simplex tableau

A-1

Module A

The Simplex Solution Method

Z06_TAYL4367_10_SE_Mod A.QXD 1/9/09 1:07 AM Page A-1

A-2 Module A The Simplex Solution Method

The simplex method is a general mathematical solution technique for solving linear pro-gramming problems. In the simplex method, the model is put into the form of a table,

and then a number of mathematical steps are performed on the table. These mathematicalsteps in effect replicate the process in graphical analysis of moving from one extremepoint on the solution boundary to another. However, unlike the graphical method, inwhich we could simply search through all the solution points to find the best one, the sim-plex method moves from one better solution to another until the best one is found, andthen it stops.

The manual solution of a linear programming model using the simplex method canbe a lengthy and tedious process.Years ago, manual application of the simplex methodwas the only means for solving a linear programming problem. Now computer solutionis certainly preferred. However, knowledge of the simplex method can greatly enhanceone’s understanding of linear programming. Computer software programs like QMfor Windows or Excel spreadsheets provide solutions to linear programming problems,but they do not convey an in-depth understanding of how those solutions are derived. Toa certain extent, graphical analysis provides an understanding of the solution process,and knowledge of the simplex method further expands on that understanding. In fact,computer solutions are usually derived using the simplex method. As a result, much ofthe terminology and notation used in computer software comes from the simplexmethod. Thus, for those students of management science who desire a more in-depthknowledge of linear programming, it is beneficial to study the simplex solution methodas provided here.

Converting the Model into Standard Form

The first step in solving a linear programming model manually with the simplex method isto convert the model into standard form. At the Beaver Creek Pottery Company NativeAmerican artisans produce bowls and mugs from labor and clay. The linear pro-gramming model is formulated as

subject to

(labor, hr.)(clay, lb.)

We convert this model into standard form by adding slack variables to each constraint asfollows:

subject to

x1, x2, s1, s2 Ú 04x1 + 3x2 + s2 = 120

x1 + 2x2 + s1 = 40

maximize Z = 40x1 + 50x2 + 0s1 + 0s2

x1, x2 Ú 04x1 + 3x2 … 120

x1 + 2x2 … 40

maximize Z = $40x1 + 50x2 (profit)

(x2)(x1)


Converting the Model into Standard Form A-3

The slack variables, and represent the amount of unused labor and clay, respectively.For example, if no bowls and mugs are produced, and and then the solutionto the problem is

and

In other words, when we start the problem and nothing is being produced, all theresources are unused. Since unused resources contribute nothing to profit, the profit iszero:

It is at this point that we begin to apply the simplex method. The model is in therequired form, with the inequality constraints converted to equations for solution with thesimplex method.

The Solution of Simultaneous EquationsOnce both model constraints have been transformed into equations, the equations shouldbe solved simultaneously to determine the values of the variables at every possible solu-tion point. However, notice that our example problem has two equations and fourunknowns (i.e., two decision variables and two slack variables), a situation that makesdirect simultaneous solution impossible. The simplex method alleviates this problem byassigning some of the variables a value of zero. The number of variables assigned values ofzero is where n equals the number of variables and m equals the number of con-straints (excluding the nonnegativity constraints). For this model, variables and

constraints; therefore, two of the variables are assigned a value of zero (i.e.,).

For example, letting and results in the following set of equations:

and

First, solve for in the first equation:

x2 = 20 2x2 = 40

x2

0 + 3x2 + s2 = 120 0 + 2x2 + 0 = 40

4x1 + 3x2 + s2 = 120 x1 + 2x2 + s1 = 40

s1 = 0x1 = 04 - 2 = 2m = 2

n = 4n - m,

Z = $0 = 40(0) + 50(0) + 0(40) + 0(120)

Z = $40x1 + 50x2 + 0s1 + 0s2

s2 = 120 lb. of clay 4(0) + 3(0) + s2 = 120

4x1 + 3x2 + s2 = 120

s1 = 40 hr. of labor 0 + 2(0) + s1 = 40 x1 + 2x2 + s1 = 40

x2 = 0,x1 = 0s2,s1Slack variables are added to

constraints and represent unused resources.

…



Then, solve for in the second equation:

This solution corresponds to point A in Figure A-1. The graph in Figure A-1 shows thatat point and the exact solution obtained by solvingsimultaneous equations. This solution is referred to as a basic feasible solution. A feasiblesolution is any solution that satisfies the constraints. A basic feasible solution satisfies theconstraints and contains as many variables with nonnegative values as there are modelconstraints—that is, m variables with nonnegative values and values set equal tozero. Typically, the m variables have positive nonzero solution values; however, when one ofthe m variables equals zero, the basic feasible solution is said to be degenerate. (The topicof degeneracy will be discussed at a later point in this module.)

n - m

s2 = 60,A, x1 = 0, x2 = 20, s1 = 0,

s2 = 60 3(20) + s2 = 120

3x2 + s2 = 120

s2

Consider a second example where and These values result in the follow-ing set of equations:

and

Solve for

Then solve for

This basic feasible solution corresponds to point C in Figure A-1, where and s2 = 0. x2 = 0, s1 = 10,

x1 = 30,

s1 = 10 30 + s1 = 40

s1:

x1 = 30 4x1 = 120

x1:

4x1 + 0 + 0 = 120 x1 + 0 + s1 = 40

4x1 + 3x2 + s2 = 120 x1 + 2x2 + s1 = 40

s2 = 0.x2 = 0

10

100

x1

x2

20 30 40

20

30

40

A

B

C

4x1 + 3x2 + s2 = 120

x1 + 2x

2 + s1 = 40 x1 = 30

x2 = 0s1 = 10s2 = 0

x1 = 0x2 = 20s1 = 0s2 = 60

x1 = 24x2 = 8s1 = 0s2 = 0

Figure A-1

Solutions at points A, B, and C

A basic feasible solution satisfiesthe model constraints and has the

same number of variables withnonnegative values as there are

constraints.


The Simplex Method A-5

Finally, consider an example where and These values result in the follow-ing set of equations:

and

These equations can be solved using row operations. In row operations, the equationscan be multiplied by constant values and then added or subtracted from each other without changing the values of the decision variables. First, multiply the top equationby 4 to get

and then subtract the second equation:

Next, substitute this value of into either one of the constraints:

This solution corresponds to point B on the graph, where andwhich is the optimal solution point.

All three of these example solutions meet our definition of basic feasible solutions.However, two specific questions are raised by the identification of these solutions.

1. In each example, how was it known which variables to set equal to zero?2. How is the optimal solution identified?

The answers to both of these questions can be found by using the simplex method. Thesimplex method is a set of mathematical steps that determines at each step which variablesshould equal zero and when an optimal solution has been reached.

The Simplex Method

The steps of the simplex method are carried out within the framework of a table, orsimplex tableau. The tableau organizes the model into a form that makes applying themathematical steps easier. The Beaver Creek Pottery Company example will be used againto demonstrate the simplex tableau and method:

subject to

x1, x2, s1, s2 Ú 04x1 + 3x2 + s2 = 120 lb.

x1 + 2x2 + s1 = 40 hr.

maximize Z = $40x1 + 50x2 + 0s1 + 0s2

s2 = 0,x1 = 24, x2 = 8, s1 = 0,

x1 = 24 x1 + 2(8) = 40

x2

x2 = 8 5x2 = 40

- 4x1 - 3x2 = -120 4x1 + 8x2 = 160

4x1 + 8x2 = 160

4x1 + 3x2 + 0 = 120 x1 + 2x2 + 0 = 40

4x1 + 3x2 + s2 = 120 x1 + 2x2 + s1 = 40

s2 = 0.s1 = 0

The simplex method is a set ofmathematical steps for solving a

linear programming problemcarried out in a table called a

simplex tableau.

Row operations are used to solvesimultaneous equations where

equations are multiplied byconstants and added or subtracted

from each other.


The initial simplex tableau for this model, with the various column and row headings, isshown in Table A-1.


Table A-1

The Simplex Tableau

Table A-2

The Basic Feasible SolutionBasic ____________________________

cj Variables Quantity x1 x2 s1 s2

s1 40 ____________________________s2 120

z1 ____________________________cj - zj

The first step in filling in Table A-1 is to record the model variables along the second rowfrom the top. The two decision variables are listed first, in order of their subscript magni-tude, followed by the slack variables, also listed in order of their subscript magnitude. Thisstep produces the row with and in Table A-1.

The next step is to determine a basic feasible solution. In other words, which two vari-ables will form the basic feasible solution and which will be assigned a value of zero? Insteadof arbitrarily selecting a point (as we did with points A, B, and C in the previous section),the simplex method selects the origin as the initial basic feasible solution because the valuesof the decision variables at the origin are always known in all linear programming prob-lems. At that point and thus, the variables in the basic feasible solution are

and

and

In other words, at the origin, where there is no production, all resources are slack, orunused. The variables and which form the initial basic feasible solution, are listed inTable A-2 under the column “Basic Variables,” and their respective values, 40 and 120, arelisted under the column “Quantity.”

s2,s1

s2 = 120 lb. 4(0) + 3(0) + s2 = 120

4x1 + 3x2 + s2 = 120

s1 = 40 hr. 0 + 2(0) + s1 = 40 x1 + 2x2 + s1 = 40

s2:s1

x2 = 0;x1 = 0

s2x1, x2, s1,The basic feasible solution in the

initial simplex tableau is the originwhere all decision variables

equal zero.

At the initial basic feasible solutionat the origin, only slack variables

have a value greater than zero.

Basic ____________________________


____________________________________________________________________

zj ____________________________

cj - zj



The initial simplex tableau always begins with the solution at the origin, where and equal zero. Thus, the basic variables at the origin are the slack variables, and Since thequantity values in the initial solution always appear as the right-hand-side values of theconstraint equations, they can be read directly from the original constraint equations.

The top two rows and bottom two rows are standard for all tableaus; however, the num-ber of middle rows is equivalent to the number of constraints in the model. For example,this problem has two constraints; therefore, it has two middle rows corresponding to and

(Recall that n variables minus m constraints equals the number of variables in the prob-lem with values of zero. This also means that the number of basic variables with valuesother than zero will be equal to m constraints.)

Similarly, the three columns on the left side of the tableau are standard, and the remain-ing columns are equivalent to the number of variables. Since there are four variables inthis model, there are four columns on the right of the tableau, corresponding to and

The next step is to fill in the values, which are the objective function coefficients, rep-resenting the contribution to profit (or cost) for each variable or in the objective func-tion. Across the top row the values 40, 50, 0, and 0 are inserted for each variable in themodel, as shown in Table A-3.

cj

sjxj

cj

s2.x1, x2, s1,

s2.s1

s2.s1

x2x1

The values for on the left side of the tableau are the contributions to profit of onlythose variables in the basic feasible solution, in this case and These values are insertedat this location in the tableau so that they can be used later to compute the values in the

row.The columns under each variable (i.e., and ) are filled in with the coefficients

of the decision variables and slack variables in the model constraint equations. The rowrepresents the first model constraint; thus, the coefficient for is 1, the coefficient for is2, the coefficient for is 1, and the coefficient for is 0. The values in the row are thesecond constraint equation coefficients, 4, 3, 0, and 1, as shown in Table A-4.

s2s2s1

x2x1

s1

s2x1, x2, s1,zj

s2.s1

cj

The quantity column values arethe solution values for the vari-

ables in the basic feasible solution.

The number of rows in a tableau isequal to the number of constraints

plus four.

The number of columns in atableau is equal to the number ofvariables (including slacks, etc.)

plus three.

Table A-3

The Simplex Tableau with cj Values

Basic 40 50 0 0


0 s1 40 ____________________________0 s2 120

zj ____________________________cj - zj

Basic 40 50 0 0


0 s1 40 1 2 1 0

0 s2 120 4 3 0 1

zj ____________________________cj - zj

The values are the contributionto profit (or cost) for each

variable.

cj

Table A-4

The Simplex Tableau withModel Constraint Coefficients



This completes the process of filling in the initial simplex tableau. The remaining valuesin the and rows, as well as subsequent tableau values, are computed mathemati-cally using simplex formulas.

The following list summarizes the steps of the simplex method (for a maximizationmodel) that have been presented so far:

1. First, transform all inequalities to equations by adding slack variables.2. Develop a simplex tableau with the number of columns equaling the number of vari-

ables plus three, and the number of rows equaling the number of constraints plus four.3. Set up table headings that list the model decision variables and slack variables.4. Insert the initial basic feasible solution, which are the slack variables and their quan-

tity values.5. Assign values for the model variables in the top row and the basic feasible solution

variables on the left side.6. Insert the model constraint coefficients into the body of the table.

Computing the zj and cj -- zj RowsSo far the simplex tableau has been set up using values taken directly from the model. Fromthis point on the values are determined by computation. First, the values in the row arecomputed by multiplying each column value (on the left side) by each column valueunder “Quantity,” and and then summing each of these sets of values. The values are shown in Table A-5.

zjs2x1, x2, s1,cj

zj

cj

cj - zjzj

The simplex method works bymoving from one solution

(extreme) point to an adjacentpoint until it locates the best

solution.

Basic 40 50 0 0


0 s1 40 1 2 1 0

0 s2 120 4 3 0 1

zj 0 0 0 0 0

cj - zj

Table A-5

The Simplex Tableau with zjRow Values

For example, the value in the row under the quantity column is found as follows:

Quantity

The value in the row under the column is found similarly:

All the other row values for this tableau will be zero when they are computed using thisformula.

zj

zj = 00 * 4 = 00 * 1 = 0

cj x1

x1zj

zq = 00 * 120 = 0

0 * 40 = 0

cj

zj

The row values are computed bymultiplying the column values

by the variable column values and summing.

cj

zj



Now the row is computed by subtracting the row values from the (top) rowvalues. For example, in the column the row value is computed as This value as well as other values are shown in Table A-6, which is the complete ini-tial simplex tableau with all values filled in. This tableau represents the solution at the ori-gin, where and The profit represented by this solution(i.e., the Z value) is given in the row under the “Quantity” column—0 in Table A-6. Thissolution is obviously not optimal because no profit is being made. Thus, we want to moveto a solution point that will give a better solution. In other words, we want to produce eithersome bowls or some mugs One of the nonbasic variables (i.e., variables not in thepresent basic feasible solution) will enter the solution and become basic.

(x2).(x1)

zj

s2 = 120.x1 = 0, x2 = 0, s1 = 40,

cj - zj

40 - 0 = 40.cj - zjx1

cjzjcj - zj

Basic 40 50 0 0


0 s1 40 1 2 1 0

0 s2 120 4 3 0 1

zj 0 0 0 0 0

cj - zj 40 50 0 0

Table A-6

The Complete Initial Simplex Tableau

The Entering Nonbasic VariableAs an example, suppose the pottery company decides to produce some bowls. With this deci-sion will become a basic variable. For every unit of (i.e., each bowl) produced, profit willbe increased by $40 because that is the profit contribution of a bowl. However, when a bowl(x1) is produced, some previously unused resources will be used. For example, if

then

and

In the labor constraint we see that with the production of one bowl, the amount ofslack, or unused, labor is decreased by 1 hour. In the clay constraint the amount of slack isdecreased by 4 pounds. Substituting these increases (for ) and decreases (for slack) intothe objective function gives

The first part of this objective function relationship represents the values in the row;the second part represents the values in the row. The function expresses the fact that toproduce some bowls, we must give up some of the profit already earned from the itemsthey replace. In this case the production of bowls replaced only slack, so no profit was lost.In general, the row values represent the net increase per unit of entering a nonbasiccj - zj

zj

cj

Z = $40Z = 40(1) + 50(0) + 0(-1) + 0(-4)

cj zj

x1

s2 = 116 lb. of clay 4(1) + 3(0) + s2 = 120

4x1 + 3x2 + s2 = 120 lb. of clay

s1 = 39 hr. of labor 1 + 2(0) + s1 = 40 x1 + 2x2 + s1 = 40 hr. of labor

x1 = 1

x1x1

The variable with the largestpositive value is the

entering variable.cj - zj



variable into the basic solution. Naturally, we want to make as much money as possible,because the objective is to maximize profit. Therefore, we enter the variable that will givethe greatest net increase in profit per unit. From Table A-7, we select variable as theentering basic variable because it has the greatest net increase in profit per unit, $50—thehighest positive value in the row.cj - zj

x2

The column, highlighted in Table A-7, is referred to as the pivot column. (The opera-tions used to solve simultaneous equations are often referred to in mathematical terminol-ogy as pivot operations.)

The selection of the entering basic variable is also demonstrated by the graph in Figure A-2.At the origin nothing is produced. In the simplex method we move from one solution pointto an adjacent point (i.e., one variable in the basic feasible solution is replaced with a variablethat was previously zero). In Figure A-2 we can move along either the axis or the axis inorder to seek a better solution. Because an increase in will result in a greater profit, wechoose x2.

x2

x2x1

x2

Table A-7

Selection of the Entering BasicVariable

10

100

x1

x2

20 30 40

20

30

40

A

B

C

Pro

duce

mug

s

Produce bowls

Figure A-2

Selection of which item toproduce—the entering basic

variable

The Leaving Basic VariableSince each basic feasible solution contains only two variables with nonzero values, one ofthe two basic variables present, or will have to leave the solution and become zero.Since we have decided to produce mugs we want to produce as many as possible or, inother words, as many as our resources will allow. First, in the labor constraint we will use all

(x2),s2,s1

The pivot column is the columncorresponding to the entering

variable.

Basic 40 50 0 0


0 s1 40 1 2 1 0

0 s2 120 4 3 0 1

zj 0 0 0 0 0

cj - zj 40 50 0 0



The leaving variable is determinedby dividing the quantity values by

the pivot column values andselecting the minimum possible

value or zero.

the labor to make mugs (because no bowls are to be produced, and because we willuse all the labor possible and unused labor resources, also):

In other words, enough labor is available to produce 20 mugs. Next, perform the sameanalysis on the constraint for clay:

This indicates that there is enough clay to produce 40 mugs. But there is enough labor to pro-duce only 20 mugs. We are limited to the production of only 20 mugs because we do not haveenough labor to produce any more than that. This analysis is shown graphically in Figure A-3.

= 40 mugs

x2 =

120 lb.

3 lb>mug

4(0) + 3x2 + 0 = 120 4x1 + 3x2 + s2 = 120 lb.

= 20 mugs

x2 =

40 hr.

2 hr>mug

1(0) + 2x2 + 0 = 40 1x1 + 2x2 + s1 = 40 hr.

s1 = 0s1 =

x1 = 0;

10

100

x1

x2

20 30 40

20

30

40

A

B

C

$50

prof

it

$40 profit

R

Figure A-3

Determination of the basicfeasible solution point

Because we are moving out the axis, we can move from the origin to either point A orpoint R. We select point A because it is the most constrained and thus feasible, whereas pointR is infeasible.

This analysis is performed in the simplex method by dividing the quantity values of thebasic solution variables by the pivot column values. For this tableau,

Basic Variables Quantity

40 the leaving basic variable120 , 3 = 40s2

, 2 = 20,s1

x2

x2



The leaving basic variable is the variable that corresponds to the minimum nonnegativequotient, which in this case is 20. (Note that a value of zero would qualify as the minimumquotient and would be the choice for the leaving variable.) Therefore, is the leavingvariable. (At point A in Figure A-3, equals zero because all the labor is used to make the20 mugs.) The row, highlighted in Table A-8, is also referred to as the pivot row.s1

s1

s1

Table A-8

Pivot Column, Pivot Row, andPivot Number

Table A-9

The Basic Variables andcj Values for the Second

Simplex Tableau

The value of 2 at the intersection of the pivot row and the pivot column is called thepivot number. The pivot number, row, and column are all instrumental in developing thenext tableau. We are now ready to proceed to the second simplex tableau and a bettersolution.

Developing a New TableauTable A-9 shows the second simplex tableau with the new basic feasible solution variablesof and and their corresponding values.cjs2x2

The various row values in the second tableau are computed using several simplex for-mulas. First, the row, called the new tableau pivot row, is computed by dividing everyvalue in the pivot row of the first (old) tableau by the pivot number. The formula for thesecomputations is

The new row values are shown in Table A-10.To compute all remaining row values (in this case there is only one other row), another

formula is used.

new tableau pivot row values =

old tableau pivot row values

pivot number

x2

Computing the new tableau pivotrow values.

Computing all remaining rowvalues.

The pivot row is the rowcorresponding to the leaving

variable.

The pivot number is the numberat the intersection of the pivot

column and row.

Basic 40 50 0 0


50 x2

0 s2

zj

cj - zj

Basic 40 50 0 0


0 s1 40 1 2 1 0

0 s2 120 4 3 0 1

zj 0 0 0 0 0

cj - zj 40 50 0 0



Thus, this formula requires the use of both the old tableau and the new one. The row val-ues are computed in Table A-11.

s2

new tableaurow values =

old tableaurow values - £corresponding

coefficients inpivot column

*

correspondingnew tableau

pivot row value≥

These values have been inserted in the simplex tableau in Table A-12.This solution corresponds to point A in the graph of this model in Figure A-3. The solu-

tion at this point is In other words, 20 mugs are producedand 60 pounds of clay are left unused. No bowls are produced and no labor hours remainunused.

x1 = 0, x2 = 20, s1 = 0, s2 = 60.

Table A-10

Computation of the New PivotRow Values

Table A-12

The Second Simplex Tableauwith Row Values

Column

Quantity 120 (3 20) 60

4 (3 1/2) 5/2

3 (3 1) 0

0 (3 1/2)

1 (3 0) 1=*-s2

-3>2=*-s1

=*-x2

=*-x1

=*-

=

New TableauRow Value

Old TableauRow Value

- £CorrespondingCoefficients inPivot Column

*

New TableauPivot

Row Value≥

Table A-11

Computation of New s2Row Values

Basic 40 50 0 0


50 x2 20 1/2 1 1/2 0

0 s2

zj

cj - zj

Basic 40 50 0 0


50 x2 20 1/2 1 1/2 0

0 s2 60 5/2 0 -3/2 1

zj

cj - zj

The second simplex tableau is completed by computing the and row values thesame way they were computed in the first tableau. The row is computed by summing theproducts of the column and all other column values.cj

zj

cj - zjzj



ColumnQuantity

The row values and the row values are added to the tableau to give the com-pleted second simplex tableau shown in Table A-13. The value of 1,000 in the row is thevalue of the objective function (profit) for this basic feasible solution.

zj

cj - zjzj

z4 = (50)(0) + (0)(1) = 0s2

z3 = (50)(1>2) + (0)(-3>2) = 25s1

z2 = (50)(1) + (0)(0) = 50x2

z1 = (50)(1>2) + (0)(5>2) = 25x1

zj = (50)(20) + (0)(60) = 1000

TIME OUT for George B. Dantzig

After developing the simplex method for solving linear pro-gramming problems, George Dantzig needed a good problemto test it on. The problem he selected was the “diet problem”formulated in 1945 by Nobel economist George Stigler. Thisproblem was to determine an adequate nutritional diet at min-imum cost (which was an important military and civilian issueduring World War II). Formulated as a linear programming

model, the diet problem consisted of 77 unknowns and 9 equa-tions. It took 9 clerks using hand-operated (mechanical) deskcalculators 120 man-days to obtain the optimal simplex solu-tion: a diet consisting primarily of wheat flour, cabbage, anddried navy beans that cost $39.69 per year (in 1939 prices). Thesolution developed by Stigler using his own numerical methodwas only 24 cents more than the optimal solution.

Table A-13

The Completed Second Simplex Tableau

Basic 40 50 0 0


50 x2 20 1/2 1 1/2 0

0 s2 60 5/2 0 -3/2 1

zj 1,000 25 50 25 0

cj - zj 15 0 -25 0

The computational steps that we followed to derive the second tableau in effect accom-plish the same thing as row operations in the solution of simultaneous equations. Thesesame steps are used to derive each subsequent tableau, called iterations.

The Optimal Simplex TableauThe steps that we followed to derive the second simplex tableau are repeated to develop thethird tableau. First, the pivot column or entering basic variable is determined. Because 15in the row represents the greatest positive net increase in profit, becomes theentering nonbasic variable. Dividing the pivot column values into the values in the quantitycolumn indicates that is the leaving basic variable and corresponds to the pivot row. Thepivot row, pivot column, and pivot number are indicated in Table A-14.

At this point you might be wondering why the net increase in profit per bowl is $15rather than the original profit of $40. It is because the production of bowls will requiresome of the resources previously used to produce mugs only. Producing some bowls meansnot producing as many mugs; thus, we are giving up some of the profit gained from producingmugs to gain even more by producing bowls. This difference is the net increase of $15.

(x2)(x1)

(x1)

s2

x1cj - zj

Each tableau is the same asperforming row operations for a

set of simultaneous equations.



The new tableau pivot row in the third simplex tableau is computed using the sameformula used previously. Thus, all old pivot row values are divided through by 5/2, thepivot number. These values are shown in Table A-16. The values for the other row arecomputed as shown in Table A-15.

(x2)

(x1)

Table A-16

The Completed Third SimplexTableau

Basic 40 50 0 0


50 x2 8 0 1 4/5 -1/5

40 x1 24 1 0 -3/5 2/5

zj 1,360 40 50 16 6

cj - zj 0 0 -16 -6

Column

Quantity 20 (1/2 24) 8

1/2 (1/2 1) 0

1 (1/2 0) 1

1/2 (1/2 -3/5) 4/5

0 (1/2 2/5) -1/5=*-s2

=*-s1

=*-x2

=*-x1

=*-

New TableauRow Value=- P

CorrespondingCoefficients inPivot Column

*

New TableauPivot

Row Value QOld Tableau

Row Value

Table A-15

Computation of the x2 Row forthe Third Simplex Tableau

These new row values, as well as the new row and row, are shown in the com-pleted third simplex tableau in Table A-16.

cj - zjzj

Observing the row to determine the entering variable, we see that a nonbasicvariable would not result in a positive net increase in profit, as all values in the roware zero or negative. This means that the optimal solution has been reached. The solution is

which corresponds to point B in Figure A-1.

Z = $1,360 profit x2 = 8 mugs x1 = 24 bowls

cj - zj

cj - zjThe solution is optimal when allvalues … 0.cj - zj

Table A-14

The Pivot Row, Pivot Column,and Pivot Number

Basic 40 50 0 0


50 x2 20 1/2 1 1/2 0

0 s2 60 5/2 0 -3/2 1

zj 1,000 25 50 25 0

cj - zj 15 0 -25 0



An additional comment should be made regarding simplex solutions in general.Although this solution resulted in integer values for the variables (i.e., 24 and 8), it is possi-ble to get a fractional solution for decision variables even though the variables reflect itemsthat should be integers, such as airplanes, television sets, bowls, and mugs. To apply thesimplex method, one must accept this limitation.

Summary of the Simplex Method

The simplex method demonstrated in the previous section consists of the followingsteps:

1. Transform the model constraint inequalities into equations.2. Set up the initial tableau for the basic feasible solution at the origin and compute the

and row values.3. Determine the pivot column (entering nonbasic solution variable) by selecting the

column with the highest positive value in the row.4. Determine the pivot row (leaving basic solution variable) by dividing the quantity

column values by the pivot column values and selecting the row with the minimumnonnegative quotient.

5. Compute the new pivot row values using the formula

6. Compute all other row values using the formula

7. Compute the new and rows.8. Determine whether the new solution is optimal by checking the row. If all

row values are zero or negative, the solution is optimal. If a positive valueexists, return to step 3 and repeat the simplex steps.

Simplex Solution of a Minimization Problem

In the previous section the simplex method for solving linear programming problems wasdemonstrated for a maximization problem. In general, the steps of the simplex methodoutlined at the end of this section are used for any type of linear programming problem.However, a minimization problem requires a few changes in the normal simplex process,which we will discuss in this section.

In addition, several exceptions to the typical linear programming problem will bepresented later in this module. These include problems with mixed constraints and ); problems with more than one optimal solution, no feasible solution, or anunbounded solution; problems with a tie for the pivot column; problems with a tie for thepivot row; and problems with constraints with negative quantity values. None of thesekinds of problems require changes in the simplex method. They are basically unusualresults in individual simplex tableaus that the reader should know how to interpret andwork with.

Ú

(= , … ,

cj - zj

cj - zj

cj - zjzj

new tableaurow values

=

old tableaurow values

- Pcorresponding

coefficients in

pivot column

*

corresponding

new tableau

pivot row valuesQ

new tableau pivot row values =

old tableau pivot row values

pivot number

cj - zj

cj - zjzj

The simplex method does notguarantee integer solutions.


Simplex Solution of a Minimization Problem A-17

Standard Form of a Minimization ModelConsider the following linear programming model for a farmer purchasing fertilizer:

subject to

of nitrogenof phosphate

whereof Super-gro fertilizerof Crop-quick fertilizer

total cost ($) of purchasing fertilizer

This model is transformed into standard form by subtracting surplus variables from thetwo constraints as follows:

subject to

The surplus variables represent the extra amount of nitrogen and phosphate that exceededthe minimum requirements specified in the constraints.

However, the simplex method requires that the initial basic feasible solution be at theorigin, where and Testing these solution values, we have

The idea of “negative excess pounds of nitrogen” is illogical and violates the nonnegativ-ity restriction of linear programming. The reason the surplus variable does not work isshown in Figure A-4. The solution at the origin is outside the feasible solution space.

s1 = -16 2(0) + 4(0) - s1 = 16

2x1 + 4x2 - s1 = 16

x2 = 0.x1 = 0

x1, x2, s1, s2 Ú 04x1 + 3x2 - s2 = 242x1 + 4x2 - s1 = 16

minimize Z = 6x1 + 3x2 + 0s1 + 0s2

Ú

Z = farmer’sx2 = bagsx1 = bags

4x1 + 3x2 Ú 24 lb.2x1 + 4x2 Ú 16 lb.

minimize Z = $6x1 + 3x2

2

4

6

8

10

12

20

4 6 8 10 12 x1

x2

x1 = 0x2 = 0s1 = –16

A

B

C

Figure A-4

Graph of the fertilizer example

Transforming a model intostandard form by subtracting

surplus variables will not work inthe simplex method.



To alleviate this difficulty and get a solution at the origin, we add an artificial variableto the constraint equation,

The artificial variable, does not have a meaning, as a slack variable or a surplus variabledoes. It is inserted into the equation simply to give a positive solution at the origin; we areartificially creating a solution:

The artificial variable is somewhat analogous to a booster rocket—its purpose is to getus off the ground; but once we get started, it has no real use and thus is discarded. The arti-ficial solution helps get the simplex process started, but we do not want it to end up in theoptimal solution, because it has no real meaning.

When a surplus variable is subtracted and an artificial variable is added, the phosphateconstraint becomes

The effect of surplus and artificial variables on the objective function must now be con-sidered. Like a slack variable, a surplus variable has no effect on the objective function interms of increasing or decreasing cost. For example, a surplus of 24 pounds of nitrogendoes not contribute to the cost of the objective function, because the cost is determinedsolely by the number of bags of fertilizer purchased (i.e., the values of and ). Thus, acoefficient of 0 is assigned to each surplus variable in the objective function.

By assigning a “cost” of $0 to each surplus variable, we are not prohibiting it from beingin the final optimal solution. It would be quite realistic to have a final solution that showedsome surplus nitrogen or phosphate. Likewise, assigning a cost of $0 to an artificial variablein the objective function would not prohibit it from being in the final optimal solution.However, if the artificial variable appeared in the solution, it would render the final solu-tion meaningless. Therefore, we must ensure that an artificial variable is not in the finalsolution.

As previously noted, the presence of a particular variable in the final solution is based onits relative profit or cost. For example, if a bag of Super-gro costs $600 instead of $6 andCrop-quick stayed at $3, it is doubtful that the farmer would purchase Super-gro (i.e.,would not be in the solution). Thus, we can prohibit a variable from being in the final solu-tion by assigning it a very large cost. Rather than assigning a dollar cost to an artificial vari-able, we will assign a value of M, which represents a large positive cost (say, $1,000,000).This operation produces the following objective function for our example:

The completely transformed minimization model can now be summarized as

subject to

x1, x2, s1, s2, A1, A2 Ú 04x1 + 3x2 - s2 + A2 = 242x1 + 4x1 - s1 + A1 = 16

minimize Z = 6x1 + 3x2 + 0s1 + 0s2 + MA1 + MA2

minimize Z = 6x1 + 3x2 + 0s1 + 0s2 + MA1 + MA2

x1

x2x1

4x1 + 3x2 - s2 + A2 = 24

A1 = 16 2(0) + 4(0) - 0 + A1 = 16

2x1 + 4x2 - s1 + A1 = 16

A1,

2x1 + 4x2 - s1 + A1 = 16

(A1)

Artificial variables are assigned alarge cost in the objective function

to eliminate them from the finalsolution.

An artificial variable allows foran initial basic feasible solution at

the origin, but it has no realmeaning.


Simplex Solution of a Minimization Problem A-19

The Simplex Tableau for a Minimization ProblemThe initial simplex tableau for a minimization model is developed the same way as one fora maximization model, except for one small difference. Rather than computing inthe bottom row of the tableau, we compute which represents the net per unitdecrease in cost, and the largest positive value is selected as the entering variable and pivotcolumn. (An alternative would be to leave the bottom row as and select the largestnegative value as the pivot column. However, to maintain a consistent rule for selecting thepivot column, we will use )

The initial simplex tableau for this model is shown in Table A-17. Notice that and form the initial solution at the origin, because that was the reason for inserting them in thefirst place — to get a solution at the origin. This is not a basic feasible solution, since theorigin is not in the feasible solution area, as shown in Figure A-4. As indicated previously, itis an artificially created solution. However, the simplex process will move toward feasibilityin subsequent tableaus. Note that across the top the decision variables are listed first, thensurplus variables, and finally artificial variables.

A2A1

zj - cj.

cj - zj

zj - cj,cj - zj

In Table A-17 the column was selected as the pivot column because is thelargest positive value in the row. was selected as the leaving basic variable (andpivot row) because the quotient of 4 for this row was the minimum positive row value.

The second simplex tableau is developed using the simplex formulas presented earlier. Itis shown in Table A-18. Notice that the column has been eliminated in the second sim-plex tableau. Once an artificial variable leaves the basic feasible solution, it will never returnbecause of its high cost, M. Thus, like the booster rocket, it can be eliminated from thetableau. However, artificial variables are the only variables that can be treated this way.

A1

A1zj - cj

7M - 3x2

The third simplex tableau, with replacing is shown in Table A-19. Both the andcolumns have been eliminated because both variables have left the solution. The rowx1A2

A1A2,x1

The row is changed toin the simplex tableau fora minimization problem.

zj - cj

cj - zj

Artificial variables are alwaysincluded as part of the initial basic

feasible solution when they exist.

Once an artificial variable isselected as the leaving variable, it

will never reenter the tableau, so itcan be eliminated.

Table A-17

The Initial Simplex Tableau Basic6 3 0 0 M M

cj Variables Quantity x1 x2 s1 s2 A1 A2

M 16 2 4 0 1 0

M 24 4 3 0 0 1

40M 6M 7M M M

0 0-M-M7M - 36M - 6zj - cj

-M-Mzj

-1A2

-1A1

Table A-18

The Second Simplex Tableau Basic6 3 0 0 M

cj Variables Quantity x1 x2 s1 s2 A2

3 4 1/2 1 0 0

M 12 5/2 0 3/4 1

3

0 0-M-3>4 + 3M>45M>2 - 9>2zj - cj

M-M-3>4 + 3M>45M>2 + 3>212M + 12zj

-1A2

-1>4x2



is selected as the pivot row because it corresponds to the minimum positive ratio of 16. Inselecting the pivot row, the value for the row was not considered because the mini-mum positive value or zero is selected. Selecting the row would result in a negative quan-tity value for in the fourth tableau, which is not feasible.s1

x2

x2-4

The fourth simplex tableau, with replacing is shown in Table A-20. Table A-20 isthe optimal simplex tableau because the row contains no positive values. The opti-mal solution is

Z = $24, total cost of purchasing fertilizer s2 = 0 extra lb. of phosphate x2 = 8 bags of Crop-quick s1 = 16 extra lb. of nitrogen x1 = 0 bags of Super-gro

zj - cj

x1,s1

Table A-20

Optimal Simplex Tableau Basic 6 3 0 0


3 x2 8 4/3 1 0 -1/3

0 s1 16 10/3 0 1 -4/3

zj 24 4 3 0 -1

zj - cj -2 0 0 -1

Simplex Adjustments for a Minimization ProblemTo summarize, the adjustments necessary to apply the simplex method to a minimizationproblem are as follows:

1. Transform all constraints to equations by subtracting a surplus variable and addingan artificial variable.

2. Assign a value of M to each artificial variable in the objective function.3. Change the row to

Although the fertilizer example model we just used included only constraints, it ispossible for a minimization problem to have and constraints in addition to con-straints. Similarly, it is possible for a maximization problem to have and constraintsin addition to constraints. Problems that contain a combination of different types ofinequality constraints are referred to as mixed constraint problems.

…

=Ú

Ú=…

Ú

zj - cj.cj - zj

cj

Ú

Table A-19

The Third Simplex Tableau Basic 6 3 0 0


3 x2 8/5 0 1 -2/5 1/5

6 x1 24/5 1 0 3/10 -2/5

zj 168/5 6 3 3/5 -9/5

zj - cj 0 0 3/5 -9/5


A Mixed Constraint Problem A-21

A Mixed Constraint Problem

So far we have discussed maximization problems with all constraints and minimiza-tion problems with all constraints. However, we have yet to solve a problem with a mix-ture of and constraints. Furthermore, we have not yet looked at a maximizationproblem with a constraint. The following is a maximization problem with andconstraints.

A leather shop makes custom-designed, hand-tooled briefcases and luggage. The shopmakes a $400 profit from each briefcase and a $200 profit from each piece of luggage. (Theprofit for briefcases is higher because briefcases require more hand tooling.) The shop hasa contract to provide a store with exactly 30 items per month. A tannery supplies the shopwith at least 80 square yards of leather per month. The shop must use at least this amountbut can order more. Each briefcase requires 2 square yards of leather; each piece of luggagerequires 8 square yards of leather. From past performance, the shop owners know they can-not make more than 20 briefcases per month. They want to know the number of briefcasesand pieces of luggage to produce in order to maximize profit.

This problem is formulated as

subject to

contracted itemsof leather

briefcases

where of luggage.The first step in the simplex method is to transform the inequalities into equations. The

first constraint for the contracted items is already an equation; therefore, it is not necessaryto add a slack variable. There can be no slack in the contract with the store because exactly30 items must be delivered. Even though this equation already appears to be in the neces-sary form for simplex solution, let us test it at the origin to see if it meets the startingrequirements:

Because zero does not equal 30, the constraint is not feasible in this form. Recall that aconstraint did not work at the origin either in an earlier problem. Therefore, an artificial vari-able was added. The same thing can be done here:

Now at the origin, where and

Any time a constraint is initially an equation, an artificial variable is added. However, theartificial variable cannot be assigned a value of M in the objective function of a maximiza-tion problem. Because the objective is to maximize profit, a positive M value wouldrepresent a large positive profit that would definitely end up in the final solution. Becausean artificial variable has no real meaning and is inserted into the model merely to create an

A1 = 30 0 + 0 + A1 = 30

x2 = 0,x1 = 0

x1 + x2 + A1 = 30

Ú

0 Z 30 0 + 0 = 30

x1 + x2 = 30

x1 = briefcases and x2 = pieces

x1, x2 Ú 0x1 … 20

2x1 + 8x2 Ú 80 yd.2x1 + x2 = 30

maximize Z = $400x1 + 200x2

=… , Ú ,Ú

=… , Ú ,Ú

…A mixed constraint problemincludes a combination of

and constraints.Ú

… , = ,

An artificial variable is added toan equality constraint for

standard form.(=)



initial solution at the origin, its existence in the final solution would render the solutionmeaningless. To prevent this from happening, we must give the artificial variable a largecost contribution, or

The constraint for leather is a inequality. It is converted to equation form by subtract-ing a surplus variable and adding an artificial variable:

As in the equality constraint, the artificial variable in this constraint must be assigned anobjective function coefficient of

The final constraint is a inequality and is transformed by adding a slack variable:

The completely transformed linear programming problem is as follows:

subject to

The initial simplex tableau for this model is shown in Table A-21. Notice that the basicsolution variables are a mix of artificial and slack variables. Note also that the third-rowquotient for determining the pivot row is an undefined value, or . Therefore,this row would never be considered as a candidate for the pivot row. The second, third, andoptimal tableaus for this problem are shown in Tables A-22, A-23, and A-24.

q(20 , 0)

x1, x2, s1, s2, A1, A2 Ú 0x1 + s2 = 20

2x1 + 8x2 - s1 + A2 = 80x1 + x2 + A1 = 30

maximize Z = 400x1 + 200x2 + 0s1 + 0s2 - MA1 - MA2

x1 + s2 = 20

…

-M.

2x1 + 8x2 - s1 + A2 = 80

Ú

-M.An artificial variable in a

maximization problem is given alarge cost contribution to drive it

out of the problem.

Table A-21

The Initial Simplex Tableau Basic400 200 0 0 -M -M

cj Variables Quantity x1 x2 s1 s2 A1 A2

-M 30 1 1 0 0 1 0

-M 80 2 8 0 0 1

0 20 1 0 0 1 0 0

-110M -3M -9M 0

0 0 0-M200 + 9M400 + 3Mcj - zj

-M-MMzj

s2

-1A2

A1

Table A-22

The Second Simplex Tableau Basic400 200 0 0 -M


-M 20 3/4 0 1/8 0 1

200 10 1/4 1 0 0

0 20 1 0 0 1 0

2,000 - 20M 50 - 3M/4 200 -25 - M/8 0

0 25 + M/8 0 0350 + 3M/4cj - zj

-Mzj

s2

-1/8x2

A1


Irregular Types of Linear Programming Problems A-23

The solution for the leather shop problem is (see Table A-24)

It is now possible to summarize a set of rules for transforming all three types of modelconstraints:

Z = $10,000 profit per month s1 = 40 extra yd.2 of leather x2 = 10 pieces of luggage x1 = 20 briefcases

Table A-24

The Optimal Simplex Tableau Basic 400 200 0 0


0 40 0 0 1

200 10 0 1 0

400 20 1 0 0 1

10,000 400 200 0 200

0 0 0 -200cj - zj

zj

x1

-1x2

-6s1

Objective Function Coefficient

Constraint Adjustment MAXIMIZATION MINIMIZATION

Add a slack variable 0 0

Add an artificial variable M

Subtract a surplus variable 0 0

and add an artificial variable M-M

Ú

-M=

…

Irregular Types of Linear Programming Problems

The basic simplex solution of typical maximization and minimization problems has beenshown in this module. However, there are several special types of atypical linear program-ming problems. Although these special cases do not occur frequently, they will be describedwithin the simplex framework so that you can recognize them when they arise.

For irregular problems the general simplex procedure does

not always apply.

Table A-23

The Third Simplex Tableau Basic400 200 0 0 -M


5 0 0 1/8 1

200 5 0 1 0

400 20 1 0 0 1 0

400 200

0 0 0-350 - 3M>425 + M>8cj - zj

-M350 + 3M>4-25 - M>89,000 - 5Mzj

x1

-1>4-1>8x2

-3>4A1-M



These special types include problems with more than one optimal solution, infeasibleproblems, problems with unbounded solutions, problems with ties for the pivot column orties for the pivot row, and problems with constraints with negative quantity values.

Multiple Optimal SolutionsConsider the Beaver Creek Pottery Company example with the objective function changedas follows:

to

subject to

The graph of this model is shown in Figure A-5. The slight change in the objective functionmakes it now parallel to the constraint line, Therefore, as the objectivefunction edge moves outward from the origin, it touches the whole line segment BC ratherthan a single extreme corner point before it leaves the feasible solution area. The endpoints ofthis line segment, B and C, are typically referred to as the alternate optimal solutions. It isunderstood that these points represent the endpoints of a range of optimal solutions.

4x1 + 3x2 = 120.

x1, x2 Ú 04x1 + 3x2 … 120

x1 + 2x2 … 40

maximize Z = 40x1 + 30x2

Z = 40x1 + 30x2

Z = 40x1 + 50x2

The optimal simplex tableau for this problem is shown in Table A-25. This correspondsto point C in Figure A-5.

The fact that this problem contains multiple optimal solutions can be determined fromthe row. Recall that the row values are the net increases in profit per unit forthe variable in each column. Thus, values of zero indicate no net increase in profitand no net loss in profit. We would expect the basic variables, and to have zero values because they are part of the basic feasible solution; they are already in the solution sothey cannot be entered again. However, the column has a value of zero and it iscj - zjx2

cj - zjx1,s1

cj - zj

cj - zjcj - zj

= 120Z

x1 = 24x2 = 8

= 120Z

x1 = 30x2 = 0

C

B

A

Point B Point C

10

100

x1

x2

20 30 40

20

30

40

Figure A-5

Graph of the Beaver CreekPottery Company example with

multiple optimal solutions

Alternate optimal solutions havethe same Z value but different

variable values.

For a multiple optimal solutionthe (or ) value for a

nonbasic variable in the finaltableau equals zero.

zj - cjcj - zj



not part of the basic feasible solution. This means that if some mugs were produced,we would have a new product mix but the same total profit. Thus, a multiple optimal solu-tion is indicated by a (or ) row value of zero for a nonbasic variable.zj - cjcj - zj

(x2)

Table A-25

The Optimal Simplex Tableau Basic 40 30 0 0


0 10 0 5/4 1

40 30 1 3/4 0

1,200 40 30 0 10

0 0 0 -10cj - zj

zj

-1>4x1

-1>4s1

Table A-26

The Alternate Optimal Tableau

Basic 40 30 0 0


30 8 0 1 4/5

40 24 1 0 2/5

1,200 40 30 0 10

0 0 0 -10cj - zj

zj

-3>5x1

-1>5x2

To determine the alternate endpoint solution, let be the entering variable (pivotcolumn) and select the pivot row as usual. This selection results in the row being thepivot row. The alternate solution corresponding to point B in Figure A-5 is shown in TableA-26.

s1

x2An alternate optimal solution isdetermined by selecting the non-

basic variable with asthe entering variable.

cj - zj = 0

An Infeasible ProblemAnother linear programming irregularity is the case in which a problem has no feasiblesolution area; thus, there is no basic feasible solution to the problem.

An example of an infeasible problem is formulated next and depicted graphically inFigure A-6:

subject to

The three constraints do not overlap to form a feasible solution area. Because no pointsatisfies all three constraints simultaneously, there is no solution to the problem. The finalsimplex tableau for this problem is shown in Table A-27.

The tableau in Table A-27 has all zero or negative values in the row, indicating thatit is optimal. However, the solution is and Because the existenceof artificial variables in the final solution makes the solution meaningless, this is not a real

A2 = 2.x2 = 4, A1 = 4,cj - zj

x1, x2 Ú 0x2 Ú 6x1 Ú 4

4x1 + 2x2 … 8


An infeasible problem has anartificial variable in the final

simplex tableau.

An infeasible problem does nothave a feasible solution space.



2

4

6

8

10

12

20

4 6 8 10 12 x1

x2

B

C

A

x1 = 4

x2 = 6

4x1 + 2x2 = 8

Figure A-6

Graph of an infeasible problem

Table A-27

The Final Simplex Tableau foran Infeasible Problem

Basic5 3 0 0 0 -M -M

cj Variables Quantity x1 x2 s1 s2 s3 A1 A2

3 4 2 1 1/2 0 0 0 0

4 1 0 0 0 1 0

2 0 0 0 1

3 M M

0 0 0-M-M-3>2 - M>2-1 - Mcj - zj

-M-M3>2 + M>26 + M12 - 6Mzj

-1-1>2-2A2-M

-1A1-M

x2

solution. In general, any time the (or ) row indicates that the solution is opti-mal but there are artificial variables in the solution, the solution is infeasible. Infeasibleproblems do not typically occur, but when they do they are usually a result of errors in defin-ing the problem or in formulating the linear programming model.

An Unbounded ProblemIn some problems the feasible solution area formed by the model constraints is not closed.In these cases it is possible for the objective function to increase indefinitely without everreaching a maximum value because it never reaches the boundary of the feasible solutionarea.

An example of this type of problem is formulated next and shown graphically in Figure A-7:

subject to

x1, x2 Ú 0x2 … 2x1 Ú 4


zj - cjcj - zj

In an unbounded problem theobjective function can increase

indefinitely because the solutionspace is not closed.



In Figure A-7 the objective function is shown to increase without bound; thus, a solution isnever reached.

The second tableau for this problem is shown in Table A-28. In this simplex tableau, ischosen as the entering nonbasic variable and pivot column. However, there is no pivot rowor leaving basic variable. One row value is and the other is undefined. This indicatesthat a “most constrained” point does not exist and that the solution is unbounded. In gen-eral, a solution is unbounded if the row value ratios are all negative or undefined.

-4

s1

2

4

6

8

10

12

20

4 6 8 10 12 x1

x2

Z = 4x1 + 2x

2

Figure A-7

An unbounded problem

Basic 4 2 0 0

cjVariables

Quantity x1 x2 s1 s2

4 4 0 0 0

0 2 1 1 0 1

16 4 0 0

0 2 4 0cj - zj

-4zj

s2

-1x1

2 , 0 = q

4 , -1 = -4

Table A-28

The Second Simplex Tableau

Unlimited profits are not possible in the real world; an unbounded solution, like aninfeasible solution, typically reflects an error in defining the problem or in formulating themodel.

Tie for the Pivot ColumnSometimes when selecting the pivot column, you may notice that the greatest positive

(or ) row values are the same; thus, there is a tie for the pivot column. Whenthis happens, one of the two tied columns should be selected arbitrarily. Even though onechoice may require fewer subsequent iterations than the other, there is no way of knowingthis beforehand.

Tie for the Pivot Row—DegeneracyIt is also possible to have a tie for the pivot row (i.e., two rows may have identical lowestnonnegative values). Like a tie for a pivot column, a tie for a pivot row should be brokenarbitrarily. However, after the tie is broken, the basic variable that was the other choice for

zj - cjcj - zj

A pivot row cannot be selected foran unbounded problem.

A tie for the pivot column isbroken arbitrarily.



the leaving basic variable will have a quantity value of zero in the next tableau. This condi-tion is commonly referred to as degeneracy because theoretically it is possible for subse-quent simplex tableau solutions to degenerate so that the objective function value neverimproves and optimality never results. This occurs infrequently, however.

In general, tableaus with ties for the pivot row should be treated normally. If the simplexsteps are carried out as usual, the solution will evolve normally.

The following is an example of a problem containing a tie for the pivot row:

subject to

For the sake of brevity we will skip the initial simplex tableau for this problem and godirectly to the second simplex tableau in Table A-29, which shows a tie for the pivot rowbetween the and rows.s3s1

x1, x2 Ú 05x1 + 10x2 … 40

x2 … 36x1 + 4x2 … 24


10 , 5 = 2

12 , 6 = 2

Table A-29

The Second Simplex Tableauwith a Tie for the Pivot Row

f Tie

The row is selected arbitrarily as the pivot row, resulting in the third simplex tableau,shown in Table A-30.

s3

Table A-30

The Third Simplex Tableauwith Degeneracy

Note that in Table A-30 a quantity value of zero now appears in the row, representingthe degenerate condition resulting from the tie for the pivot row. However, the simplexprocess should be continued as usual: should be selected as the entering basic variableand the row should be selected as the pivot row. (Recall that the pivot row value of zero isthe minimum nonnegative quotient.) The final optimal tableau is shown in Table A-31.

s1

s2

s1

A tie for the pivot row is brokenarbitrarily and can lead to

degeneracy.

Basic 4 6 0 0 0

cj Variables Quantity x1 x2 s1 s2 s3

0 12 6 0 1 0

6 3 0 1 0 1 0

0 10 5 0 0 1

18 0 6 0 6 0

4 0 0 0-6cj - zj

zj

-10s3

x2

-4s1

Basic4 6 0 0 0


0 0 0 0 1 8

6 3 0 1 0 1 0

4 2 1 0 0 1/5

26 4 6 0 4/5

0 0 0 2 -4>5cj - zj

-2zj

-2x1

x2

-6>5s1



Notice that the optimal solution did not change from the third to the optimal simplextableau. The graphical analysis of this problem shown in Figure A-8 reveals the reason forthis.

Table A-31

The Optimal Simplex Tableaufor a Degenerate Problem

Basic4 6 0 0 0


0 0 0 0 1/8 1

6 3 0 1 0 3/20

4 2 1 0 1/4 0

26 4 6 1/4 0 1/2

0 0 0 -1>2-1>4cj - zj

zj

-1>10x1

-1>8x2

-3>20s2

2

20

x1

x2

4 6 8

4

6

8

C

BA

6x1 + 4x2 = 24

5x1 + 10x2 = 40

x2 = 3

Figure A-8

Graph of a degenerate solution

Notice that in the third tableau (Table A-30) the simplex process went to point B, whereall three constraint lines intersect. This is, in fact, what caused the tie for the pivot row andthe degeneracy. Subsequently, the simplex process stayed at point B in the optimal tableau(Table A-31). The two tableaus represent two different basic feasible solutions correspond-ing to two different sets of model constraint equations.

Negative Quantity ValuesOccasionally a model constraint is formulated with a negative quantity value on the rightside of the inequality sign—for example,

This is an improper condition for the simplex method, because for the method to work, allquantity values must be positive or zero.

This difficulty can be alleviated by multiplying the inequality by which also changesthe direction of the inequality:

6x1 - 2x2 … 30(-1)(-6x1 + 2x2 Ú -30)

-1,

-6x1 + 2x2 Ú -30

Degeneracy occurs in a simplexproblem when a problem

continually loops back to the samesolution or tableau.

Standard form for simplexsolution requires positive right-

hand-side values.

A negative right-hand-side valueis changed to a positive by

multiplying the constraint by ,which changes the direction of the

inequality.

-1



Now the model constraint is in proper form to be transformed into an equation and solvedby the simplex method.

Summary of Simplex IrregularitiesMultiple optimal solutions are identified by for a nonbasic variable.To determine the alternate solution(s), enter the nonbasic variable(s) with a valueequal to zero.

An infeasible problem is identified in the simplex procedure when an optimal solution isachieved (i.e., when all ) and one or more of the basic variables are artificial.

An unbounded problem is identified in the simplex procedure when it is not possible toselect a pivot row—that is, when the values obtained by dividing the quantity values by thecorresponding pivot column values are negative or undefined.

The Dual

Every linear programming model has two forms: the primal and the dual. The originalform of a linear programming model is called the primal. All the examples in this moduleare primal models. The dual is an alternative model form derived completely from the pri-mal. The dual is useful because it provides the decision maker with an alternative way oflooking at a problem. Whereas the primal gives solution results in terms of the amount ofprofit gained from producing products, the dual provides information on the value of theconstrained resources in achieving that profit.

The following example will demonstrate how the dual form of a model is derived andwhat it means. The Hickory Furniture Company produces tables and chairs on a dailybasis. Each table produced results in $160 in profit; each chair results in $200 in profit. Theproduction of tables and chairs is dependent on the availability of limited resources—labor, wood, and storage space. The resource requirements for the production of tables andchairs and the total resources available are as follows:

cj - zj … 0

cj - zj

cj - zj (or zj - cj) = 0

Resource Requirements

Resource TABLE CHAIR Total Available per Day

Labor (hr.) 2 4 40

Wood (bd. ft.) 18 18 216

Storage 24 12 240(ft.2)

The dual solution variablesprovide the value of the resources,

that is, shadow prices.

The original linear programmingmodel is called the primal, andthe alternative form is the dual.

The company wants to know the number of tables and chairs to produce per day tomaximize profit. The model for this problem is formulated as follows:

subject to

of laborof wood

of storage space x1, x2 Ú 0

24x1 + 12x2 … 240 ft.2 18x1 + 18x2 … 216 bd. ft.

2x1 + 4x2 … 40 hr.



The Dual A-31

where

of tables producedof chairs produced

This model represents the primal form. For a primal maximization model, the dual form isa minimization model. The dual form of this example model is

minimize Zd

subject to

The specific relationships between the primal and the dual demonstrated in this exam-ple are as follows:

1. The dual variables, and correspond to the model constraints in the primal.For every constraint in the primal there will be a variable in the dual. For example,in this case the primal has three constraints; therefore, the dual has three decisionvariables.

2. The quantity values on the right-hand side of the primal inequality constraintsare the objective function coefficients in the dual. The constraint quantity values in the primal, 40, 216, and 240, form the dual objective function:

3. The model constraint coefficients in the primal are the decision variable coefficientsin the dual. For example, the labor constraint in the primal has the coefficients 2 and4. These values are the variable coefficients in the model constraints of the dual:

and 4. The objective function coefficients in the primal, 160 and 200, represent the model

constraint requirements (quantity values on the right-hand side of the constraint) inthe dual.

5. Whereas the maximization primal model has constraints, the minimization dualmodel has constraints.

The primal–dual relationships can be observed by comparing the two model forms shownin Figure A-9.

Now that we have developed the dual form of the model, the next step is determiningwhat the dual means. In other words, what do the decision variables and mean,what do the model constraints mean, and what is being minimized in the dual objectivefunction?

Interpreting the Dual ModelThe dual model can be interpreted by observing the simplex solution to the primal form ofthe model. The simplex solution to the primal model is shown in Table A-32.

Interpreting this primal solution, we have

Z = $2,240 profit s3 = 48 ft.2 of storage space x2 = 8 chairs x1 = 4 tables

Ú

y3y1, y2,

Ú

…

4y1.2y1

y1

Z = 40y1 + 216y2 + 240y3.

y3,y1, y2,

y1, y2, y3 Ú 04y1 + 18y2 + 12y3 Ú 2002y1 + 18y2 + 24y3 Ú 160

= 40y1 + 216y2 + 240y3

x2 = numberx1 = number

The dual is formulated entirelyfrom the primal.

A primal maximization modelwith constraints converts to

a dual minimization model with constraints, and

vice versa.Ú

…



This optimal primal tableau also contains information about the dual. In the row ofTable A-32, the negative values and under the and columns indicate that ifone unit of either or was entered into the solution, profit would decrease by $20 or$6.67 (i.e., 20/3), respectively.

Recall that represents unused labor and represents unused wood. In the presentsolution and are not basic variables, so they both equal zero. This means that all thematerial and labor are being used to make tables and chairs, and there are no excess (slack)labor hours or board feet of material left over. Thus, if we enter or into the solution,then or no longer equals zero, and the use of labor or wood is decreased. If, for exam-ple, one unit of is entered into the solution, then one unit of labor previously used is notused, and profit is reduced by $20.

Let us assume that one unit of has been entered into the solution, so that we have onehour of unused labor Now let us remove this unused hour of labor from the solu-tion so that all labor is again being used. We previously noted that profit was decreased by$20 by entering one hour of unused labor; thus, it can be expected that if we take this hourback (and use it again), profit will be increased by $20. This is analogous to saying that if wecould get one more hour of labor, we could increase profit by $20. Therefore, if we could

(s1 = 1).s1

s1

s2s1

s2s1

s2s1

s2s1

s2s1

s2s1-20>3-20cj - zj

cj qi

160

200

160 200 40 216 240

40

216

240

4

18

12

2

18

24

2y1 + 18y2 + 24y3

4y1 + 18y2 + 12y3

x2 ≤

x2 ≤

x2 ≤

y1, y2, y3 0≥

subject to:subject to:

Primal Dual

minimize Zd = y1 + y2 + y3maximize Zp = x1 + x2

x1 +

x1 +

x1 +

≥

≥

x1, x2 ≥ 0

Figure A-9

The primal–dual relationships

160 200 0 0 0

Quantity

200 8 0 1 1/2 0

160 4 1 0 1/9 0

0 48 0 0 6 1

2,240 160 200 20 20/3 0

0 0 0-20>3-20cj - zj

zj

-2s3

-1>2x1

-1>18x2

s3s2s1x2x1

BasicVariablescj

Table A-32

The Optimal Simplex Solutionfor the Primal Model


The Dual A-33

purchase one hour of labor, we would be willing to pay up to $20 for it because that is theamount by which it would increase profit.

The negative row values of $20 and $6.67 are the marginal values of labor and wood respectively. These dual values are also often referred to as shadow prices,since they reflect the maximum “price” one would be willing to pay to obtain one more unitof the resource.

What happened to the third resource, storage space? The answer can be seen in Table A-32. Notice that the row value for (which represents unused storage space) iszero. This means that storage space has a marginal value of zero; that is, we would not bewilling to pay anything for an extra foot of storage space.

The reason more storage space has no marginal value is because storage space was nota limitation in the production of tables and chairs. Table A-32 shows that 48 square feet ofstorage space were left unused (i.e., ) after the 4 tables and 8 chairs were produced.Since the company already has 48 square feet of storage space left over, an extra squarefoot has no additional value; the company cannot even use all the storage space it hasavailable.

We need to consider one additional aspect of these marginal values. In our discus-sion of the marginal value of these resources, we have indicated that the marginal value(or shadow price) is the maximum amount that would be paid for additional resources.The marginal value of $60 for one hour of labor is not necessarily what the HickoryFurniture Company would pay for an hour of labor. This depends on how the objectivefunction is defined. In this example we are assuming that all the resources available, 40hours of labor, 216 board feet of wood, and 240 square feet of storage space, are alreadypaid for. Even if the company does not use all the resources, it still must pay for them.They are sunk costs. In other words, the cost of any additional resources secured areincluded in the objective function coefficients. As such, the profit values in the objec-tive function for each product are unaffected by how much of a resource is actuallyused; the profit is independent of the resources used. If the cost of the resources is notincluded in the profit function, then securing additional resources will reduce the mar-ginal value.

Continuing our analysis, we note that the profit in the primal model was shown tobe $2,240. For the furniture company, the value of the resources used to produce tablesand chairs must be in terms of this profit. In other words, the value of the labor andwood resources is determined by their contribution toward the $2,240 profit. Thus, ifthe company wanted to assign a value to the resources it used, it could not assign anamount greater than the profit earned by the resources. Conversely, using the samelogic, the total value of the resources must also be at least as much as the profit theyearn. Thus, the value of all the resources must exactly equal the profit earned by theoptimal solution.

s3 = 48

s3cj - zj

(s2),(s1)cj - zj

TIME OUT for John Von Neumann

John Von Neumann, the famous Hungarian mathematician, iscredited with many contributions in science and mathematics,including crucial work on the development of the atomic bombduring World War II and the development of the computer fol-lowing the war. In 1947 George Dantzig visited Von Neumann at

the Institute for Advanced Study at Princeton and described thelinear programming technique to him. Von Neumann graspedthe technique immediately, because he had been working onsimilar concepts himself, and went on to explain duality toDantzig for the first time.

The shadow price is themaximum amount that should be

paid for one additional unit of aresource.

The total marginal value of theresources equals the optimal profit.

The values for slackvariables are the marginal values

of the constraint resources, i.e.,shadow prices.

cj - zj

If a resource is not completelyused, i.e., there is slack, its

marginal value is zero.



Now let us look again at the dual form of the model:

subject to

Given the previous discussion on the value of the model resources, we can now define thedecision variables of the dual, and to represent the marginal values of the resources:

Use of the DualThe importance of the dual to the decision maker lies in the information it provides aboutthe model resources. Often the manager is less concerned about profit than about the use ofresources because the manager often has more control over the use of resources than overthe accumulation of profits. The dual solution informs the manager of the value of theresources, which is important in deciding whether to secure more resources and how muchto pay for these additional resources.

If the manager secures more resources, the next question is, How does this affect theoriginal solution? The feasible solution area is determined by the values forming the modelconstraints, and if those values are changed, it is possible for the feasible solution area tochange. The effect on the solution of changes to the model is the subject of sensitivityanalysis, the next topic to be presented here.

Sensitivity Analysis

In this section we will show how sensitivity ranges are mathematically determined usingthe simplex method. While this is not as efficient or quick as using the computer, closeexamination of the simplex method for performing sensitivity analysis can provide a morethorough understanding of the topic.

Changes in Objective Function CoefficientsTo demonstrate sensitivity analysis for the coefficients in the objective function, we will usethe Hickory Furniture Company example developed in the previous section. The model forthis example was formulated as

subject to

of labor of wood

of storage space

wherenumber of tables producednumber of chairs producedx2 =

x1 =

x1, x2 Ú 024x1 + 12x2 … 240 ft.218x1 + 18x2 … 216 bd. ft.

2x1 + 4x2 … 40 hr.


y3 = marginal value of 1 ft.2 of storage space = $0 y2 = marginal value of 1 bd. ft. of wood = $6.67 y1 = marginal value of 1 hr. of labor = $20

y3,y1, y2,

y1, y2, y3, Ú 0 4y1 + 18y2 + 12y3 Ú 200 2y1 + 18y2 + 24y3 Ú 160

minimize Zd = 40y1 + 16y2 + 240y3

The dual provides the decisionmaker with a basis for deciding

how much to pay for moreresources.

The dual variables equal themarginal value of the resources,

the shadow prices.


Sensitivity Analysis A-35

The coefficients in the objective function will be represented symbolically as (the samenotation used in the simplex tableau). Thus, and Now, let us consider achange in one of the values by an amount . For example, let us change by

In other words, we are changing from $160 to $250. The effect of this change onthe solution of this model is shown graphically in Figure A-10.

c1¢ = 90.c1 = 160¢cj

c2 = 200.c1 = 160cj

Originally, the solution to this problem was located at point B in Figure A-10, whereand However, increasing from $160 to $250 shifts the slope of the objec-

tive function so that point becomes the optimal solution. This demon-strates that a change in one of the coefficients of the objective function can change theoptimal solution. Therefore, sensitivity analysis is performed to determine the range overwhich can be changed without altering the optimal solution.

The range of that will maintain the optimal solution can be determined directly fromthe optimal simplex tableau. The optimal simplex tableau for our furniture companyexample is shown in Table A-33.

cj

cj

C (x1 = 8, x2 = 4)c1x2 = 8.x1 = 4

First, consider a change for This will change the value from toas shown in Table A-34. Notice that when is changed to the

new value is included not only in the top row but also in the left-hand column. This is because is a basic solution variable. Since is in the left-hand column, it 160 + ¢x1

cjcj

160 + ¢,c1c1 = 160 + ¢,c1 = 160c1c1.¢

The sensitivity range for a valueis the range of values over whichthe current optimal solution will

remain optimal.

cj

is added to in the optimalsimplex tableau.cj¢

5

50

x1

x2

10 15 20

10

15

20

AB

CD

c1 = 250

c1 = 160

Figure A-10

A change in c1

160 200 0 0 0

Quantity

200 8 0 1 1/2 0

160 4 1 0 1/9 0

0 48 0 0 6 1

2,240 160 200 20 20/3 0

0 0 0-20>3-20cj - zj

zj

-2s3

-1>2x1

-1>18x2

s3s2s1x2x1

BasicVariablescj

Table A-33

The Optimal Simplex Tableau



becomes a multiple of the column values when the new row values and the subsequentrow values, also shown in Table A-34, are computed.cj - zj

zj

The solution shown in Table A-34 will remain optimal as long as the row valuesremain negative. (If becomes positive, the product mix will change, and if it becomeszero, there will be an alternative solution.) Thus, for the solution to remain optimal,

and

Both of these inequalities must be solved for :

and

Thus, and Now recall that therefore,Substituting the amount for in these inequalities,

and

Therefore, the range of values for over which the solution basis will remain optimal(although the value of the objective function may change) is

Next, consider a change in so that The effect of this change in thefinal simplex tableau is shown in Table A-35.

c2 = 200 + ¢.c2¢

100 … c1 … 200

c1

c1 Ú 100 c1 - 160 Ú -60

¢ Ú -60

c1 … 200 c1 - 160 … 40

¢ … 40

¢c1 - 160¢ = c1 - 160.c1 = 160 + ¢;¢ Ú - 60.¢ … 40

¢ Ú -60 - ¢ … 60

- ¢>9 … 20>3 -20>3 - ¢>9 … 0

¢ … 40 ¢>2 … 20

-20 + ¢>2 … 0

¢

-20>3 - ¢>9 … 0

-20 + ¢>2 … 0

cj - zj

cj - zjFor the solution to remain optimalall values in the row must

be … 0.cj - zj

160 200 0 0 0

Quantity

200 8 0 1 1/2 0

160 4 1 0 1/9 0

0 48 0 0 6 1

2,240 160 200 20 20/3 0

0 0 0- ¢/9-20>3+ ¢/2-20cj - zj

+ ¢/9- ¢/2+ ¢+ 4¢zj

-2s3

-1>2x1+ ¢

-1>18x2

s3s2s1x2x1

BasicVariablescj

+ ¢

Table A-34

The Optimal Simplex Tableauwith c1 160 + ¢=



As before, the solution shown in Table A-35 will remain optimal as long as the row values remain negative or zero. Thus, for the solution to remain optimal, we must have

and

Solving these inequalities for gives

and

Thus, and Since we have Substi-tuting this value for in the inequalities yields

and

Therefore, the range of values for over which the solution will remain optimal is

The ranges for both objective function coefficients are as follows:

However, these ranges reflect a possible change in either or not simultaneouschanges in both and Both of the objective function coefficients in this example werefor basic solution variables. Determining the sensitivity range for a decision variable that is not basic is much simpler. Because it is not in the basic variable column, the change ¢

cj

c2.c1

c2,c1

160 … c2 … 320 100 … c1 … 200

160 … c2 … 320

c2

c2 … 320 c2 - 200 … 120

¢ … 120

c2 Ú 160 c2 - 200 Ú -40

¢ Ú -40

¢

¢ = c2 - 200.c2 = 200 + ¢,¢ … 120.¢ Ú -40

¢ … 120 ¢>18 … 20>3

-20>3 + ¢>18 … 0

¢ Ú -40 - ¢>2 … 20

-20 - ¢>2 … 0

¢

-20>3 + ¢>18 … 0

-20 - ¢>2 … 0

cj - zj

160 200 0 0 0

Quantity

200 8 0 1 1/2 0

160 4 1 0 1/9 0

0 48 0 0 6 1

2,240 160 200 20 20/3 0

0 0 0+ ¢/18-20>3- ¢/2-20cj - zj

- ¢/18+ ¢/2+ ¢+ 8¢zj

-2s3

-1>2x1

-1>18x2+ ¢

s3s2s1x2x1

BasicVariablescj

+ ¢

Table A-35

The Optimal Simplex Tableauwith c2 200 + ¢=



does not become a multiple of the row. Thus, the change will show up in only one col-umn in the row.

Changes in Constraint Quantity ValuesTo demonstrate the effect of a change in the quantity values of the model constraints, wewill again use the Hickory Furniture Company example:

subject to

of laborof wood

of storage space

The quantity values 40, 216, and 240 will be represented symbolically as Thus,and Now consider a change in For example, let us

change by In other words, is changed from 216 board feet to 234board feet. The effect of this change is shown graphically in Figure A-11.

q2¢ = 18.q2 = 216q2.¢q3 = 240.q1 = 40, q2 = 216,

qi.

x1, x2 Ú 024x1 + 12x2 … 240 ft.218x1 + 18x2 … 216 bd. ft.

2x1 + 4x2 … 40 hr.


cj - zj

¢zj

In Figure A-11 a change in is shown to have the effect of changing the feasible solu-tion area from 0ABCD to Originally, the optimal solution point was B; however,the change in causes to be the new optimal solution point. At point B the optimalsolution is

At point the new optimal solution is

x2 = 7 x1 = 6

B¿

Z = $2,240 s1 and s2 = 0

s3 = 48 x2 = 8 x1 = 4

B¿q2

0AB¿C¿D.q2

5

50

x1

x2

10 15 20

10

15

20

AB

CD

B'C'

Original optimal solution

New optimal solutionwith change in q2

Figure A-11

A change in q2¢



Thus, a change in a value can change the values of the optimal solution. At some point anincrease or decrease in will change the variables in the optimal solution mix, includingthe slack variables. For example, if q2 increases to 240 board feet, then the optimal solutionpoint will be at

Notice that has left the solution; thus, the optimal solution mix has changed. At thispoint, where which is the upper limit of the sensitivity range for the shadowprice will also change. Therefore, the purpose of sensitivity analysis is to determine therange for over which the optimal variable mix will remain the same and the shadow pricewill remain the same.

As in the case of the values, the range for can be determined directly from the opti-mal simplex tableau. As an example, consider a increase in the number of labor hours.The model constraints become

Notice in the initial simplex tableau for our example (Table A-36) that the changes inthe quantity column are the same as the coefficients in the column.s1

24x1 + 12x2 … 240 + 0¢

18x1 + 18x2 … 216 + 0¢

2x1 + 4x2 … 40 + 1¢

¢

qicj

qi

q2,q2 = 240,s3

Z = $2,400 s1 and s2 = 0

s3 = 0 x2 = 6.67 x1 = 6.67

qi

qi

Z = $2,360 s1 and s2 = 0

s3 = 12

This duplication will carry through each subsequent tableau, so the column valueswill duplicate the changes in the quantity column in the final tableau (Table A-37).

In effect, the changes form a separate column identical to the column. Therefore, todetermine the change, we need only observe the slack column corresponding to themodel constraint quantity being changed.

Recall that a requirement of the simplex method is that the quantity values not be nega-tive. If any value becomes negative, the current solution will no longer be feasible and anew variable will enter the solution. Thus, the inequalities

qi

(qi)(si)¢

s1¢

¢

s1

The sensitivity range for a value is the range of values over

which the right-hand-side valuescan vary without changing the

solution variable mix, includingslack variables and the shadow

prices.

qi

A in a value is duplicated inthe column in the final tableau.si

qi¢

Table A-36

The Initial Simplex Tableau160 200 0 0 0

Quantity

0 2 4 1 0 0

0 18 18 0 1 0

0 24 12 0 0 1

0 0 0 0 0 0

160 200 0 0 0cj - zj

zj

240 + 0¢s3

216 + 0¢s2

40 + 1¢s1

s3s2s1x2x1

BasicVariablescj



are solved for

and

and

Since

then

These values are substituted into the inequalities and asfollows:

Summarizing these inequalities, we have

24 … 32 … q1 … 48

q1 Ú 32 q1 - 40 Ú -8

¢ Ú -8 q1 … 48

q1 - 40 … 8 ¢ … 8 q1 Ú 24

q1 - 40 Ú -16 ¢ Ú -16

¢ Ú -8¢ Ú -16, ¢ … 8,

¢ = q1 - 40

q1 = 40 + ¢

¢ Ú -8 6¢ Ú -48

48 + 6¢ Ú 0

¢ … 8 - ¢>2 Ú -4

4 - ¢>2 Ú 0

¢ Ú -16 ¢>2 Ú -8

8 + ¢>2 Ú 0

¢:

48 + 6¢ Ú 0 4 - ¢>2 Ú 0 8 + ¢>2 Ú 0

160 200 0 0 0

Quantity

200 x2 0 1 1/2 0

160 1 0 1/9 0

0 0 0 6 1

2,240 160 200 20 0

0 0 0-20/3-20cj - zj

20/3+ 20¢zj

-248 + 6¢s3

-1/24 - ¢/2x1

-1/188 + ¢/2

s3s2s1x2x1

BasicVariablescj

Table A-37

The Final Simplex Tableau



The value 24 can be eliminated, since must be greater than 32; thus,

As long as remains in this range, the present basic solution variables will remain thesame and feasible. However, the quantity values of those basic variables may change. Inother words, although the variables in the basis remain the same, their values canchange.

To determine the range for (where ), the column values are used todevelop the inequalities:

The inequalities are solved as follows:

Since

we have

Substituting this value into the inequalities and gives arange of possible values for

That is,

180 … q2 … 240 … 360

q2 … 240 q2 - 216 … 24

¢ … 24

q2 Ú 180 q2 - 216 Ú -36

¢ Ú -36

q2 … 360 q2 - 216 … 144

¢ … 144

q2:¢ … 24¢ … 144, ¢ Ú -36,

¢ = q2 - 216

q2 = 216 + ¢

¢ … 24 -2¢ Ú -48

48 - 2¢ Ú 0

¢ Ú -36 ¢>9 Ú -4

4 + ¢>9 Ú 0

¢ … 144 - ¢>18 Ú -8

8 - ¢>18 Ú 0

48 - 2¢ Ú 0 4 + ¢>9 Ú 0

8 - ¢>18 Ú 0

¢

s2q2 = 216 + ¢q2

q1

32 … q1 … 48

q1



The value 360 can be eliminated, because cannot exceed 240. Thus, the range overwhich the basic solution variables will remain the same is

The range for is

The upper limit of means that can increase indefinitely (without limit) withoutchanging the optimal variable solution mix in the shadow price.

Sensitivity analysis of constraint quantity values can be used in conjunction with thedual solution to make decisions regarding model resources. Recall from our analysis of thedual solution of the Hickory Furniture Company example that

Because the resource with the greatest marginal value is labor, the manager might desireto secure some additional hours of labor. How many hours should the manager get? Giventhat the range for is the manager could secure up to an additional 8hours of labor (i.e., 48 total hours) before the solution basis changes and the shadow pricealso changes. If the manager did purchase 8 more hours, the solution values could be foundby observing the quantity values in Table A-37:

Since

Total profit would be increased by $20 for each extra hour of labor:

In this example for the Hickory Furniture Company, we considered only constraintsin determining the sensitivity ranges for values. To compute the sensitivity range,we observed the slack column, since a change in was reflected in the column.However, recall that with a constraint we subtract a surplus variable rather than adding aslack variable to form an equality (in addition to adding an artificial variable). Thus, for a constraint we must consider a change in in order to use the (surplus) columnto perform sensitivity analysis. In that case sensitivity analysis would be performed exactly

siqi- ¢Ú

Ú

siqi¢si,qiqi

…

= $2,400 = 2,240 + 160 = 2,240 + 20(8)

Z = $2,240 + 20¢

= 96 s3 = 48 + 6(8)

= 0 x1 = 4 - (8)>2

= 12 x2 = 8 + (8)>2

¢ = 8,

s3 = 48 + 6¢

x1 = 4 - ¢>2 x2 = 8 + ¢>2

32 … q1 … 48,q1

y3 = $0, marginal value of storage space y2 = $6.67, marginal value of wood y1 = $20, marginal value of labor

q3q

192 … q3 6 q

q3

180 … q2 … 240

q2

The shadow prices are valid onlywithin the sensitivity range for the

right-hand-side values.


Problems A-43

as shown in this example, except that the value of would be used instead ofwhen computing the sensitivity range for

Problems

1. Following is a simplex tableau for a linear programming model:

qi.qi + ¢qi - ¢

a. What is the solution given in this tableau?b. Is the solution in this tableau optimal? Why?c. What does equal in this tableau? ?d. Write out the original objective function for the linear programming model, using only

decision variables.e. How many constraints are in the linear programming model?f. Explain briefly why it would have been difficult to solve this problem graphically.

2. The following is a simplex tableau for a linear programming model:

s2x3

Basic10 2 6 0 0 0

cj Variables Quantity x1 x2 x3 s1 s2 s3

10 10 1 0 1 0

2 40 0 1 2 0 1/2 0

0 30 0 0 8 3/2 1

420 10 2 16 2 4 0

0 0 0-4-2-10cj - zj

zj

-3s3

x2

-1>2-2x1

Basic6 20 12 0 0

cj Variables Quantity x1 x2 x3 s1 s2

6 20 1 1 0 0 0

12 10 0 1/3 1 0

0 10 0 1/3 0 1

240 6 10 12 0

0 0 0 -2-10zj - cj

-2zj

-1>6s1

-1>6x3

x1

a. Is this a maximization or a minimization problem? Why?b. What is the solution given in this tableau?c. Is the solution given in this tableau optimal? Why?



d. Write out the original objective function for the linear programming model using onlydecision variables.

e. How many constraints are in the linear programming model?f. Were any of the constraints originally equations? Why?g. What is the value of in this tableau?

3. The following is a simplex tableau for a linear programming problem:

x2

Basic60 50 45 50 0 0 0 0

cj Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4

0 20 0 1 0 0 1 0 0 0

50 15 0 0 0 1 0 1 0 0

60 12 1 1/2 0 0 0 0 1/10 0

0 45 0 0 8 6 0 0 1

1,470 60 30 0 50 0 50 6 0

0 20 45 0 0 0-6-50cj - zj

zj

-6s4

x1

x4

s1

Basic8 10 4 0 0 0 M M

cj Variables Quantity x1 x2 x3 s1 s2 s3 A1 A2

M 30 2/3 0 1 1/6 0 0 0

10 10 1/3 1 0 0 0 1 0

M 100 0 0 1 0 0 0 1

10 M M

0 0 0-MM>6 - 5>3-M2M - 42M>3 - 14>3zj - cj

-MM>6 - 5>3-M2M2M>3 + 10>3130M + 100zj

-1A3

-1>6x2

-1A1

a. Is this a maximization or a minimization problem?b. What are the values of the decision variables in this tableau?c. What are the values of the slack variables in this tableau?d. What does the value of “20” in the “ ” column mean?e. Is this solution optimal? Why? If the solution is not optimal, determine the optimal solution.

4. The following is a simplex tableau for a linear programming problem:

x2cj - zj

a. Is this a maximization or a minimization problem?b. What is the value of in this tableau?c. What does the value “ ” in the “ ” column of the row mean?d. What is the minimum number of additional simplex iterations that this problem must go

through to determine a feasible optimal solution?e. Is this solution optimal? Why? If the solution is not optimal, compute the optimal solution.

zj - cjs2M>6 - 5>3x3


Problems A-45

a. Is this a maximization or a minimization problem? Why?b. What are the values of the decision variables in this tableau?c. Were any of the constraints in this problem originally equations? Why?d. What is the value of in this tableau?e. Is this solution optimal? Why? If the solution is not optimal, complete the next iteration

(tableau) and indicate if it is optimal.

s2

Basic4 6 0 0 M


M 2 0 1/2 1/2 1

4 6 1 1/2 0 0

4 M

0 0M>2 - 2-MM>2 - 4zj - cj

M>2 - 2-MM>2 + 224 + 2Mzj

-1>2x1

-1A1

Basic10 5 0 0 --M


10 5 1 1/2 0 0

4 0 1 0 0 1

0 15 0 7/2 1/2 1 0

10 0

0 M 5 0 0cj - zj

-M-5-M + 5-4M + 50zj

s2

A2-M

-1>2x1

a. Is this a maximization or a minimization problem? Why?b. What is the value of in this tableau?c. Does the fact that has a value equal to “0” in this tableau mean that multiple

optimal solutions exist? Why?d. What does the value for the column mean?e. Is this solution optimal? Why? If not, solve this problem and indicate if multiple optimal

solutions exist.

s1cj - zj

cj - zjx1

x2

5. Given is the following simplex tableau for a linear programming problem:

6. Following is a simplex tableau for a linear programming problem:

7. The Munchies Cereal Company makes a cereal from several ingredients. Two of the ingredients,oats and rice, provide vitamins A and B. The company wants to know how many ounces of oats andrice it should include in each box of cereal to meet the minimum requirements of 48 milligrams ofvitamin A and 12 milligrams of vitamin B while minimizing cost. An ounce of oats contributes8 milligrams of vitamin A and 1 milligram of vitamin B, whereas an ounce of rice contributes



6 milligrams of vitamin A and 2 milligrams of vitamin B. An ounce of oats costs $0.05, and anounce of rice costs $0.03. Formulate a linear programming model for this problem and solve usingthe simplex method.

8. A company makes product 1 and product 2 from two resources. The linear programming modelfor determining the amounts of product 1 and 2 to produce ( and ) is

subject to

(resource 1, lb.)(resource 2, lb.)

Solve this model using the simplex method.

9. A company produces two products that are processed on two assembly lines. Assembly line 1 has100 available hours, and assembly line 2 has 42 available hours. Each product requires 10 hours ofprocessing time on line 1, while on line 2 product 1 requires 7 hours and product 2 requires 3hours. The profit for product 1 is $6 per unit, and the profit for product 2 is $4 per unit. Formulatea linear programming model for this problem and solve using the simplex method.

10. The Pinewood Furniture Company produces chairs and tables from two resources—labor andwood. The company has 80 hours of labor and 36 board feet of wood available each day. Demandfor chairs is limited to 6 per day. Each chair requires 8 hours of labor and 2 board feet of wood toproduce, while a table requires 10 hours of labor and 6 board feet of wood. The profit derived fromeach chair is $400 and from each table, $100. The company wants to determine the number ofchairs and tables to produce each day to maximize profit. Formulate a linear programming modelfor this problem and solve using the simplex method.

11. The Crumb and Custard Bakery makes both coffee cakes and Danish in large pans. The main ingre-dients are flour and sugar. There are 25 pounds of flour and 16 pounds of sugar available and thedemand for coffee cakes is 8. Five pounds of flour and 2 pounds of sugar are required to make onepan of coffee cake, and 5 pounds of flour and 4 pounds of sugar are required to make one pan ofDanish. One pan of coffee cake has a profit of $1, and one pan of Danish has a profit of $5.Determine the number of pans of cake and Danish that the bakery must produce each day so thatprofit will be maximized. Formulate a linear programming model for this problem and solve usingthe simplex method.

12. The Kalo Fertilizer Company makes a fertilizer using two chemicals that provide nitrogen, phos-phate, and potassium. A pound of ingredient 1 contributes 10 ounces of nitrogen and 6 ounces ofphosphate, whereas a pound of ingredient 2 contributes 2 ounces of nitrogen, 6 ounces of phos-phate, and 1 ounce of potassium. Ingredient 1 costs $3 per pound, and ingredient 2 costs $5 perpound. The company wants to know how many pounds of each chemical ingredient to put into abag of fertilizer to meet minimum requirements of 20 ounces of nitrogen, 36 ounces of phosphate,and 2 ounces of potassium while minimizing cost. Formulate a linear programming model for thisproblem and solve using the simplex method.

13. Solve the following model using the simplex method:

subject to

minimize Z = 0.06x1 + 0.10x2

x1, x2 Ú 02x1 + 6x2 … 184x1 + 5x2 … 20

maximize Z = 8x1 + 2x2 (profit, $)

x2x1


Problems A-47

14. The Copperfield Mining Company owns two mines, both of which produce three grades of ore—high, medium, and low. The company has a contract to supply a smelting company with at least 12tons of high-grade ore, 8 tons of medium-grade ore, and 24 tons of low-grade ore. Each mine pro-duces a certain amount of each type of ore each hour it is in operation. Mine 1 produces 6 tons ofhigh-grade, 2 tons of medium-grade, and 4 tons of low-grade ore per hour. Mine 2 produces 2 tonsof high-grade, 2 tons of medium-grade, and 12 tons of low-grade ore per hour. It costs $200 perhour to mine each ton of ore from mine 1, and it costs $160 per hour to mine a ton of ore frommine 2. The company wants to determine the number of hours it needs to operate each mine sothat contractual obligations can be met at the lowest cost. Formulate a linear programming modelfor this problem and solve using the simplex method.

15. A marketing firm has contracted to do a survey on a political issue for a Spokane television station.The firm conducts interviews during the day and at night, by telephone and in person. Each houran interviewer works at each type of interview results in an average number of interviews. In orderto have a representative survey, the firm has determined that there must be at least 400 day inter-views, 100 personal interviews, and 1,200 interviews overall. The company has developed thefollowing linear programming model to determine the number of hours of telephone interviewsduring the day telephone interviews at night personal interviews during the day andpersonal interviews at night that should be conducted to minimize cost:

subject to

(day interviews)(personal interviews)

(total interviews)


16. A jewelry store makes both necklaces and bracelets from gold and platinum. The store has devel-oped the following linear programming model for determining the number of necklaces andbracelets ( and ) that it needs to make to maximize profit:

subject to

(gold, oz.)(platinum, oz.)

(demand, bracelets)


17. A sporting goods company makes baseballs and softballs on a daily basis from leather and yarn.The company has developed the following linear programming model for determining the numberof baseballs and softballs to produce ( and ) to maximize profits:x2x1

x1, x2 Ú 0x2 … 4

2x1 + 4x2 … 203x1 + 2x2 … 18


x2x1

x1, x2, x3, x4 Ú 0x1 + x2 + x3 + x4 Ú 1,200

4x3 + 5x4 Ú 10010x1 + 4x3 Ú 400

minimize Z = 2x1 + 3x2 + 5x3 + 7x4 (cost, $)

(x4)(x3),(x2),(x1),

x1, x2 Ú 05x1 + 2x2 Ú 103x1 + 6x2 Ú 124x1 + 3x2 Ú 12



subject to

(yarn, yd.)


18. A clothing shop makes suits and blazers. Three main resources are used: material, rack space, andlabor. The shop has developed this linear programming model for determining the number of suitsand blazers to make ( and ) to maximize profits:

subject to

(rack space)(labor, hr.)


19. Solve the following linear programming model using the simplex method:

subject to

20. The following is a simplex tableau for a linear programming model:

x1, x2, x3 Ú 0x3 … 40

2x1 + 2x2 + 2x3 … 1003x1 + 5x2 … 60

maximize Z = 100x1 + 20x2 + 60x3

x1, x2 Ú 010x1 + 20x2 … 300

x1 + x2 … 2010x1 + 4x2 … 160 (material, yd.2)


x2x1

x1, x2 Ú 010x1 + 4x2 … 2,000

0.3x1 + 0.5x2 … 150 (leather, ft.2)


Basic1 2 --1 0 0 0


2 10 0 1 1/4 1/4 0 0

0 20 0 0 3/4 1

1 10 1 0 1 0 1/2

30 1 2 3/2 0 0 1/2

0 0 0 0 -1>2-5>2cj - zj

zj

-1>2x1

-1>2-3>4s2

x2

a. Is this a maximization or a minimization problem? Why?b. What is the solution given in this tableau?c. Write out the original objective function for the linear programming model, using only

decision variables.d. How many constraints are in the linear programming model?e. Were any of the constraints originally equations? Why?f. What does equal in this tableau?s1


Problems A-49

g. This solution is optimal. Are there multiple optimal solutions? Why?h. If there are multiple optimal solutions, identify the alternate solutions.

21. A wood products firm in Oregon plants three types of trees—white pines, spruce, and ponderosapines—to produce pulp for paper products and wood for lumber. The company wants to plantenough acres of each type of tree to produce at least 27 tons of pulp and 30 tons of lumber. Thecompany has developed the following linear programming model to determine the number ofacres of white pines spruce and ponderosa pines to plant to minimize cost:

subject to

(pulp, tons)(lumber, tons)


22. A baby products firm produces a strained baby food containing liver and milk, each of which con-tribute protein and iron to the baby food. Each jar of baby food must have 36 milligrams of proteinand 50 milligrams of iron. The company has developed the following linear programming model todetermine the number of ounces of liver and milk to include in each jar of baby food tomeet the requirements for protein and iron at the minimum cost:

subject to

(protein, mg)(iron, mg)


23. Solve the linear programming model in Problem 22 graphically, and identify the points on thegraph that correspond to each simplex tableau.

24. The Cookie Monster Store at South Acres Mall makes three types of cookies—chocolate chip,pecan chip, and pecan sandies. Three primary ingredients are chocolate chips, pecans, and sugar.The store has 120 pounds of chocolate chips, 40 pounds of pecans, and 300 pounds of sugar. Thefollowing linear programming model has been developed for determining the number of batchesof chocolate chip cookies pecan chip cookies and pecan sandies to make to maxi-mize profit:

subject to

(sugar, lb.)(chocolate chips, lb.)

(pecans, lb.)


x1, x2, x3 Ú 0x1 + 2x3 … 40

10x1 + 5x2 … 12020x1 + 15x2 + 10x3 … 300

maximize Z = 10x1 + 12x2 + 7x3 (profit, $)

(x3)(x2),(x1),

x1, x2 Ú 05x1 + 5x2 Ú 506x1 + 2x2 Ú 36

minimize Z = 0.05x1 + 0.10x2 (cost, $)

(x2)(x1)

x1, x2, x3 Ú 02x1 + 6x2 + 3x3 Ú 30

4x1 + x2 + 3x3 Ú 27

minimize Z = 120x1 + 40x2 + 240x3 (cost, $)

(x3)(x2),(x1),



25. The Eastern Iron and Steel Company makes nails, bolts, and washers from leftover steel and coatsthem with zinc. The company has 24 tons of steel and 30 tons of zinc. The following linear pro-gramming model has been developed for determining the number of batches of nails bolts

and washers to produce to maximize profit:

subject to

(steel, tons)(zinc, tons)



subject to


subject to

28. Solve the linear programming model in Problem 27 graphically, and identify the points on thegraph that correspond to each simplex tableau.

29. Transform the following linear programming model into proper form for solution by the simplexmethod:

subject to

30. Transform the following linear programming model into proper form for solution by the simplexmethod:

x1, x2, x3 Ú 0x1 + 2x2 Ú 20

4x1 + x2 + 2x3 … 503x2 + 4x3 Ú 60

2x1 + 6x2 + x3 = 30

minimize Z = 8x1 + 2x2 + 7x3

x1, x2 Ú 02x1 + 6x2 Ú 12

x1 + x2 Ú 43x1 + x2 Ú 6

minimize Z = 20x1 + 16x2

x1, x2, x3, x4 Ú 0100x1 + 200x4 … 2,200200x1 + 250x3 … 2,000

4x3 + x4 … 253x1 + 2x2 … 40

maximize Z = 100x1 + 75x2 + 90x3 + 95x4

x1, x2, x3 Ú 02x1 + 6x2 + 3x3 … 30

4x1 + x2 + 3x3 … 24

maximize Z = 6x1 + 2x2 + 12x3 (profit, $1,000s)

(x3)(x2),(x1),


Problems A-51

subject to

31. A manufacturing firm produces two products using labor and material. The company has a con-tract to produce 5 of product 1 and 12 of product 2. The company has developed the following lin-ear programming model to determine the number of units of product and product toproduce to maximize profit:

subject to

(material, lb.)(labor, hr.)

(contract, product 1)(contract, product 2)


32. A custom tailor makes pants and jackets from imported Irish wool cloth. To get any cloth at all, thetailor must purchase at least 25 square feet each week. Each pair of pants and each jacket requires 5 square feet of material. The tailor has 16 hours available each week to make pants and jackets. Thedemand for pants is never more than 5 pairs per week. The tailor has developed the following lin-ear programming model to determine the number of pants and jackets to make each weekto maximize profit:

subject to

(labor, hr.)(demand, pants)


33. A sawmill in Tennessee produces cherry and oak boards for a large furniture manufacturer. Eachmonth the sawmill must deliver at least 5 tons of wood to the manufacturer. It takes the sawmill 3days to produce a ton of cherry and 2 days to produce a ton of oak, and the sawmill can allocate 18days out of a month for this contract. The sawmill can get enough cherry to make 4 tons of woodand enough oak to make 7 tons of wood. The sawmill owner has developed the following linearprogramming model to determine the number of tons of cherry and oak to produce tominimize cost:

(x2)(x1)

x1, x2 Ú 0x1 … 5

2x1 + 4x2 … 165x1 + 5x2 Ú 25 (wool, ft.2)

maximize Z = x1 + 5x2 (profit, $100s)

(x2)(x1)

x1, x2 Ú 0x2 Ú 12x1 Ú 5

4x1 + 4x2 … 72x1 + 2x2 … 30


2 (x2)1 (x1)

x1, x2, x3 Ú 05x2 + 3x3 Ú 100

x1 + 3x2 + x3 Ú 502x1 + x2 + x3 = 40

x1 + 2x2 + 3x3 … 60

minimize Z = 40x1 + 55x2 + 30x3



subject to

(production time, days)(contract, tons)(cherry, tons)(oak, tons)



subject to

35. Solve the following linear programming problem using the simplex method:

subject to

36. Solve the following linear programming problem using the simplex method:

subject to

37. A farmer has a 40-acre farm in Georgia. The farmer is trying to determine how many acres of corn,peanuts, and cotton to plant. Each crop requires labor, fertilizer, and insecticide. The farmer hasdeveloped the following linear programming model to determine the number of acres of corn peanuts and cotton to plant to maximize profit:

subject to

(labor, hr.)(fertilizer, tons)4x1 + 3x2 + x3 … 160

2x1 + 3x2 + 2x3 … 120


(x3)(x2),(x1),

x1, x2, x3 Ú 0x1 + x2 - x3 Ú 8

2x1 + x2 + 5x3 = 20x1 + x2 + 2x3 … 12

maximize Z = x1 + 2x2 + 2x3

x1, x2, x3 Ú 02x1 + 4x2 + 3x3 … 60

x1 - x2 … 204x2 + x3 … 40

maximize Z = x1 + 2x2 - x3

x1, x2 Ú 0x1 + 4x2 … 20

x2 = 42x1 + x2 Ú 10


x1, x2 Ú 0x2 … 7x1 … 4

x1 + x2 Ú 53x1 + 2x2 … 18

minimize Z = 3x1 + 6x2 (cost, $)


(insecticide, tons)(acres)


38. Solve the following linear programming model (a) graphically and (b) using the simplex method:

subject to

39. Solve the following linear programming model (a) graphically and (b) using the simplex method:

subject to


subject to


subject to

42. The Old English Metal Crafters Company makes brass trays and buckets. The number of trays and buckets that can be produced daily is constrained by the availability of brass and labor, asreflected in the following linear programming model:

(x2)(x1)

x1, x2 Ú 03x1 + 2x2 … 92x1 + x2 Ú 6

3x1 + 4x2 Ú 12

minimize Z = 15x1 + 25x2

x1, x2, x3 Ú 0x3 … 6

x1 + x2 … 252x1 + x2 + x3 … 40

x1 + x2 + x3 … 25

maximize Z = 7x1 + 5x2 + 5x3

x1, x2 Ú 0-x1 + 2x2 … 4

x1 - x2 Ú -1

maximize Z = x1 + x2

x1, x2 Ú 0x1 + x2 Ú 2x1 + x2 … 1


x1, x2, x3 Ú 0x1 + x2 + x3 … 40

3x1 + 2x2 + 4x3 … 100

Problems A-53


subject to

(brass, lb.)(labor, hr.)

The final optimal simplex tableau for this model is as follows:

x1, x2 Ú 02x1 + 2x2 … 60

x1 + 4x2 … 90



Basic 6 10 0 0

cjVariables

Quantity x1 x2 s1 s2

10 20 0 1

6 10 1 0

260 6 10

0 0 -7>3-4>3cj - zj

7>34>3zj

2>3-1>3x1

-1>61>3x2

Basic200 300 0 0


300 6 0 1 17/750

200 20 1 0 1/30

5,800 200 300 310/75 70/15

0 0 -70>15-310>75cj - zj

zj

-1>75x1

-1>150x2

a. Formulate the dual of this model.b. Define the dual variables and indicate their value.c. Determine the optimal ranges for and d. Determine the feasible ranges for (pounds of brass) and (labor hours).e. What is the maximum price the company would be willing to pay for additional labor

hours, and how many hours could be purchased at that price?

43. The Southwest Foods Company produces two brands of chili—Razorback and Longhorn—fromseveral ingredients, including chili beans and ground beef. The number of 100-gallon batches ofRazorback chili and Longhorn chili that can be produced daily is constrained by the avail-ability of chili beans and ground beef, as shown in the following linear programming model:

subject to

(chili beans, lb.)(ground beef, lb.)


x1, x2 Ú 034x1 + 20x2 … 80010x1 + 50x2 … 500


(x2)(x1)

q2q1

c2.c1


a. Formulate the dual of this model and indicate what the dual variables equal.b. What profit for Razorback chili will result in no production of Longhorn chili? What will

the new optimal solution values be?c. Determine what the effect will be of changing the amount of beans in Razorback chili from

10 pounds per batch to 15 pounds per batch.d. Determine the optimal ranges for and e. Determine the feasible ranges for (pounds of beans) and (pounds of ground beef).f. What is the maximum price the company would be willing to pay for additional pounds of

chili beans, and how many pounds could be purchased at that price?g. If the company could secure an additional 100 pounds of only one of the ingredients, beans

or ground beef, which should it be?h. If the company changed the selling price of Longhorn chili so that the profit was $400

instead of $300, would the optimal solution be affected?

44. The Agrimaster Company produces two kinds of fertilizer spreaders—regular and cyclone.Each spreader must undergo two production processes. Letting the number of regularspreaders produced and the number of cyclone spreaders produced, the problem can beformulated as follows:

subject to

(process 1, production hr.)(process 2, production hr.)

The final optimal simplex tableau for this problem is as follows:

x1, x2 Ú 04x1 + 8x2 … 40

12x1 + 4x2 … 60


x2 =

x1 =

q2q1

c2.c1

Problems A-55

Basic9 7 0 0


9 4 1 0

7 3 0 1

57 9 7

0 0 -12>20-11>20cj - zj

12>2011>20zj

3>20-1>20x2

-1>201>10x1

a. Formulate the dual for this problem.b. Define the dual variables and indicate their values.c. Determine the optimal ranges for and d. Determine the feasible ranges for and (production hours for processes 1 and 2,

respectively).e. What is the maximum price the Agrimaster Company would be willing to pay for

additional hours of process 1 production time, and how many hours could be purchased atthat price?

q2q1

c2.c1


45. The Stratford House Furniture Company makes two kinds of tables—end tables and coffeetables The manufacturer is restricted by material and labor constraints, as shown in the fol-lowing linear programming formulation:

subject to

(labor, hr.)(wood, bd. ft.)

The final optimal simplex tableau for this problem is as follows:

x1, x2 Ú 03x1 + 3x2 … 1352x1 + 5x2 … 180


(x2).(x1)


Basic200 300 0 0


300 30 0 1

200 15 1 0

12,000 200 300

0 0 -400>9-100>3cj - zj

400>9100>3zj

5>9-1>3x1

-2>91>3x2

a. Formulate the dual for this problem.b. Define the dual variables and indicate their values.c. What profit for coffee tables will result in no production of end tables, and what will the

new optimal solution values be?d. What will be the effect on the optimal solution if the available wood is increased from 135

to 165 board feet?e. Determine the optimal ranges for and f. Determine the feasible ranges for (labor hours) and (board feet of wood).g. What is the maximum price the Stratford House Furniture Company would be willing to pay

for additional wood, and how many board feet of wood could be purchased at that price?h. If the furniture company wanted to secure additional units of only one of the resources,

labor or wood, which should it be?

46. A manufacturing firm produces electric motors for washing machines and vacuum cleaners. Thefirm has resource constraints for production time, steel, and wire. The linear programming modelfor determining the number of washing machine motors and vacuum cleaner motors toproduce has been formulated as follows:

subject to

(production, hr.)(steel, lb.)(wire, ft.)


x1, x2 Ú 0x1 + 2x2 … 20

x1 + x2 … 142x1 + x2 … 19


(x2)(x1)

q2q1

c2.c1


a. Formulate the dual for this problem.b. What do the dual variables equal, and what do they mean?c. Determine the optimal ranges for and d. Determine the feasible ranges for (production hours), (pounds of steel), and (feet

of wire).e. Managers at the firm have determined that the firm can purchase a new production

machine that will increase available production time from 19 to 25 hours. Will this changeaffect the optimal solution?

47. A manufacturer produces products 1 and 2, for which profits are $9 and $12, respectively. Eachproduct must undergo two production processes that have labor constraints. There are also mater-ial constraints and storage limitations. The linear programming model for determining the num-ber of product 1 to produce and the number of product 2 to product is given as follows:

subject to

(process 1, labor hr.)(process 2, labor hr.)

(material, lb.)


x1, x2 Ú 0x2 … 7 (storage space, ft.2)x1 … 7 (storage space, ft.2)

15x1 + 8x2 … 1205x1 + 5x2 … 504x1 + 8x2 … 64


(x2)(x1)

q3q2q1

c2.c1

Problems A-57

Basic70 80 0 0 0


70 6 1 0 0

0 1 0 0 1

80 7 0 1 0

980 70 80 20 0 30

0 0 0 -30-20cj - zj

zj

2>3-1>3x2

-1>3-1>3s2

-1>32>3x1

Basic9 12 0 0 0 0 0

cj Variables Quantity x1 x2 s1 s2 s3 s4 s5

9 4 1 0 0 0 0

0 1 0 0 0 0 1

0 12 0 0 1 0 0

0 3 0 0 0 1 0

12 6 0 1 0 0 0

108 9 12 0 0 0

0 0 0 0 0-6>5-3>4cj - zj

6>53>4zj

-1>51>4x2

-2>51>4s4

-22>57>4s3

1>5-1>4s5

2>5-1>4x1



a. Formulate the dual for this problem.b. What do the dual variables equal, and what does this dual solution mean?c. Determine the optimal ranges for and d. Determine the range for (process 1, labor hr.).e. Due to a problem with a supplier, only 100 pounds of material will be available for

production instead of 120 pounds. Will this affect the optimal solution mix?

48. A manufacturer produces products 1, 2, and 3 daily. The three products are processed throughthree production operations that have time constraints, and the finished products are then stored.The following linear programming model has been formulated to determine the number of prod-uct product and product to produce:

subject to

(operation 1, hr.)(operation 2, hr.)(operation 3, hr.)


x1, x2, x3 Ú 0x1 + x2 + x3 … 40 (storage, ft.2)

3x1 + 2x2 + 4x3 … 1004x1 + 3x2 + x3 … 160

2x1 + 3x2 + 2x3 … 120


3 (x3)2 (x2),1 (x1),

q1

c2.c1

Basic40 35 45 0 0 0 0

cj Variables Quantity x1 x2 x3 s1 s2 s3 s4

0 10 0 0 1 0

0 60 2 0 0 0 1 1

45 10 1/2 0 1 0 0 1/2

35 30 1/2 1 0 0 0 2

1,500 40 35 45 0 0 5 25

0 0 0 0 0 -25-5cj - zj

zj

-1>2x2

-1x3

-5s2

-41>2-1>2s1

a. Formulate the dual for this problem.b. What do the dual variables equal, and what do they mean?c. How does the fact that this problem has multiple optimum solutions affect the

interpretation of the dual solution values?d. Determine the optimal range for e. Determine the feasible range for (square feet of storage space).f. What is the maximum price the manufacturer would be willing to pay to lease additional

storage space, and how many additional square feet could be leased at that price?

49. A school dietitian is attempting to plan a lunch menu that will minimize cost and meet certainminimum dietary requirements. The two staples in the meal are meat and potatoes, which provideprotein, iron, and carbohydrates. The following linear programming model has been formulated todetermine how many ounces of meat and ounces of potatoes to put in a lunch:

minimize Z = 0.03x1 + 0.02x2 (cost, $)

(x2)(x1)

q4

c2.


subject to

(protein, mg)(iron, mg)(carbohydrates, mg)


x1, x2 Ú 03x1 + 2x2 Ú 12

12x1 + 3x2 Ú 304x1 + 5x2 Ú 20

Problems A-59

Basic0.03 0.02 0 0 0


0.02 3.6 0 1 0 0.20

0.03 1.6 1 0 0 0.20

0 4.4 0 0 1 0.47

0.12 0.03 0.02 0 0

0 0 0 0 -0.01zj - cj

-0.01zj

-3.2s1

-0.13x1

-0.80x2

a. Formulate the dual for this problem.b. What do the dual variables equal, and what do they mean?c. Determine the optimal ranges for and d. Determine the ranges for and (milligrams of protein, iron, and carbohydrates,

respectively).e. What would it be worth for the school dietitian to be able to reduce the requirement for

carbohydrates, and what is the smallest number of milligrams of carbohydrates that wouldbe required at that value?

50. The Overnight Food Processing Company prepares sandwiches (among other processed fooditems) for vending machines, markets, and business canteens around the city. The sandwiches aremade at night and delivered early the following morning. Any sandwiches not purchased duringthe previous day are thrown away. Three kinds of sandwiches are made each night, a basic cheesesandwich a ham salad sandwich and a pimento cheese sandwich The profits are$1.25, $2.00, and $1.75, respectively. It takes 0.5 minutes to make a cheese sandwich, 1.2 minutes tomake a ham salad sandwich, and 0.8 minutes to make a pimento cheese sandwich. The companyhas 20 hours of labor available to produce the sandwiches each night. The demand for ham saladsandwiches is at least as great as the demand for the two types of cheese sandwiches combined.However, the company has only enough ham salad to produce 500 sandwiches per night. The com-pany has formulated the following linear programming model in order to determine how many ofeach type of sandwich to make to maximize profit:

subject to

(production time, min.)(demand for ham salad sandwiches)

(ham salad sandwich limit)x1, x2, x3 Ú 0

x2 … 500x1 + x3 … x2

0.5x1 + 1.2x2 + 0.8x3 … 1,200

maximize Z = $1.25x1 + 2.00x2 + 1.75x3

(x3).(x2),(x1),

q3q1, q2,c2.c1


The optimal simplex tableau follows:


Basic1.25 2.00 1.75 0 0 0


0 200 0 0 1

1.75 500 1 0 1 0 1 1

2.00 500 0 1 0 0 0 1

1,875 1.75 2.00 1.75 0 1.75 3.75

0 0 0 -3.75-1.75- .5cj - zj

zj

x2

x3

-2-0.8-0.3s1

Basic3 5 2 0 0 0


0 15 0 0 1

3 5 1 0 0

5 30 0 1 0

165 3 5 0

0 0 0 -1-4-13zj - cj

-1-4-11zj

-1>2-1>2-1x2

1>2-1>2-2x1

-1>2-3>2-4s2

a. Formulate the dual for this problem and define the dual variables.b. Determine the optimal ranges for and c. Determine the range for (ham salad sandwiches).d. Overnight Foods is considering advertising its cheese sandwiches to increase demand. The

company estimates that spending $100 on some leaflets that would be packaged with allother sandwiches would increase the demand for both kinds of cheese sandwiches by 200.Should it make this expenditure?

51. Given the linear programming model,

subject to

and its optimal simplex tableau,

x1, x2, x3 Ú 0-x1 + x2 Ú 25x1 + 2x2 Ú 50

x1 + x2 - 3x3 Ú 35

minimize Z = 3x1 + 5x2 + 2x3

q3

c3.c1, c2,


a. Find the optimal ranges for all values.b. Find the feasible ranges for all values.

52. The Sunshine Food Processing Company produces three canned fruit products—mixed fruit fruit cocktail and fruit delight The main ingredients in each product are pears andpeaches. Each product is produced in lots and must undergo three processes—mixing, canning,and packaging. The resource requirements for each product and each process are shown in the fol-lowing linear programming formulation:

subject to

(pears, lb.)(peaches, lb.)

(mixing, hr.)(canning, hr.)(packaging, hr.)

The optimal simplex tableau is as follows:

x1, x2, x3 Ú 02x1 + x2 + x3 … 40

x1 + x2 + x3 … 60x1 + 2x2 + 2x3 … 43

10x1 + 20x2 + 16x3 … 40020x1 + 10x2 + 16x3 … 320


(x3).(x2),(x1),

qi

cj

Problems A-61

Basic10 6 8 0 0 0 0 0

cj Variables Quantity x1 x2 x3 s1 s2 s3 s4 s5

10 8 1 0 8/15 1/15 1/30 0 0 0

6 16 0 1 8/15 1/15 0 0 0

0 3 0 0 2/5 0 1 0 0

0 36 0 0 0 1 0

0 8 0 0 0 0 0 1

176 10 6 128/15 7/15 1/15 0 0 0

0 0 0 0 0-1>15-7>15-8>15cj - zj

zj

-1-3>5s5

-1>30-1>30-1>15s4

-1>10s3

-1>30x2

x1

a. What is the maximum price the company would be willing to pay for additional pears?How much could be purchased at that price?

b. What is the marginal value of peaches? Over what range is this price valid?c. The company can purchase a new mixing machine that would increase the available mixing

time from 43 to 60 hours. Would this affect the optimal solution?d. The company can also purchase a new packaging machine that would increase the available

packaging time from 40 to 50 hours. Would this affect the optimal solution?e. If the manager was to attempt to secure additional units of only one of the resources, which

should it be?

53. The Evergreen Products Firm produces three types of pressed paneling from pine and spruce. Thethree types of paneling are Western Old English and Colonial Each sheet must be cutand pressed. The resource requirements are given in the following linear programming formulation:

(x3).(x2),(x1),


subject to

(pine, lb.)(spruce, lb.)

(cutting, hr.)(pressing, hr.)


x1, x2, x3 Ú 02x1 + 4x2 + 2x3 … 80

x1 + x2 + 2x3 … 502x1 + 5x2 + 2x3 … 1605x1 + 4x2 + 4x3 … 200



Basic4 10 8 0 0 0 0

cj Variables Quantity x1 x2 x3 s1 s2 s3 s4

0 80 0 0 1 0

0 70 0 0 0 1

8 20 0 1 0 0

10 10 1 0 0 0 1/3

260 6 10 8 0 0 2 2

0 0 0 0 -2-2-2cj - zj

zj

-1>31>3x2

-1>62>31>3x3

-4>31>3-1>3s2

-2>3-4>37>3s1

a. What is the marginal value of an additional pound of spruce? Over what range is this valuevalid?

b. What is the marginal value of an additional hour of cutting? Over what range is this value valid?c. Given a choice between securing more cutting hours or more pressing hours, which should

management select? Why?d. If the amount of spruce available to the firm was decreased from 160 to 100 pounds, would

this reduction affect the solution?e. What unit profit would have to be made from Western paneling before management would

consider producing it?f. Management is considering changing the profit of Colonial paneling from $8 to $13. Would

this change affect the solution?

54. A manufacturing firm produces four products. Each product requires material and machine pro-cessing. The linear programming model formulated to determine the number of product product product and product to produce is as follows:

subject to

(material, lb.)(machine processing, hr.)

x1, x2, x3, x4 Ú 0x1 + 2x2 + 2x3 + x4 … 160

2x1 + x2 + 4x3 + 2x4 … 200

maximize Z = 2x1 + 8x2 + 10x3 + 6x4 (profit, $)

4 (x4)3 (x3),2 (x2),1 (x1),



Problems A-63

Basic2 8 10 6 0 0

cj Variables Quantity x1 x2 x3 x4 s1 s2

6 80 1 0 2 1 2/3

8 40 0 1 0 0 2/3

800 6 8 12 6 4/3 10/3

0 0 -10>3-4>3-2-4cj - zj

zj

-1>3x2

-1>3x4

What is the marginal value of an additional pound of material? Over what range is this value valid?


simplex tableau

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