Module 2
Lesson 4
Linear
Programming:
Method (Maximization)
Simplex
4.1 Introduction
The procedure for solving a Linear Programming Problem by the Simplex Method involves an
iteration of one basic feasible solution to another. An initial simplex tableau will be constructed
out of the different kinds of variables together with their coefficients from the modified set of
equality constraints which is called the new program of equality constraints. This tabular
presentation of the solution is derived from the concept of reducing the augmented and
elementary matrices to its echelon form.
In any type of linear programming model, whether maximization or minimization, the following
assumptions should be taken into consideration.
1. A set of variables: basic, slack, and artificial variables are constrained to be non-negative,
since these variables would represent measurable quantities.
2. A set of linear constraints, each of which contains a ≤, ≥ or = signs, is converted to a new
program of equality constraints for the purpose of transferring its coefficients and variables
to the simplex tableau.
3. A linear objective function is either maximization or minimization.
4. The basic variables, usually denoted by X1, X2, …., Xn, represent the product mix; or i.e.,
the desired units that will yield an optimum or feasible solution.
5. The slack variables, denoted by S1, S2, …. Sj, represent the unused resources or the
surplus and, therefore, these will demonstrate negligible or zero profit or cost contribution.
6. The artificial variables, denoted by A1, A2, …., Aj, represent the most expensive
resources substitutes, thus they will demonstrate the highest cost contribution to the
objective function. It is suggested that students will use any of the powers of 10 (e.g., 10,
100, 1,000, 10,000…) for the coefficient of A, in the objective function as long as it is the
highest among the given coefficients of the variables.
4.2. Steps in Solving Linear Programming Problems by the Simplex Method:
Maximization Problems
Step 1: Represent the unknowns by the basic variables X1, X2,…….Xn (any other letters can be
used) and indicate the units for measuring these unknown quantities.
Step 2: State the objective function (which is usually maximizing profit or minimizing cost) and set
up the algebraic expressions describing the problem constraints with proper inequality or
equality signs.
Step 3: Convert the constraints to a new program of equality constraints with these rules:
i)
If the symbol is “≤”, then add slack variable Sj. to the left side then change the sign
to equality “=”.
ii)
If the symbol is “=’, then add slack variable Sj to the left side and retain the equality
“=” sign.
iii)
If the symbol is “≥”, then multiply both sides by (-1), add slack variable, Sj to the left
side and change the sign to equality “=” sign.
Step 4:
Construct the initial simplex tableau by entering the variables and their coefficients
in the following standard form.
Contribution to
Feasible
Constant
Basic
Slack
profit
product
variable
Variable
Variable
Pj
Objective
coefficient row
Product
Qty.
B1
B2
B3
Zj
Pj - Zj
X1
X2
S1, S2,..Sn
Variable row
C11 C12
C21
C31
C22
C32
Constraints
coefficient row
Total
profits
row
Net profit row
For a simplex maximization tableau, it is just logical to start entering the slack variables in the
Product column since slack variable Sj has zero profit contributions. It is a good start to have zero
total profit along the Zj row so that the subsequent tableau will demonstrate an increasing total
profit along the Zj row as we develop the solution of the previous simplex tableau until we reach
the so-called optimum table that shows the maximum profit.
Step 5: Determine the entries for Zj row and (Pj – Zj) for a maximization problem and every variable
column entry. Zj row entries are obtained by taking the sum of the products of Pj column
entries and every variable column entries.
Step 6: Determine the entering variable of the optimum column, i.e., the column with the highest
positive entry of Pj – Zj row. Encircle the optimum column.
Step 7: Determine the outgoing variable of the Pivot row, i.e., the row with the least positive
quotient between the quantity column entries and optimum column entries. (Disregard
24 −3 0
ratios like 0 , 4 . 5). Encircle the Pivot row.
Step 8: Determine the New Pivotal Row by dividing each pivotal row entries with the pivot
element. This pivot element is the entry along intersection of pivot row and optimum
column. Encircle the New Pivot Row.
*New Pivot Row = Pivot Row ÷ Pivot element
Step 9: Determine the New Remaining Rows. This is done by taking the sum of the entries from
an old row and new pivot row multiplied by the negative of an element along the
intersection of Optimum column and the Old row which is being evaluated. Notice that in
Steps 8 and 9, the concept of reducing matrix to echelon form is partially performed
whenever we convert the column entries to zero and one.
*New Rj = Old Rj + (-e)(NPR)
Step 10:
Construct the next simplex tableau by entering the value obtained from Step 8 and
9 and replacing the so-called outgoing variable of the pivot with the entering variable of
the Optimum column from the previous simplex tableau. Complete this simplex tableau by
recalculating the Zj row and Pj – Zj row. If there is still positive entry found on the Pj – Zj row
then go back to Step 6 to 10 in reconstructing the simplex tableau. Otherwise, if there is
no positive entry found on the Pj – Zj row, then the process stops for the final or optimum
table has been obtained. Notice that in this Optimum Table, the product column entries
and the Quantity Column entries reveal the optimum solution, together with the desired
objective. State the decision.
Example 1:
Maximize: P = 10x + 10y + 5z
New Objective function: 10x + 10y + 5z + 0S1 + 0S2 + 0S3 + 0S4
Subject to:
Explicit Constraints:
1. 2x + y + 4z ≥ 15
2.
2x ≤ 50
3.
3y ≤ 40
4.
y + 2z ≤ 30
New Program of constraints: (see Step 3)
-2x - y – 4z + S1 = -15
2x + S2 = 50
3y + S3 = 40
.
y + 2z + S4= 30
Implicit constraints:
x + y+ z ≥ 0
x + y + z + S1 + S2 + S3 + S4 ≥ 0
Step 4 to 7: See attached excel file.
DO-IT-YOURSELF EXERCISES
Example 2:
Maximize P=10x + 8y
Subject to:
1) 7x + y ≤ 630
2) 2x + 5y ≤ 600
3) x + 2y ≥ 708
Implicit constraints:
x≥0
y≥0
Example 3:
An appliance manufacturer produces two models of microwave oven: x and y. Both models
require fabrication and assembly work. Each unit of model x uses four hours of fabrication and
two hours of assembly. On the other hand, each unit of model y uses two hours of fabrication and
four hours of assembly. There are 60 fabrication hours and 48 assembly hours available this
week. Each model x contributes P8 to profit and each model y contributes P6 to profit. What mix
of model x and y will maximize profit?