advertisement

(User.programs) Item: 437 by akcs.rmarimon@hpcvbbs.cv.hp.com [Ricardo Jose Marimon] Subj: Solving Linear Programming Problems with SIMPLEX3 Date: 14 Nov 1992 This solves linear programs using a "pseudo revised" simplex method. It works very fast. Take a look at it!!! [Note: "Linear Programming Problems" is mathematical jargon for what is simply a system of linear inequalities with the goal of maximizing a particular variable. Concrete example: Suppose that the Sierra Cement Company has several cement trucks, drivers, delivery routes, and types of cement. It is known that the sum of the costs of the delivery routes plus the drivers' pay must fall below some particular maximum. And the other variables are similarly tied to given maximum values. These inequalities are called "constraints". If Sierra Cement wants to maximize their profit, they could spend all day doing "what-if" tinkering on a spreadsheet on a computer, or they could solve it in a few seconds using the Simplex method, which kicks out the absolute maximum possible profit, and the exact values of all the variables required to produce that maximum. Nice, eh? Every business should use it, but few have even heard about it. Complete details can be found in any textbook about "Finite Mathematics". -jkh-] SIMPLEX3 can solve a problem with 8 variables and 5 constraints in less than 15 seconds performing 5 iterations on the tableau. The problem this program is intended to solve is of the form, min st c.x A.x = b x >= 0 (* b must be greater than zero *) The way the problem is entered is as a matrix of the following form, [ c | 0 | 0 ] [---+---+---] [ A | I | b ] Where I is the matrix formed by adding the slack and excess variables. I is NOT necessarily the identity matrix. For example to solve the following problem max x1 + 2 x2 st x1 + x2 x1 - x2 + x2 + x1 + x2 + + x3 <= 3 x3 <= 3 x3 <= 3 x3 >= 3 x2 <= 2 x1, x2, x3 >=0 We convert it to the standard form min - x1 - 2 x2 - x3 st x1 + x2 + x4 x1 - x2 + x3 + x5 x2 + x3 + x6 x1 + x2 + x3 - x7 x2 + x8 x1, x2, x3 >= 0 = = = = = 3 3 3 3 2 And write it as a matrix to enter it in the program. [[ -1 -2 -1 0 0 0 0 0 [ 1 1 0 1 0 0 0 0 [ 1 -1 1 0 1 0 0 0 [ 0 1 1 0 0 1 0 0 [ 1 1 1 0 0 0 -1 0 [ 0 1 0 0 0 0 0 1 0 3 3 3 3 2 ] ] ] ] ] ]] This example matrix is included in the SIMPLEX3 directory under the name "P". To solve this problem right away, enter the matrix [just press {alpha} P {enter} -jkh-], press the [INIT] key which initializes some of the variables used by the program and then press the [SOLVE] key which iterates and gives the solution in three parts as follows, 3: 2: 1: :X: [ 2 .999999999998 2 0 0 0 2 1 ] :\Gl: [ -1 0 -1 0 0 ] :Z: -6 Where x is the actual point that minimizes the problem, the \Gl (lambda) is the solution to the dual problem or shadow prices of each of the constraints, and Z is the objective function value at the point found to be the minimum. This is the end of the quick start procedure to solve linear programming problems with my program. Now lets go to some details... The solutions of LP (linear programming problems) can be classified into: unfeasible, when there is no point satisfying the set of constraints unbounded, when the set of constraints can not restrict the decrease of the objective function (c.x) feasible & bounded, when there is a solution to the set of constraints which minimizes the objective function value. The problem can handle all three cases, giving the solution when it is feasible and bounded, providing a ray (which is a vector through which you can infinitely decrease the objective function while remaining in the feasible region) when the solution is unbounded, or flagging that there is no feasible solution. To achieve the different solutions, the program goes through the two phase simplex method automatically (you don't have to include artificial variables). Phase 1 is used to derive a starting basic feasible solution by introducing a set of artificial variables whith the objective of minimizing its sum; when this is achieved, the set of artificial variables is removed and Phase 2 starts with the original objective function. (refer to any LP book for more details into the 2 phase simplex method) One important thing to notice, is that the columns corresponding to the artificial variables, although not taken into account on phase 2, are not removed from the problem because it is sometimes usefull to have these columns at the end of the problem in order to perform sensitivity analysis. The basic algorithm that I used to solve the problem is a pseudo revised simplex method, where all that is maintained is the inverse of the base, the set of basic and non basic variables, and the rest of the data provided at the initialization process. I think this is the most effective procedure in terms of time because it requires less computations... Here's a brief explanation of what the programs in the directory do. This first set of programs are the ones intended to be used directly to solve the problem: [INIT] Initializes the set of variables that are needed by the program. [TABL] Provides the current tableau. [SOLU] Extracts the solution from the current tableau. [SRAY] Extracts a ray from an unbounded LP. [TRACE] Performs one iteration at a time; it is equivalent to solve, but you can take a look at the partial tableaus that are generated by the procedure. [Great for homework or exams in Finite Mathematics courses, where you must show "your" work! -jkh-] [SOLVE] Gives the answer to the LP if it exists. This second set of programs should not be used directly unless you REALLY know what you're doing. There are some system-RPL programs that can crash your system... [ITER] Given which variable enters and which variable exits, this program performs the necessary iterations. Be careful because the numbers that this programs expects are the indices into the BVAR and NVAR variables corresponding to the basic and non-basic variables. [INVAR] Calculates which variable enters the base. It returns 0 if it is an optimum. [OUTVAR] Given which variable enters the base, it gives which variable must exit the base. It returns 0 if the solution is unbounded. [PH1] Performs all the calculations necessary to start phase 1. [PH2] Same but for phase 2. [MCHOP] Given a matrix and the contents of the 'ZERO' variable if an element of the matrix is less than 'ZERO' in absolute value it changes it to 0. [MVIN] Calculates the min element such that it is strictly less than zero. It returns the index into the array corresponding to the element. It returns 0 if no negative elements are found. [MVOUT] It performs the ratio test among two vectors. 2: v2 1: v1 Returns i s.t. min over all i [ v1[i]/v2[i] for v2[i]>0 ] Returns 0 if all v2[i] are less or equal to 0. [MJOC] Given two matrices it joins them in the standard form 2: [ A ] 1: [ B ] It returns [ A | B ] [MSBC] It takes the columns of a column as specified by a list with the numbers of the columns. 2: [ A ] 1: { 3 2 2 1} It returns [ A3 A2 A2 A1 ] where Ai represents the ith column of A. [MELM] This program is used to calculate the matrix that when multiplied by the inverse of the base, gives the new inverse. 2: [1 2 4] 1: 1 It returns [[ 1 0 0] [-2 1 0] [-4 0 1] Multyplying by this matrix is equivalent to performing row reduction in row 1 (as specified by element in level 1) given that the column involved is specified by the element in level 2. [MBASE] It quickly calculates wether there is a base in the system and if there is, which one it is. It returns a list such that its elements correspond to the elements of the original matrix that corresponds to the the columns of the identity. [MSPL] It just separates the original problem form, [ c | 0 | 0 ] [---+---+---] [ A | I | b ] Into the matrices [ A | I ], [ b ], and [ c ]. [CLR14] Clears the top 14 rows of the display. Here are the variables used by the above procedures. Do not play with these variables during a run of the problem because you might get incorrect answers. [ZERO] [BINV] Threshold after which numbers are considered to be zero. This in important because due to numeric problems, results tend to show some rounding errors. This is the inverse of the base. [ACST] [NVAR] [BVAR] [COST] [DATA] [RSRC] [ARTI] [COLS] [ROWS] Actual cost function being used, it changes from phase 1 to phase 2. Non basic variables. Basic variables, in the order in which they are in the inverse of the base. The original objective function coefficients of the problem. The matrix [ c ] The matrix [ A | I ] The matrix [ b ] Number of artificial variables included. Number of variables in the problem, excluding artificials. Number of constraints in the problem. The custom menu provides a quick access to all of these variables. I hope that most of the question will be answered by the above brief description, but you can always contact me, and if I have time I will get back yo you fairly soon. Ricardo Marimon 28-D Escondido Village Stanford CA 94305 Email: rmarimon@leland.stanford.edu Most of the bugs have already been worked out, but as Murphy says, there is always one more bug... Please send any bug reports to my email address... Of course, I am not responsible for any damage that this program might cause to your calculator, your job, yout studies, or your life... :-) But again, I haven't had any troubles for months.