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228   CHAPTER 5 LINEAR PROGRAMMING: THE SIMPLEX METHOD


                                     Interpreting the Optimal Solution
                                     Using the final simplex tableau, we find the optimal solution to the HighTech
                                     problem consists of the basic variables x 1 , x 2 and s 2 and nonbasic variables s 1 and
                                     s 3 with:
                                                                      x 1 ¼ 30
                                                                      x 2 ¼ 12
                                                                     s 1 ¼ 0
                                                                     s 2 ¼ 8
                                                                     s 3 ¼ 0

                                     The value of the objective function is E1980. If management wants to maximize the
                                     total profit contribution, HighTech should produce 30 units of the Deskpro and 12
                                     units of the UltraPortable. When s 2 ¼ 8, management should note that there will be
                                     eight unused UltraPortable display units. Moreover, because s 1 ¼ 0 and s 3 ¼ 0, no
                                     slack is associated with the assembly time constraint and the warehouse capacity
                                     constraint; in other words, these constraints are both binding. Consequently, if it is
                                     possible to obtain additional assembly time and/or additional warehouse space,
                    The first computer-based  management should consider doing so.
                    simplex application was  Figure 5.3 shows the computer solution to the HighTech problem using Excel.
                    in 1951 when the US Air  The optimal solution with x 1 ¼ 30 and x 2 ¼ 12 is showntohave anobjective
                    Force looked at the
                    deployment and support  function value of E1980. The values of the slack variables complete the optimal
                    of aircraft. The problem  solution with s 1 ¼ 0, s 2 ¼ 8and s 3 ¼ 0. The values in the Reduced Costs column
                    had 71 variables and 48
                                     are from the net evaluation row of the final simplex tableau. Note that the c j – z j
                    constraints. There were
                    73 simplex iterations  values in columns corresponding to x 1 and x 2 are both 0. The shadow prices are
                    needed to find the  the z j values for the three slack variables in the final simplex tableau. Referring
                    optimal solution and this  to the final tableau, we see that the shadow price for constraint 1 is the z j value
                    took the computer 18  corresponding to s 1 ¼ 2.8. Similarly, the shadow price for constraint 2 is 0, and
                    hours.
                                     the dual price for constraint 3 ¼ 5.2. The use of the Simplex method to calculate
                                     shadow (dual) prices will be discussed further when we cover sensitivity analysis
                                     in Chapter 6.


                                     Summary of the Simplex Method
                                     Let us now summarize the steps followed to solve a linear programme using the
                                     Simplex method. We assume that the problem has all less-than-or-equal-to con-
                                     straints and involves maximization.
                                       Step 1. Formulate a linear programming model of the problem.
                                       Step 2. Add slack variables to each constraint to obtain standard form.
                                       Step 3. Set up the initial simplex tableau.
                                       Step 4. Choose the nonbasic variable with the largest entry in the net evaluation
                                              row to bring into the basis. This variable identifies the pivot column: the
                                              column associated with the incoming variable.
                                       Step 5. Choose as the pivot row that row with the smallest ratio of b i /a ij for a ij > 0
                                             where j is the pivot column. This pivot row is the row of the variables
                                             leaving the basis when variable j enters.
                                       Step 6. Perform the necessary elementary row operations.
                                              a. Divide each element of the pivot row by the pivot element (the
                                                element in the pivot row and pivot column).







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