1T.T. Narendran Department of Management Studies Indian Institute of Technology Madras TTN – DoMS, IIT Madras, 17-Apr-13 2 Example Problem No. 8 : A furniture shop manufactures tables and chairs. The operations take place sequentially in two work centers. The associated profits and the man hours required by each product at each work center are shown in the table below : 52 TTN – DoMS, IIT Madras, 17-Apr-13 53 3 Furniture Shop Tables Chairs Available man hours Profit / unit Work Center-I Work Center-II 8 4 2 6 2 4 60 48 Formulate a linear programme to determine the optimal number of tables and chairs to be manufactured so as to maximize the profit. 52 TTN – DoMS, IIT Madras, 17-Apr-13 53 4 Solution to Example Problem No.8 The following linear programme is formulated This problem can be solved graphically. The following figure shows the constraints ; 50 TTN – DoMS, IIT Madras, 17-Apr-13 51 17-Apr-13 51 .12) B(12.5 The shaded region is the feasible solution set to the problem X2 A(0.0) X1 C(15. IIT Madras.6) O(0.0) 50 TTN – DoMS. 6 Before we indicate the procedure to solve this problem. IIT Madras. a few concepts are introduced. 17-Apr-13 . TTN – DoMS. IIT Madras. 17-Apr-13 .7 Convex Set: If there exists a set in which a straight line joining any two points in the set is also contained in the set. then such a set is called a convex set TTN – DoMS. IIT Madras. 17-Apr-13 . This property is used for determining the solution to the given linear programme. The expression TTN – DoMS.8 The extreme of a convex set will always lie at a corner point. 17-Apr-13 . gives the optimal solution. These are called Isoprofit lines. The corner point through which the last isoprofit line passes. The value of Z is seen to increase as the isoprofit lines move farther away from the origin. In the figure OABC. IIT Madras. the corner points are evaluated as follows: TTN – DoMS.9 A series of parallel lines for various assumed values of Z can be drawn. The profit Z along each line is the same. 12) B(12.0) X1 TTN – DoMS.Iso Profit Line 10 X2 A(0.0) .6) O(0. 17-Apr-13 C(15. IIT Madras. . the optimal solution occurs at corner point B (12.e.11 Z0 = 0 ZA = 72 ZB = 132 ZC = 120 For the given problem. 17-Apr-13 . TTN – DoMS.6) i. X2 = 6 with the corresponding profit of 132 which is the maximum. X1 = 12. IIT Madras. IIT Madras. is for a method that can solve problems of realistic size. For this purpose.12 Note: The graphical method. The need. 17-Apr-13 . however. TTN – DoMS.B. Dantzig. can solve problems with just two variables. there exists a method called the Simplex Algorithm developed by George. obviously. IIT Madras. 17-Apr-13 . Basic Solution: Consider the following set of equations : TTN – DoMS.13 A few basic concepts must be learnt before the algorithm is introduced. 17-Apr-13 . it is possible to find a set of solutions called basic solutions which are obtained as follows In the given system of equations. Hence this cannot be solved as simultaneous equations. TTN – DoMS.14 There are four variables and two equations. IIT Madras. we can solve for the remaining variables. However. if we set the 'extra' variables = 0. 15 For instance. IIT Madras. the solutions that also satisfy the non-negativity condition are called basic feasible solutions TTN – DoMS. Of these. if we set We get This is called a basic solution. . We can see that there are as many as solutions. 17-Apr-13 . IIT Madras. There will be basic solutions TTN – DoMS.m) variables = 0 and solving for the remaining variables. 17-Apr-13 .16 In general. we can obtain basic solutions by setting (n . if we have a system of m equations with n variables (where n > m). 17-Apr-13 . TTN – DoMS. The given system is seen to be in Canonical form since satisfy this condition. such a system is said to be in Canonical form. IIT Madras.17 Canonical form : If there exists a system of equations such that each equation has one variable with coefficient of 1 in that equation and a coefficient of 0 in all the other equations. IIT Madras. 17-Apr-13 .18 Other Definitions : The variables that are set = 0 are called non basic variables while the remaining variables are called basic variables. Now let us consider the same example: TTN – DoMS. 17-Apr-13 . expressing the basic variables in terms of non-basic variables. Choose the variables in canonical form. Rewrite the equations. i.. IIT Madras. TTN – DoMS.19 Remove the inequalities and rewrite the constraints as equations by introducing 'slack variables' are shown below This is a system of two equations with four variables. as the basic variables for the initial solution.e. 20 Both the non basic variables have positive coefficients in the objective function and hence have the potential to increase the value of Z . IIT Madras. 17-Apr-13 . TTN – DoMS. one of the existing basic variables must become non basic since there can be only two basic variables at any stage. To do this. To find out which variable is to be replaced. 17-Apr-13 .21 Let us now consider making X1 a basic variable. TTN – DoMS. we find the maximum possible value for X1 in equations (1) and (2). IIT Madras. IIT Madras.22 TTN – DoMS. 17-Apr-13 . only the lower value will satisfy both the constraints.23 Of these. IIT Madras. Hence TTN – DoMS. 17-Apr-13 . IIT Madras.24 TTN – DoMS. 17-Apr-13 . we see that X3 displaces X4 as basic variable. 17-Apr-13 . IIT Madras.25 When we examine for the maximum possible value of X2 in (3) and (4). TTN – DoMS. 26 Here. IIT Madras. there is no further scope for increasing the value of Z. the coefficients of both the nonbasic variables in the objective function are negative. So the final solution is TTN – DoMS. 17-Apr-13 . Hence. IIT Madras.T. Narendran Department of Management Studies Indian Institute of Technology Madras TTN – DoMS.27 T. 17-Apr-13 . The key features of the tabular form are as follows: TTN – DoMS. IIT Madras.28 Simplex Algorithm : Tabular form The algorithm explained above can be implemented in a tabular form for greater working convenience. 17-Apr-13 . This form is also easier for coding and automation purposes. 17-Apr-13 .29 Each variable is represented as column. CB = Coefficient of Basic Variable in the objective function It can be seen that the tabular form is just another way of representing the same set of equations used above TTN – DoMS. IIT Madras. The corresponding coefficients appear in the boxes of the table. 17-Apr-13 .30 TTN – DoMS. IIT Madras. X4 = 48. 17-Apr-13 . IIT Madras. Z = 0 TTN – DoMS.31 The solution at this stage reads as X3 = 60. We have to find the departing variable. TTN – DoMS. We choose X1to become the basic variable. IIT Madras.32 In the row Cj – Zj . 17-Apr-13 . This requires one of the existing basic variables to become non basic. X1is the entering variable. the positive coefficients for the non basic variables X1 and X2 indicate that the value of Z will increase if one of these variables becomes a basic variable. In other words. The ratios are 60 / 4 = 15 and 48 / 2 = 24 The minimum ratio determines the departing variable. IIT Madras. The coefficient 4 corresponding operation is called the pivot.33 Now. evaluate the ratio bi / aij for all aij >0 in this column. to this TTN – DoMS. Hence. 17-Apr-13 . in this case X3 is the departing variable. 15) We get (1 ½ ¼ 0 | TTN – DoMS. In the 'basis' column X1 replaces X3. IIT Madras. 17-Apr-13 . Divide the pivotal row (4 2 1 0 | 60) by the pivot (4).34 The next table is obtained by performing the following operations. do the following : Multiply the new row just obtained by the coefficient in the pivotal column (2). 17-Apr-13 .35 To get the X4 row. We get (2 1 ½ 0 | 30) Subtract corresponding elements from the old X4 row. That is TTN – DoMS. IIT Madras. 17-Apr-13 . IIT Madras. The Cj – Zj row is evaluated as follows : Zj = CB aij For example. Z2 = 8 (½) + 0 (3) = 4 and C2-Z2 = 6-4 = 2 In this manner. all the values of Cj – Zj are computed TTN – DoMS.36 This constitutes the new X4 row in the table. 37 TTN – DoMS. 17-Apr-13 . IIT Madras. Z = 120. the solution reads as X1 = 15. we find 18 / 6 = 3 is less than 15 / (½) = 30 TTN – DoMS. 17-Apr-13 . we now bring in X2 as the entering variable. Proceeding as before. IIT Madras. Since the X2 column in the Cj – Zj row shows a positive value.38 At the end of this iteration. X4 = 18. 3 is the pivot and X4 is the departing variable. IIT Madras. The next table is obtained using the same steps described earlier TTN – DoMS. 17-Apr-13 .39 Therefore. 17-Apr-13 . IIT Madras.40 TTN – DoMS. The given solution is the optimal solution TTN – DoMS.41 Now the solution reads X1 = 12. Therefore. 17-Apr-13 . there is no further scope for increasing the value of Z. Hence the algorithm stops at this stage. IIT Madras. X2 = 6. Z = 132 The coefficients of the non basic variables in the Cj – Zj row are all non positive. P in tabular form. IIT Madras. 17-Apr-13 . Generally.42 The steps involved in the simplex algorithm are given below : Rewrite inequalities as equations using slack variables. TTN – DoMS. Determine the entering variable as one with a positive Cj – Zj coefficient . the variable with the maximum Cj – Zj is chosen as the entering variable though this is not a strict criterion. Set up L. 17-Apr-13 .43 Determine the departing variable as the row in which the aij coefficient yields the minimum positive bi / aij (aij >0). In order to obtain the next table. IIT Madras. Divide the pivotal row throughout by the pivot. The aij corresponding to the departing variable is called the pivot. TTN – DoMS. perform the following steps. If not. Check if there is any Cj – Zj > 0. If so. TTN – DoMS.44 Multiply this new row by coefficients in the pivotal column and subtract the product from the corresponding old row. repeat steps (3) . Repeat this process till the table is complete. 17-Apr-13 .(7). read off the optimal solution from the last table. IIT Madras.