1 Lecture 11 Resource Allocation Part1 (involving Continuous Variables- Linear Programming) 1.040/1.401/ESD.018 Project Management Samuel Labi and Fred.

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Lecture 11 Resource Allocation Part1

(involving Continuous Variables-

Linear Programming)

1.040/1.401/ESD.018Project Management

Samuel Labi and Fred Moavenzadeh

Massachusetts Institute of Technology

April 2, 2007

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Linear Programming

This Lecture

Part 1: Basics of Linear Programming

Part 2: Methods for Linear Programming

Part 3: Linear Programming Applications

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Linear Programming

Part 1: Basics of Linear Programming

- The link to resource allocation in project management

- What is a “feasible region”?

- How to sketch a feasible region on a 2-D Cartesian axis

- Vertices of a feasible region

- Some standard terminology

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The link to resource allocation in project management

Project output = f(Resource 1, Resource 2, Resource 3, … Resource n)

The goal is to determine the levels of each resource that would maximize project output.

Assume only 1 resource variable: X

Project output

Amount of Resource X

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Assume only 2 resources: X and Y

Linear Programming

WW

W

X

XX

YY

Y

Examples of W =f(X,Y) response surfaces

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Assume only 2 resources: X and Y (consider simplified cross section of response surface)

Resource Y

Output, W

Resource X

Linear Programming

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Resource Y

Output, W

Resource X

Linear Programming

Local space

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Resource Y

Output, W

Resource X

Linear Programming

Local space

Local maximum

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Resource Y

Output, W

Resource X

Linear Programming

Global Space

Local maximum

Local space

Global Maximum

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Resource YOutput, W

Resource X

Linear Programming

Local space

Local maximum

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In the real world, there are more than 2 resource types (variables)- equipment types- labor types or crew types- money

Therefore, in project management, resource allocation can be a multi-dimensional linear programming problem.

Linear Programming

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Linear Programming

Example 1: Sketch the following region:y – 2 > 0

SolutionFirst, make y the subjectWrite the equation of the critical boundarySketch the critical boundaryIndicate the region of interest

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Linear Programming

Sketch of the region: y > 2

2

- 1

1

3

4

5

- 2

y = 2

x

y

Critical Boundary

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Linear Programming

Example 2: Sketch of the region: x - 5 < 0

21 3 4 5

x = 5

x

y

(Critical Boundary)

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Linear Programming

Linear Programming

Example 3: Sketch of the region: y > 2

2

- 1

1

3

4

5

- 2

y = 1

x

y

(Critical Boundary)

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Linear Programming

Example 4: Sketch of the region: 1 – x ≤ 0

21 3 4 5

x = 1

x

y

(Critical Boundary)

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Linear Programming

Example 5: Sketch of the region: y > 0

2

- 1

1

3

4

5

- 2

x axis, or y = 0

y

(Critical Boundary)

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Linear Programming

Example 6: Sketch of the region: y - 3 ≤ 0

2

- 1

1

3

4

5

- 2

x axis

y

(Critical Boundary)

y = 3

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Mean and Variance

Linear Programming

Example 7: Sketch of the region: x + 1 ≤ 0

21-1-2-3

x = -1

x

y

(Critical Boundary)

3

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Linear Programming

Mean and Variance

Linear Programming

Example 8: Sketch of the region: 2 - x ≤ 0

21-1-2-3

x = -2

x

y

(Critical Boundary)

3

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Linear Programming

How to Sketch a Region whose Critical Boundary is a bi-variate Function

First, make y the subject of the inequality

Write the equation of the critical boundary

Sketch the critical boundary (often a sloping line)

Indicate the region of interest

Note that …

- the sign < means the region below the sloping line- the sign > means the region above the sloping line)

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Linear Programming

Example 9: Sketch of the region: y ≤ x

y = x

x

y (Critical Boundary)y x

Thus, the critical boundary is:

y = x

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Linear Programming

Example 10: Sketch of the region: y < x

y = x

x

y (Critical Boundary)y < x


y = x

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Linear Programming

Linear Programming

Example 11: Sketch of the region: x – y ≤ 0

y = x

x

y

(Critical Boundary)

x – y ≤ 0

Making y the subject yields:

y x


y = x

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Linear Programming

Linear Programming

Example 12: Sketch of the region: y > 2x + 1

y = 2x+1

x

y

(Critical Boundary)

y > 2x +1


y = 2x+1

When x = 0, y = -0.5

CB passes thru (0,-0.5)

When y = 0, x = 1

CB passes thru (1,0)

1

-3

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Linear Programming

Example 13: Sketch of the region: y < 4x - 3

y = 4x- 3

x

y

(Critical Boundary)y < 4x - 3


y = 4x - 3

When x = 0, y = -3

CB passes thru (0, -3)

When y = 0, x = 3/4

CB passes thru (0.75, 0)

0.75

-3

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Linear Programming

Example 14: Sketch of the region: y ≤ -3.8x + 13

y = 4x- 3

x

y

(Critical Boundary)

y < -3.8x + 3


y = - 3.8x +3

When x = 0, y = 13

CB passes thru (0, 13)

When y = 0, x = 13/3.8

CB passes thru (13/3.8, 0)

13/3.8

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How to sketch a region bounded by two or more critical boundaries

First make y the subject of each inequality

Write the equation of the critical boundary

Sketch the critical boundaries for each inequality

Indicate the overlapping region of interest

Linear Programming

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Linear Programming

Example 15: Sketch the region bounded (or constrained) by the following functions

y > 0x > 0y < -3.5x + 5

y=0

y

(Critical Boundary)

y > 0

Its critical boundary is:

y = 0

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Linear Programming


y > 0x > 0y < -3.5x + 5

y=0

y

(Critical Boundary)

x > 0

Its critical boundary is:

x = 0

x=0

(Critical Boundary)

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Linear Programming


y > 0x > 0y < -3x + 5

y=0

y

(Critical Boundary)

y < -3x + 5

Thus, the critical boundary is:y = -3x + 5When x = 0, y = 5CB passes thru (0, 5)

When y = 0, x = 5/3CB passes thru (5/3, 0)

x=0

(Critical Boundary)

(Critical Boundary)

y= -3x + 5

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Linear Programming


y > 0x > 0y < -3x + 5

y=0

y

(Critical Boundary)

This is the FEASIBLE region.

All points in this region satisfy all the three constraining functions.

x=0

(Critical Boundary)

(Critical Boundary)

y= -3x + 5

Feasible Region

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Linear Programming


y > 0y> - 0.2x + 5y < -0.5x + 5

x

y

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Linear Programming


y > 0y> - 0.2x + 5y < -0.5x + 5

y=0

y

(Critical Boundary)



(Critical Boundary)

y = -0.2x + 5y= -0.5x + 5

(Critical Boundary)

Feasible Region

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Linear Programming


y > 3y < -2x + 6y < x + 1

y

x

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Linear Programming


y > 3y < -2x + 6y < x + 1

y=3

y

(Critical Boundary)



(Critical Boundary)

y = -0.2x + 6

y= x + 1

(Critical Boundary)

Feasible Region

x3-1

6

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Linear Programming


y > 0x > 0y < -x + 5y < x+2

y

x

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Linear Programming


y > 0x > 0y < -x + 5y < x+2

y=3

x=0

(Critical Boundary)



(Critical Boundary)

y = -x + 5y= x + 2

(Critical Boundary)

Feasible Region

y=03-2

5

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Linear Programming


y > 0x > 0y < -0.33x + 1y > 2x - 5

y

x

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Linear Programming


y > 0x > 0y < -0.33x + 1y > 2x - 5

y=3

y

(Critical Boundary)



(Critical Boundary)

y = 2x - 5

y= 0.33x + 1

(Critical Boundary)

Feasible Region

x

5/2

1

(Critical Boundary)

(Critical Boundary)

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What are the “vertices” of a feasible region?

Simply refers to the corner points

How do we determine the vertices of a feasible region?- Plot the boundary conditions carefully on a graph sheet and read off the values at the corners, OR- Solve the equations simultaneously

Linear Programming

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Linear Programming


y > 0x > 0y < -0.33x + 1y > 2x - 5

y=3

y

(Critical Boundary)



(Critical Boundary)

y = 2x - 5

y= 0.33x + 1

(Critical Boundary)

Feasible Region

x

(0, 1)

(3.6, 2.2)

(0, 0)(2.5, 0)

(Critical Boundary)

(Critical Boundary)

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Why are vertices important?

They often represent points at which certain combinations of X and Y is either a maximum or minimum.

Certain combination … ? Yes!For example: W = x + y

W = 2x + 3y W = x2 + y W = x0.5 + 3y2 W = (x + y)2

etc., etc.So we typically seek to optimize (maximize or minimize) the value

of W. In other words, W is the objective function.

Linear Programming

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W is also referred to as the OBJECTIVE

FUNCTION or project performance output.

(It is our objective to maximize or minimize W

x and y can be referred to as Project CONTROL VARIABLES or DECISION VARIABLES

Linear Programming

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Symbols for decision variables

In some books, (x1, x2) is used instead of (x,y)

(x1, x2, x3) is used instead of (x, y, z)

(x1, x2, x3 , x4) is used instead of (x, y, z, v) etc.

x1

x2

x3

x2

x1

Linear Programming

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Dimensionality of Optimization Problems

An optimization problem with n decision variables n-dimensional

Linear Programming

W=f(x1)

1 Decision Variable

x1

1-dimensional

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x1

x2

2-dimensional

2 Decision Variables

Intersecting lines yield vertices (problem solutions)

Linear Programming

W=f(x1 , x2)W=f(x1)

1 Decision Variable

x1

1-dimensional

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x1

x2

x2

x1

2-dimensional 3-dimensional

2 Decision Variables 3 Decision Variables


Intersecting planes yield vertices (problem solutions)

x3

Linear Programming

W=f(x1 , x2)W=f(x1) W=f(x1 , x2, x3)

1 Decision Variable

x1

1-dimensional

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x1

x2

x2

x1

2-dimensional 3-dimensional

2 Decision Variables 3 Decision Variables

n-dimensional

n Decision Variables

Sorry! Cannot

be visualize

d


Intersecting planes yield vertices (problem solutions)

Intersecting objects yield vertices (problem solutions)

x3

Linear Programming

W=f(x1 , x2)W=f(x1) W=f(x1 , x2, x3) W=f(x1 , x2, …, xn)

1 Decision Variable

x1

1-dimensional

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Example of 2-dimensional problem

Given that W = 8x + 5y

Find the maximum value of Z subject to the following:

y > 0

x > 0

y < -0.33x + 1

y < 2x - 5

Linear Programming

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Solution

The objective function is: W = 8x + 5y

The constraints are:y > 0x > 0y < -0.33x + 1y < 2x – 5

The control values are x and y.

Linear Programming

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Linear Programming

y=3

y

(Critical Boundary)

(Critical Boundary)

y = 2x - 5y= 0.33x + 1

(Critical Boundary)

Feasible Region

x

(0, 1)

(3.6, 2.2)

(0, 0)(2.5, 0)

Vertices of Feasible Region

x y W = 8x+5y

(0, 0) 0 0 = 8(0) + 5(0) = 0

(0, 1) 0 1 = 8(0) + 5(1) = 5

(2.5, 0) 2.5 0 = 8(2.5) + 5(0) = 20

(3.6, 2.2) 3.6 2.2 = 8(3.6) + 5(2.2) = 36

Solution (cont’d)

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Solution (continued)

Therefore, the maximum value of W is 36,And this happens when x = 3.6 and y = 2.2

That is: Wopt = 36 units

yopt = 3.6 units

xopt = 2.2 units

This set of answers represents the “optimal solution”.

Linear Programming

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What if there are several variables and constraints?

- In project management resource allocation, a typical problem may have tens, hundreds, or even thousands of variables and several constraints.

- Solutions methods - Graphical method- Simultaneous equations- Vector algebra (matrices)- Software packages

Linear Programming

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Next Lecture

Common Methods for Solving Linear Programming Problems

Graphical Methods- The “Z-substitution” Method- The “Z-vector” Method

Various Software Programs: - GAMS- CPLEX- SOLVER

Linear Programming

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Questions?

Linear Programming

1 Lecture 11 Resource Allocation Part1 (involving Continuous Variables- Linear Programming) 1.040/1.401/ESD.018 Project Management Samuel Labi and Fred.

Documents

resource n

resource x slide

resource variable

y resource y output

resource allocation

x project output

basics of linear programming

project management samuel