Optimum Design Problem Formulation
Cheng-Liang Chen
PSE
LABORATORY
Department of Chemical Engineering National TAIWAN University
Chen CL
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Optimum Design Problem Formulation Introduction
Formul rmulat atio ion n of an opti optimu mum m desi design gn probl roblem em invo involv lves es tran transc scri ribi bing ng a verb verbal al desc descri ript ptio ion n of a probl roblem em into into a well-defined mathematical statement
The correct formulation of a problem takes roughly 50% of the total effort needed to solve it
Chen CL
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Optimum Design Problem Formulation Introduction
Formul rmulat atio ion n of an opti optimu mum m desi design gn probl roblem em invo involv lves es tran transc scri ribi bing ng a verb verbal al desc descri ript ptio ion n of a probl roblem em into into a well-defined mathematical statement
The correct formulation of a problem takes roughly 50% of the total effort needed to solve it
Chen CL
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Optimum Design Problem Formulation Problem Formulation Process Step Step 1: Projec Project/P t/Prob roblem lem Stat Statemen ementt Are the project goals clear ?
Step Step 2: Data Data and and Info Inform rmat atio ion n Coll Collec ecti tion on Is all the information available to solve the problem ?
Step Step 3: Identifi Identificat cation ion/De /Defini finitio tion n of Design Varia Variables bles What are these variables for describing the system ? How do I identify them ?
Step Step 4: Iden Identi tific ficat atio ion n of a Crite Criteri rion on (Obje (Object ctiv ive e Funct unctio ion) n) to Be Optimized How do I know that my design is the best ? (to judge whether or not a given design is better than others)
Step Step 5: Identifi Identificat cation ion of of Constr Constraint aintss What restrictions do I have on my design (the system) ?
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A Two-Bar Structure Design Problem Description
Problem:
desi desiggn a twoo-m membe emberr
bracket to support a force W W without structural failure
(given material) material)
Objective: minimize its mass while also satisfying certain fabrication and space limitations
Given: bracket material; σa; W, θ;
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A Two-Bar Structure Design Design Variables
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 ?
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Notes:
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 do, di , t case (2): do, r ≡
di do
There is a minimum number of DVs required to formulate a design problem properly 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 Vector of Design Variables: x
x
=
h s d d d o1 i1
o2
di2
x x x x x 1 2
=
3 4 5
x6
=
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 Cost (Objective) Function
Cost (Objective) Function: a criterion to compare various designs (to be minimized )
Cost (Objective) Function should be dependent on Design Variables
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
f (x) = =
ρπ + x + − − 4x + x x + x − x − x (total mass) x3 2 2
πρ 8
2 1
x4 2 2
2 2
2 3
x5 2 2
2 5
x6 2 2
2 4
2 6
2 1
1/2 x2 2 2
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Design Constraints (Restrictions)
Feasible (Acceptable) Design: a design that meets all requirements
Infeasible (Unacceptable) Design
Implicit Constraints
Linear and Nonlinear Constraints
Equality and Inequality Constraints
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A Two-Bar Structure Design Design Constraints −F 1 sin α + F 2 sin α = W cos θ −F 1 cos α − F 2 cos α = W sin θ
= h
2
+ (0.5s)2
F 1 = F 2 =
⇓ F < 0 for compression
− W2 sinh θ − W2 sinh θ
stress σ1 ≡
IF
sin θ x1
sin θ x1
+ 2 coss θ − 2 coss θ
−F 1
A1
=
2W π (x23−x24 )
sin θ x1
θ + 2 cos x2
≤ σa
π 2 4 (x3 π 2 4 (x5
•
θ ≥ 2 cos → F 2 is a compressive force x2
σ2 ≡
IF
A1 = A2 =
−F 2
A2
=
2W π (x25−x26 )
sin θ x1
≤ σa
•
≤ σa
•
θ − 2 cos x2
θ → F 2 is a tensile force < 2 cos x2
σ2 ≡
F 2 A2
=
−2W
π (x25−x26 )
sin θ x1
θ − 2 cos x2
xi ≤ xi ≤ xiu
i = 1, . . . , 6
•
− x24) − x26)
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A Two-Bar Structure Design Optimum Design Problem
min f (x) x∈Ω
x
= [x1 x2 . . . x6]T
Ω = {x | all • are satisfied} Find values of design variables x1, x2, x3, x4, x5 and x6 to minimize the cost function f (x) subject to all of the above-mentioned constraints
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Major Steps for Problem Formulation Summary
Identify and define design variables
Identify the 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)
x
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Optimum Design Problem Formulation Design of A Beer Can
Fabrication, handling, asthenic, shipping considerations impose following restrictions on size:
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
The can is required to hold at least 400
m
Design objective: to minimize total surface area of sheet metal
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DVs
:
D, H
f (D, H ) = πDH + 2 π4 D2 π 2 4 D H
≥ 400
3.5 ≤ D ≤ 8 8 ≤ H ≤ 18
min f (x)
⇒
x∈Ω x
= {All Var.s}
Ω = {x| All Constraints}
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Optimum Design Problem Formulation Minimum Cost Cylindrical Tank Design
Problem: Design a minimum cost (surface area) cylindrical tank closed at both ends to contain a fixed volume of fluid V . DV: R, H A = 2πR 2 + 2πRH
⇒ 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
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Optimum Design Problem Formulation Saw Mill Operation
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.
Distance (km)
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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DVs:
x1 : number of logs shipped from Forest 1 to Mill A x2 : number of logs shipped from Forest 2 to Mill A x3 : number of logs shipped from Forest 1 to Mill B x4 : number of logs shipped from Forest 2 to Mill B
Cost Function:
f (x) = 24(0.15)x1 + 20.5(0.15)x2 + 17.2(0.15)x3 + 18(0.15)x4 = 3.5x1 + 3.075x2 + 2.58x3 + 2.7x4
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Constraints: x1 + x2 ≤ 240 (Mill A handling limitation) x3 + x4 ≤ 300 (Mill B handling limitation) x1 + x3 ≤ 200 (Forest 1 supply limitation) x2 + x4 ≤ 200 (Forest 2 supply limitation) x1 + x2 + x3 + x4 ≥ 300 (Market minimum demand) xi ≥ 0
i = 1, 2, 3, 4
min f (x)
⇒
x∈Ω x
= {All Var.s}
Ω = {x| All Constraints}
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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
gasoline (x3) kerosene (x4) fuel oil (x5) residual (x6) processing cost
vol % yield #1 #2 80 44 5 10 10 36 5 10 0.5 1.0
max allowable prod (bbl/day)
24, 000 2, 000 6, 000 −
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Let x1, x2 denote (bbl/day) of crude #1 and #2, respectively profit
income
Max: f ( ) = 36x + 24x + 21x + 10x − (24x + 15x )− (0.5x + x ) X
3
4
1
5
2
6
1
2
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
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
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 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
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) = 0.75(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 2:
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 + x4 =
8 × 5 × 100 = 4000 (for C 1)
x2 + x5 =
5 × 6 × 100 = 3000 (for C 2)
x3 + x6 = 15 × 3 × 100 = 4500 (for C 3) x1 + x2 + x3 ≤ 6000 (bolting capacity) x4 + x5 + x6 ≤ 8000 (riveting capacity) xi ≥ 0
i = 1 to 6
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Formulation 3:
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 + x2 = 8 (for C 1) x3 + x4 = 5 (for C 2) x5 + x6 = 15 (for C 3) 100(5x1 + 6x3 + 3x5) ≤ 6000 (bolting capacity) 100(5x2 + 6x4 + 3x6) ≤ 8000 (riveting capacity) xi ≥ 0
i = 1 to 6
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Optimum Design Problem Formulation Insulated Spherical Tank 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.
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 annual running cost
= c3 G = c4 G
total cost f (t) = c14πr 2t + cA2t (c3 + uspwf (0.1, 10)c4) (365)(24)(∆T ) = at + bt t ≥ tmin uspwf : uniform series present worth factor
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Optimum Design Problem Formulation Minimum Weight Tubular Column Design
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 π2EI/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 1: R
:
mean radius
R >> t A
=
2πRt
I
=
πR 3t
mass
=
ρ(A)
=
2ρπRt
P P Material crushing const: ≤ σa = 2πRt A π 2EI π 3ER3t Buckling load constraint: P ≤ P cr = = 2 4 42 Rmin ≤ R ≤ Rmax tmin ≤ t ≤ tmax
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Formulation 2: Ri
:
inner radius
Ro
:
outer radius
A = π(Ro2 − Ri2) I =
π 4 (R o 4
− Ri4)
mass = ρ(A) = ρπ(Ro2 − Ri2) P P Material crushing const: ≤ σa = 2 2 A π(Ro − Ri ) π 2EI π 3E 4 4 − Buckling load constraint: P ≤ = (R R ) o i 2 2 4 16 Romin ≤ Ro ≤ Romax Rimin ≤ Ri ≤ Rimax Ro ≥ Ri
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Optimization Applications 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
+ (M + 11.4 − 93.8) error of 1st measure A
error of 3rd measure
⇒
M A = 82.4 = M B − M C
2
error of 2nd measure
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Optimization Applications VLE Data Fitting
VLE Data for Water-1,4-Dioxane, 20oC Low Pressure: ⇒ yiP = γ ixiP isat
van Laar Model for γ i ln γ 1 = A12 ln γ 2 = A21
A21 x2 A12x1 +A21 x2 A12 x1 A12x1 +A21 x2
2 2
= f 1 (A12, A21) = f 2 (A12, A21)
⇒ P = y1P + y2P = γ 1x1P 1sat + γ 2x2P 2sat = ef 1(•)x1P 1sat + ef 2(•)x2P 2sat
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i Antoine Equation for Saturated Pressure log P isat = Ai − T B +C i
Water 1,4-Dioxane
Ai
Bi
C i
8.071 7.432
1730.630 1554.679
223.426 240.337 x1
0. 28.1
.1 34.4
··· ···
1.0 17.5
Experimental VLE Data
Find A12, A21 to minimize sum of errors between P exp and P pred
P (mmHg)
11
f (A12, A21
)= P
exp k
k=1
−
P pred k
2
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Optimization Applications Project Selection in Manufacturing
A micro-electronics manufacturing facility is considering six projects to improve operations as well as profitability
Project
Description
1
modify existing production line
2
build existing production line automate new production line install plating line build waste recovery plant sub-contract waste disposal
3 4 5 6
1st year expenditure
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
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Resource Limitations: First year expenditure: $450, 000 Second year expenditure: $400, 000 Engineering hours:
$10, 000
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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Formulation: x j = 1 ⇒ project j is selected
Max:
f = 100000x1 + 150000x2 + 35000x3 + 75000x4 + 125000x5 +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}
(0 ≤ g 6 ≤ 1)
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Solution: x1
x2
x3
x4
x5
x6
f
I .88 .12 .12 .40 1.0 0.0 265, 200 II
1
0
0
0
1
0
225, 000
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Standard Design Optimization Model
min f (x) = f (x1, x2, · · · , xn) x
subject to: h j (x) = h j (x1, x2, · · · , xn) = 0; j = 1 to p gk (x) = gk (x1, x2, · · · , xn)≤0; xi ≤ xi ≤ xiu ;
f (x): objective function hj (x): equality constraints gk (x): inequality constraints x ∈ R1: single variable
x
∈ Rn: multi-variable
k = 1 to m i = 1 to n
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Linear Program if all f, hj , gk are linear to
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
Mixed-Integer Linear Program
Mixed-Integer Nonlinear Program
x
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Maximization Problem Treatment
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 | h j (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 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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x1 = # of A machine manufactured each day x2 = # of B machine manufactured each day profit = 400x1 + 600x2 ⇒ f (x) = −(400x1 + 600x2) g1 : x1 + x2 ≤ 16 (shipping and handling constraint) g2 : x281 + x142 ≤ 1 (manufacturing constraint) g3 : x141 + x242 ≤ 1 (limitation on sales department) g4 : −x1 ≤ 0; g5 : − x2 ≤ 0 ∗ ⇒ x = (4, 12); f (x∗) = − 8800
41
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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)
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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)
43
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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
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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
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Graphical Optimization Design Problem with Unbounded Solution
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max F (x) = x1 − 2x2 s.t. : 2x1 − x2 ≥ 0
−2x1 + 3x2 ≤ 6 x1 ≥ 0; x2 ≥ 0
⇓ min f (x) = −x1 + 2x2 s.t. :
− 2x1 + x2 ≤ 0 −2x1 + 3x2 ≤ 6 −x1 ≤ 0 −x2 ≤ 0
