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Unformatted text preview: PERFORMANCE EXCELLENCE
IN THE WOOD PRODUCTS INDUSTRY EM 8720-E
$3.00 Using the Simplex Method to Solve Linear
Programming Maximization Problems
J. Reeb and S. Leavengood
A key problem faced by managers is how to allocate scarce resources
among activities or projects. Linear programming, or LP, is a method of
allocating resources in an optimal way. It is one of the most widely used
operations research (OR) tools. It has been used successfully as a decisionmaking aid in almost all industries, and in financial and service organizations.
Programming refers to mathematical programming. In this context, it
refers to a planning process that allocates resources—labor, materials,
machines, and capital—in the best possible (optimal) way so that costs are
minimized or profits are maximized. In LP, these resources are known as
decision variables. The criterion for selecting the best
About this series
values of the decision variAccording to the Operations Research Society of
ables (e.g., to maximize profits
America, “Operations research [OR] is concerned with
or minimize costs) is known as
scientifically deciding how to best design and operate
the objective function. The
man-machine systems, usually under conditions
limitations on resource availrequiring the allocation of scarce resources.”
ability form what is known as
This publication, part of a series, should be useful
a constraint set.
for supervisors, lead people, middle managers, and
For example, let’s say a
anyone who has planning responsibility for either a
furniture manufacturer prosingle manufacturing facility or for corporate planning
duces wooden tables and
over multiple facilities. Although managers and
chairs. Unit profit for tables is
planners in other industries can learn about OR
$6, and unit profit for chairs is
techniques through this series, practical examples are
$8. To simplify our discussion,
geared toward the wood products industry.
let’s assume the only two
resources the company uses to
produce tables and chairs are wood (board feet) and labor (hours). It takes
30 bf and 5 hours to make a table, and 20 bf and 10 hours to make a chair. James E. Reeb, Extension forest products manufacturing specialist; and Scott Leavengood, Extension
agent, Klamath County; Oregon State University. OPERATIONS RESEARCH Decision variables. . . “The resources
Constraint set. . . “The limitations on
Objective function. . . “The criterion for
selecting the best
values of the decision
variables.” 2 There are 300 bf of wood available and 110 hours of labor available. The company wishes to maximize profit, so profit maximization becomes the objective function. The resources (wood and
labor) are the decision variables. The limitations on resource
availability (300 bf of wood and 110 hours of labor) form the
constraint set, or operating rules that govern the process. Using LP,
management can decide how to allocate the limited resources to
The “linear” part of the name refers to the following:
• The objective function (i.e., maximization or minimization) can
be described by a linear function of the decision variables, that
is, a mathematical function involving only the first powers of the
variables with no cross products. For example, 23X2 and 4X16 are
valid decision variable terms, while 23X22, 4X163, and (4X1 * 2X1)
are not. The entire problem can be expressed as straight lines,
planes, or similar geometrical figures.
• The constraint set can be expressed as a set of linear equations.
In addition to the linear requirements, nonnegativity conditions
state that the variables cannot assume negative values. It is not
possible to have negative resources. Without these conditions, it
would be mathematically possible to use more resources than are
In EM 8719, Using the Graphical Method to Solve Linear
Programs, we use the graphical method to solve an LP problem
involving resource allocation and profit maximization for a furniture manufacturer. In that example, there were only two variables
(wood and labor), which made it possible to solve the problem
Problems with three variables also can be graphed, but threedimensional graphs quickly become cumbersome. Problems with
more than three variables cannot be graphed. Most real-world
problems contain numerous objective criteria and resources, so
they’re too complicated to represent with only two or three variables. Thus, for all practical purposes, the graphical method for
solving LP problems is used only to help students better understand how other LP solution procedures work.
This publication will build on the example of the furniture
company by introducing a way to solve a more complex LP problem. The method we will use is the simplex method. SIMPLEX METHOD Overview of the simplex method
The simplex method is the most common way to solve large LP
problems. Simplex is a mathematical term. In one dimension, a
simplex is a line segment connecting two points. In two dimensions, a simplex is a triangle formed by joining the points. A threedimensional simplex is a four-sided pyramid having four corners.
The underlying concepts are geometrical, but the solution algorithm, developed by George Dantzig in 1947, is an algebraic
As with the graphical method, the simplex method finds the
most attractive corner of the feasible region to solve the LP problem. Remember, any LP problem having a solution must have an
optimal solution that corresponds to a corner, although there may
be multiple or alternative optimal solutions.
Simplex usually starts at the corner that represents doing nothing. It moves to the neighboring corner that best improves the
solution. It does this over and over again, making the greatest
possible improvement each time. When no more improvements
can be made, the most attractive corner corresponding to the
optimal solution has been found.
A moderately sized LP with 10 products and 10 resource constraints would involve nearly 200,000 corners. An LP problem
10 times this size would have more than a trillion corners. Fortunately, the search procedure for the simplex method is efficient
enough that only about 20 of the 200,000 corners are searched to
find the optimal solution.
In the real world, computer software is used to solve LP problems using the simplex method, but you will better understand the
results if you understand how the simplex method works. The
example in this publication will help you do so. PROFITS Review of the graphical method
First, let’s quickly review the graphical procedure for solving an
LP problem, which is presented in EM 8719, Using the Graphical
Method to Solve Linear Programs. Let’s say a furniture manufacturer wishes to maximize profit. Information about available
resources (board feet of wood and hours of labor) and the objective criterion is presented in Table 1. For a complete, step-by-step
review of the graphical method, see EM 8719 or one of the textbooks listed in the “For more information” section. 3 OPERATIONS RESEARCH Tip . .
In our example,
X1 refers to tables,
X2 refers to chairs, and
Z refers to profit. Table 1.—Information for the wooden tables and chairs linear programming
Labor (hr) Table (X1)
5 Chair (X2)
10 Unit profit $6 Available
110 $8 Maximize: Z = 6X1 + 8X2 (objective function) Subject to: 30X1 + 20X2 < 300 (wood constraint: 300 bf available) 5X1 + 10X2 < 110 (labor constraint: 110 hours available) X1, X2 > 0 (nonnegativity conditions) Based on the above information, graphically solve the LP
(Figure 1). Graph the two constraint equation lines. Then plot two
objective function lines by arbitrarily setting Z = 48 and Z = 72 to
find the direction to move to determine the most attractive corner.
The coordinates for the most attractive corner (where the wood and
labor constraint equations intersect) can be found by simultaneously solving the constraint equations with two unknowns. Chairs Wood Labor
region Tables Figure 1.—Determining the most attractive corner corresponding to the optimal
solution. 4 SIMPLEX METHOD To simultaneously solve the two constraint equations, first
multiply the labor equation by -2, and add it to the wood equation:
30X1 + 20X2 = 300 (wood)
-2(5X1 + 10X2 = 110) (labor)
20X1 + 0
X1 = 80
= 4 tables Next, substitute into either of the constraint equations to find the
number of chairs. We can substitute into both equations to illustrate
that the same value is found.
Wood constraint Labor constraint 30(4) + 20X2 = 300 5(4) + 10X2 = 110 120 + 20X2 = 300 20 + 10X2 = 110 20X2 = 300 - 120 10X2 = 110 - 20 20X2 = 180 10X2 = 90 X2 = 180/20 X2 = 90/10 X2 = 9 chairs X2 = 9 chairs Now, determine the value of the objective function for the
optimal solution. Substitute into the equation the number of tables
and chairs, and solve for Z.
Z = $6(4) + $8(9) = $96 The optimal solution is to manufacture four tables and nine
chairs for a profit of $96. 5 OPERATIONS RESEARCH Using the simplex method
By introducing the idea of slack variables (unused resources) to
the tables and chairs problem, we can add two more variables to
the problem. With four variables, we can’t solve the LP problem
graphically. We’ll need to use the simplex method to solve this
more complex problem.
We’ll briefly present the steps involved in using the simplex
method before working through an example. Table 2 shows an
example of a simplex tableau. Although these steps will give you a
general overview of the procedure, you’ll probably find that they
become much more understandable as you work through the
A list of shortcuts is found on page 23. You can refer to the six
steps and shortcuts while working through the example. Step 1. Formulate the LP and construct a simplex tableau. Add
slack variables to represent unused resources, thus eliminating
inequality constraints. Construct the simplex tableau—a table that
allows you to evaluate various combinations of resources to
determine which mix will most improve your solution. Use the
slack variables in the starting basic variable mix.
Step 2. Find the sacrifice and improvement rows. Values in the
sacrifice row indicate what will be lost in per-unit profit by making
a change in the resource allocation mix. Values in the improvement
row indicate what will be gained in per-unit profit by making a
Table 2.—Example of a simplex tableau.
0 0 Basic mix
0 X1 X2 SW SL Solution SW
Improvement 6 0 ➝ Current profit
— SIMPLEX METHOD Step 3. Apply the entry criteria. Find the entering variable and
mark the top of its column with an arrow pointing down. The
entering variable is defined as the current non-basic variable that
will most improve the objective if its value is increased from 0. If
ties occur, arbitrarily choose one as the entering variable. When no
improvement can be found, the optimal solution is represented by
the current tableau.
If no positive number appears in the entering variable’s column,
this indicates that one or more constraints are unbounded. Since it
is impossible to have an unlimited supply of a resource, an
unbounded solution indicates that the LP problem was formulated
Step 4. Apply the exit criteria. Using the current tableau’s
exchange coefficient from the entering variable column, calculate
the following exchange ratio for each row as:
Solution value/Exchange coefficient The exchange ratio tells you which variable is the limiting
resource, i.e., the resource that would run out first.
Find the lowest nonzero and nonnegative value. This variable is
the limiting resource. The basic variable in this row becomes the
exiting variable. In case of identical alternatives, arbitrarily choose
one. Mark the exiting variable row with an arrow pointing left. Step 5. Construct a new simplex tableau. Constructing a new
tableau is a way to evaluate a new corner. One variable will enter
the basic mix (entering variable), and one variable will leave the
basic mix and become a non-basic variable (exiting variable). The
operation of an entering variable and an exiting variable is called a
pivot operation. The simplex method is made up of a sequence of
such pivots. The pivot identifies the next corner to be evaluated.
The new basic mix always differs from the previous basic mix by
one variable (exiting variable being replaced by the entering
To construct the new tableau, replace the exiting variable in the
basic mix column with the new entering variable. Other basic mix
variables remain unchanged. Change the unit profit or unit loss
column with the value for the new entering variable. Compute the
new row values to obtain a new set of exchange coefficients
applicable to each basic variable.
Step 6. Repeat steps 2 through 5 until you no longer can improve
7 OPERATIONS RESEARCH Slack variables. . .
surplus resources.” In
real-life problems, it’s
unlikely that all
resources will be used
completely, so there
usually are unused
resources. A simplex method example:
Production of wooden tables and chairs
Step 1. Formulate the LP and construct
a simplex tableau.
From the information in Table 3, we can formulate the LP
problem as before.
Table 3.—Information for the wooden tables and chairs linear programming
Labor (hr) Table (X1)
5 Chair (X2)
10 Unit profit $6 Available
110 $8 Maximize: Z = 6X1 + 8X2 (objective function) Subject to: 30X1 + 20X2 < 300 (wood constraint: 300 bf available) 5X1 + 10X2 < 110 (labor constraint: 110 hours available) X1, X2 > 0 (nonnegativity conditions) Slack variables
Using the simplex method, the first step is to recognize surplus
resources, represented in the problem as slack variables. In most
real-life problems, it’s unlikely that all resources (usually a large
mix of many different resources) will be used completely. While
some might be used completely, others will have some unused
capacity. Also, slack variables allow us to change the inequalities
in the constraint equations to equalities, which are easier to solve
algebraically. Slack variables represent the unused resources
between the left-hand side and right-hand side of each inequality;
in other words, they allow us to put the LP problem into the standard form so it can be solved using the simplex method.
The first step is to convert the inequalities into equalities by
adding slack variables to the two constraint inequalities. With SW
representing surplus wood, and SL representing surplus labor, the
constraint equations can be written as:
30X1 + 20X2 + SW = 300
5X1 + 10X2 + SL = 110 8 (wood constraint: 300 bf)
(labor constraint: 110 hours) SIMPLEX METHOD All variables need to be represented in all equations. Add slack
variables to the other equations and give them coefficients of 0 in
those equations. Rewrite the objective function and constraint
Maximize: Z = 6X1 + 8X2 + 0SW + 0SL (objective function) Subject to: 30X1 + 20X2 + SW + 0SL = 300 (wood constraint: 300 bf) 5X1 + 10X2 + 0SW + SL = 110 (labor constraint: 110 hours) X1, X2, SW, SL > 0 (nonnegativity conditions) We can think of slack or surplus as unused resources that don’t
add any value to the objective function. Thus, their coefficients are
0 in the objective function equation. Basic variable mix and non-basic variables
Since there are more unknown variables (four) than equations
(two), we can’t solve for unique values for the X and S variables
using algebraic methods. Whenever the number of variables is
greater than the number of equations, the values of the extra
variables must be set arbitrarily, and then the other variables can be
solved for algebraically.
First we’ll choose which variables to solve for algebraically.
These variables are defined to be in the basic variable mix. We can
solve for these variables after we fix the other variables at some
The fixed-value variables are identified as not being in the basic
mix and are called non-basic variables. We’ll arbitrarily give the
non-basic variables the value of 0. The algebraic solution of the
constraint equations, with non-basic variables set to 0, represents a
For any given set of variables, there are several possible combinations of basic variables and non-basic variables. For illustration,
Table 4 contains the six basic mix pairs and the corresponding nonbasic variables for the tables and chairs LP problem. Figure 2
illustrates where each corner (A through F in Table 4) lies on a
graph. Basic variable mix. . .
“The variables that we
choose to solve for
algebraically.” Non-basic variables. . .
“Variables that are
arbitrarily given a value
of 0 so that we can
solve for the variables
in the basic variable
mix.” 9 OPERATIONS RESEARCH Table 4.—Basic variable mix combinations and algebraic solutions.
F Chairs E
D Labor C A F Tables Figure 2.—Corners corresponding to Table 4 data. We can evaluate each corner to find the values of the basic
variables and of Z:
Set X1 = 0 and X2 = 0
30(0) + 20(0) + SW + 0SL = 300
5(0) + 10(0) + 0SW + SL = 110
Therefore, SW = 300
Solution: X1 = 0, X2 = 0, SW = 300, SL = 110 Profit: 10 SL = 110 Z = $6(0) + $8(0) + 0(300) + 0(110) = 0 SIMPLEX METHOD Corner B:
Set X1 = 0 and SL = 0
30(0) + 20X2 + SW + 0(0) = 300
5(0) + 10X2 + 0SW + 0 = 110 Therefore, 20X2 + SW = 300
and 10X2 = 110 Solution: X1 = 0, X2 = 11, SW = 80, SL = 0 Profit: Z = $6(0) + $8(11) + 0(80) + 0(0) = $88 Corner C:
Set X2 = 0 and SW = 0
30X1 + 20(0) + 0 + 0SL = 300 5X1 +10(0) + 0(0) + SL = 110
Therefore, 30X1 = 300
and 5X1 + SL = 110 Solution: X1 = 10, X2 = 0, SW = 0, SL = 60 Profit: Z = $6(10) + $8(0) + 0(0) + 0(60) = $60 Corner D:
Set SW = 0 and SL = 0
30X1 + 20X2 + 0 + 0(0) = 300 5X1 + 10X2 + 0(0) + 0 = 110 Therefore, 30X1 + 20X2 = 300
and 5X1 + 10X2 = 110 Solution: X1 = 4, X2 = 9, SW = 0, SL = 0 Profit: Z = $6(4) + $8(9) + 0(0) + 0(0) = $96 Point E is infeasible because it violates the labor constraint, and
point F is infeasible because it violates the wood constraint. The
simplex algorithm never evaluates infeasible corners.
Remember, with slack variables added, the tables and chairs LP
is now four-dimensional and is not represented by Figure 2. Points
on the constraint lines in Figure 2 represent 0 slack for both wood The simplex algorithm
never evaluates infeasible corners, i.e.,
those that violate one
of the constraint
equations. 11 OPERATIONS RESEARCH Most real-world problems are too complex
to solve graphically.
They have too many
corners to evaluate,
and the algebraic
solutions are lengthy. A
simplex tableau is a
way to systematically
evaluate variable mixes
in order to find the best
one. and labor resources. A feasible point off a constraint line represents
positive slack and cannot be read off the two-dimensional graph.
In Table 4, all corners in the LP were identified, and all feasible
corners were algebraically evaluated to find the optimum solution.
You can see that a graph wasn’t necessary to list all variable mixes
and that each variable-mix pair corresponded to a corner solution.
One reason we can’t use this procedure to solve most LP problems is that the number of corners for real-life LP problems usually
is very large. Another reason is that each corner evaluation requires
a lengthy algebraic solution. To obtain each corner solution for a
10-constraint linear program, 10 equations with 10 unknowns must
be solved, which is not a simple arithmetic task. Many LP problems are formulated with many more than 10 constraints. Simplex tableau
As you’ll recall, we formulated the tables and chairs LP in
standard form as:
Maximize: Z = 6X1 + 8X2 + 0SW + 0SL (objective function) Subject to: 30X1 + 20X2 + SW + 0SL = 300 (wood constraint: 300 bf) 5X1 + 10X2 + 0SW + SL = 110 (labor constraint: 110 hours) X1, X2, SW, SL > 0 (nonnegativity conditions) The information for the tables and chairs example can be incorporated into a simplex tableau (Table 5). A tableau is a table that
allows you to evaluate the values of variables at a given corner to
determine which variable should be changed to most improve the
Table 5.—Tables and chairs simplex tableau.
0 0 Basic mix X1 X2 SW SL Solution SW
110 The top of the tableau lists the per-unit profit for the objective
function. The rows in the body of the tableau indicate the basic
variable mix for the corner point being evaluated. The first row in
the body of the tableau lists the coefficients of the first constraint 12 SIMPLEX METHOD equation (the wood constraint) in their original order. The second
row lists the coefficients of the second constraint equation (the
labor constraint) in their original order.
The basic mix column lists the slack variables. All variables not
listed in this column are designated as non-basic variables and will
be arbitrarily fixed at a value of 0 when we plug them into the
The solution column lists the values of the basic variables,
SW = 300 and SL= 110. Thus, the solution mix shows that all of the
resources (wood and labor) remain unused. The solutions to the
two constraint equations after zeroing out X1 and X2 are as follows:
30(0) + 20(0) + SW + 0SL = 300 or SW = 300 5(0) + 10(0) + 0SW + SL = 110 or SL = 110 The original constraint coefficients, highlighted in the simplex
tableau, are called exchange coefficients. They indicate how many
units of the variable listed on the left (basic mix column) must be
given up to achieve a unit increase in the variable listed at the top
of the tableau. The 30 indicates that 30 board feet of unused wood
can be exchanged for one table, and the 20 indicates that 20 board
feet of wood can be exchanged for one chair. Likewise, 5 hours of
labor can be exchanged for one table, and 10 hours of labor can be
exchanged for one chair.
The exchange coefficients are 0 or 1 for the basic mix variables.
These numbers are not very meaningful. For example, the 1 in the
first row indicates that 1 board foot of wood can be exchanged for
1 board foot of wood. The 0 in row one indicates that no unused
wood is required to accommodate more unused labor. ▼ Step 2. Find the sacrifice and improvement rows.
The next step is to expand the simplex tableau as in Table 6.
Table 6.—Expanded simplex tableau.
0 0 Basic mix
0 X1 X2 SW SL Solution SW
Improvement ➝ Current profit 13 OPERATIONS RESEARCH Sacrifice row. . .
“Indicates what will be
lost in per-unit profit
by making a change in
the resource allocation
mix.” Improvement row. . .
“Indicates what will be
gained in per-unit
profit by making a
change in the resource
allocation mix.” In the expanded tableau, list the per-unit profit for the basic
variables in the far left-hand column (Unit profit). The per-unit
profit for both slack variables is 0.
Values in the sacrifice row indicate what will be lost in per-unit
profit by making a change in resource allocation. Values in the
improvement row indicate what will be gained in per-unit profit by
making a change in resource allocation. The sacrifice and improvement rows help you decide what corner to move to next. Sacrifice row
Values for the sacrifice row are determined by:
Unit sacrifice = Unit profit column * Exchange coefficient column To obtain the first sacrifice row value, calculations are:
(Unit profit column value) * (X1 column value)
+ * 30 =0 0 * 5 = 0 Add the products
0 First value in the
sacrifice row The first product is the unit profit of unused wood multiplied by
the amount needed to make one table. The second product is the
unit profit of unused labor multiplied by the amount needed to
make one table. Together, these products are the profit that is
sacrificed by the basic mix variables for producing one more table.
Since both basic mix variables are slack variables, and slack refers
to unused resources, zero profit is sacrificed by producing another
The next sacrifice row value is calculated in the same manner:
(Unit profit column value) * (X2 column value)
+ 14 * 20 =0 0 * 10 = 0 Add the products
0 Second value in
the sacrifice row SIMPLEX METHOD The other sacrifice row values are calculated in the same manner. In the case of our example, the unit profit values are 0, so the
sacrifice row values are all 0.
The values in the solution column of the sacrifice row are
(Unit profit column value) * (Solution column value)
+ * 300 =0 0 * 110 = 0 Add the products
0 The sum of these products represents the current profit (Z). Improvement row
Improvement row values are calculated by subtracting each
value in the sacrifice row from the value found above it in the unit
profit row. Therefore:
Unit improvement = Unit profit - Unit sacrifice For example, the improvement for X1 is calculated as:
- Unit sacrifice = 0
$6 (first value in the improvement row) Since all of the sacrifice values are 0 in this example, all of the
improvement row values are the same as those found in the unit
profit row. Step 3. Apply the entry criterion.
The next step is to apply the entry criterion, that is, to determine
the entering variable. The entering variable is defined as the
current non-basic variable that will most improve the objective if it
is increased from 0. It’s called the entering variable because it will
enter the basic mix when you construct your next tableau to evaluate a new corner. For profit maximization problems, you determine
the entering variable by finding the largest value in the improvement row. Entering variable. . .
“The current non-basic
variable that will most
improve the objective if
its value is increased
from 0.” 15 OPERATIONS RESEARCH In our example, the largest value in the improvement row is 8.
Thus, we can increase profit (improve the current solution) by $8
per unit for each chair made. Increasing the value of X2 from 0 to
$8 is the best improvement that can be made. If we increase the
value of X1, our solution improves by only $6. Therefore, X2 is the
entering variable. The entering variable is marked by placing a
downward facing arrow in the X2 (chair) column (Table 7). ➝ Table 7.—Entering variable, exchange ratios, exiting variable, and pivot
SL ➝ 30
1 Improvement 0
0 Sacrifice 300
110 300/20 = 15
110/10 = 11 0 ➝ Current profit
— Step 4. Apply the exit criterion. Limiting resource. . .
“The resource that
would run out first.”
Exchange ratio. . .
“Tells you which
variable is the limiting
The lowest nonzero
exchange ratio denotes
the limiting resource.
The basic variable in
this row becomes the
exiting variable. 16 The next step is to determine the exiting variable. The exiting
variable is the variable that will exit the basic mix when you
construct your next simplex tableau.
We’ll find the exiting variable by calculating the exchange ratio
for each basic variable. The exchange ratio tells us how many
tables or chairs can be made by using all of the resource for the
current respective basic variable. To find the exchange ratio, divide
the solution value by the corresponding exchange coefficient in the
entering variable column. The exchange ratios are:
300/20 = 15 (SW basic mix row) and 110/10 = 11 (SL basic mix row) By using all 300 board feet of wood, we can make 15 chairs
because it takes 20 board feet of wood to make a chair. By using
all 110 hours of labor, we can make 11 chairs because it takes
10 hours of labor per chair. Thus, it’s easy to see the plant can’t
manufacture 15 chairs. We have enough wood for 15 chairs but
only enough labor for 11. In this case, labor is the limiting
resource. If all the labor were used, there would be leftover wood. SIMPLEX METHOD The exit criterion requires that the limiting resource (the basic
mix variable with the smallest exchange ratio) exit the basic mix.
In this case, the exiting variable is SL. Because of this, wood (SW)
remains in the basic mix. Indicate the exiting variable by placing a
small arrow pointing toward the SL (Table 7).
Next circle the pivot element—the value found at the intersection of the entering variable column and the exiting variable row.
In this case, the value 10 (X2 column and SL row) is the pivot
element. We’ll use this value to evaluate the next corner point
represented by exchanging X2 and SL. Step 5. Construct a new simplex tableau.
The next step is to create a new simplex tableau. First, let’s look
at the old constraint equations that represented the X1 and X2 rows
in our original tableau:
30X1 + 20X2 + SW + 0SL = 300 (wood constraint)
5X1 + 10X2 + 0SW + SL = 110 (labor constraint) Since X2 is to replace SL, we need to transform the second equation so that X2 will have a coefficient of 1. This requires some
algebraic manipulation. Although the resulting equations will look
different, they will be equivalent to the original constraints of the
First, we’ll multiply the labor constraint equation by 1⁄10 (the
same as dividing each of the variables by 10). We get the following
equivalent equation: Constructing a new
tableau is a way to
evaluate a new corner
point. One variable will
enter the basic mix
(entering variable), and
one variable will leave
the basic mix and
become a non-basic
variable (exiting variable). The operation of
an entering variable
and an exiting variable
is called a pivot operation. The simplex
method is made up of
a sequence of such
pivots. 1 ⁄2X1 + X2 + 0SW + 1⁄10SL = 11 This now becomes the new X2 row. By setting the non-basic
variables, X1 and SL, both to 0, we get:
1 ⁄2(0) + X2 + 0SW + 1⁄10(0) = 11
X2 = 11 We want the solution, X2 = 11, to satisfy both constraint equations. We can do this by zeroing the X2 term in the first equation
(the wood constraint). We’ll do this by multiplying the second
equation by -20 and adding it to the first equation:
Multiply times -20:
-20(1⁄2X1 + X2 + 0SW + 1⁄10SL) = 11 -10X1 - 20X2 - 0SW - 2SL = -220 17 OPERATIONS RESEARCH Add to the first equation:
30X1 + 20X2 + SW + 0SL = 300
-10X1 - 20X2 - 0SW - 2SL = -220
20X1 + 0X2 + SW - 2SL = 80 This equation becomes the new SW row. Thus, our new constraint equations are:
20X1 + 0X2 + SW 1 SL = 80 (wood) ⁄2X1 + X2 + 0SW + 1⁄10SL = 11 (labor) When the non-basic variables X1 and SL are set to 0, the solution
20(0) + 0 + SW - 2(0) = 80
SW = 80 and 1⁄2(0) + X2 + 0 + 1⁄10(0) = 11
X2 = 11 These two new equations give us some information. Remember,
SW represents the amount of surplus or slack wood, that is, the
amount of wood not used. When 11 chairs are manufactured,
80 board feet of surplus or slack wood will remain.
The new simplex tableau is shown in Table 8c. An easier method
It’s easier to find a
solution by using the
simplex tableau than by
doing the algebra. The above equations can be calculated much more easily
directly from the original simplex tableau than by doing the
algebra. Refer to Tables 8a through 8c as we work through the
example. 1. Fill in the new X2 row.
Referring to Table 8a, divide all values in the exiting variable
row, SL, by the pivot element, 10. The calculations are 5/10, 10/10
(pivot element divided by itself), 0/10, and 1/10. Place the new
values in the same location in the new tableau (Table 8b). 18 SIMPLEX METHOD Place the unit profit row value for X2, the new entering variable
(8), into the unit profit column.
Table 8a.—Original simplex tableau.
8 0 0 ➝ Basic mix
0 X1 X2 SW SL SW
0 ➝ Sacrifice
Improvement Table 8b.—Second simplex tableau—X2 row.
110 300/20 = 15
110/10 = 11 0 ➝ Current profit
— 0 X1 X2 SW SL Solution 0.5 1 0 0.1 11 Sacrifice
Improvement 2. Fill in the new SW row.
Now we’ll find the values for the SW row. Referring to Table 8a,
find the value in the SW row in the old tableau in the pivot element
column (20). Multiply it times the first value in the new X2 row
(0.5 from Table 8b). Subtract your answer from the value in the
first position of the old SW row.
Thus, for the first value (to replace the 30 in the first tableau):
(20 * 0.5) = 10
30 - 10 = 20 For the second value (to replace 20 in the first tableau):
(20 * 1) = 20
20 - 20 = 0 For the third value in this row:
20 * 0 = 0
1 - 0 = 1 (Stays the same in the new tableau.) 19 OPERATIONS RESEARCH For the fourth value in this row:
20 * 0.1 = 2
0 - 2 = -2 For the solution column value for this row:
20 * 11 = 220
300 - 220 = 80 The new SW row is shown in Table 8c.
Table 8c.—Second simplex tableau—SW row.
0 0 X1 X2 SW SL Solution 20
11 Basic mix 0
Improvement 3. Find the sacrifice and improvement rows.
Find the sacrifice and improvement rows using the same method
as in the first tableau. See Table 8d.
Table 8d.—Second simplex tableau—sacrifice and improvement rows.
Basic mix 0
8 X1 X2 SW SL Solution SW
Improvement ➝ Current profit We now see that profit has been improved from 0 to $88. 4. Complete the pivot operation (entering and
Recall that the pivot operation results in new entering
and exiting variables. The greatest per-unit improvement is
20 SIMPLEX METHOD 2 (X1 column). The others offer no improvement (either 0 or a
negative number). X1 becomes the new entering variable. Mark the
top of its column with an arrow (Table 8e). Remember, when no
improvement can be found at this step, the current tableau represents the optimal solution.
Now determine the exiting variable. To do so, first determine the
80/20 = 4
and 11/0.5 = 22 Now choose the smallest nonnegative exchange ratio (4 versus
22). SW becomes the exiting variable. Mark that row with an arrow.
Draw a circle around the pivot element, 20. (Table 8e).
Table 8e.—Second simplex tableau—pivot operation.
X2 ➝ X1 X2 SW SL Solution Exchange
11 80/20 = 4
11/0.5 = 22 8
— ➝ Basic mix 4
Sacrifice 5. Construct the third tableau from the second tableau.
Replace the entering variable in the basic mix where the exiting
variable left. Bring over the unit profit from the top row of the old
table to the new table. Fill in the pivot element row by dividing
through by the pivot element (Table 8f).
Table 8f.—Third simplex tableau—X1 row.
X2 0 X1 X2 SW SL 1 0 0.05 -0.1 Solution
Improvement 21 OPERATIONS RESEARCH Fill in the first value in the X2 row as follows. First, multiply the
previous tableau’s X2 pivot value (0.5) times the first value in the
new tableau’s X1 row (1):
0.5 * 1 = 0.5 Now subtract this number from the first value in the previous
tableau’s X2 row (0.5):
0.5 - 0.5 = 0 Place this value in the first position of the new tableau’s X2 row.
Repeat this process to fill in the remaining values in the new X2
row (Table 8g).
Table 8g.—Third simplex tableau—X2 row.
0 0 Basic mix
8 X1 X2 SW SL Solution X1
Improvement Fill in the sacrifice row (Table 8h). The first value is (6 * 1) +
(8 * 0) = 6.
Fill in the improvement row. The first value is 6 - 6 = 0.
Table 8h.—Third simplex tableau—sacrifice and improvement rows.
Basic mix X2 SW SL Solution X1
9 Sacrifice 6
8 X1 6
— Improvement 22 SIMPLEX METHOD There are no positive numbers in the new improvement row.
Thus, we no longer can improve the solution to the problem. This
simplex tableau represents the optimal solution to the LP problem
and is interpreted as:
X1 = 4, X2 = 9, SW = 0, SL = 0, and profit or Z = $96 The optimal solution (maximum profit to be made) is to manufacture four tables and nine chairs for a profit of $96. Shortcuts
Several shortcuts can make the construction of simplex
• In the new tableau, only the columns for the non-basic
and exiting variables change. Move the values for all other
basic variables directly into the new tableau.
• When 0 is found in the pivot column, that row always is
the same in the new tableau. When 0 is found in the pivot
row, that column always is the same in the new tableau.
• With two exceptions, the newly entered basic variable’s
column will contain a zero in all locations. A 1 will go in
the same location as the pivot element in the preceding
tableau, and the per-unit profit or loss for this variable will
appear in the sacrifice row. 23 OPERATIONS RESEARCH Summary
In this publication, the simplex method was used to solve a
maximization problem with constraints of the form of < (less than
or equal to). In the next publication in this series, we will discuss
how to handle > (greater than or equal to) and = (equal to) constraints. We’ll also explore how to use LP as an economic tool with
sensitivity analysis. Quick review of the simplex method
Step 1. Formulate the LP and construct a simplex tableau.
Add slack variables to represent unused resources.
Step 2. Find the sacrifice and improvement rows. Values in
the sacrifice row indicate what will be lost in per-unit profit
by making a change in the resource allocation mix. Values in
the improvement row indicate what will be gained in per-unit
profit by making a change.
Step 3. Apply the entry criteria. The entering variable is
defined as the current non-basic variable that will most
improve the objective if its value is increased from 0.
Step 4. Apply the exit criteria. Using the current tableau’s
exchange coefficient from the entering variable column,
calculate the exchange ratio for each row. Find the lowest
nonzero and nonnegative value. The basic variable in this
row becomes the exiting variable.
Step 5. Construct a new simplex tableau. Replace the
exiting variable in the basic mix column with the new entering variable. Change the unit profit or unit loss column with
the value for the new entering variable. Compute the new
row values to obtain a new set of exchange coefficients
applicable to each basic variable.
Step 6. Repeat steps 2 through 5 until you no longer can
improve the solution. 24 SIMPLEX METHOD For more information
Bierman, H., C.P. Bonini, and W.H. Hausman. Quantitative Analysis for Business Decisions (Richard D. Irwin, Inc., Homewood,
IL, 1977). 642 pp.
Dykstra, D.P. Mathematical Programming for Natural Resource
Management (McGraw-Hill, Inc., New York, 1984). 318 pp.
Hillier, F.S., and G.J. Lieberman. Introduction to Operations
Research, sixth edition (McGraw-Hill, Inc., New York, 1995).
Ignizio, J.P., J.N.D. Gupta, and G.R. McNichols. Operations
Research in Decision Making (Crane, Russak & Company, Inc.,
New York, 1975). 343 pp.
Lapin, L.L. Quantitative Methods for Business Decisions with
Cases, third edition (Harcourt Brace, Jovanovich, Publishers,
San Diego, 1985). 780 pp.
Ravindran, A., D.T. Phillips, and J.J. Solberg. Operations
Research: Principles and Practice, second edition (John Wiley
& Sons, New York, 1987). 637 pp. 25 OPERATIONS RESEARCH PERFORMANCE EXCELLENCE
IN THE WOOD PRODUCTS INDUSTRY ABOUT THIS SERIES
This publication is part of a series, Performance
Excellence in the Wood Products Industry. The various
publications address topics under the headings of wood
technology, marketing and business management,
production management, quality and process control,
and operations research.
To view and download any of the other titles in the
series, visit the OSU Extension Web site at http://
eesc.oregonstate.edu/ then “Publications & Videos” then
“Forestry” then “Wood Processing” and “Business
Management”. Or, visit the OSU Wood Products Extension
Web site at http://wood.oregonstate.edu/ 26 SIMPLEX METHOD 27 © 1998 Oregon State University
This publication was produced and distributed in furtherance of the Acts of Congress of May 8 and June 30, 1914. Extension
work is a cooperative program of Oregon State University, the U.S. Department of Agriculture, and Oregon counties. Oregon
State University Extension Service offers educational programs, activities, and materials—without regard to race, color, religion,
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Rehabilitation Act of 1973. Oregon State University Extension Service is an Equal Opportunity Employer.
Published October 1998. ...
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