1. 5.2 Iteration  We are at an extreme point of the feasible region  We want to move to an adjacent extreme point.  We want to move to a better extreme point.  Observation:  A pair of basic feasible solutions which differ only in that a basic and non-basic variable are interchanged corresponds to adjacent feasible extreme points.

2. Geometry Two hyperplane (constraints) in common Only one hyperplane in common (adjacent) (not adjacent)

3. Moving to an adjacent extreme point  Step 1: – Select which nonbasic variable becomes basic  Step 2: – Determine which basic variable becomes nonbasic  Step 3: – Reconstruct a new canonical form reflecting this change

4. Simplex Tableau  It is convenient to describe how the Simplex Method works using a table (=tableau).  There are a number of different layouts for these tables.  All of us shall use the layout specified in the lecture notes.

5. Observation  It is convenient to incorporate the objective function into the formulation as a functional constraint.  We can do this by viewing z, the value of the objective function, as a decision variable, and introduce the additional constraint  z =  j=1,...,n c j x j or equivalently  z - c 1 x 1 - c 2 x 2 -... - c n x n = 0

7.  Terminology: We refer to the last row as the Z-row, and to the coefficient of x their as reduced costs. For example, the reduced cost of x 1 is . Tableau (5.10)

8. Step 1: Selecting a new basic variable  Issue: – which one of the current non-basic variables should add to the basis?  Observation: The Z-row tells us how the value of the objective function (Z) changes as we change the decision variables: z - c 1 x 1 - c 2 x 2 -... - c n x n = 0

9. Because all the nonbasic variables are equal to zero, if we decide to add x j to the basis we must have z - c j x j = 0 namely z = c j x j

10.  Since we try to maximize the objective function, it would be better to select a non- basic variable with a large (positive) cost coefficient (large cj).  Thus, if we do the selection via the reduced costs, we will prefer a variable with a negative reduced cost.

11. ConclusionConclusion  If we maximize the objective function, to improve (increase) the value of the objective function we have to select a non- basic variable whose reduced cost is negative.

12. Greedy Rule Select the non-basic variable with the most negative reduced cost

13. Example (Continued) The most negative reduced cost in the Z-row is  4, corresponding to j=1. Thus, we select x 1 as the new basic variable.

14. Step 2: Determining the new nonbasic variable  Suppose we decided to select x j as a new basic variable.  Since the number of basic variables is fixed (m), we have to take one variable out of the basis.  Which one?

15. Observation  As we increase x j from zero, sooner or later one or more of the basic variables will become negative.  We can thus take the first such variable out of the basis and set it to zero.

16. Example (continued)  Suppose we select x 1 as the new basic variable.  Since x 2 is a nonbasic variable, its value is zero. Thus the above system can be simplified!

17.  Each equation involves only two variables: – The new basic variable (x 1 ) – The old basic variable associated with the respective constraint.  We can thus express the old basic variables in terms of the new one! x 3 = 40 - 2x 1 x 4 = 30 - x 1 x 5 = 15 - x 1

18.  We can now compute the critical values of the new basic variable (x 1 ), namely the values for which the old basic variables will reach zero:  We had: x 3 = 40 - 2x 1 x 4 = 30 - x 1 x 5 = 15 - x 1 We take x 5 out of the basis.

19. thus the critical values are obtained from: 0 = 40 - 2x 1 (x 1 *=20) 0 = 30 - x 1 (x 1 *=30) 0 = 15 - x 1 (x 1 *=15) Conclusions: The critical value of x 1 is 15. We take x 5 out of the basis.

20. More generally....  If we select x j as the new basic variable, then for each of the functional constraints we have  a ij x j + x i = b i (i=1,2,...,m)  where x i is the old basic variable associated with constraint i.

21.  Thus, x i = b i - a ij x j so that the critical values of x j are determined by setting the x i ’s to zero: 0 = b i - a ij x j (i=1,2,...,m)

22. Bottom Line  Question:  Why aren’t we interested in rows for which a ij  0 ??? > 0,

23. Ratio Test Given the new basic variable x j, take out of the basis the old basic variable corresponding to row i where the following ratio attains its smallest value: ( > 0 )

24. Example (Continued)  Take x 5 out of the basis.

25. Step 3: Restore The Canonical Form  We interchanged a basic variable with a nonbasic variable  We have a new basis  We have to construct the simplex tableau for the new set-up  This is done by one pivot operation

26. Example (Continued) 1 Old New

27. How do we “read” a Simplex Tableau ?  New basis: (x 3,x 4,x 1 )  New basic feasible solution: x = (15,0,10,15,0)

28.  New value of objective function: z = 60