Ch13 - NLP,DP,GP2005
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Transcript of Ch13 - NLP,DP,GP2005
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13.1 Introduction to Nonlinear Programming
Most real-life situations are more accurately depicted by nonlinear models than by linear models.
Nonlinear models formulated in this chapter fall into three broad categories: Dynamic programming models (DP). Goal programming models (GP). General nonlinear programming models
(NLP).
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13.2 Dynamic Programming Dynamic programming problems can be thought
of as multistage problems, in which decisions are made “in sequence.”
Known applications of dynamic programming in business: Resource allocation, Equipment replacement, Production and inventory control, Reliability.
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13.2 Dynamic Programming At stage n a system is found to be in a certain
state. A decision made at stage n takes the system to
stage n+1, and leaves it in a new state for a stage-state related cost/profit.
The decision maker’s challenge is to find a set of optimal decisions for the entire process.
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13.5 Goal Programming In real life decision situations, virtually all
problems have several objectives. When objectives are conflicting, the
optimal decision is not obvious. Goal programming is an approach that
seeks to simultaneously take into account several objectives.
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Goals are prioritized in some sense, and their level of aspiration is stated.
An optimal solution is attained when all the goals are reached as close as possible to their aspiration level, while satisfying a set of constraints.
There are two types of goal programming models: Nonpreemtive goal programming - no goal is pre-determined to
dominate any other goal. Preemtive goal programming - goals are assigned different priority
levels. Level 1 goal dominates level 2 goal, and so on.
13.5 Goal Programming
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13.6 Nonlinear Programming
A nonlinear programming problem (NLP) is one in which the objective function, F, and or one or more constraint functions, Gi, possess nonlinear terms.
There is no a universal algorithm that can find the optimal solution to every NLP.
One class of NLPs “Convex Programming Problems” can be solved by algorithms that are guaranteed to converge to the optimal solution.
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The objective is to maximize a concave function or to minimize a convex function.
The set of constraints form a convex set.
Properties of Convex Programming Problems
9
A smooth function (no sharp points, no discontinuities)
One global maximum (minimum). A line drawn between any two points on the
curve of the function will lie below (above) the curve or on the curve.
A One Variable Concave (Convex) Function
X
A Concave function
A convex function
X
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If a straight line that joins any two points in the set lies within the set, then the set is called a convex set.
Convex set Non-convex set
Convex Sets
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In a NLP model all the constraints are of the “less than or equal to” form Gi(X) B.
If all the functions Gi are convex, the set of constraints forms a convex set.
NLP and Convex Sets
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Dynamic ProgrammingUS Department of Labour
Goal ProgrammingAdvertisement example
Non Linear ProgrammingToshi CameraPBI
Examples for NLP Problems
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The U.S. Department of Labor has made up to $5 million available to the city of Houston for job creation.
Requests for funding are to be prepared by four Requests for funding are to be prepared by four departments.departments.
The Department of Labor would like to allocate funds to maximize the number of jobs created.
THE U.S. DEPARTMENT OF LABOR
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Data
Department Cost $ millions Jobs CreatedHousinng 4 225Employment 1 45
2 1253 190
Highway 50 jobs per $1 million fundingLaw Enforcement 2 75
3 1554 220
Estimated new jobs created for proposal
THE U.S. DEPARTMENT OF LABOR
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SOLUTION
The U.S. Department of Labor wants to:
Maximize the total number of new jobs.
Limit funding to $5 million.
Fund at most one proposal from each department
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Notation Dj = the amount allocated to department j, where j is:1 - housing, 2 - Employment, 3 - Highway, 4 - Law Enforcement.
R(Dj) = the number of new jobs created by funding department j with $Dj million.
The model
Max R1(D1) + R2(D2) + R3(D3) + R4(D4) STD1 + D2 + D3 + D4 <= 5D1, D2, D3 , D4 >= 0
Non-linear functions
SOLUTION
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Define Fj(Xj) as the maximum number of new jobs created by departments (stages) j, j+1,…, 4, given that there is $Xj million in funding available for departments j (state) through 4.
The Backward Dynamic Programming
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Stage 4: The Law Enforcement,(LED) Start with the last stage j=4 (Law Enforcement
department, LED).Allocate to this department funds that maximize the number of new jobs created
Obviously, the optimal solution for the last department is to use all the amount available at this stage).
The optimal solution for the last stage is called “The boundary condition
The Backward Dynamic Programming
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Stage 4: The Law Enforcement table
Available Optimal Funding Maximum jobsFunding For Stage 4 F4(X4)
0 0 01 0 02 2 753 3 1554 4 2205 5 220
States
Cost JobsLaw Enforcement 2 75
3 1554 220
Recall
SOLUTION
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Stage 3: The Highway Department, (HD) At this stage we consider the funding of both the
Highway department and the Law Enforcement department.
For a given amount of funds available for funding of these two departments, the decision regarding the funds allocated to the HD affects the funds available for the LED).
SOLUTION
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Available Possible Remaining Max. New Jobs Optimal F3(X3)Funding Funding Funds for When Allocatingfor Stage 3,4 for Stage 3 Stage 4 D3 to Stage 3 Optimal D3
(X3) (D3) (X3-D3) R3(D3)+F4(X3-D3)0 0 0 0 + 0 F3(0) = 0; D3(0) = 01 0 1 0+0
1 0 50+0 = 50 F3(1) = 50; D3 = 1 2 0 2 0+75 = 75
1 1 50+0 = 502 0 100+0 F3(3) = 100; D3 = 2
3 0 3 0+155 = 1551 2 50+75 = 1252 1 100+0 = 1003 0 150+0 = 150 F3(3) = 155; D3 = 0
Stage 3: The Highway Department table
SOLUTION
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Available Possible Remaining Max. New Jobs Optimal F3(X3)Funding Funding Funds for When Allocatingfor Stage 3,4 for Stage 3 Stage 4 D3 to Stage 3 Optimal D3
(X3) (D3) (X3-D3) R3(D3)+F4(X3-D3)4 0 4 0+220 = 220
1 3 50+155 = 2052 2 100+75 = 1753 1 150+ 0 = 1504 0 200+ 0 = 220 F3(4) = 220; D3 = 0
5 0 5 0+220 = 2201 4 50+220 = 2702 3 100+155 = 2553 2 150+75 = 2254 1 200+0 = 2005 0 250+0 = 250 F3(5) = 270; D3 = 1
SOLUTION
Stage 3: The Highway Department table
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Stage 2: The Employment Department, (ED) At this stage we consider the funding of both the
Employment Department and the previous departments (HD and the LED)
For a given state (the amount of funds available for funding of these three departments), the decision regarding the funds allocated to the ED affects the funds available for the HD and LED. (the state at the stage j=3)
SOLUTION
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Funding Funding Funds for When Allocatingfor Stage 2,3,4 for Stage 2 Stage 3,4 D2 to Stage 2 Optimal D2
(X2) (D2) (X2-D2) R2(D2)+F3(X2-X2)0 0 0 0 + 0 F2(0) = 0; D2 = 01 0 1 0+50 = 50
1 0 45+0 = 45 F2(1) = 50; D2 = 0 2 0 2 0+100 = 100
1 1 45+50 = 952 0 125+0 = 125 F2(2) = 125; D2 = 2
3 0 3 0+155 = 1551 2 45+100 = 1452 1 125+50 = 1753 0 190+0 = 190 F2(3) = 190; D2 = 3
4 0 4 0+220 = 2201 3 45+155 = 1952 2 125+100 = 2253 1 190+50 = 240 F2(4) = 240; D2 = 3
5 0 5 0+270 = 2701 4 45+220 = 2652 3 125+155 = 2803 2 190+100 = 290 F2(5) = 290; D2 = 3
SOLUTION
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Stage 1: The Housing Department
At this stage we consider the funding for the Housing Department and all previous departments.
Note that at stage 1 we know there is exactly $5 million to allocate (X1 = 5).
SOLUTION
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Stage 1: The Housing Department submitted only one proposal of $4 million, thus (D1 = 0, 4).
Available Possible Remaining Max. New Jobs Optimal F1(X1)Funding Funding Funds for When Allocatingfor Stage 1,2,3,4 for Stage 1 Stage 2,3,4 D3 to Stage 3 Optimal D1
(X1) (D1) (X1-D1) R1(D1)+F2(X1-D1)5 0 5 0+290 = 290
4 1 225+50 = 275 F1(5) = 290; D1 = 0
Do not fund Fund the project
SOLUTION
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Stage State0 1 2 3 4 5
1 D1 = 02 D2 = 0 D2 = 0 D2 = 2 D2 = 3 D2 = 3 D2 = 33 D3 = 0 D3 = 1 D3 = 2 D3 = 0 D3 = 0 D3 = 14 D4 = 0 D4 = 1 D4 = 2 D4 = 3 D4 = 4 D4 = 5
In summary, the optimal funds allocation to maximize the number of jobs created is:
Housing = $0Employment = $3 millionHighways = $2 millionLaw Enforcement = $0
Maximum number of jobs created = 290
SOLUTION
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Notation Description ExampleStage Variable j A decision point The four departmentsState Variable Xj The amount of resource Funds left to allocate
left to be allocated to departmentsDecision Variable Dj A possible decision Funds that could have
at stage j been give to department jStage Return Values Rj(Dj) The stage return for Number of jobs created
making decision Dj at department j if funding DjOptimal Value Fj(Xj) The best cumulative Maximum number of new
return for stages j and jobs created for dept j,…,4.remaining stages at state Xj.
Boundary Condition F(XN) A set of optimal values The new jobs created fromfor the last stage (N) funding Dept 4 with $X4 million.
Optimal Solution F1(T) The best cumulative The maximum total number Value return of new jobs given $Tmillion to
allocate to all departments
Components of Dynamic Programming
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Bellman’s principle of optimality. From a given state at a given stage, the optimal solution for the
remainder of the process is independent of any previous decisions made to that point.
Dynamic programming is a recursive process. The following recursive relationship describes the process for the
U.S. Department of Labor problem. Define Fj(Xj) as the maximum number of new jobs created by
departments (stages) j, j+1,…, 4, given that there is $Xj million in funding available for departments j (state) through 4.
Fj(Xj) = Max {(Rj(Xj) + Fj+1(Xj - Dj)} all feasible Xj
Dynamic Recursive Relationship
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The form of the recursion relation differs from problem to problem, but the general idea is the same: Do the best you can for the remaining stages with the remaining resources available.
Dynamic Recursive Relationship
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If these were constraints rather than goals we would have:3000X1 + 800X2 + 250X3 25,0001000X1 + 500X2 + 200X3 30,000
X1 10 No feasible solution exists that satisfies all
the constraints. When these constraints are simply goals they
are to be reached as close as possible.
An Advertisement Example(Goal Programming)
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Detrimental variablesUi = the amount by which the left hand side
falls short of (under) its right had side value.
Ei = the amount by which the left side exceeds its right had side value.
The goal equations3000X1 + 800X2 + 250X3 + U1 – E1 = 25,0001000X1 + 500X2 + 200X3 + U2 – E2 = 30,000
X1 + U3 – E3 = 10
An Advertisement Example
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The objective is to minimize the penalty of not meeting the goals, represented by the detrimental variables
E1, U2, U3.
An Advertisement Example
25,000 30,000 10
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The penalties are estimated to be as follows:
Each extra dollar spent on advertisement above $25,000 cost the company $1.
There is a loss of $5 to the company for each customer not being reached, below the goal of 30,000.
Each television spot below 10 is worth 100 times each dollar over budget.
An Advertisement Example
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It is assumed that no advantage is gained by overachieving a goal.
Minimize 1E1 + 5U2 + 100U3
s.t.3000X1 + 800X2 + 250X3 + U1 – E1 = 25,0001000X1 + 500X2 + 200X3 + U2 – E2 = 30,000
X1 + U3 – E3 = 10
All variables are non-negative.
An Advertisement Example – The goal programming model
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Non Preemptive Goal Prog Minimize (the total weighted deviation)
Subject to • Expresion of the goal Equation• Functional Constraint• Non Negativity Constraint
Minimize E1 + 5U2 +100 U3 ST
3000X1 + 800 X2 + 250 X3 + U1 – E1 =25000 1000X1 + 500 X2 + 200 X3 +U2 – E2 =
30000 X1 +U3 – E3=10
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Preemptive Goal Prog 7 Goals 2 in priority level 1 3 in priority level 2 2 in priority level 3 Detrimental deviation
E1, U2,E3,U4,E5,U6 and E7 Weights
W1, W2 Level 1 Goals W3,W4,W5 Level 2 Goals W6, W7 Level 3 Goals
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Preemptive Goal Prog(Problem Formulation)
Priority 1 Minimize W1E1 + W2U2ST Goal Equation
Functional ConstraintNon Negativity Constraint
Priority 2 Minimize W3E3 + W4U4 + W5E5ST Goal Equation
Functional ConstraintW1E1 + W2U2 = V1 (new Constraint)Non Negativity Constraint
Priority 2 Minimize W6U6 + W7E7 ST Goal Equation
Functional ConstraintW1E1 + W2U2 = V1 W3E3 + W4U4 + W5E5 = V2 (New Constraint)Non Negativity Constraint
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The NECC is planning next month production of its two bicycles B2 and S10.
Data Both models use the same seats and tires. 2000 seats are available; 2400 tires are available. 1000 gear assembly are available (used only in the
S10 model). Production time per unit: 2 hours for B2; 3 hours for
S10. Profit: $40 for each B2; 10$ for each S10.
PREEMTIVE GOAL PROGRAMMING - New England Cycle Company
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Priority 1: Fulfill a contract for 400 B2 bicycles to be delivered next month.
NECC – Prioritized Goals
Priority 4: At least 200 tires left
over at the end of the month.
At least 100 gear assemblies left over at the end of the month.
Priority 2: Produce at least 1000 total bicycles during the month.
Priority 3: Achieve at least
$100,000 profit for the month.
Use no more than 1600 labor-hours during the month.
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Management wants to determine the production schedule that best meets its prioritized schedule.
New England Cycle Company Example
45
NECC - SOLUTION Decision variables
X1 = The number of B2s to be produced next month
X2 = The number of S10s to be produced next month Functional / nonnegativity constraints
0X,XTires2400X2X2
assembliesGear1000X
Seats2000XX2
21
21
2
21
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Goal constraints Priority 1 (goal 1): Production of at least 400 B2sPriority 1 (goal 1): Production of at least 400 B2s
XX11 + U+ U11 - E - E11 = 400 = 400
Priority 2 (goal 2): Production of at least 1000 total cyclesPriority 2 (goal 2): Production of at least 1000 total cyclesXX11 + X + X22 + U+ U22 - E - E22 = 1000 = 1000
Priority 3 (goal 3) Profit of at least $100,000Priority 3 (goal 3) Profit of at least $100,000 .04X .04X11 + .10X + .10X22 + U+ U33 - E - E33 = 100 (in $1000) = 100 (in $1000)Priority 3 (goal 4) Use a maximum of 1600 labor hoursPriority 3 (goal 4) Use a maximum of 1600 labor hours 2X 2X11 + 3X + 3X22 + U+ U44 - E - E44 = 1600 = 1600
Priority 4 (goal 5) At least 200 leftover tiresPriority 4 (goal 5) At least 200 leftover tires 2X 2X11 + 2X + 2X22 + U+ U55 - E - E55 = 2200 = 2200Priority 4 (goal 6) At least 100 leftover gear assemblyPriority 4 (goal 6) At least 100 leftover gear assemblyXX22 + U+ U66 - E - E66 = 900 = 900
NECC - SOLUTION
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Priority level objectives
Priority 1: Underachieving a production of 400 B2s:Minimize U1
Priority 2: Underachieving a total production of 1000: Minimize U2
NECC - SOLUTION
48
Priority level objectives Priority 3: Underachieving a $100,000 profit Using
more than 1600 labor-hoursMinimize 30U3 + E4
NECC - SOLUTION
Each $1,000 short of the $100,000 goal is considered 30 times as important as utilizing an extra labor-hour.
49
Priority level objectives Priority 4: Using more than 2200 tires
Using more than 900 gear assembliesMinimize E5 + 2E6
NECC - SOLUTION
Each leftover gear assembly is deemed twice as important as leftover tire.
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Solve the linear goal programming for priority 1 objective, under the set of regular constraints and goal constraint as shown below Minimize U1
ST
NECC - The solution procedure
X1 + U1 - E1 = 400Tires2400X2X2
Gear1000XSeats2000XX2
21
2
21
X1 = 400, thus
U1 = 0, and priority 1 goal is fully achieved.
51
Solve the linear goal programming for priority 2 level objective, under the set of original constraints plus the constraint X1 400 (maintain the level of achievement of the priority 1 goal).
Minimize U2
ST
NECC - The solution procedure
Every point that satisfies X1 + X2 1000 yields U2 = 0, and therefore, priority 2 goal is fully achieved.
Tires2400X2X2
Gear1000XSeats2000XX2
21
2
21
X1 400X1 + X2 + U2 - E2 = 1000
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Solve the linear goal programming for priority 3 level objective, under the set of original constraints plus the constraint X1 400 (maintain the level of achievement of the priority 1 goal), plus the constraint X1 + X2 1000 (maintain the level of achievement of the priority 2 goal).
• Every point in the range X1 = 400 and 600 X2 800 is optimal for this model; 30U3 + E4 = 1720 is the level of achievement for the priority 3 goal.
NECC - The solution procedure
53
Solve the linear goal programming for priority 4 level objective, and notice that after the previous step the feasible region is reduced to a segment of a straight line between the points (400,600) and (400,800).
• X1 =400; 600 X2 700 and E5 + 2E6 = 0
NECC - The solution procedure
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In summary NECC should produce 400 B2 model Between 600 and 700 S10 model
NECC - Solution Summary
55
Unconstraint Models• Single Variable
• Necessary & Sufficient Conditions (df/dx = 0 , D2f/dx2 <> 0))• Multi Variable
• Partial Derivative Constraint Models
• Single Variable• Shadow price• Optimality Criteria
• Multi Variable• Kuhn Tucker Condition
Global View on NLP problems
56
13.7 Unconstrained Nonlinear Programming
One-variable unconstrained problems are demonstrated by the Toshi Camera problem.
The inverse relationships between demand for an item and its value (price) are utilized in this problem.
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TOSHI CAMERA (example for unconstrained NLP) Toshi camera of Japan has just
developed a new product, the Zoomcam.
It is believed that demand for the initial product will be linearly related to the price.
Price EstimatedP ($) Demand (X)100 350,000150 300,000200 250,000250 200,000300 150,000350 100,000
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Unit production cost is estimated to be $50. What is the production quantity that
maximizes the total profit from the initial production run?
SOLUTION
Total profit = Revenue - Production cost
F(X) = PX - 50X
TOSHI CAMERA
59
From the Price / Demand table it can be verified that P = 450 - .001X
The Profit function becomesF(X) = (450 - .001X)X = 400X - .001X2
This is a concave function.400,0000
TOSHI CAMERA
60
To obtain an optimal solution (maximum profit), two conditions must be satisfied: A necessary condition dF/dX = 0 A sufficient condition d2F/dX2 < 0.
The necessary condition is satisfied at:dF/dX = 400 - 2(.001)X = 0; X = 200,000.
The sufficient condition is satisfied since d2F/dX2 = -.002.
The optimal solution: Produce 200,000 cameras. The profit is F(200,000) = $40,000,000.
TOSHI CAMERA
61
If a function is known to be concave or convex at all points, the following condition is both a necessary and sufficient condition for optimality:
The point X* gives the maximum value for a concave function, or the minimum value for a convex function, F(X), if at X*
dF/dX = 0
Optimal solutions for concave/convex functions with one variable
62
Determining whether or not a multivariate function is concave or convex requires analysis of the second derivatives of the function.
A point X* is optimal for a concave (convex) function if all its partial derivatives are equal to zero at X*.
For example, in the three variable case:
FX
FX
FX
11
22
33
0 0 0 ; ; ;
Optimal solutions for concave/convex functions with more than one variable
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13.8 Constrained Nonlinear Programming Problems – one variable
The feasible region for a one variable problem is a segment on a straight line (X a or X b).
When the objective function is nonlinear the optimal solution must not be at an extreme point.
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TOSHI CAMERA - revisited
Toshi Camera needs to determine the optimal production level from among the following three alternatives: 150,000 X 300,000 50,000 X 175,000 150,000 X 350,000
654000
X 250,000X 350,000
X* = 250,000
40004000
Maximize F(X) = 400X - .001X2
X 150,000X 300,000
150
X 50,000X 175,000
50 175X*=200,000 X* = 175,000
The objective function does not change:
TOSHI CAMERA – solution
66
Constrained Nonlinear Programming Problems – m variables
mn21
2n21
1n21
n21
B)X..., ,X ,Gm(X...
B)X..., ,X ,G2(XB)X..., ,X ,G1(X
ST)X..., ,X,X(FMaximize
Let us define Y1, Y2, …,Ym as the instantaneous improvementin the value of F for one unitincrease in B1, B2, …Bm
respectively.
67
This is a set of “necessary conditions” for optimality of most nonlinear problems.
If the problem is convex, the K-T conditions are also sufficient for a point X* to be optimal.
mm
211
mm2
11
1
mm2211
m21
XGmY...X
2G2YX1GYXn
F
.
..
XGmY...X
2GYX1GYX
F.4
0SY...,,0SY,0SY.30Y..., , Y,Y .2
.feasibleis*X.1
2
S1, S2, …,Smare defined as the slack variablesin each constraint.
Kuhn-Tucker optimality conditions
68
PBI INDUSTRIES
PBI wants to determine an optimal production schedule for its two CD players during the month of April.
Data Unit production cost for the portable CD player = $50.
Unit production cost for the deluxe table player = $90.
There is additional “intermix” cost of $0.01(the number of portable CD’s)(the number of deluxe CD’s).
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Forecasts indicate that unit selling price for each CD player is related to the number of units sold as follows: Portable CD player unit price = 150 - .01X1 Deluxe CD player unit price = 350 - .02X2
PBI INDUSTRIES
70
Resource usage Each portable CD player uses 1 unit of a particular
electrical component, and .1 labor hour. Each deluxe CD player uses 2 units of the
electrical component, and .3 labor hour. Resource availabilty
10,000 units of the electrical component units; 1,500 labor hours.
PBI INDUSTRIES
71
PBI INDUSTRIES – SOLUTION
Decision variablesX1 - the number of portable CD players to produceX2 - the number of deluxe CD players to produce
The modelMax (150-.
X X X X XST
X X
X
01X1)X1+ (350-.02X2)X2 - 50X1- 90X2-.01X1X2 =
-.01X1
.1X1 +.3X2 1,500 - X1
2
. .
,
02 2 01 1 2 100 1 260 2
1 2 2 10 000
02 0
2
Production cannot be negative
Resource constraints
72
For a point X1, X2 to be optimal, the K-T conditions require that:Y1S1 = 0; Y2S2 = 0; Y3S3 = 0; Y4S4 = 0, and
)1(Y)0(Y)3(.Y)2(Y2602X04.X01.or
XGYX
GYXGYX
GYXF
)0(Y)1(Y)1(.Y)1(Y100X01.X02.or
XG4YX
GYXGYX
GYXF
43211
2
44
2
33
2
22
2
11
2
432121
1
4
1
33
1
22
1
11
1
PBI INDUSTRIES – SOLUTION
73
Finding an optimal production plan. Assume X1>0 and X2>0.
• The assumption implies S3>0 and S4>0.• Thus, Y3 = 0 and Y4 = 0.
Add the assumption that S1 = 0 and S2 = 0.• From the first two constraints we have X1 = 0 and
X2 = 5000.X1= 0
A contradictionX1>0
AS A RESULT THE SECOND ASSUMPTION CANNOT BE TRUE
PBI INDUSTRIES – SOLUTION
74
Change the second assumption. Assume that S1 = 0 and S2 > 0.
• As before, from the first assumption S3 > 0 and S4 > 0.• Thus, Y3 = 0 and Y4 = 0.• From the second assumption Y2 = 0.• Substituting the values of all the Ys in the partial derivative
equations we get the following two equations:-.02X1 - .01X2 + 100 = Y1-.01X1 - .04X2 + 260 = 2Y1
• Also, since S1 = 0 (by the second assumption)X1 + 2X2 = 10,000
• Solving the set of three equations in three unknowns we get:
PBI INDUSTRIES – SOLUTION
75
X1 = 1000, X2 = 4500, Y1 = 35 This solution is a feasible point (check the constraints).
• X1 and X2 are positive.• 1X1 + 2X2 <= 10,000 [1000 + 2(4500) = 10,000]• .1X1 + .3X2 <= 1,500 [.1(1000) + .3(4500) = 1450]
This problem represents a convex program since• It can be shown that the objective function F is concave.• All the constraints are linear, thus, form a convex set.
The K-T conditions yielded an optimal solution
PBI INDUSTRIES – SOLUTION