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Linear programing problem

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Linear programing problem

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Linear programing problem

  1. 1. Page | 1 Linear Programming Optimization Problem: Problems which seek to maximize or minimize a numerical function of a number of finite variables subject to certain constraints are called optimization problems Programming Problem: Programming problems deal with determining optimal allocations of limited resources to meet given objectives. The constraints or limited resources are given by linear or non-linear inequalities or equations. The given objective may be to maximize or minimize certain function of finite variables. Linear Programming and Linear Programming Problem: Suppose we have given m linear inequalities or equations in n variables and we wish to find non-negative values of these variables which will satisfy the constraints and maximize or minimize some linear functions of these variables (objective functions), then this procedure is known as linear programming and the problem which is described is known as linear programming problem. Mathematically it can be described as, suppose we have m linear inequalities or equations in n unknown variables of the form n ij j i j=1 a x { ,=, }b (i= 1, 2,....,m)  where for each constraint one and only one of the signs ,=,  holds. Now we wish to find the non-negative values of jx , j = 1, 2,………,n. which will satisfy the constraints and maximize of minimize a linear function n j j j=1 z = c x . Here ija , ib and jc are known constant. Application: (i) Linear programming problem is widely applicable in business and economic activities (ii) It is also applicable in government, military and industrial operations (iii) It is also extensively used in development of planning. Objective Function: In a linear programming problem, a linear function n j j j=1 z = c x of the variables jx , j = 1, 2,………,n. which is to be optimized is called objective function. In a objective function no constant term will be appeared. i. e. we cannot write the objective function of the type n j j j=1 z = c x +k Example of Linear Programming Problem: Machine Type Products 1 2 3 4 Total Time Available Per Week A B C 1.5 1 2.4 1 1 5 1 3.5 1.5 3 3.5 1 2000 8000 5000 Unit Profits 5.24 7.30 8.34 4.18
  2. 2. Page | 2 Suppose three types of machines A, B and C turns out four products 1, 2, 3, 4. The above table shows (i) the hours required on each machine type to produce per unit of each product (ii) total available machine hours per week, and (iii) per unit profit on sale of each of the product. Suppose jx (j = 1, 2, 3, 4) is the no. of units of product j produced per week. So we have the following linear constraints; 1 2 3 4 1 2 3 4 1 2 3 4 1.5x +x +2.4x +x 2000 x +5x +x +3.5x 8000 1.5x +3x +3.5x +x 5000    Since the amount of production cannot be negative so, jx 0 (j = 1, 2, 3, 4) . The weekly profit is given by 1 2 3 4z= 5.24x +7.3x +8.34x +4.18x . Now we wish to determine the values of the variables jx 's for which (i), (ii), (iii) and (iv) will be satisfied and (v) will be maximized Formulation of Linear Programming Problem (i) Transportation Problem: Suppose given amount of uniform product are available at each of a no. of origins say warehouse. We wish to send specified amount of the products to each of a no. of different destinations say retail stores. We are interested in determining the minimum cost-routing from warehouse to the retail stores. Let use define m = no. of warehouses n = no. of retail stores ijx the amount of product shipped from the ith warehouse to the jth retail store. Since negative amounts cannot be shipped so we have ijx 0 i, j  ia = total no. of units of the products available for shipment at the ith (i= 1, 2,………m)warehouse. jb = the no. of units of the product required at the jth retail store. Since we cannot supply more than the available amount of the product from ith warehouse to the different retail stores, therefore we have i1 i2 in ix +x +............+x a i= 1, 2,……..,m We must supply at each retail store with the no. of units desired, therefore 1j 2j mj jx +x +.............+x =b ; j = 1, 2,………….,n The total amount received at any retail store is the sum over the amounts received from each warehouse. The needs of the retail stores can be satisfied m n i i=1 j=1 a b j 
  3. 3. Page | 3 Let us define ijc is the per unit cost of shifting from ith warehouse to the jth retial store, then the total cost of shifting m n ij ij i=1 j=1 z= c x Now we wish to determine ijx which minimize the cost m n ij ij i=1 j=1 z= c x subject to the constraints i1 i2 in ix +x +............+x a 1j 2j mj jx +x +.............+x =b It is a linear programming problem in mn variables with (m+n) constraints. (2) The Diet Problem Suppose we have given the nutrient content of a no. of different foods. We have also given the minimum daily requirement for each nutrient and quantities of nutrient contained in one of each food being considered. Since we know the cost per ounce of food, the problem is to determine the diet that satisfy the minimum daily requirement of nutrient and also the minimum cost diet. Let us define m = the no. of nutrients n = the no. of foods ija = the quantity (mg) of ith nutrient per (oz) of the jith food ib = the minimum quantity of ith nutrient jc = the cost per (oz) of the jth food jx = the quantity of jth food to be purchased The total amount of ith nutrient contained in all the purchased foods cannot be less than the minimum daily requirements Therefore we have n i1 1 i2 2 in n ij j i j=1 a x +a x +............+a x = a x b The total cost for all purchased foods is given by; n j j j=1 z = c x Now our problem is to minimize cost n j j j=1 z = c x subject to the constraints n i1 1 i2 2 in n ij j i j=1 a x +a x +............+a x = a x b and jx 0 This is called the linear programming problem.
  4. 4. Page | 4 Feasible Solution: Any set of values of the variables jx which satisfies the constraints n ij j i j=1 a x { , , b   , where ija and ib are constant is called a solution to the linear programming problem and any solution which satisfies the non-negative restrictions i. e. jx 0 is called a feasible solution. Optimal Feasible Solution In a linear programming problem there is an infinite no. of feasible solutions and out of all these solutions we must find one feasible solution which optimize the objective function n j j j=1 z = c x is called optimal feasible solution In other words, any feasible solution which satisfies the following conditions; (i) n ij j i j=1 a x { , , b   (ii) jx 0 (iii) optimize objective function n j j j=1 z = c x , is called a optimal feasible solution. Corner Point Feasible Solution: A feasible solution which does not lie on the line segment, connection any other two feasible solution is called a corner point feasible solution. Properties: (i) If there is a exactly one optimal solution of the linear programming problem, then it is a corner point feasible solution. (ii) If there are more than two optimal solutions of the given problem, then at least two of them are adjacent corner points. (iii) In a linear programming problem there are a finite number of corner points (iv) If a corner point feasible solution is better than its adjacent corner point solution, then it is better than all other feasible solutions. Methods for SolvingLinear ProgrammingProblems (1) Graphical Method (2) Algebraic Method (3) SimplexMethod Graphical Method: The graphical method to solve a linear programming problem involves two basic steps (1) At the first step we have to determine the feasible solution space. We represent the values of the variable 1x to the X axis and the their corresponding values of the variable 2x to the Y axis. Any point lying in the first quadrant satisfies 1x > 0 and 2x 0 . The easiest way of accounting for the remaining constraints for optimization objective function is to replace inequalities with equations and then plot the resulting straight lines Give an example:
  5. 5. Page | 5 Next we consider the effect of the inequality. All the inequality does is to divide the 1 2(x , x ) -plane into two spaces that occur on both sides of the plotted line: one side satisfies the inequality and the other one dies not. Any point lying on or below the line satisfies the inequality. A procedure to determine the feasible side is to use the origin (0, 0) as a reference point. Step 2: At the second step we have to determine the optimal solution. Problem: Find the non-negative value of the variables 1 2x and x which satisfies the constraints 1 23x +5x 15 1 25x +2x 10 And which maximize the objective function 1 2z = 5x +3x Solution: We introduce an 1 2x x co-ordinate system. Any point lying in the first quadrant has 1 2x ,x 0 . Now we show the straight lines 1 23x +5x =15 and 1 25x +2x =10 on the graph. Any point lying on or below the line 1 23x +5x =15 satisfies the 1 23x +5x 15 . Similarly any point lying on or below the line 1 25x +2x =10 satisfies the constraint 1 25x +2x 10 B(0,3) A(1.053, 2.368) 1 23x +5x =15 O C (5,0) 1 2z = 5x +3x 1 25x +2x =10 So, the region ABOC containing the set of points satisfying both the constraints and the non negative restriction. So, the points in this region are the feasible solution. Now we wish to find the line with the largest value of 1 2z = 5x +3x which has at least one point in common with the region of feasible solution. The line is drawn in the graph above. It shows that the value of 1x and 2x at the point A are the required solution. Here 1x =1.053 and 2x 2.368 approximate. Now from the objective function we get the maximum value of z which is given by z = 5 1.053+3 2.368=12.37 
  6. 6. Page | 6 Algebraic Method: In LP problems, generally the constraints are not all equations. Since equations are easy to handle as compared to inequalities, a simple conversion is needed to make the inequalities into equality. Let us consider first, the constraints having less than or equal signs (). Any constraint of this category can be written as h1 1 h2 2 hn n ha x +a x +..............+a x b (1) Let us introduce a new variable n+hx which satisfies that n+hx 0 where n n+h h hj j j=1 x b a x 0   , to convert the inequalities to the equality such that h1 1 h2 2 hn n n+h ha x +a x +..............+a x +x b (2) The new variable n+hx is the difference between the amount available of resource and the amount actually used and it is called the slack variables. Next we consider the constraints having signs greater than or equal ( ). A typical inequality in this set can be written as; k1 1 k2 2 kn n ka x +a x +..............+a x b (3) Introducing a new variable n+kx 0 , the inequality can be written as equality which is given by; k1 1 k2 2 kn n n+k ka x +a x +..............+a x -x b (4) Here the variable n+kx is called the surplus variable, because it is the difference between resources used and the minimum amount to be produced is called the surplus. Therefore using algebraic method for solving a linear programming problem, the linear programming problem with original constraints can be transformed into a LP problem with constraints of simultaneously linear equation form by using slack and surplus variable Example: Considering the LP problem 1 2Min: -x -3x St 1 2x -2x 4 1 2-x +x 3 1 2x , x > 0 Now introducing two new variables 3 4x and x , the problem can be written as; 1 2 3 4Min: -x -3x +0.x +0.x St: 1 2 3 1 2 4 1 2 3 4 x -2x x =4 -x +x x = 3 x , x , x , x > 0   Here 3x is the slack variable and 4x is the surplus variable. Effect of Introducing Slack and Surplus Variables Suppose we have a linear programming problem 1P such that Optimize 1 1 2 2 n nZ= c x +c x +..............+c x (1) Subject to the condition h1 1 h2 2 hn n ha x +a x +..............+a x { , , }b   (2) Where one and only one of the signs in the bracket hold for each constraint The problem is converted to another linear programming problem 2P such that 1 1 2 2 n n n+1 mZ= c x +c x +..............+c x +0.x +............+0.x (3)
  7. 7. Page | 7 Subject to the condition h1 1 h2 2 hn n hn+1 n+1 hm m hAx = a x +a x +..............+a x a x ........... a x b    (4) Where  ij n m A= a  and ja ( j = 1, 2,…….,m) is the jth column of A. We claim that optimizing (3) subject to (4) with jx 0 is completely equivalent to optimizing (1) subject to (2) with jx 0 To prove this, we first note that if we have any feasible solution to the original constraints, then our method of introducing slack or surplus variables will yield a set of non-negative slack or surplus variables such that equation (4) is satisfied with all variables non-negative consequently if we have a feasible solution to (4) with all variables non-negative, then its first n components will yield a feasible solution to (2) .Thus there exist one –to-one correspondence between the feasible solutions to the original set of constraints and the feasible solution to the set of simultaneous linear equations. Now if * * * * 1 2 mX = (x x ,........,x ) 0 is a feasible optimal solution to linear programming 2P then the first n components of * X that is * * * 1 2 n(x x ,........,x ) is an optimal solution by annexing the slack and surplus variables to any optimal solution to 1P we obtain an optimal solution to 2P Therefore, wet may conclude that if slack and surplus variables having a zero cost are introduced to convert the original set of constraint into a set of simultaneous linear equations, so the resulting problem is equivalent to the original problem. Existence of Extreme Basic Feasible Solution: Reduction of any feasible solution to a basic feasible solution Let us consider a linear programming problem with m linear equations in n unknowns such that AX = b X 0 Which has at least one basic feasible solution without loss of generality suppose that Rank(A) = m and let 1 2 nX=(x , x ,......,x )be as feasible solution. Further suppose that 1 2 px , x ,......,x >0 and that p+1 P+2 nx , x ,......,x =0 . And let 1 2 pa , a ,......,a be the respective columns of A corresponding to the variables 1 2 px , x ,......,x . If 1 2 pa , a ,......,a are linearly independent then X is a basic feasible solution. in such case p m . If p=m from the theory of system of linear equation, the solution is non-degeneracy basic feasible solution. If p<m, the system have a degenerate basic feasible solution with (m-p) of the basic variables are equal to zero. If 1 2 pa , a ,......,a are dependent then there exist scalars 1 2, ,......, p   with at least one positive
  8. 8. Page | 8 j such that 1 0 p j j j a    Considering the following point X with j 0 j x ; 1,2,...., x = 0; 1, 2,..... j j p j p p n         where j k o j=1,2,....,p k x x =Minimum ; 0 = >0j j             If j 0  , then jx >0 , since both jx and 0 are positive. If 0j  , then by the definition of 0 we have j o j 0 x x j j        . Thus jx >0 Furthermore k k k 0 k k x x = x - =x - =0k k     . Hence x has at most (p-1) positive components. Also, n j j j=1 n j j 0 j=1 n n j j 0 j j=1 j=1 Ax = a x = a (x ) = a x a = b j j             Thus we have a constructed feasible solution x since Ax =b , x 0  with at most (p-1) positive components. If the columns of A corresponding to these positive components are linearly independent then x is basic feasible solution. Otherwise the process is repeated. Eventually a basic feasible solution (BFS) will be obtained. Example: Consider the following inequalities 1 2 2 1 2 x +x 6 x 3 x , x 0    Find basic solution, BFS and extreme points. Solution. By introducing slack variables 3 4x and x , the problem is put into the following standard format 1 2 3 2 4 1 2 3 4 x +x x =6 x x =3 x , x ,x ,x 0    So, the constraint matrix A is given by; 1 1 1 0 A = 0 1 0 1       = 1 2 3 4(a , a , a , a ) , 6 b= 3       Rank(A) = 2
  9. 9. Page | 9 Therefore, the basic solutions corresponding to finding a 2 2 basis B. Following are the possible ways of extracting B out of A (i) 1 2 1 1 B=(a , a ) = 0 1       , -1 1 -1 B = 0 1       , -1 B 1 -1 6 3 x =B b= = 0 1 3 3                , 3 n 4 x 0 x = = x 0            (ii) 1 3 1 1 B=(a , a )= 0 0       , Since |B|=0, it is not possible to find -1 B and hence Bx (iii) 1 4 1 0 B=(a , a )= 0 1       ; -1 1 0 B = 0 1       21 -1 B n 34 xx 1 0 6 6 0 x = =B b= = x = = xx 0 1 3 3 0                               (iv) 2 3 1 1 B=(a , a )= 1 0       -1 0 1 B = 1 1       2 1-1 B n 3 4 x x0 1 6 3 0 x = =B b= = x = = x x1 1 3 3 0                              (v) 2 4 1 0 B=(a , a )= 1 1       ; -1 1 0 B = -1 1       ; 12 -1 B n 34 xx 1 0 6 6 0 x = =B b= = x = = xx -1 1 3 -3 0                               (vi) 3 4 1 0 B=(a , a )= 0 1       ; -1 1 0 B = 0 1       ; 3 1-1 B n 4 2 x x1 0 6 6 0 x = =B b= = x = = x x0 1 3 3 0                               Hence we have the following five basic solutions 1 3 3 x = 0 0              ; 2 6 0 x = 0 3              ; 3 0 3 x = 3 0              ; 4 0 6 x = 0 -3              ; 5 0 0 x = 6 3              Of which except 4x are BFS because it violates non-negativity restrictions. The BFS belong to a four dimensional space. These basic feasible solutions are projected in the 1 2(x , x ) space gives rise to the following four points. 3 6 0 0 0 , , , 3 0 3 6 0                            From the graphical representation the extreme points are (0, 0), (0, 3), (3, 3) and (6,0) which are the same as the BFSs. Therefore the extreme points are precisely the BFS. The no. of BFS is 4 less than 6. The SimplexMethod: General Mathematical Formulation for Linear Programming Let us define the objective function which to be optimized 1 1 2 2 n nz = c x +c x +...................+c x We have to find the values of the decision variables 1 2 nx , x ,.........,x on the basis of the following m constraints;
  10. 10. Page | 10 11 1 12 2 1n n 1 21 1 22 2 2n n 2 m1 1 m2 2 mn n m a x +a x +.........+a x ( ,=, )b a x +a x +.........+a x ( ,=, )b a x +a x +.........+a x ( ,=, )b       and jx 0; j = 1, 2,.......,n The above formulation can be written as the following compact form by using the summation sign; Optimize (maximize or minimize) n j j j=1 z = c x Subject to the conditions; n ij j i j=1 a x ( ,=, )b ;i=1, 2,.......,m  and jx 0; j = 1, 2,.......,n The constants jc ; j =1, 2,......,n are called the cost coefficients; the constants ib ; i =1, 2,.......,m are called stipulations and the constants ija ; i =1, 2,.....,m; j=1,2,.....,n are called structural coefficients. In matrix notation the above equations can be written as; Optimize z = CX Subject to the conditions AX( ,=, )B  Where  1 2 n 1 n C= c c ... ... c  ; 1 2 n n 1 x x . X= . . . x                        ; 11 12 1n 21 22 2n m1 m2 mn m n a a ...... a a a ...... a A= . . . . . . . . a a ...... a                  ; 1 2 m m n b b . B= . . . b                        Where, A is called the coefficient matrix, X is called the decision vector, B is called the requirement vector and C is called the cost vector of linear programming problem The Standard Form of LP Problem The use of basic solutions to solve the general LP models requires putting the problem in standard form. The followings are the characteristics of the standard form (i) All the constraints are expressed in the form of equations except the non-negative restrictions on the decision variables which remain inequalities (ii) The right hand side of each constraint equation is non-negative
  11. 11. Page | 11 (iii) All the decision variables are non-negative (iv) The objective function may be of the maximization or the minimization type Conversionof Inequalities into Equations: The inequality constraint of the type ,( )  can be converted to an equation by adding or subtracting a variable from the left-hand sides of such constraints. These new variables are called the slack variables or simply slacks. They are added if the constraints are of the  types and subtracted if the constraints are of the  types. Since in the cases of  type the subtracted variables represent the surplus of the left-hand side over right-hand side, it is commonly known as the surplus variables and is in fact a negative slack. For example 1 2 1x +x b Is equivalent to 1 2 1 1x +x s = b If 1 2 2x +x b Is equivalent to 1 2 1 2x +x =bs The general LP problem that discussed above can be expressed as the following standard form; n j j j=1 z = c x Subject to the conditions n ij j i i j=1 a x s =b ;i=1, 2,.......,m jx 0; j = 1, 2,.......,n And is 0; i = 1, 2,.....,m In the matrix notation, the general LP problem can be written as the following standard form; Optimize z = CX Subject to the conditions AX S = B X 0 S 0 Example: Express the following LP problem in a standard form; Maximize 1 2z= 3x +2x Subject to the conditions; 1 2 1 2 1 2 2x +x 2 3x +4x 12 x ,x 0    Solution: Introducing slack and surplus variables, the problem can be expressed as the standard form and is given below; Maximize 1 2z= 3x +2x Subject to the conditions;
  12. 12. Page | 12 1 2 1 1 2 2 1 2 1 2 2x +x =2 3x +4x =12 x ,x , , 0 s s s s    Conversionof Unrestricted Variable into Non-negative Variables An unrestricted variable jx can be expressed in terms of two non-negative variables by using substitution such that + - j j jx = x -x ; + - j jx ,x 0 For example, if jx = -10 , then + - j jx =0, and x = 10. If jx = 10 , then + - j jx =10, and x = 0. The substitution is effected in all constraints and in the objective function. After solving the problem in terms of + jx and - jx , the value of the original variable jx is then determined through back substitution. Example: Express the following linear programming problem in the standard form; Maximize, 1 2 3z= 3x +2x +5x Subject to 1 2 1 2 3 1 3 2x -3x 3 x +2x 3x 5 3x +2x 2     Solution: Here 1x and 2x are restricted to be non-negative while 3x is unrestricted. Let us express as, + - 3 3 3x = x -x where, + - 3 3x 0 and x 0  . Now introducing slack and surplus variable the problem can be written as the standard form which is given by; Maximize, + - 1 2 3 3z= 3x +2x +5(x -x ) Subject to the conditions; 1 2 1 + - 1 2 3 3 2 + - 1 3 3 3 + - 1 2 3 3 1 2 3 2x -3x 3 x +2x 3x -3x =5 3x +2x -2x =2 x ,x ,x , x , , , 0 s s s s s s       Conversionof Maximization to Minimization: The maximization of a function 1 2 nf(x , x ,.....,x ) is equivalent to the minimization of 1 2 n-f(x , x ,.....,x ) in the sense that both problems yield the same optimal values of 1 2x ,x ,......, and nx

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