a simplex

PIVOT Constraint Objective Function Simplex Tableau Standard Maximizing Problem
Hopping phase I Objective row Ratios (Quotients) Identity Sub-Matrix (ISM)
Indicator Z-column basic solution Final Tableau pivoting , pivot operation

A linear programming problem (or linear program) is a set of (linear) inequalities (with a solution set S) and a (linear) function (often cost or profit) whose value (within S) is to be maximized or minimized.
A STANDARD MAXIMIZING PROBLEM is a linear programming problem
which satisfies all of the following 4 CONDITIONS (Rolf, pg.263) :
[C1] The objective function is to be maximized.
[C2] All inequalities are of the type.
[C3] All right hand constants are non-negative.
[C4] All variables are non-negative.
A NON-STANDARD PROBLEM is simply a problem which is not standard, and hence fails to satisfy at least one of [C1] through [C4] above. In our text, [C4] is always true, which simplifies discussions.

Reference : An example of how to apply the following procedure to a non-standard problem is available, with abundant comments and cross-references.

Reference : Many EXERCIZES are available for each step of this method.

Step NS-1
Convert your problem into one satisfying conditions C1,C2, and C4 described above. How to do this is explained in Rolf (pg 307-308).
see example
of this step?

Step NS-2
Convert all constraints to equations with slack variables, and then write the problem as a tableau with some negative right sides, with or without a Z-COLUMN.

The basic solution for a tableau with some negative right sides is a point like A or B in the figure above : it will not be a corner of the RED shaded solution set, but rather will be an intersection of extended boundaries of that set. Our first task will be to locate a corner point of the actual solution set : this task might be called PHASE I and is described here : it differs from the procedure for a standard problem only in STEPS NS-1 and NS-3.
After locating some corner of the solution set (shaded above), we must set out to locate the BEST corner : this second and final stage of the solution is called PHASE II, and is identical to the procedure for a standard problem.
PHASE I, as described in Rolf (Step 6, Pg 323), allows the freedom to choose any number as pivot. On tests, you must follow the more restrictive PHASE I procedure given below, not the unrestricted procedure in Rolf.

FINDING THE PIVOT (steps NS-3 through NS-6)

Step NS-3
Locate the row containing the most negative right-hand number. This row will be called the INDICATOR ROW. If no right-hand number is negative, then your tableau is STANDARD.

Step NS-4
INDICTORS will consist of all numbers in the row found in NS-3, except the right-most number; the PIVOT COLUMN will contain the most negative of these indicators. If no indicator is negative, then there is no pivot column, and the problem is unsolvable.

Step NS-5
Form RATIOS or QUOTIENTS for all (non-objective) rows : for each row, divide the right-most number by the number in the pivot column.

Step NS-6
The PIVOT will be in the ROW with smallest non-negative ratio. Note that 0(+1) and 0(-1) are both numerically zero, but in calculating RATIOS, consider 0(+1) as positive (OK), and 0(-1) as negative (not OK).

Step NS-7
Perform a pivot transformation on the above pivot (Rolf, Pg 92). Check out the PIVOT ENGINE to speed pactice.

Step NS-8
If all right-most (non-objective row) entries are non-negative, then phase I is ended. PHASE II now consists of applying steps 3-9 of the standard maximizing procedure to the new tableau obtained in step NS-7 above. Note: PHASE II is described as step #8 on Page 323 of our text, Rolf.

Step NS-9
Otherwise, if some right-most (non-objective row) entry is negative, apply steps NS-3 through NS-9 to the new tableau obtained in NS-7.