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.

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

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

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.

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 c n x n = 0

 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)

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 c n x n = 0

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

 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.

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.

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

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.

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?

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.

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!

 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 = x 1 x 4 = 30 - x 1 x 5 = 15 - x 1

 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 = x 1 x 4 = 30 - x 1 x 5 = 15 - x 1 We take x 5 out of the basis.

thus the critical values are obtained from: 0 = x 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.

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.

 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)

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

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 )

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

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

Example (Continued) 1 Old New

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)

 New value of objective function: z = 60