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MESM543

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MESM543

Operations Research OPTIMIZATION

Operations Research (OR) is the field of how to form mathematical models of complex management decision problems and how to analyze the models to gain insight about possible solutions.

Applications grouped by type of organizational client

- Business
- Government and Non-Profit
- Health Care
- Military

Applications grouped by function

- Planning, Strategic Decision-Making
- Production
- Distribution, Logistics, Transportation
- Supply Chain Management
- Marketing Engineering
- Financial Engineering

Build Your Knowledge to increase your success in practice

- Linear Programming
- Non-linear Programming
- Dynamic Programming
- Markov Decision Processes
- Multiple Criteria Decision Making
- Queuing Models
- General Simulation

Decisions

Production System

7

Production Systems

8

OR Process

Assessment

Real world problem

Real world solution

Abstraction

Interpretation

Analysis

Model

Model solution

9

Operations Research Techniques

- Linear Programming
- Dynamic Programming
- Integer Programming
- Nonlinear programming
- Goal Programming
- Network Programming

10

26.10.2014

The principal phases for implementing OR in practice

- Definition of the problem
- Decision alternative definition
- Objective determination
- Operational limitations specification

- Construction of the model
Entails translating the problem definition into mathematical relationships

- Solution of the model
It entails the use of well-defined optimization algoritms

- Validation of the model
Does the model provide a reasonable prediction of the system’s behavior?

- Implementation of the solution
Translation of the model’s results into operating instructions

Linear Programming (LP)

An important topic of Deterministic Operations Research

- Agenda
- Modeling problems
- Examples of models and some classical problems
- Graphical interpretation of LP
- Solving LP by Simplex using MS Excel
- Some theoretical ideas behind LP and Simplex

Example 1: Product Mix Problem

Fertilizer manufacturing company, 2 types of fertilizer

Type A: high phosphorus

Type B: low phosphorus

Product Mix Problem: Modeling

Step 1. The decision variables

Daily production ofType A: x tons

Type B: y tons

Step 2. The objective function (maximize profit)

z =15x + 10y

Product Mix Problem: Modeling..

Step 3. The constraints

Limited supply of raw materials per day:

Urea:2x + y ≤ 1500

Potash:x + y ≤ 1200

Rock Phosphate:x ≤ 500

Product Mix Problem: Complete model

Maximizez( x, y) = 15 x + 10y

subject to2x + y ≤ 1500

x + y ≤ 1200

x ≤ 500

x ≥ 0,

y ≥ 0

Interesting Aspects:

Linearity, Inequalities

Feasible solutions: (0, 0), (1, 1), …

Infeasible solutions: (600, 500), …

Background: Petroleum Refinery

Example 2. Blending Problem

Three types of petrol (minimum Octane rating: 85, 90, 95)

Four types of oils (Octane rating: 68, 86, 91, 99)

Blending oils petrol, with proportional Octane rating

Objective: best product mix [how much of each petrol, oil to sell]

Example 2. Blending Problem, the data

68x11 + 86x21 + 91x31 + 99x41

Its Octane Rating:

≥ 95,

x11 + x21 + x31 + x41

Blending Problem: Modeling

Step 1. The decision variables

xij = barrels/day of oil i( i = 1, 2, 3, or 4)

to make petrol j (j = 1, 2, or 3)

Total premium petrol per day = x11 + x21 + x31 + x41

68x11 + 86x21 + 91x31 + 99x41 - 95(x11 + x21 + x31 + x41) ≥ 0.

premium

45.15(x11 + x21 + x31 + x41) +

42.95(x12 + x22 + x32 + x42) +

40.99(x13 + x23 + x33 + x43) +

36.85 (4000 – (x11 + x12 + x13)) +

36.85 (5050 – (x21 + x22 + x23)) +

38.95 (7100 –(x31 + x32 + x33)) +

38.95 (4300 – (x41 + x42 + x43))

super

regular

Oil 1

Oil 2

Oil 3

Oil 4

Blending Problem: Modeling..

Step 2.The objective function

Maximize profit Maximize revenue

Blending Problem: Modeling...

Step 3.The constraints

(a) The OcR constraints:

68x11 + 86x21 + 91x31 + 99x41 - 95(x11 + x21 + x31 + x41) ≥ 0

68x12 + 86x22 + 91x32 + 99x42 - 90(x12 + x22 + x32 + x42) ≥ 0

68x13 + 86x23 + 91x33 + 99x43 - 85(x13 + x23 + x33 + x43) ≥ 0

Blending Problem: Modeling....

Step 3.The constraints..

(b) Can’t use more oil than we have:

x11 + x12 + x13 ≤ 4000

x21 + x22 + x23 ≤ 5050

x31 + x32 + x33 ≤ 7100

x41 + x42 + x43 ≤ 4300

Blending Problem: Modeling…..

Step 3.The constraints...

(c) The demand constraints:

x11 + x21 + x31 + x41 ≤ 10,000

x13 + x23 + x33 + x43 ≥ 15,000

(d) Allowed values of variables

xij ≥ 0 for i = 1, 2, 3, 4, and j = 1, 2, 3.

Blending Problem: complete model

Maximize:

45.15(x11 + x21 + x31 + x41) + 42.95(x12 + x22 + x32 + x42) + 40.99(x13 + x23 + x33 + x43) +

36.85(4000 – (x11 + x12 + x13)) + 36.85 (5050 – (x21 + x22 + x23)) +

38.95 (7100 –(x31 + x32 + x33)) + 38.95 (4300 – (x41 + x42 + x43))

Subject to:

68x11 + 86x21 + 91x31 + 99x41 - 95(x11 + x21 + x31 + x41) ≥ 0

68x12 + 86x22 + 91x32 + 99x42 - 90(x12 + x22 + x32 + x42) ≥ 0

68x13 + 86x23 + 91x33 + 99x43 - 85(x13 + x23 + x33 + x43) ≥ 0

x11 + x12 + x13 ≤ 4000

x21 + x22 + x23 ≤ 5050

x31 + x32 + x33 ≤ 7100

x41 + x42 + x43 ≤ 4300

x11 + x21 + x31 + x41 ≤ 10,000

x13 + x23 + x33 + x43 ≥ 15,000

xij ≥ 0 for I = 1, 2, 3, 4, and j = 1, 2, 3.

Octane

rating

Supply

Demand

Example 3: Transportation problem

Background:

Company has several factories (sinks), and

several suppliers (sources)

Objective: Minimize the cost of transportation

Example 3. Transportation problem, the data

Transportation problem: the model

Step 1. The decision variables

xij= amount of ore shipped from mine i to plant j per day.

Step 2:The objective function

Minimize the transportation costs:

Minimize: 11x11 + 8x12 + 2x13 + 7x21 + 5x22 + 4x23

Transportation problem: the model..

Step 3.The constraints

(a) Shipment from each mine less than daily production

x11 + x12 + x13 ≤ 800[capacity of mine 1]

x21 + x22 + x23 ≤ 300[capacity of mine 2]

(b) Demand of each plant must be met

x11 +x21≥ 400[demand at plant 1]

x12 + x22≥ 500[demand at plant 2]

x13+ x23≥ 200[demand at plant 3]

(c) Decision variables can’t be negative

xij ≥ 0, for all i= 1, 2, j = 1, 2, 3.

Transportation problem: historical note

Kantorovich in USSR in the 1930’s,

Koopmans in 1940’s

Dantzig in 1950’s Simplex method

Kantorovich and Koopmans, Nobel prize (Economics) in 1975

Point in a 1D space:x = c

c

0

y

3

2x+3y = 9

2x+3y = 3

2

2x+3y = 6

2x+3y = 0

1

x

3

4.5

(0,0)

1.5

The Geometry of Linear Programs

Line in 2D:ax + by = c

z

1

Plane

x + y + z = 1

Plane:

x + y + z = 2

z

Plane:

x + y + z = 0

1

x

1

1

1

y

y

1

x

The Geometry of Linear Programs

Plane in 3D:ax + by + cz = d

y

3

2x+3y > 6

2

2x+3y < 6

1

x

4.5

3

1.5

(0,0)

2x+3y = 6

The Geometry of Linear Programs

Hyper-plane in n-Dimensions: a1x1 + a2x2 + … + anxn = c

??

2-D Half spaces:

y

y ≥0

1500

1000

500

x ≥0

x

500

Feasible set

1000

(0,0)

1500

x + y ≤ 1200

x ≤ 500

2x+y ≤ 1500

The Geometry of LP: Product Mix revisited

maxz( x, y) = 15 x + 10y

ST2x + y ≤ 1500

x + y ≤ 1200

x ≤ 500

x ≥ 0,

y ≥ 0

y

y ≥0

15x + 10y = 5000

1500

15x + 10y = 13,500

1000

15x + 10y = 0

(300, 900)

500

x ≥0

x

500

Feasible set

1000

(0,0)

1500

x + y ≤ 1200

x ≤ 500

2x+y ≤ 1500

The Geometry of LP: Product Mix revisited

maxz( x, y) = 15 x + 10y

ST2x + y ≤ 1500

x + y ≤ 1200

x ≤ 500

x ≥ 0,

y ≥ 0

Try point: x = 0, y = 0:

Summary

1. LP formulations are very common in modern industry

2. Beautiful connection between Algebra and Geometry

3. Geometry not useful for > 3 variables

4. Practical problems: 1000’s of variables (see next slide)

5. Need Algebraic method !

Dynamic Programming – Recursive formula (forward, Backward)

A state k is the information that is needed at any stage to make an optimal decision.

A problem can be decomposed into stages, where each stage j represents an operation.

Define fj(xj) as fj(xj) = maximum return for stage i given state yj

Forward Recursive:

Forward vs. Backward Recursion

Backward recursion formula:

Budgeting Problem –FirstExample1.m

A corporation has $5 million to allocate to its three plants for possible expansion Each plant has submitted a number of proposals on how it intends to spend the money.

Each proposal gives the cost of the expansion (c) and the total revenue expected (r). The following table gives the proposals generated:

Problem formulation

Define state x[6]=(0 1 2 3 4 5); available resources

Define Stage j[3]=(1 2 3); The operation or processing of the problem takes place

Define Policy or Proposal of problem m[4]=(1 2 3 4)

%Dynamic Programming 1st example, CMU

% F. Nagi 5/2/12

clear

clc

Alloc = 6; %Allocation 7 million +1 for index '0'

r=[0 5 6 0;

0 8 9 12;

0 4 0 0]; %revenue of Proposal

c=[0 1 2 0;

0 2 3 4;

0 1 0 0]; %cost of proposal

x =[0 ;

1 ;

2 ;

3 ;

4 ;

5 ];

f2(1:Alloc,1:Alloc)=0; %initialze Answer function f2

x=x';

for j=1 %state

for m=1:4 %proposal

for k=1:Alloc %state

if c(j,m)<=x(k)

f1(m,k)=r(j,m);

end

end

end

end

f1=max(f1)

for j =2:3 %stage

for m=1:4 %proposal

for k=1:Alloc %states

if c(j,m)<=x(k)

f1(j,k)=max(r(j,m)+f1(j-1,x(k) c(j,m)+1), f2(j,k))

f2(j,k)=f1(j,k);

end

end

end

end

f2=max(f1)

WorkStation Production Problem

Ans, f1 =

0 5 6 6 6 6

0 5 8 13 14 17

0 5 9 13 17 18

Ans, f2 =

0 5 9 13 17 18

Problem formulation

Resource Alloc = 5; %Allocation 4 cycle time +1 for index '0'

r=[0 31 62 0 0;

0 47 0 0 0;

0 14 28 42 56]; %revenue of Products

c=[0 2 4 0 0;

0 3 0 0 0;

0 1 2 3 4]; %cycle time of products

x =[0 ;

1 ;

2 ;

3 ;

4 ]; % cycle time

The cycle of a Genetic Algorithms is presented below

Each cycle in Genetic Algorithms produces a new generation of possible solutions for a given problem

Genetic Algorithm Steps

- The elements of the population are encoded into bit-strings, called chromosomes.

- The performance of the strings, often called fitness, is then evaluated with the help of some functions f(x) , representing the constraints of the problem.

- The crossover operation that recombines the bits (genes) of each two selected strings (chromosomes) . The crossover helps to span over the solution space

Genetic Algorithm Steps

In Mutation the bits at one or more randomly selected positions of the chromosomes are altered. The mutation process helps to overcome trapping at local maxima

Mutation of a chromosome at

the 5th bit position.

Example: The Genetic Algorithms cycle is illustrated in this example for maximizing a function f(x) = x2in the interval 0 = x = 31.

In this example the fitness function is f (x) itself. Starts with 4 initial strings. The fitness value of the strings and the percentage fitness of the total are estimated in Table A.

Example

Table A

- Since fitness of the second string is large, we select 2 copies of the second string and one each for the first and fourth string in the mating pool. The selection of the partners in the mating pool is also done randomly

- Here in table B, we selected partner of string 1 to be the 2-nd string and partner of 4-th string to be the 2nd string. The crossover points for the first-second and second-fourth strings have been selected after 0-th and 2-nd bit positions respectively in table B.

Table B

Example

The second generation of the population without mutation in the first generation is presented in table C.

Table C

gaex.m % Example in >>gatool

0-4, five bits

25 =32 (0-31)

Fitness f(x)

x -variable

GA- Problems

- Use Gatool and maximize the quadratic equation
- f(x) = x2 +4x-1 within the range
- −5 ≤x≤0.

2. Use Gatool and maximize the function

f(x1, x2, x3)= −5 sin(x1) sin(x2) sin(x3) + (- sin(5x1) sin(5x2)sin(x3))

where 0 <= xi <= pi, for 1 <= i <= 3.

3. Create a “gatool” to minimize the function

f(x) = cos(2x) within the range 0≤x≤3.14