1.040/1.401/ESD.018 Project Management

transcript

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

Linear Programming

This Lecture

Part 1: Basics of Linear Programming

Part 2: Methods for Linear Programming

Part 3: Linear Programming Applications

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

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

Assume only 2 resources: X and Y

Linear Programming

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

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

Resource Y

Output, W

Resource X

Linear Programming

Resource Y

Output, W

Resource X

Linear Programming

Local space

Resource Y

Output, W

Resource X

Linear Programming

Local space

Local maximum

Resource Y

Output, W

Resource X

Linear Programming

Global Space

Local maximum

Local space

Global Maximum

Resource YOutput, W

Resource X

Linear Programming

Local space

Local maximum

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

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

Linear Programming

Sketch of the region: y > 2

Critical Boundary

Linear Programming

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

21 3 4 5

(Critical Boundary)

Linear Programming

Example 3: Sketch of the region: y > 2

(Critical Boundary)

Linear Programming

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

21 3 4 5

(Critical Boundary)

Linear Programming

Example 5: Sketch of the region: y > 0

x axis, or y = 0

(Critical Boundary)

Linear Programming

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

x axis

(Critical Boundary)

Mean and Variance

Linear Programming

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

21-1-2-3

x = -1

(Critical Boundary)

Linear Programming

Mean and Variance

Linear Programming

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

21-1-2-3

x = -2

(Critical Boundary)

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)

Linear Programming

Example 9: Sketch of the region: y ≤ x

y (Critical Boundary)y x

Thus, the critical boundary is:

Linear Programming

Example 10: Sketch of the region: y < x

y (Critical Boundary)y < x

Linear Programming

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

(Critical Boundary)

x – y ≤ 0

Making y the subject yields:

Linear Programming

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

y = 2x+1

(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)

Linear Programming

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

y = 4x- 3

(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)

Linear Programming

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

y = 4x- 3

(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

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

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

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

(Critical Boundary)

Its critical boundary is:

Linear Programming

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

(Critical Boundary)

Its critical boundary is:

(Critical Boundary)

Linear Programming

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

(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)

(Critical Boundary)

y= -3x + 5

Linear Programming

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

(Critical Boundary)

This is the FEASIBLE region.

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

(Critical Boundary)

y= -3x + 5

Feasible Region

Linear Programming

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

Linear Programming

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

(Critical Boundary)

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

(Critical Boundary)

Feasible Region

Linear Programming

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

Linear Programming

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

(Critical Boundary)

y = -0.2x + 6

y= x + 1

(Critical Boundary)

Feasible Region

Linear Programming

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

Linear Programming

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

(Critical Boundary)

y = -x + 5y= x + 2

(Critical Boundary)

Feasible Region

y=03-2

Linear Programming

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

Linear Programming

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

(Critical Boundary)

y = 2x - 5

y= 0.33x + 1

(Critical Boundary)

Feasible Region

(Critical Boundary)

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

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

(Critical Boundary)

y = 2x - 5

y= 0.33x + 1

(Critical Boundary)

Feasible Region

(0, 1)

(3.6, 2.2)

(0, 0)(2.5, 0)

(Critical Boundary)

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

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

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.

Linear Programming

Dimensionality of Optimization Problems

An optimization problem with n decision variables n-dimensional

Linear Programming

W=f(x1)

1 Decision Variable

1-dimensional

2-dimensional

2 Decision Variables

Intersecting lines yield vertices (problem solutions)

Linear Programming

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

1 Decision Variable

1-dimensional

2-dimensional 3-dimensional

2 Decision Variables 3 Decision Variables

Intersecting planes yield vertices (problem solutions)

Linear Programming

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

1 Decision Variable

1-dimensional

2-dimensional 3-dimensional

2 Decision Variables 3 Decision Variables

n-dimensional

n Decision Variables

Sorry! Cannot

be visualize

Intersecting planes yield vertices (problem solutions)

Intersecting objects yield vertices (problem solutions)

Linear Programming

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

1 Decision Variable

1-dimensional

Example of 2-dimensional problem

Given that W = 8x + 5y

Find the maximum value of Z subject to the following:

y < -0.33x + 1

y < 2x - 5

Linear Programming

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

(Critical Boundary)

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

(Critical Boundary)

Feasible Region

(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)

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

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

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

Questions?

Linear Programming

1.040/1.401/ESD.018 Project Management

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