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

Transcription

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

2 Today s Learning Objectives Simplex Method 2

3 Simplex : Example 2 Max Z = 3x1+5x2 Subject to x1 4 2x2 12 3x1+2x2 18 x1 0,x2 0 3

4 Solution Augmented form Max Z = 3x1+5x2 Subject to x1 + x3 =4 2x2 +x4 =12 3x1+2x2 +x5 =18 x1 0,x2 0 Simplex Table x1 x2 x3 x4 x5 Solution Z x x x

5 Solution x1 x2 x3 x4 x5 Solution Z x x x Solution (0,0,4,12,18) and Z =0 Do we have an optimal solution? No What is the entering variable? x2 5

6 Solution Entering variable x1 x2 X3 x4 x5 Solution Ratio Z x x /2 x /2 Pivot Element Now compute the different ratios We can see that row corresponding to x4 has the minimum ratio Hence x4 is the leaving variable Leaving variable 6

7 Solution Start making the changes to the table Step 1: for pivot row divide the pivot row elements by pivot element (in the example 2) Step 2: for every other row: New row = current row Pivot column coefficient * new pivot row x1 x2 x3 x4 x5 Solution Z / x x4x2 0/2 2/2 0/2 1/2 0/2 12/2 x

8 Solution Consolidate x1 x2 x3 x4 x5 Solution Z / x x /2 0 6 x Solution (0,6,4,0,6) and Z = 30; Are we optimal yet? No x1 will enter Leaving variable Min(4/1, 6/0, 6/3) => x5 leaves 8

9 Solution Do the operations x1 x2 x3 x4 x5 Solution Z / x /3 0-1/3 4-2 x ½ x5x1 3/3 0/3 0/3-1/3 1/3 6/3 Consolidate x1 x2 x3 x4 x5 Solution Z / x /3-1/3 2 x ½ 0 6 x /3 1/3 2 Optimal????? Yes Solution (2,6,2,0,0) 9

10 Summary Subject to x1 4 2x2 12 3x1+2x2 18 x1 0,x2 0 x1 = 2, x2 = 6 So Eqn 1 has abundant resources Eqn 2 and Eqn 3 lead to scarce resources 10

11 Minimum Ratio - Notes In the simplex table computing minimum ratio has two components Coefficients for the entering variable Have to be >0 RHS 0 Hence Minimum Ratio 0 given you meet the above constraints So if RHS is 0 and coefficient is ive that is not a valid ratio to consider 11

12 Insights on simplex What if there is a tie for entering variable? At a juncture in the simplex tableau you can have two variable ive and of the same magnitude How do we determine what enters Choose arbitrarily! Eventually you will reach the solution What if tie in the Minimum ratio test What does it imply? Two constraints are such that they yield a same lower limit on entering variable i.e. two current basic variables go to 0 simultaneously Referred to as degeneracy How to address it Break tie arbitrarily You might enter a loop, if so, then the time of the tie use the other variable as leaving variable 12

13 Degeneracy example Max Z = 3x1+9x2 x1+4x2 8 x1+2x2 4 x1, x2 0 x1 x2 x3 x4 Solution Z x x Entering variable? x2 Leaving variable : Min(8/4, 4/2) Lets decide as x4 13

14 Degeneracy example x2 enters and x4 leaves x1 x2 x3 x4 Solution Z -3+9/ / x x4x2 1/2 2/2 0/2 1/2 4/2 x1 x2 x3 x4 Solution Z 3/ /2 18 x x2 ½ 1 0 1/2 2 So we reached optimal value 14

15 Degeneracy example What if we picked the other variable Push x3 out in the first iteration x1 x2 x3 x4 Solution Z -3+9/ / x3x2 ¼ 4/4 1/4 0/4 8/4 x4 1-2/ / x1 x2 x3 x4 Solution Z -3/4 0 9/ x2 ¼ 1 1/4 0 2 x4 1/2 0-1/2 1 0 Now x1 enters What leaves? x4 15

16 Degeneracy example x1 x2 x3 x4 Solution Z -3/4+3/ /4+3/4 0+3/ x2 ¼-1/4 1-0 ¼+1/4 0-1/2 2-0 x4x1 ½*2 0*2-1/2*2 1*2 0*2 x1 x2 x3 x4 Solution Z /2 18 X /2-1/2 2 X Optimal? 16

17 Insights on simplex Just as you can have two possible leaving variables you can have 0 variables for leaving Happens when Z is unbounded The basic variable entering can be increased indefinitely 17

18 Unbounded Z Max z =2x1 + x2 x1-x2 10 2x1 40 x1,x2 0 x1 x2 x3 x4 Solution Z x x Entering variable x1, leaving variable x3 x1 x2 x3 x4 Soluti on x1 x2 x3 x4 Solu on Z x3x x Z x x

19 Unbounded Z Entering variable x2 Leaving variable x4 x1 x2 x3 x4 Soluti on Z / x / x4x2 0/2 2/2-2/2 1/2 20/2 x1 x2 x3 x4 Soluti on Z /2 50 x ½ 20 x ½ 10 Now x3 entering.. But no leaving variable.. 19

20 Multiple Optimal Solutions Max Z = 2x1+4x2 x1+2x2 5 x1+x2 4 x1,x2 0 Basic x2 enters leaving variable x3 Z x1 x2 x3 x4 Solut ion Z x x Basic Z x1 x2 x3 x4 Solut ion Z / x3x2 x4 0/2 ½ 2/2 1/2 0/2 5/ / / /2 Z x1 x2 x3 x4 Solu tion Z x2 x4 0 ½ 1 ½ 0 5/2 0 ½ 0-1/2 1 3/2 Z = 10 and (0,5/2,0,3/2) 20

21 Multiple Optimal Solutions We reached optimal but see that x1 and x2 have 0 coefficients Lets try to look at z We notice that one of the non-basic variables has a coefficient of 0.. So without changing z we can have x1 enter If x1 enters x4 leaves 21 Z x1 x2 x3 x4 Solu tion Z x2 0 ½-1/2 1-0 x4 x1 0 ½ *2 0*2 ½+1/ 2-1/2* /2-3/2 1*2 3/2*2 Z x1 x2 x3 x4 Solu tion Z x3 x1 Z = 10 and (3,1,0,0)

22 Summary of special cases Entering ties Degeneracy Results from multiple leaving options Possibility of a loop Unbounded Z Simplex is unable to find the corner point Multiple optimal solutions Results when an edge is the optimal solution 22

23 Simplex assumptions So far we examined simplex.. But implicitly we made the following assumptions for the standard problem constraints Slack variables easily provide basic feasible solution All the RHS values are non-negative Ensure the variables are not <0 Maximization We have ive values in simplex (these enter) But we need to know how to adapt the simplex for other forms also! 23

24 References Hillier F.S and G. J. Lieberman. Introduction to Operations Research, Ninth Edition, McGraw- Hill, 2010 Revelle C.S, E.E. Whitlatch and J. R. Wright. Civil and Environmental Systems Engineering 24