civil engineering systems analysis class...

Civil Engineering Systems Analysis

Lecture XII

Instructor: Prof. Naveen Eluru

Department of Civil Engineering and Applied Mechanics

Today’s Learning Objectives

Dual

Midterm

10/16/2012 2 CIVE 208: CIVIL ENGINEERING SYSTEMS ANALYSIS

Let us look at a complex case

10/16/2012 CIVE 208: CIVIL ENGINEERING SYSTEMS ANALYSIS 3

Minimize 8x1 + 5x2 + 4x3

Subject to

4x1 + 2x2 + 8x3 = 12

7x1 + 5x2 + 6x3 ≥ 9

8x1 + 5x2 + 4x3 ≤ 10

3x1 + 7x2 + 9x3 ≥ 7

x1 ≥ 0, x2 unrestricted, x3 ≤ 0

We know how to write the dual if we have maximization and ≤ constraints, and non-negative variables

Solution


Convert Min to Max multiply with -1

Now convert x2 into a non-negative variable combination x2 = x4 – x5

Now convert x3 into a non-negative variable combination x3 = -x6

Max -8x1 - 5x4 + 5x5 + 4x6

4x1 + 2x4 – 2x5 - 8x6 = 12

7x1 + 5x4 – 5x5 - 6x6 ≥ 9

8x1 + 5x4 – 5x5 - 4x6 ≤ 10

3x1 + 7x4 – 7x5 - 9x6 ≥ 7

x1, x4, x5, x6 ≥ 0

Convert constraints to ≤


4x1 + 2x4 – 2x5 - 8x6 = 12

Write it as two equations

4x1 + 2x4 – 2x5 - 8x6 ≤ 12 (good!)

4x1 + 2x4 – 2x5 - 8x6 ≥ 12 (change this by

multiplying with -1)

-4x1 - 2x4 + 2x5 + 8x6 ≤ -12 (good!)

7x1 + 5x4 – 5x5 - 6x6 ≥ 9

-7x1 - 5x4 + 5x5 + 6x6 ≤ -9 (good!)

8x1 + 5x4 – 5x5 - 4x6 ≤ 10 (good!)

3x1 + 7x4 – 7x5 - 9x6 ≥ 7

-3x1 - 7x4 + 7x5 + 9x6 ≤ -7

In standard form


Maximize -8x1 - 5x4 + 5x5 + 4x6

4x1 + 2x4 – 2x5 - 8x6 ≤ 12 (good!)

-4x1 - 2x4 + 2x5 + 8x6 ≤ -12 (good!)

-7x1 - 5x4 + 5x5 + 6x6 ≤ -9 (good!)

8x1 + 5x4 – 5x5 - 4x6 ≤ 10

-3x1 - 7x4 + 7x5 + 9x6 ≤ -7

x1, x4, x5, x6 ≥ 0

5 constraints and 4 variables

So, dual will have 5 variables and 4 constraints

Lets write the dual

Dual


Minimize 12y1 – 12y2 -9y3 +10y4 -7y5

4y1 – 4y2 – 7y3 + 8y4 - 3y5 ≥ -8

2y1 – 2y2 – 5y3 + 5y4 – 7y5 ≥ -5

-2y1 + 2y2 + 5y3 – 5y4 + 7y5 ≥ 5

-8y1 + 8y2 + 6y3 – 4y4 + 9y5 ≥ 4

y1, y2, y3, y4, y5 ≥ 0

Reconvert to match with the original problem

Convert minimize to maximize

Max -12y1 + 12y2 +9y3 -10y4 +7y5

Match variables with original problem


Original problem has 4 constraints so only 4

variables!

Can we make some changes to the variables

Set y6 = y2-y1

–4y6 – 7y3 + 8y4 - 3y5 ≥ -8

-2y6 – 5y3 + 5y4 – 7y5 ≥ -5

2y6 + 5y3 – 5y4 + 7y5 ≥ 5

8y6 + 6y3 – 4y4 + 9y5 ≥ 4

Max 12y6 +9y3 -10y4 +7y5

Match variables with original problem


Also, the dual should have objective function

values from RHS of primal.. So all terms

positive

Set y7 = -y4

–4y6 – 7y3 - 8y7 - 3y5 ≥ -8

-2y6 – 5y3 - 5y7 – 7y5 ≥ -5

2y6 + 5y3 + 5y7 + 7y5 ≥ 5

8y6 + 6y3 + 4y7 + 9y5 ≥ 4

Max 12y6 +9y3 +10y7 +7y5

Match no. of constraints


Lets make all constraints have positive RHS

4y6 + 7y3 + 8y7 + 3y5 ≤ 8

2y6 + 5y3 + 5y7 + 7y5 ≤ 5

2y6 + 5y3 + 5y7 + 7y5 ≥ 5

8y6 + 6y3 + 4y7 + 9y5 ≥ 4

We had 3 variables in the primal, so 3 constraints

We can join equations 2 and 3 as =

4y6 + 7y3 + 8y7 + 3y5 ≤ 8

2y6 + 5y3 + 5y7 + 7y5 = 5

8y6 + 6y3 + 4y7 + 9y5 ≥ 4

Max 12y6 +9y3 +10y7 +7y5

For simplicity replace y6 with y1, y3 with y2 and so on

Solution


Primal

Minimize 8x1 + 5x2 + 4x3

4x1 + 2x2 + 8x3 = 12

7x1 + 5x2 + 6x3 ≥ 9

8x1 + 5x2 + 4x3 ≤ 10

3x1 + 7x2 + 9x3 ≥ 7

x1 ≥ 0, x2 unrestricted, x3 ≤ 0

Dual

Max 12y1 +9y2 + 10y3 +7y4

4y1 + 7y2 + 8y3 + 3y4≤ 8

2y1 + 5y2 + 5y3 + 7y4 = 5

8y1 + 6y2 + 4y3 + 9y4 ≥ 4

y1 unrestricted, y2, y4 ≥ 0 and y3 ≤ 0

Points to Remember


“=” constraint gives an unrestricted variable and vice versa

Desirable constraints yield desirable variables and vice-versa

In a minimization ≥ is desirable hence the variable corresponding to that will be ≥ 0 (look at constraints 2 and 4 in primal)

In a minimization ≤ is undesirable and hence it gives us ≤0 variable (look at constraint 3 in primal)

In any problem non-negativity is desirable,

non-negative variable yield desirable constraint signs (see x1 in primal)

Negative variables yield non-desirable constraint signs (see x3 in primal

Another example


Primal

Max 3x1 + 7x2 + 5x3 + 3x4

2x1 + 2x2 + 8x3 = 12

5x1 + 2x2 + 4x3 + x4 ≥ 9

6x1 + 3x2 + 5x3 + 2x4 ≤ 10

6x1 + 5x2 + 7x3 + 5x4 ≥ 7

x1 ≥ 0, x2 unrestricted, x3 ≤ 0, x4 ≥0

Dual

Min 12y1+9y2+10y3+7y4

2y1+5y2+6y3+6y4 ≥ 3

2y1+2y2+3y3+5y4 = 7

8y1+4y2+5y3+7y4 ≤ 5

y2+2y3+5y4 ≥ 3

y1 unrestricted, y2 ≤0, y3 ≥0, y4 ≤0

Relationship summary


MIDTERM REVIEW


Mid-term


Location

Trottier 100 [Even number student ids]

MC13 [Odd number student ids]

If you turn up in the wrong room [-10 points]

Time October 18th 1.05 – 2.25

Late comers will not be provided extra time!

Syllabus covered up to today

Cheat sheet on one side only

Important Components*


LP problem formulation (8-18)

Simplex (8-16)

Algebraic

Graphical

Table

Sensitivity analysis

Simplex theory

Matrix Simplex (8-15)

Revised Simplex (8–15)

Dual (4–8)

*I have not yet prepared the exam; so these mark allocations are for indication only

LP problem formulation


1-2 problems on formulating an LP

Simplex


Approach to solve LP problems Solves based on the corner point approach… Simplex moves from one

corner point to another with the objective of maximizing the rate of increase of the objective function

It does not consider the interior of the feasible space

For two dimensions graphical approach

For more than two dimensions move to the table approach

Requirements Create an identity matrix – so that we have an easy starting solution

Then , from there we set about increasing the objective function

At any iteration we only consider exchanging one element (highest –ive coefficient in z row)

So one variable enters the basis and on leaves the basis (based on minimum ratio)

Then do gauss-jordan transformations to get the new solution

Important elements to note Entering variable

Leaving variable

Simplex


Important points to note Tie in the leaving variable

Unbounded z

Multiple optima

Adapting simplex to other forms Artificial variables (M)

= artificial variable with penalty M in the objective function

≥ surplus + artificial variable with penalty M in the objective function

Minimization Max –Z

Also most cases will have a M based variable (so change the obj. function accordingly)

Theory Ensuring simplex works

Matrix and Revised Simplex forms


In this we just compute the below expression to replace for the table

1 cBB

−1𝐴 − 𝑐 cBB−1

0 B−1𝐴 B−1

𝑍xxs

= cBB

−1bB−1b

B-1 is expensive operation

So we improve the mechanism to compute B-1

Revised Simplex Method ((𝐵−1)𝑛𝑒𝑤 )𝑖𝑗 = (𝐵−1

𝑜𝑙𝑑)𝑖𝑗 - (aik/ark) (𝐵−1𝑜𝑙𝑑)𝑟𝑗 if i ≠r

= (1/ark) (𝐵−1𝑜𝑙𝑑)𝑖𝑗 if i = r

In matrix notation

(𝐵−1)𝑛𝑒𝑤= E (𝐵−1)𝑜𝑙𝑑 where E is an identity matrix except that its rth

column is replaced by vector

𝜂 =

𝜂1

𝜂2

⋮𝜂𝑚

where 𝜂I = -(aik/ark) if i ≠r

= (1/ark) if i = r

Dual


A dual problem is formed by Maximization ←→ Minimization

Constraints ←→ Variables No. of constraints in primal = no. of variables in dual

Column of coefficients ←→row of coefficients

Resources ←→ objective function values

Rules for conversion “=” constraint gives an unrestricted variable and vice versa

Desirable constraints yield desirable variables and vice-versa In a minimization ≥ is desirable hence the variable corresponding to

that will be ≥ 0

In a minimization ≤ is undesirable and hence it gives us ≤0 variable

In any problem non-negativity is desirable, Non-negative variable yield desirable constraint signs

Negative variables yield non-desirable constraint signs


Last year mid-term solutions

References


Hillier F.S and G. J. Lieberman. Introduction to

Operations Research, Ninth Edition, McGraw-

Hill, 2010

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