LINEAR PROGRAMMING PROBLEM

Transcription

1 LINEAR PROGRAMMING PROBLEM INTRODUCTION Linear Programming deals with the optimization (maximum or minimum) of a function of variables, known as objective function, subject to a set of linear equations and/or inequalities known as constraints. The objective function may be profit, cost, production capacity or any other measure of effectiveness, which is to be obtained in the best possible or optimal manner. The constraints may be imposed by different resources such as market demand, production process and equipment, storage capacity, raw material availability, etc. By linearity is meant a mathematical expression in which the expression among the variables are linear equation, the expression a 1 x 1 + a 2 x a n x n is linear. Higher powers of variables or their products do not appear in the expressions for the objective functions as well as the constraints. The variables obey the properties of proportionality and additivity. It was in 1947 that George Dantzig and his associates found out a technique for solving military planning problems while they were working on a project for U.S. Air Force. This technique consisted of representing the various activities of an organization as a Linear Programming model and arriving at the optimal programme by minimizing a linear objective function. Afterwards, Dantzig suggested this approach for solving business and industrial problems. He also developed the most powerful mathematical tool known as Simplex Method to solve Linear Programming problems. [ 42 ]

2 2.2 MATHEMATICAL FORMULATION OF LINEAR PROGRAMMING PROBLEM Mathematically, the Linear Programming Problem can be formulated as: Maximize Z = CX (Objective Function) Subject to AX = b, b 0 (Constraint Equation) and X = (x 1,x 2, x n ) 0 (Non-negative restrictions) where C = Cost Vector = (c 1 c c n ) b = Requirement Vector = (b 1 b b n ) T a 11 a 12.. a 1n and A = Activity Matrix = a 21 a 22.. a 2n a n1 a n2.. a nn 2.3 VARIOUS METHODS FOR SOVLING LINEAR PROGRAMMING PROBLEMS Graphical Method Graphical Method can be used only for Linear Programming Problem with only two variables. The graphical method provides a pictorial representation of the solution process and a great deal of insight into the basic concepts used in solving large Linear Programming Problems. The method consists of the following steps:- (a) Represent, the given problem in mathematical form i.e., formulate the mathematical model for the given problem. (b) Draw the x 1 and x 2 axis. The non-negativity restriction x 1 0 and x 2 0 i.e. x 1, x 2 lies in the first quadrant. This eliminates a number of infeasible (c) alternatives that lie in second, third and fourth quadrants. Plot each of the constraints on the graph. The constraints, whether equations or inequalities are plotted as equations. For each constraint, assign an arbitrary value to one of the variable and get the value of other variable. Plot these two points and connect them by a straight line. Thus each constraint is plotted as line in the first quadrant. [ 43 ]

3 (d) (e) Identify the feasible region that satisfies all the constraints simultaneously. For type constraint, the area on or above the constraint line i.e., towards away from origin and for type constraint, the area on or above the constraint line i.e., towards from the origin will be considered. They are common to all the constraints is called feasible region and is shown shaded. Any point on or within the shaded region represents a feasible solution to the given problem. Use iso-profit (cost) function line approach. For this, plot the objective function by assuming. This will be a line passing through the origin. As the value of z is increased from zero, the line starts moving to the right, parallel to itself. Draw line parallel to this line till the line is farthest away from the origin. For a minimization problem, the line will be nearest to the origin. The point of the feasible region through which this line passes will be the optimal point. It is possible that this line may coincide with one of the edges of the feasible region. In that case every point on that edge will give the same max/min value of the objective function and will be the optimal point Simplex Method The graphical method cannot be applied when the number of variables involved in Linear Programming Problem is more than two. The Simplex Method developed by Prof. George B. Dantzig can be used to solve any Linear Programming Problem consisting any number of variables and constraints. The computational procedure in the Simplex Method is based on the fundamental property that the optimal solution to an Linear Programming Problem, if it exists, occurs only at one of the course points of the feasible region. The method involves the following steps:- (i) Express the problem in standard form: If the problem is of minimization, then convert it into a maximization problem. If any one of the b i is negative, multiply the corresponding constraints by 1 in order to make that b i positive. Add slack and surplus [ 44 ]

4 (ii) variables if needed. Introduce artificial variables, if needed, and then write the constraints in the form Ax = b. Construct the first simplex tableau: As we always start with a B.F.S. with B=I, first simplex tableau can easily be constructed. We have y j = B -1 = I -1 α j = I α j = α j x B = B -1 b=ib=b Thus initially always y j s are nothing but α j s and the column of values in the tableau considered with the requirement vector b. (iii) Calculate z j c j for all the columns of A. for as initially B=I we have z j c j = c B α j - c j Note that for all the basic vectors, necessarily z j c j = 0 (iv) If all the z j c j 0 then the solution of this iteration is optimum. Further if atleast one z j c j = 0 for a non-basic vector and at least one y ij > 0 for this vector, then by introducing that vector into the basis, one can get an alternative optimum basic feasible solution. If atleast one z j c j = 0 for a non-basic vector and for this vector all y ij 0 then one can obtain an alternative non-basic optimal solution. If alternative BFS exists then as any convex combination of these two BFS s will also provide an optimum solution, the problem has in this case infinite number of optimal solutions. All convex combinations will provide non-basic optimum solutions. Lastly if for all the non-basic vector z j c j > 0 strictly, then the problem admits a unique solution. (v) If at least one z j c j < 0 and corresponding to some α j with z j c j < 0, y j < 0, i.e. y ij 0 problem admits an unbounded solution. A i = 1,2,---m then it is to be concluded that the given (vi) If (iv) and (v) fail, then choose the minimum most of all z j c j. [ 45 ]

5 Suppose Min (z j c j ) = z K -c K. Find the vector α k as the entering vector. Note that at least one y ik > 0, otherwise the solution will be unbounded. Furthermore if Min (z j -c j ) is not unique i.e. it occurs for more than one α j s (vii) can be selected as the entering vector arbitrarily. Mark ( ) below the entering vector in tableau. Now fix up the departing vector: x Bi y ik Compute Min, y ik > 0 Where k is the number of marked column. If this minimum is unique and occurs at i = r then delete B r i.e. r th vector of the basis, from the basis. Mark ( ) below the departing vector B r in tableau. In the next BFS x Br will become zero. Now construct the next simplex tableau by using the transformations. On the other hand if this minimum is not unique then more than one basic variable of this solution will become zero in the next solution.therefore the next solution will become a degenerate BFS. (viii) If the problem involves the artificial variables, then first remove all the artificial vector from the basis if possible by using the simple criterions and get BFS in terms of original variables. 2.4 ABS METHOD ABS algorithm have been introduced by Abaffy, Broyden and Spedicato to solve determined or undetermined linear systems and have been later extended to linear least squares, nonlinear equations, optimization problems and integers (Diophantine) equation and linear programming problems. The class of ABS method unifies most existing method for solving linear system and provide variety of alternative ways of implementing a specific algorithm. Extensive computational experience has shown that ABS methods are implementable in a stable way, being often more accurate than the corresponding traditional algorithm and the linear programming is one of the most important problem in optimization. [ 46 ]

6 2.4.1 Preliminaries Let the linear programming problem under consideration is of the form; Max Z = CX Subject to AX = b, X R n, b R m, m n Where A is said to be activity matrix, b is a requirement vector and C is cost vector ABS Algorithm 1. Let x 1 R n be an arbitrary vector. Let H 1 R n,n be an arbitrary nonsingular matrix. 2. Cycle for i = 1,2,. n (a) Let z i R n, be a vector arbitrary save for the condition; T i 3. z H i a i 0 Compute search vector p i : T i 4. p i = H z i (b) Compute step size α i : 5. α i = a i T x i - b i p i T a i Which is well defined with regard to (3) to (4). (c) Compute the new approximation of the solution. 6. X i +1 = X i α i p i If i = n stop; x n+1 solve the system. (d) Let w i R n be a vector arbitrary for the condition; T i 7. w H i a i = 1 [ 47 ]

7 and to update the matrix H 1 : T i 8. H i+1 = H i - H i a i w H i There are three eligible parameters in the general version of the ABS algorithm, matrix H1 and two systems of vectors z i and w i. The new algorithms or a new formulation of the classic algorithms can be created by a suitable choice of these parameters. Abaffy etal have studied the above system for a variety of choices of z i and w i calculating the storage and arithmetic operations are required to solve the system with various kind of matrices. [ 48 ]

8 2.4.3 ABS Algorithm in C //Program to solve LPP Problems using ABS Method. #include<stdio.h> #include<conio.h> #include<math.h> //Prototype of Multiply() void multiply(float a[][10],float b[][10], //Prototype of Transpose() float c[][10],int r1,int c1,int c2); void trans(float a[][10],float t[][10],int r,int c); //Prototype of ColArr() void colarr(float a[][10],float col[][10], //Prototype of Identity() void identity(float idt[][10],int size); int main() int x,int r,int c); float A[10][10],b[10][10],x[10][10],H[10][10], s[10][10],a[10][10],z[10][10],p[10][10], HT[10][10],ZT[10][10],Y[10][10],X[10][10], P[10][10]; float Q,T,p1[10][10],PT[10][10],PT1[10][10], PT2[10][10],AT[10][10],BU[10][10],AU[10][10]; int r1,c1,i,j,k,l,rank; // clrscr(); printf("\n\n\t\t\tbasic ABS Algorithm for LPP"); printf("\n\ninput Section : "); //Input of Matrix A printf("\n\nenter Size of Matrix A : "); printf("\n\n\t\t Rows : "); scanf("%d",&r1); [ 49 ]

9 printf("\n\n\t\t Cols : "); scanf("%d",&c1); rank=(r1<c1)?r1:c1; printf("\n\nenter Elements of Matrix A : "); for(i=0;i<r1;i++) for(j=0;j<c1;j++) printf("\n\n\telement[%d][%d]: ",i+1,j+1); //Input of Matrix b scanf("%f",&a[i][j]); printf("\n\nenter Elements of Matrix b : "); j=0; for(i=0;i<r1;i++) printf("\n\n\telement[ %d ][ %d ] : ",i+1,j+1); scanf("%f",&b[i][0]); if(r1>c1) trans(a,at,r1,c1); else multiply(at,b,bu,c1,r1,1); //asigning BU to b for updation for(k=0;k<rank;k++) b[k][0]=bu[k][0]; trans(a,au,r1,c1); //asigning AU to A for updation for(k=0;k<c1;k++) for(j=0;j<r1;j++) [ 50 ]

10 A[k][j]=AU[k][j]; //Intialization of Matrix x to zero. for(i=0;i<c1;i++) x[i][0]=0; //Generating Identity Matrix identity(h,c1); //Computing Vector p and s i=0; printf("\n\noutput Section : "); while(i<rank) colarr(a,a,i+1,c1,r1); //gained matrix a multiply(h,a,s,c1,c1,1); //gained matrix s trans(h,ht,c1,c1); //gained matrix HT //Assigning a to z for(k=0;k<c1;k++) z[k][0]=a[k][0]; //gained matrix p (search direction) multiply(ht,a,p,c1,c1,1); trans(z,zt,c1,1); (matrix ZT) //gained transpose of z [ 51 ]

11 multiply(zt,s,y,1,c1,1); //gained matrix Y multiply(zt,x,x,1,c1,1); T=X[0][0]-b[i][0]; if(s[i][0]==0 && T==0) continue; else if(s[i][0]==0 && T!=0) printf("\n\nsystem is Incompatible..."); else if(y[0][0]!=0) //Updating x multiply(zt,x,x,1,c1,1); multiply(zt,p,p,1,c1,1); Q=(X[0][0]-b[i][0])/P[0][0]; for(k=0;k<c1;k++) p1[k][0]=q*p[k][0]; for(k=0;k<c1;k++) x[k][0]-=p1[k][0]; //Updating H trans(p,pt,c1,1); multiply(p,pt,pt1,c1,1,c1); multiply(pt,p,pt2,1,c1,1); for(k=0;k<c1;k++) for(l=0;l<c1;l++) PT1[k][l]/=PT2[0][0]; [ 52 ]

12 for(k=0;k<c1;k++) for(l=0;l<c1;l++) H[k][l]-=PT1[k][l]; for(k=0;k<c1;k++) printf("\t"); for(l=0;l<c1;l++) printf("%f\n ",H[k][l]); //Printing X printf("\n\nvalue of x [%d]: ",i+1); for(k=0;k<c1;k++) printf("%f ",x[k][0]); i++; printf("\n\n Value of x : "); for(i=0;i<c1;i++) printf("%f ",x[i][0]); getch(); return 0; [ 53 ]

13 //Definition of Multiply() void multiply(float a[][10],float b[][10], float c[][10],int r1,int c1,int c2) int i,j,k; for(i=0;i<r1;i++) for(j=0;j<c2;j++) c[i][j]=0; for(i=0;i<r1;i++) for(j=0;j<c2;j++) for(k=0;k<c1;k++) c[i][j]=c[i][j]+a[i][k]*b[k][j]; //Definition of Transpose() void trans(float a[][10],float t[][10],int r,int c) int i,j; for(i=0;i<c;i++) for(j=0;j<r;j++) t[i][j]=a[j][i]; printf("\n"); [ 54 ]

14 //Definition of ColArr() void colarr(float a[][10],float col[][10],int x, int r,int c) int i,j=x-1; for(i=0;i<r;i++) col[i][0]=a[i][j]; //Definition of Identity() void identity(float idt[][10],int size) int i,j; for(i=0;i<size;i++) for(j=0;j<size;j++) if(i==j) idt[i][j]=1; else idt[i][j]=0; //End of Source Code [ 55 ]

15 2.5 EXAMPLE Max z = 3x + 2y Subject to x+y 4 x-y x, y Solution by Graphical Method The solution space satisfying the constraints x+y 4, x y 2 and the non-negative restriction x,y 0 is shown in fig. below. The coordinates of the vertices of the convex polygon are O(0,0), C(2,0), E(3,1) and B(0,4). Graph I [ 56 ]

16 Values of the objective function or the feasible solution at these vertices are O: x = 0, y=0 => z = 3 (0) + 2(0) z=0 C: x = 2, y=0 => z = 3 (2) + 2(0) z=6 E: x = 3, y=1 => z = 3 (3) + 2(1) z=11 B: x = 0, y=4 => z = 3 (0) + 2(4) z=8 The maximum value of the objective function occurs at point E. Therefore the optimal solution of the above Linear Programming Problem comes out to be x=3, y=1 and Max z= Solution by Simplex Method Max z = 3x + 2y Subject to x + y 4 x - y 2 x, y 0 Introducing slack variables s 1 and s 2, the problem can be expressed in the following standard form:- Max z = 3x + 2y + 0s 1 + 0s 2 Subject x + y + s 1 = 4 x - y + s 2 = x, y, s 1, s 2 0 Find the initial basic feasible solution. [ 57 ]

17 Let x 1 = x 2 = 0. Substituting these values in the constraints we have the initial Simplex Table as : Table First Iteration table c j FR C B Basic x y s 1 s 2 b 1 0 s s 2 (1) Perform the optimality test z j c j - z j Entering Variable Leaving Variable Since c j - z j is positive under x and y.initial basic feasible solution is not optimal and can be improved. Mark the key column and key row as x and s 2 are incoming/entering and leaving variables respectively. Key element is 1. Table Second Iteration table c j FR C B Basic x y s 1 s 2 b 0 s 1 0 (2) Leaving Variable 3 s z j c j - z j Entering Variable Since all c j - z j under y is still positive. Therefore this solution is not optimal and can be improved. [ 58 ]

18 Mark the key column and key row as y and s 1 are entering and leaving variables respectively. Key element is 2. Table Third Iteration table c j FR C B Basic x y s 1 s 1 b 2 y x 1 0 z j 3 2 c j - z j (11) Since all c j - z j elements are either zero or negative, the table represents an optimal solution. The optimal solution is x = 3, y =1 and Max z = Solution by ABS Method Max z = 3x + 2y Subject to x + y 4 x - y 2 x, y 0 Let the identity matrix be [ 59 ]

19 Compute the search vector T Now for i = 2, solve the system again, update H i.e., [ 60 ]

20 Compute the next search vector After the final iteration. [ 61 ]

21 2.5.4 Solution by ABS Method by C-Approach Input Section: Enter the size of Matrix A : Row : 2 Cols : 2 Enter elements of Matrix A : Enter elements of Matrix b : Output Section: Element [1][1] : 1 Element [1][2] : 1 Element [2][1] : 1 Element [2][2] : 1 Element [1][1] : 4 Element [2][1] : 2 Value of x[1] : Value of x[2] : Value of x : Max z = APPLICATIONS OF LINEAR PROGRAMMING PROBLEMS: The Diet Problem: Linear programming has been applied to know the nutrient contents such as vitamins, proteins, fats, carbohydrates, starch etc. in each of the food stuffs. The method is also used to find the minimum cost diet that satisfies the minimum daily requirements of nutrients when minimum cost of each type of food stuff and minimum daily requirement of each nutrient in the diet is given. Agriculture Problem: These problems are concerned with maximizing net revenue by allocating the input resources such as average of land water, labour, fertilizers and capital to various crops. [ 62 ]

22 Flight Scheduling Problem: To utilize the aircrafts and crew efficiently by determining the most economical patterns and timing of flight. Product mix Problem: The problem is concerned with determining the product mix that will maximize the total profit, when to manufacture various products, a certain production capacity (men, machines, money, material, markets, etc.) are available on various manufacturing process. As different products will have different selling prices, will require different amounts of production capacity at the several processes and will, therefore have different unit profit, there may also be conditions on maximum and / or minimum product levels. Blending Problems: These problems arises when manufacturing process involves blending of some of the various raw material available of various composition and prices, in varying qualities to make a product of desired specification for example different grades of gasoline are used in aviation purpose. About the input ingredients, prices and specifications, such as octane raling, tetra ethyl lead concentrations, maximum vapour pressure, etc. and problem is to decide the proportions of these ingredients to make the desired grades of gasoline so that (i) maximum output is obtained and (ii) storage capacity restrictions are satisfied. The technique of linear programming is used to handle many similar situations such preparation of chemicals, fertilizers and alloys etc. Production Scheduling Problem: Such problem involves the determination of optimum production schedule to meet the fluctuating demand. The objective is to fulfill the demand, minimizing the total cost of production and inventory while keeping employment and inventory at reasonable minimum levels. Trim Loss Problem: They are applicable to paper, sheet metal and glass manufacturing industries where according to customers requirements, items of standard sizes have to be cut to smaller sizes with the objective of minimizing the waste produced. Media Selection Problem: They have the objective to maximize the public exposure to company s product by selecting the proper advertising mix among the [ 63 ]

23 different advertising media such as T.V., Radio, Magazines and Newspapers. Total advertising budget, maximum expenditure in each media, maximum number of insertions in each media and etc. are issues on which the constraints may be. Portfolio Selection Problem: The banks, financial companies, insurance companies, investment services etc. frequently encounters such problems. Amount is distributed among several investment alternatives such as bonds, saving certificates, common stock, mutual funds, real estate etc. to minimize the expected risk and maximize the expected returns. Transportation Problem: When n sources situated at different locations the problem is to transport the products to m different destinations. The problem is to design the optimum transportation plan that minimizes the total transportation cost (distance or time) when the supply position at the sources demand at destinations, freight charges and storages costs etc. are known. Man Power Scheduling Problem: The big hospitals, restaurants and companies operating in number of shifts face such problems. Problem is to minimize the overtime cost by allocating man-power in each shift. 2.7 LIMITATIONS OF LINEAR PROGRAMMING Although Linear Programming is an improvement over conventional theory of production yet it is not free from limitations Linear Programming is an important technique of planning. To specify an objective function in mathematical form is not an easy task. Even if objective function is determined it is difficult to determine social, institutional, financial and other constraints. It is also possible that the objective function and constraints may not be directly specified by linear in equality equations. To determine the relevant values of the co-efficient of constraints involved in Linear Programming is the main problem. The assumptions of Linear Programming are also unrealistic. It assumes that factory proportion remain constant. In addition for it, the [ 64 ]

24 relationship between input and output, production and cost and production and total revenue are assumed to be linear. All these assumptions imply constant returns to scale and perfect competition in the market. It is a very complex method as it uses mathematical techniques extensively Linear Programming models presents the trial and error solutions and it is difficult to find out really optimal solutions to business problems. Under Linear Programming to increase production by a single process the quantity of all inputs is to be increased in a fixed proportion. But the production of a number of goods can be increased to some, extend by increasing only one or two inputs. It means that production can be increased to some extent by varying factors proportion. 2.8 CONCLUSION In this chapter, four different methods are considered to solve Linear Programming Problems. They are:- (i) (ii) (iii) (iv) Graphical Method, Simplex Method, ABS Method, and ABS Algorithm via C-Approach. Our main focus is to develop and represent ABS method and ABS algorithm via C-approach to solve Linear Programming Problems when degeneracy has been treated properly. All the methods provide the same optimal solution.the last two methods viz. ABS method and ABS Algorithm via C-approach are the new methods which are yet to be established in the field of Mathematical Programming / Operations Research. [ 65 ]