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Lecture 1. Pages In Book:1, 16, 23-39 Operations Research = Management Science The scientific approach to management decision making. The science of better decision making. 1. Decisions 2. Constraints 3. Objective(s). Section 1.6. Deterministic Models – all data are known

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Lecture 1

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### Lecture 1

Pages In Book:1, 16, 23-39

Operations Research = Management Science

The scientific approach to management decision making.

The science of better decision making.

1. Decisions

2. Constraints

3. Objective(s)

EMIS 8360

### Section 1.6

Deterministic Models – all data are known

Stochastic Models – quantities are known only by probability distributions

Stochastic models are much more difficult to solve!

(Example: point-to-point demand for service in a telecommunication data network)

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### Chapter 2

Deterministic Optimization Models equal Mathematical Programs

Composed Of The Following:

1. Subscripts

2. Constants and Sets

3. Decision Variables

4. Constraints

5. Objective Function

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### Example 2.1 – Page 24

Subscripts (we only have one for this example)

j – denotes the source of crude petroleum

j = 1 implies Saudi Arabia

j = 2 implies Venezuela

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### Constants

Conversion From Crude To Products

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### Demand – Must Be Met Each Day

Units are barrels/day

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### Decision Variables & Constraints

X1 – denotes the number of barrels of Saudi crude processed each day

X2 – denotes the number of barrels of Venezuela crude processed each day

(Demand On Gasoline)

0.3X1 + 0.4X2> 2000

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### Constraints Continued

(Demand On Jet Fuel)

0.4X1 + 0.2X2 > 1500

(Demand On Lubricants)

0.2X1 + 0.3X2> 500

(Max Supply From Saudi Arabia)

X1< 9000

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### Constraints Continued

(Max Supply From Venezuela)

X2< 6000

(Non-negativity)

X1> 0

X2 > 0

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### Objective Function

(Minimize Cost)

Minimize 20X1 + 15X2

Units = (\$/barrel)(barrels)

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### The Model

Minimize 20X1 + 15X2

Subject To

0.3X1 + 0.4X2> 2000

0.4X1 + 0.2X2> 1500

0.2X1 + 0.3X2> 6000

0 < X1< 9000

0 < X2< 6000

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### The Graph

X2 (000s)

0.3X1 + 0.4X2 = 2000

(6666.7,0)

(0,5000)

0.3X1 + 0.4X2> 2000

8

6

4

2

X1 (000s)

2

4

6

8

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### The Graph - Continued

X2 (000s)

0.4X1 + 0.2X2 = 1500

(3750,0)

(0,7500)

0.4X1 + 0.2X2> 1500

8

6

4

2

X1 (000s)

2

4

6

8

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### The Graph - Continued

X2 (000s)

0.2X1 + 0.3X2 = 500

(2500,0)

(0,1666.7)

0.2X1 + 0.3X2> 500

Redundant!

8

6

4

2

X1 (000s)

2

4

6

8

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X2 (000s)

0 < X1< 9000

0 < X2< 6000

Feasible Region

8

6

4

2

X1 (000s)

2

4

6

8

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C = 20X1 + 15X2

X2 (000s)

8

6

4

2

X1 (000s)

2

4

6

8

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### The Graph - Continued

X2 (000s)

Optimum

(2000, 3500)

8

6

0.3X1+0.4X2 = 2000

0.4X1+0.2X2 = 1500

Cost = 92,500

4

2

X1 (000s)

2

4

6

8

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### AMPL Software

http://www.ampl.com

Try AMPL

2. (for Windows Users)

amplcml.zip

place this file in some directory – such as 8360/AMPL/

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### AMPL Software - Continued

extract all

to run double click sw

Note: My model is in a:ex1.txt

type

sw: ampl

ampl: model a:ex1.txt;

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### Example in a:ex1.txt

# AMPL Model Ex2.1 – a:ex1.txt

option solver cplex;

printf"AMPL Model Ex2.1 - run on PC\n\n";

var x1 >= 0, <= 9000;

var x2 >= 0, <= 6000;

minimize cost: 20*x1 + 15*x2;

subject to Gas: 0.3*x1 + 0.4*x2 >= 2000;

subject to Jet_Fuel: 0.4*x1 + 0.2*x2 >= 1500;

subject to Lubricants: 0.2*x1 + 0.3*x2 >= 500;

expand cost; expand Gas; expand Jet_Fuel;

expand Lubricants;

solve;

display x1; display x2;

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### Solution Obtained

ampl: model a:ex1.txt;

AMPL Model Ex2.1 - run on PC

minimize cost:

20*x1 + 15*x2;

subject to Gas:

0.3*x1 + 0.4*x2 >= 2000;

subject to Jet_Fuel:

0.4*x1 + 0.2*x2 >= 1500;

subject to Lubricants:

0.2*x1 + 0.3*x2 >= 500;

CPLEX 8.0.0: optimal solution; objective 92500

2 dual simplex iterations (0 in phase I)

x1 = 2000

x2 = 3500

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### To Save Solution

While still in sw:

use edit to select all

use edit to copy

then paste to Word or notepad document

then you can turn in your output

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### Generic Model

The generic model does not have the data with the model.

The data is in a separate file.

Hence we can solve any problem of this type.

We simply provide the proper data in a data file.

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# Ex2.1 AMPL Model - Generic - a:ex2.txt

option solver cplex;

set S; # sourses - sa = Saudi Arabia, v = Venezuela

set P; # products - g = gas, j = jet fuel, l = lubs

param a {S}; # availability of crude from sources

param c {S}; # cost/barrel

param d {P}; # demand for products

param cf {S,P}; # conversion factor - crude to product

data a:data.txt;

var x {s in S} >= 0, <= a[s];

minimize cost: sum {s in S} c[s]*x[s];

subject to Demand {p in P}:

sum {s in S} cf[s,p]*x[s] >= d[p];

solve;

display x;

expand cost; expand Demand;

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# Ex2.1 AMPL Model - Generic - a:data.txt

set S := sa v;

set P := g j l;

param a := sa 9000 v 6000;

param c := sa 20 v 15;

param d := g 2000 j 1500 l 500;

param cf :=

sa g 0.3

v g 0.4

sa j 0.4

v j 0.2

sa l 0.2

v l 0.3;

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ampl: model a:ex2.txt;

CPLEX 8.0.0: optimal solution; objective 92500

2 dual simplex iterations (0 in phase I)

x [*] :=

sa 2000

v 3500;

minimize cost:

20*x['sa'] + 15*x['v'];

subject to Demand['g']:

0.3*x['sa'] + 0.4*x['v'] >= 2000;

subject to Demand['j']:

0.4*x['sa'] + 0.2*x['v'] >= 1500;

subject to Demand['l']:

0.2*x['sa'] + 0.3*x['v'] >= 500;

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### Optimal Solutions For Linear Programs (LPs)

There will be either a unique optimum or an infinite number of optimal solutions.

There cannot be exactly 2 optimal solutions to a LP.

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Max 2X + Y

Subject To

X + Y < 5

0 < X < 3

0 < Y < 4

optimum

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### Infinite Number Of Solutions

All points are optimal

Max 2X + 2Y

Subject To

X + Y < 5

0 < X < 3

0 < Y < 4

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### Output From AMPL For An Infinite Number Of Solutions

# Example With An Infinite Number Of Solutions

# runL1c.txt

option solver cplex;

var X >= 0, <= 3; var Y >= 0, <= 4;

maximize profit: 2*X + 2*Y;

subject to Constraint: X + Y <= 5;

expand profit; expand Constraint;

solve;

display X, Y;

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### Output Is A Single Solution

sw: ampl

ampl: model runL1c.txt;

maximize profit:

2*X + 2*Y;

s.t. Constraint:

X + Y <= 5;

CPLEX 7.1.0: optimal solution; objective 10

0 simplex iterations (0 in phase I)

X = 3

Y = 2

ampl:

Other Optimal Solutions:

(1,4), (2,3),(1.5,3.5),

Etc.

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max X

subject To:

X + Y < 2

X + Y > 4

X > 0, Y > 0

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### Example Of An Unbounded Problem

Max X

Subject To:

X + Y > 10

X > 0

0 < Y < 10

Try (100,5), (1000, 7) (10000, 10)

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