)
x4 + x5 + x6 ≤ 8000
(riveting capacity)
100(5x2 + 6x4 + 3x6) ≤ 8000
(riveting capacity)
xi ≥ 0
i = 1 to 6
xi ≥ 0
i = 1 to 6
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Optimum Design Problem Formulation
Optimum Design Problem Formulation
Insulated Spherical Tank Design
Minimum Weight Tubular Column Design
Problem: The goal is to choose insulation thickness t to minimize the cooling cost for a spherical tank. The cooling costs include the cost of installing and running the refrigeration equipment, and the cost of installing the insulation. Assume a 10-year life, 10% annual interest rate and no salvage value.
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DV: t (meter) A = 4πr 2, t << r ⇒ insulation cost = c1At = c14πr 2t annual heat gain = (365)(24)(∆T )A Watt-hours c2 t
purchasing cost for ref. equip
= c3G
annual running cost = c4G total cost f (t) = c14πr 2t + cA2 t (c3 + uspwf (0.1, 10)c4) (365)(24)(∆T ) = at + bt t ≥ tmin uspwf : uniform series present worth factor
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➢ Formulation 1:
➢
Problem: Straight columns as structural elements are used in many civil, mechanical, aerospace, agricultural and automotive structures. The problem is to design a minimum weight tubular column of length supporting a load P without buckling or over-stressing. The column is fixed at base and free at the top (a cantilevel column). Buckling load for such a column is given as π 2EI/42. Here I is the moment of inertia for the cross-section of the column and E is the material property called modulus of elasticity (Young’s modulus). The material stress σ for the column is defined as P /A, where A is the cross-sectional area. The material allowable stress under axial load is σa, and material mass density is ρ. Formulate the design problem.
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➢ Formulation 2:
R
:
mean radius
R >> t
Ri
:
inner radius
Ro
:
outer radius
A
=
2πRt
A = π(Ro2 − Ri2)
I
=
πR 3t
I =
mass
=
ρ(A)
=
2ρπRt
P P ≤ σa = A 2πRt π 2EI π 3ER 3t Buckling load constraint: P ≤ P cr = = 2 4 42 Rmin ≤ R ≤ Rmax Material crushing const:
tmin ≤ t ≤ tmax
π 4 4 (Ro
− Ri4)
mass = ρ(A) = ρπ(Ro2 − Ri2) P P ≤ σa = A π(R2o − Ri2) 2 3 π EI π E 4 = (R − R4i ) Buckling load constraint: P ≤ 2 4 162 o Romin ≤ Ro ≤ Romax Material crushing const:
Rimin ≤ Ri ≤ Rimax Ro ≥ Ri
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Optimization Applications
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Material Balance Reconciliation
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Three measurements ⇒ find M A to minimize deviations between input and output
MB: M A + M C = M B Obj:
f (M A) = (M A + 11.1 − 92.4)2 + (M A + 10.8 − 94.3)2
error of 1st measure
error of 2nd measure
+ (M A + 11.4 − 93.8)2
⇒
error of 3rd measure
M A = 82.4
= M B − M C
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Optimization Applications
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Antoine Equation for Saturated Pressure
VLE Data Fitting ➢
VLE Data for Water-1,4-Dioxane,
Low Pressure: ⇒ yiP = γ ixiP isat ➢
Water 1,4-Dioxane
20oC ➢
ln γ 1 = A12 ln γ 2 = A21
A21x2 A12x1 +A21x2 A12x1 A12x1 +A21x2
2 2
➢
= f 1 (A12, A21)
Ai
Bi
C i
8.071 7.432
1730.630 1554.679
223.426 240.337
0. 28.1
.1 34.4
··· ···
1.0 17.5
Find A12, A21 to minimize sum of errors between P exp and P pred 11
= f 2 (A12, A21)
⇒ P = y1P + y2P = γ 1x1P 1sat + γ 2x2P 2sat = ef 1(•)x1P 1sat + ef 2(•)x2P 2sat
Bi = A i − T +C i
Experimental VLE Data x1 P (mmHg)
van Laar Model for γ i
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log P isat
f (A12, A21) =
P kexp − P kpred
k=1
2
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Optimization Applications
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Resource Limitations:
Project Selection in Manufacturing
First year expenditure: $450, 000 Second year expenditure: $400, 000
➢
A micro-electronics manufacturing facility is considering six projects to improve operations as well as profitability
Project
Description
1
2nd year expenditure
Engineering hours
Net present value
$300, 000
0
4, 000
$100, 000
$100, 000
$300, 000
7, 000
$150, 000
0
$200, 000
2, 000
$35, 000
$50, 000 $50, 000
$100, 000 $300, 000
6, 000 3, 000
$75, 000 $125, 000
$100, 000
$200, 000
600
$60, 000
modify existing production line build existing production line
2 3
automate new production line
4 5
install plating line build waste recovery plant
6
1st year expenditure
sub-contract waste disposal
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➢ Formulation: xj = 1 ⇒ project j is selected
Other Limitations:
a new or modernized production line must be provided (project 1 or 2) ☞ automation is feasible only for new line ☞ either project 5 or project 6 can be selected, but not both ➢
Objective: determine which projects maximize the net present value subject to various constraints
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Solution: x1
f = 100000x1 + 150000x2 + 35000x3 + 75000x4 + 125000x5 I
+60000x6 s.t. g1 = 300000x1 + 100000x2 + 0x3 + 50000x4 + 50000x5 +100000x6 ≤ 450000 g2 = 0x1 + 300000x2 + 200000x3 + 100000x4 + 300000x5 +200000x6 ≤ 400000 g3 = 4000x1 + 7000x2 + 2000x3 + 6000x4 + 3000x5 +600x6 ≤ 10000 g4 = x1 + x2 ≥ 1
(g4 ≥ 1)
g5 = x2 − x3 ≥ 0 g6 = x5 + x6 ≤ 1 xj ∈ {0, 1}
$10, 000
☞
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Max:
Engineering hours:
(0 ≤ g6 ≤ 1)
II
x2
x3
x4
x5
x6
.88 .12 .12 .40 1.0 0.0 1
0
0
0
1
0
f 265, 200 225, 000
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Model for Binary Distillation Design
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Superstructure
➢
Variables ☞
➢
Objective
Consider the design of a binary distillation column which separates saturated liquid feed mixture into distillate and bottom product of specified purity ☞ Objectives are the determination of the number of trays, reflux ratio, flow rates, and compositions in the distillation column that minimize the total annual cost (TAC) ☞
➢
Assumptions Equi-molar flow, constant relative volatility, total condenser, and partial boiler
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Binary variables zi denote the existence of trays in the column, zi = N (total # of trays)
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Overall Material and Component Balance
☞
Continuous variables represent the liquid flow rates Li and compositions xi, vapor flow rates V i and compositions yi, the reflux Ri and vapor boilup V Bi, and the column diameter Di
☞
T R = {1, . . . , N } , AF = {N f + 1, . . . , N } , BF = {2, . . . , Nf − 1}
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Phase Equilibrium
D + B − F = 0
αxi − yi[1 + xi(α − 1)] = 0
DxD + Bx B − F z = 0
αxB − yB [1 + xB (α − 1)] = 0 i = 2, . . . , N
➢
Total Condenser V N −
Ri − D = 0
➢
i∈AF
➢
yN − xD = 0
Partial Boiler
B +
V Bi − L2 = 0
➢
Component Balances Lixi + V iyi − Li+1xi+1 − V i−1yi−1 − RixD = 0
i ∈ AF
Lixi + V iyi − Li+1xi+1 − V i−1yi−1 − F z = 0
i = N f
Lixi + V iyi − Li+1xi+1 − V i−1yi−1 − V BiyB = 0
i ∈ BF
Equimolar Flow
i∈BF
Bx D +
V Bi yB − L2x2 = 0
i∈BF
V i − V i−1 = 0
i ∈ AF
Li − Li+1 = 0
i ∈ BF
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Diameter
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v − kv · f f ·
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Product Specifications
(ρL − ρV )/ρV = 0 4 · V Nf · M W Di2 − = 0 π · v · ρV
➢
− xD ≤ 0 xspec D ≤ 0 xB − xspec B ➢
Reflux and Holdup Constraints Ri − F max(zi − zi+1) ≤ 0
i ∈ AF
V Bi − F max(zi − zi−1) ≤ 0
i ∈ BF
In rectifying section, all trays above the tray on which the reflux enters have no liquid flows (no mass transfers) ☞ In stripping section, all trays below the tray on which the boilup enters have no vapor flows ☞ The reflux and reboiler constraints ensure that the reflux and boilup enter on one tray ☞
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➢
min
i ∈ AF
zi − zi+1 ≤ 0
i ∈ BF
Economic Objective Function
f (N,Di) = 12.3[615 + 324Di2 + 486(6 + 0.76N )Di] + 245N (0.7 + 1.5Di2)
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Standard Design Optimization Model
Cost min f (x) = f (x1, x2, · · · , xn) x
x = all continuous design variables
subject to:
y = all integer (or binary) design variables Ω
zi − zi−1 ≤ 0
Cost = β tax(C LP S ∆H vap + C CW ∆H cond)V + f (N,Di)/β pay
A Standard Optimization Problem x,y ∈Ω
Sequential Tray Constraints
hj (x) = hj (x1, x2, · · · , xn) = 0; j = 1 to p
= all constraints mentioned above
gk (x) = gk (x1, x2, · · · , xn) ≤ 0; k = 1 to m xi ≤ xi ≤ xiu; ➢
➢
i = 1 to n
f (x): objective function hj (x): equality constraints gk (x): inequality constraints x ∈ R1: single variable
x ∈ Rn: multi-variable
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Standard Design Optimization Model
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Maximization Problem Treatment
min f (x) = f (x1, x2, · · · , xn) x
subject to: hj (x) = hj (x1, x2, · · · , xn) = 0; j = 1 to p gk (x) = gk (x1, x2, · · · , xn) ≤ 0; k = 1 to m xi ≤ xi ≤ xiu;
i = 1 to n
➢
Linear Program if all f, hj , gk are linear to x
➢
Nonlinear Program if any one of f, hj , gk is nonlinear
➢
Integer Program if xi’s are integer values
➢
Mixed-Integer Program if some xi’s are integer values
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Mixed-Integer Linear Program
➢
Mixed-Integer Nonlinear Program
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max F (x) ⇓ min f (x) = −F (x)
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Constraints ➢
➢
Treatment of Greater Than Type Constraints Gk (x)
≥
0
gk (x)
=
−Gk(x) ≤ 0
Constraint Set (Feasible Region) S = {x | hj (x) = 0, j = 1, . . . , p; gk (x) ≤ 0, k = 1, . . . , m}
⇓ min f (x)
x∈S
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Active/Inactive/Violated Constraints ➢
An inequality constraint gk(x) ≤ 0 is said to be active at a design point x∗ if it is satisfied at equality, i.e. gk (x) = 0
➢
An inequality constraint gk(x) ≤ 0 is said to be inactive at a design point x∗ if it is strictly satisfied, i.e. gk (x) < 0
➢
An inequality constraint gk (x) ≤ 0 is said to be violated at a design point x∗ if its value is positive, i.e. gk (x) > 0
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Graphical Optimization Profit Maximization Problem
profit f (x) ⇒ g1 : x1 + x2 x1 x2 g2 : 28 + 14 x1 x2 g3 : 14 + 24 −x1 g4 : ⇒ x∗
A company manufactures two machines, A and B . Using available resources either 28 A or 14 B machines can be manufactured each day. The sales department can sell up to 14 A machines or 24 B machines. The shipping facility can handle no more than 16 machines each day. The company makes a profit of 400 on each A machine and 600 on each B machine. How many A and B machines should the company manufacture every day to maximize profit ?
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% ProfitMax.m % for Profit Maximization Problem % [x1,x2] = meshgrid(0:0.5:25, 0:0.5:25); prof = 400*x1+600*x2; g1 = x1+x2-16; g2 = x1/28+x2/14-1; g3 = x1/14+x2/24-1; g4 = -x1; g5 = -x2; cla reset; axis([0 25 0 25]) set(gca,’XTick’,[0,5,10,15,20,25],... ’YTick’,[0,5,10,15,20,25],... ’XTickLabel’,{’0’,’5’,’10’,’15’,’20’,’25’},... ’YTickLabel’,{’0’,’5’,’10’,’15’,’20’,’25’},... ’FontSize’,16,’LineWidth’,3)
x1 = # of A machine manufactured each day x2 = # of B machine manufactured each day = = ≤ ≤ ≤ ≤ =
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400x1 + 600x2 −(400x1 + 600x2) 16 (shipping and handling constraint) 1 (manufacturing constraint) 1 (limitation on sales department) 0; g5 : − x2 ≤ 0 (4, 12); f (x∗) = − 8800
Chen CL
xLabel(’\bf x_1’,’FontSize’,16,’Color’,[0,0,0]) yLabel(’\bf x_2’,’FontSize’,16,’Color’,[0,0,0]) title(’\bf Profit Maximization Problem’,... ’FontSize’,16,’Color’,[0,0,0]) hold on cv = [0 0]; profv = [2000, 5000, 8800, 12000, 15000]; const1 = contour(x1,x2,g1,cv,’r’,’LineWidth’,3); const2 = contour(x1,x2,g2,cv,’m’,’LineWidth’,3); const3 = contour(x1,x2,g3,cv,’b’,’LineWidth’,3); const4 = contour(x1,x2,g4,cv,’k’,’LineWidth’,3); const5 = contour(x1,x2,g5,cv,’k’,’LineWidth’,3); profs = contour(x1,x2,prof,profv,’k-’); clabel(profs)
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Chen CL
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text( 1.1, 15,’\bf g_1’,’FontSize’,12,’Color’,[1,0,0]) text( 15, 7,’\bf g_2’,’FontSize’,12,’Color’,[1,0,1]) text( 1.1, 22,’\bf g_3’,’FontSize’,12,’Color’,[0,0,1]) text(-0.05,20.75,’\bf g_4’,’FontSize’,12) text( 18, 0.85,’\bf g_5’,’FontSize’,12) plot(4,12,’g’,’Marker’,’o’,’MarkerSize’,6) text( 1, 5,’\it\bf Feasible Region’,’FontSize’,14) text( 15, 20,’\it\bf Infeasible Region’,’FontSize’,14) hold off
Chen CL
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Graphical Optimization Design Problem with Multiple Solutions min f (x) = −x1 − 0.5x2 subject to: 2x1 + 3x2 ≤ 12 2x1 + x2 ≤ 8
−x1 ≤ 0 −x2 ≤ 0
Chen CL
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Graphical Optimization
Design Problem with Unbounded Solution
Minimum Tubular Weight Column
2x1 − x2 ≥ 0
−2x1 + 3x2 ≤ 6 x1 ≥ 0; x2 ≥ 0
⇓ min f (x) = −x1 + 2x2 s.t. :
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Graphical Optimization max F (x) = x1 − 2x2 s.t. :
Chen CL
− 2x1 + x2 ≤ 0 −2x1 + 3x2 ≤ 6 −x1 ≤ 0 −x2 ≤ 0
P = 10 MN, E = 207 GPa, ρ = 7833 kg/m3, = 5 m, σa = 248 MPa f (R, t) = 2πρRt = 2π(7833)(5)Rt = 2.4608 × 105Rt kg g1 (R, t) =
P 2πRt
g2 (R, t) = P −
6
×10 ) − σa = 10(1.0 − 248(1.0 × 106) ≤ 0 2πRt
π3 ER3t 42
g3 (R, t) = −R ≤ 0,
= 10(1.0 × 106) − π
3
(207.0×109 )R3 t 4(5)(5)
g4(R, t) = − t ≤ 0
≤0
Chen CL
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Thank You for Your Attention Questions Are Welcome
