Water Resources Systems: Modeling Techniques and Analysis

Transcription

1 INDIAN INSTITUTE OF SCIENCE Water Resources Systems: Modeling Techniques and Analysis Lecture - 9 Course Instructor : Prof. P. P. MUJUMDAR Department of Civil Engg., IISc.

2 Summary of the previous lecture Graphical method General form of LP Motivation for the simple method

3 LP Simple Method Motivation for the Simple method Ma Z = 6 8 s.t

4 LP Simple Method B A C Feasible region O D E 4

5 LP Simple Method In the general form of LP, the constraints are converted as equations and 4 unknowns 0 0; 0 0 0; , 4 are slack variables 5

6 LP Simple Method In general, m equations in n unknowns (n > m), including slack and surplus variables It is possible to solve for m variables in terms of the other (n m) variables Basic solution: A basic solution is a solution obtained by setting (n m) variables to zero The m variables whose solution is sought by setting the remaining (n m) variables to zero are called the basic variables The (n m) variables which are set to zero are called the non-basic variables. It may so happen that some variables may take a value of zero arising out of the solution of the m equations, but they are not nonbasic variables as they are not initialized to zero. 6

7 LP Simple Method Since out of n variables, any of the (n m) variables may be set to zero, the no. of basic solutions will correspond to the no. of ways in which m variables can be selected out of n variables i.e., nc m = n! m! n m! For the problem being considered, with n = 4 and m =, the no. of basic solutions will be 4!! 4! 6 7

8 LP Simple Method B A C Feasible region O D E 8

9 LP Simple Method In the eample, the basic solutions are 3 4 Point on graph Feasible O Y A Y B N C Y D Y E N 9

10 LP Simple Method All the basic solutions need not (and, in general, will not) be feasible. A basic solution which is also feasible is called as the Basic Feasible Solution. All the corner points of the feasible space are basic feasible solutions. 0

11 LP Simple Method As the size of a LP problem increases, the no. of basic solutions also increases. The possible no. of basic feasible solutions can be too large to be enumerated completely. The goal would be, starting with an initial basic feasible solution, to generate better and better basic feasible solutions until the optimal basic feasible solution is obtained.

12 LP Simple Method Algebraic approach (Simple algorithm): Maimize s.t ; 0 0; Z To solve the problem algebraically, we must be able to sequentially generate a set of basic feasible solutions

13 LP Simple Method. Determine an initial basic feasible solution: The best way to obtain an initial basic feasible solution would be to put all the decision variables (n m in no., if we have one slack/surplus variable associated with each constraint) to zero. In the eample, choose the slack variables 3 and 4 to be basic and and to be non-basic (i.e., set = 0 and = 0 ) Basis :

14 LP Simple Method Solve the m (=) equations for m basic variables Point O on the graph Initial basic feasible solution is Z =

15 LP Simple Method Now we have to generate another basic feasible solution from the initial basic feasible solution such that there is an improvement in the objective function value, Z. The new solution is obtained as follows ONE presently non-basic variable ( or ) must be selected to enter the current basis, thus becoming basic. ONE presently basic variable ( 3 or 4 ) must be selected to leave the current basis, thus becoming non-basic. 5

16 First operation: LP Simple Method Z Current basis : Either or can enter the basis. The question now is, out of and which variable should enter the basis? Since the problem is to maimize the objective function (OF), we must look for the variable which will increase the OF value the fastest. Because the coefficient of in the OF is higher than that of, increases the OF value faster than does. Hence should enter the basis

17 LP Simple Method Second operation: As the value of newly selected basic variable is increased ( still being zero), the other two variables, 3 and 4 (which are currently in the basis) keep reducing. One of these will reach its lower limit (zero) earlier than the other = 0 when 0 = 60 or = 6 4 = 0 when 4 = 40 or =

18 LP Simple Method To increase the OF value, we would be interested in increasing to the greatest etent possible. As soon as reaches 6, 3 becomes zero ( 4 is still non-zero) Therefore for entering the basis, 3 must leave the basis. Therefore new basis is 4 The new basic solution is obtained by Guass Jordan elimination method. The process continues until the maimum value of the function is arrived at. 8

19 Eample Maimize Z 3 5 s.t Constraints Decision variables 9

20 Eample (Contd.) The problem is converted into standard LP form s.t. Maimize Z ; 0; ; n = no. of variables = 5; m = no. of constraints =3 0

21 Eample (Contd.) Seek solution from m variables by setting (n m) variables to zero and are non basic variables (whose values are set to zero) 3, 4 and 5 are basic variables (whose solution is sought) The solution is obtained by creating a table with coefficients of variables.

22 Iteration- Eample (Contd.) Z Basis Row Z b i Z