Civil Engineering Systems Analysis Lecture XII. Instructor: Prof. Naveen Eluru Department of Civil Engineering and Applied Mechanics

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Conrad Cook
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1 Civil Engineering Systems Analysis Lecture XII Instructor: Prof. Naveen Eluru Department of Civil Engineering and Applied Mechanics

2 Today s Learning Objectives Dual Midterm 2

3 Let us look at a complex case 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 nonnegative variables 3

4 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 4

5 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 5

6 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 6

7 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 7

8 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 8

9 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 9

10 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 10

11 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 11

12 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 12

13 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 13

14 Relationship summary 14

15 MIDTERM REVIEW 15

16 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 18 th Late comers will not be provided extra time! Syllabus covered up to today Cheat sheet on one side only 16

17 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 17

18 LP problem formulation 1-2 problems on formulating an LP 18

19 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 19

20 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 20

21 Matrix and Revised Simplex forms In this we just compute the below expression to replace for the table 1 c B B 1 A c c B B 1 Z 0 B 1 A B 1 x = cbb 1 b B x 1 b s 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 - (a ik /a rk ) (B 1 old) rj if i r = (1/a rk ) (B 1 old) ij if i = r In matrix notation (B 1 ) new = E (B 1 ) old where E is an identity matrix except that its r th column is replaced by vector η = η 1 η 2 η m where ηi = -(a ik/a rk) if i r = (1/a rk ) if i = r 21

22 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 viceversa 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 22

23 Last year mid-term solutions 23

24 References Hillier F.S and G. J. Lieberman. Introduction to Operations Research, Ninth Edition, McGraw- Hill,

Linear Programming, Lecture 4

Linear Programming, Lecture 4 Corbett Redden October 3, 2016 Simplex Form Conventions Examples Simplex Method To run the simplex method, we start from a Linear Program (LP) in the following standard simplex