Introduction to Optimum Design phần 4 pdf

We identify a column having negative reduced cost coefficient because the nonbasic variable associatedwith this column can become basic to reduce the cost function from its current value

TABLE 6-5 Pivot Step to Interchange Basic Variable x4with Nonbasic Variable xl for Example 6.5

Initial canonical form.

To interchange x1with x4 , choose row 2 as the pivot row and column 1 as the pivot column.

Perform elimination using a21 as the pivot element.

Result of the pivot operation: second canonical form.

Solution. The given canonical form can be written in a tableau as shown in Table

6-5; x1and x2are nonbasic and x3and x4are basic, i.e., x1= x2= 0, x3= 4, x4.= 6 Thiscorresponds to point A in Fig 6-2 In the tableau, the basic variables are identified inthe leftmost column and the rightmost column gives their values Also, the basic vari-ables can be identified by examining columns of the tableau The variables associated

with the columns of the identity matrix are basic; e.g., variables x3and x4in Table

6-5 Location of the positive unit element in a basic column identifies the row whose

right side parameter biis the current value of the basic variable associated with that

column For example, the basic column x3has unit element in the first row, and so x3

is the basic variable associated with the first row Similarly, x4is the basic variableassociated with row 2

To make x1basic and x4a nonbasic variable, one would like to make a¢21= 1 and

a11¢ = 0 This will replace x1with x4as the basic variable and a new canonical form

will be obtained The second row is treated as the pivot row, i.e., a21= 1 ( p = 2, q =

1) is the pivot element Performing Gauss-Jordan elimination in the first column with

a21= 1 as the pivot element, we obtain the second canonical form as shown in Table

6-5 For this canonical form, x2= x4= 0 are the nonbasic variables and x1= 6 and x3

= 10 are the basic variables Thus, referring to Fig 6-2, this pivot step results in amove from the extreme point A(0, 0) to an adjacent extreme point D(6, 0)

6.3.5 Basic Steps of the Simplex Method

In this section, we shall illustrate the basic steps of the Simplex method with an exampleproblem In the next subsection, we shall explain the basis for these steps and summarizethem in a step-by-step general algorithm The method starts with a basic feasible solution,i.e., at a vertex of the convex polyhedron A move is then made to an adjacent vertex while

maintaining feasibility of the new solution as well as reducing the cost function This isaccomplished by replacing a basic variable with a nonbasic variable In the Simplex method,movements are to the adjacent vertices only Since there may be several points adjacent tothe current vertex, we naturally wish to choose the one that makes the greatest improvement

in the cost function f If adjacent points make identical improvements in f, the choice becomes

arbitrary An improvement at each step ensures no backtracking Two basic questions nowarise:

1 How to choose a current nonbasic variable that should become basic?

2 Which variable from the current basic set should become nonbasic?

The Simplex method answers these questions based on some theoretical considerationswhich shall be discussed in Chapter 7 Here, we consider an example to illustrate the basicsteps of the Simplex method that answer the foregoing two questions Before presentation ofthe example problem, an important requirement of the Simplex method is discussed In this

method the cost function must always be given in terms of the nonbasic variables only To

accomplish this, the cost function expression cT

x= f is written as another linear equation in the Simplex tableau; for example, the (m+ l)th row One then performs the pivot step on the

entire set of (m + 1) equations so that x1, x2, , xm and f are the basic variables This way

the last row of the tableau representing the cost function expression is automatically given

in terms of the nonbasic variables after each pivot step The coefficients in the nonbasic

columns of the last row are called the reduced cost coefficients written as c¢ j Example 6.6

describes the steps of the Simplex method in a systematic way

EXAMPLE 6.6 Steps of the Simplex Method

Solve the following LP problem: maximize

Solution. The graphical solution for the problem is given in Fig 6-3 It can be seen

that the problem has an infinite number of solutions along the line C–D (z* = 4)because the objective function is parallel to the second constraint The Simplex method

is illustrated in the following steps:

1 Convert the problem to the standard form We write the problem in the

standard LP form by transforming the maximization of z to minimization of

f = -2x1- x2, and adding slack variables x3, x4, and x5to the constraints Thus,the problem becomes

(a)subject to

(b)(c)

2x1+x2+x4=4

4x1+3x2+x3=12

minimize f = -2x1-x2

z=2x1+x2 subject to 4x1+3x2£12 2, x1+x2£4,x1+2x2£4,x x1, 2≥0

We use the tableau and notation of Table 6-3 which will be augmented withthe cost function expression as the last row The initial tableau for theproblem is shown in Table 6-6, where the cost function expression -2x1- x2

= f is written as the last row Note also that the cost function is in terms of only the nonbasic variables x1and x2 This is one of the basic requirements of

the Simplex method—that the cost function always be in terms of the nonbasic variables When the cost function is only in terms of the nonbasic

variables, then the cost coefficients in the last row are the reduced cost

coefficients, written as c¢j.

2 Initial basic feasible solution.

To initiate the Simplex method, a basic feasible solution is needed This isalready available in Table 6-6 which is given as:

basic variables: x3 = 12, x4= 4, x5= 4

nonbasic variables: x1 = 0, x2= 0

cost function: f = 0Note that the cost row gives 0 = f after substituting for x1and x2 Thissolution represents point A in Fig 6-3 where none of the constraints is activeexcept the nonnegativity constraints on the variables

3 Optimality check We scan the cost row, which should have nonzero entries only in the nonbasic columns, i.e., x1and x2 If all the nonzero entries are

nonnegative, then we have an optimum solution because the cost function

x i≥0; i=1 to 5

x1+2x2+x5=4

cannot be reduced any further and the Simplex method is terminated Thereare negative entries in the cost row so the current basic feasible solution is

not optimum (see Chapter 7 for further explanation).

4 Choice of a nonbasic variable to become basic We select a nonbasic column

having a negative cost coefficient; i.e., -2 in the x1column This identifies a

nonbasic variable (x1) that should become basic Thus, eliminations will be

performed in the x1column This answers question 1 posed earlier: “How tochoose a current nonbasic variable that should become basic?” Note also thatwhen there is more than one negative entry in the cost row, the variable

tapped to become basic is arbitrary among the indicated possibilities The

usual convention is to select a variable associated with the smallest value inthe cost row (or, negative element with the largest absolute value)

Notation. The boxed negative reduced cost coefficient in Table 6-6 indicatesthe nonbasic variable associated with that column selected to become basic, a nota-tion that is used throughout

5 Selection of a basic variable to become nonbasic To identify which current basic variable should become nonbasic (i.e., to select the pivot row), we take

ratios of the right side parameters with the positive elements in the x1 column

as shown in Table 6-7 We identify the row having the smallest positive ratio, i.e., the second row This will make x4nonbasic The pivot element is a21= 2(the intersection of pivot row and pivot column) This answers the question 2posed earlier: “Which variable from the current basic set should become

nonbasic?” Selection of the row with the smallest ratio as the pivot row

maintains feasibility of the new basic solution This is justified in Chapter 7

TABLE 6-6 Initial Tableau for the LP Problem of Example 6.6

Notation The reduced cost coefficients in the nonbasic columns are boldfaced The selected

negative reduced cost coefficient is boxed.

TABLE 6-7 Selection of Pivot Column and Pivot Row for Example 6.6

The selected pivot element is boxed Selected pivot row and column are shaded x1 should

become basic (pivot column) x4row has the smallest ratio, and so x4 should become nonbasic.

Notation. The selected pivot element is also boxed, and the pivot column androw are shaded throughout.

6 Pivot operation We perform eliminations in column x1using row 2 as the

pivot row and Eqs (6.15) to (6.17) to eliminate x1from rows 1, 3, and thecost row as follows:

• divide row 2 by 2, the pivot element

• multiply new row 2 by 4 and subtract from row 1 to eliminate x1from row 1

• subtract new row 2 from row 3 to eliminate x1from row 3

• multiply new row 2 by 2 and add to the cost row to eliminate x1

As a result of this elimination step, a new tableau is obtained as shown inTable 6-8 The new basic feasible solution is given as

nonnegative

Note that the cost coefficients corresponding to the nonbasic variable x2in the last

row is zero in the final tableau This is an indication of multiple solutions for the

problem In general, when the reduced cost coefficient in the last row corresponding

to a nonbasic variable is zero, the problem may have multiple solutions We shalldiscuss this point later in more detail

Let us see what happens if we do not select a row with the least ratio as the pivot row Let a31= 1 in the third row be the pivot element in Table 6-6 This will inter-

change nonbasic variable x1with the basic variable x5 Performing the eliminationsteps in the first column as explained earlier, we obtain the new tableau given in Table6-9 From the tableau, we have

The foregoing solution corresponds to point G in Fig 6-3 We see that this basic

solution is not feasible because x3and x4have negative values Thus, we conclude that

if a row with the smallest ratio (of right sides with positive elements in the pivot column) is not selected, the new basic solution is not feasible Note that a spreadsheet program, such as Excel, can be used to carry out the pivot step Such a program can

facilitate learning of the Simplex method without getting bogged down with themanual elimination process

TABLE 6-9 Result of Improper Pivoting in Simplex Method for LP Problem of

that in general the reduced cost coefficients c¢ jof the nonbasic variables may be positive,

neg-ative, or zero Let one of c¢ jbe negative, then if a positive value is assigned to the associated

nonbasic variable (i.e., it is made basic), the value of f will decrease If more than one ative c¢ jis present, a widely used rule of thumb is to choose the nonbasic variable associated

neg-with the smallest c¢ j (i.e., the most negative c¢ j) to become basic Thus, if any c¢ j for (m+ 1) £

j £ n (for nonbasic variables) is negative, then it is possible to find a new basic feasible tion (if one exists) that will further reduce the cost function If a c¢ jis zero, then the associ-

solu-ated nonbasic variable can be made basic without affecting the cost function value If all c¢ j

are nonnegative, then it is not possible to reduce the cost function any further, and the currentbasic feasible solution is optimum These ideas are summarized in the following Theorems6.3 and 6.4

Theorem 6.3 Improvement of Basic Feasible Solution Given a nondegenerate basic

fea-sible solution with the corresponding cost function f0, suppose that c¢ j < 0 for some j Then, there is a feasible solution with f < f0 If the jth nonbasic column associated with c¢ j can be substituted for some column in the original basis, the new basic feasible solution will have

f < f0 If the jth column cannot be substituted to yield a basic feasible solution (i.e., there is

no positive element in the jth column), then the feasible set is unbounded and the cost

func-tion can be made arbitrarily small (toward negative infinity)

Theorem 6.4 Optimum Solution for LP Problems If a basic feasible solution has reduced

cost coefficients c¢ j ≥ 0 for all j, then it is optimum.

According to Theorem 6.3, the basic procedure of the Simplex method is to start with aninitial basic feasible solution, i.e., at the vertex of the convex polyhedron If this solution is

not optimum according to Theorem 6.4, then a move is made to an adjacent vertex to reduce

the cost function The procedure is continued until the optimum is reached The steps of the

Simplex method illustrated in the previous subsection in Example 6.6 are summarized as follows assuming that the initial tableau has been set up as described earlier:

Step 1 Initial Basic Feasible Solution This is readily obtained if all constraints are “£

type” because the slack variables can be selected as basic and the real variables as nonbasic

If there are equality or “≥ type” constraints, then the two-phase simplex procedure explained

in the next section must be used

Step 2 The Cost Function Must be in Terms of Only the Nonbasic Variables This is

readily available when there are only “£ type” constraints and slack variables are added intothem to convert the inequalities to equalities The slack variables are basic, and they do notappear in the cost function In subsequent iterations, eliminations must also be performed inthe cost row

Step 3 If All the Reduced Cost Coefficients for Nonbasic Variables Are Nonnegative

(≥0), We Have the Optimum Solution Otherwise, there is a possibility of improving the

cost function We need to select a nonbasic variable that should become basic We identify

a column having negative reduced cost coefficient because the nonbasic variable associatedwith this column can become basic to reduce the cost function from its current value This

is called the pivot column.

Step 4 If All Elements in the Pivot Column Are Negative, Then We Have an Unbounded Problem Design problem formulation should be examined to correct the situation If there

are positive elements in the pivot column, then we take ratios of the right side parameters withthe positive elements in the pivot column and identify a row with the smallest positive ratio Inthe case of a tie, any row among the tying ratios can be selected The basic variable associated

with this row should become nonbasic (i.e., become zero) The selected row is called the pivot

row, and its intersection with the pivot column identifies the pivot element.

Step 5 Complete the Pivot Step Use the Gauss-Jordan elimination procedure and the

pivot row identified in Step 4 Elimination must also be performed in the cost function row

so that it is only in terms of nonbasic variables in the next tableau This step eliminates thenonbasic variable identified in Step 3 from all the rows except the pivot row

Step 6 Identify Basic and Nonbasic Variables, and Their Values Identify the cost

func-tion value and go to Step 3

Note that when all the reduced cost coefficient c¢j in the nonbasic columns are strictly itive, the optimum solution is unique If at least one c¢j is zero in a nonbasic column, then there is a possibility of an alternate optimum If the nonbasic variable associated with a zero

pos-reduced cost coefficient can be made basic by using the foregoing procedure, the extremepoint (vertex) corresponding to an alternate optimum is obtained Since the reduced cost coef-ficient is zero, the optimum cost function value will not change Any point on the line segmentjoining the optimum extreme points also corresponds to an optimum Note that all these

optima are global as opposed to local, although there is no distinct global optimum metrically, multiple optima for an LP problem imply that the cost function hyperplane is par-

Geo-allel to one of the constraint hyperplanes Example 6.7 shows how to obtain a solution for

an LP problem using the Simplex method

EXAMPLE 6.7 Solution by the Simplex Method

Using the Simplex method, find the optimum (if one exists) for the LP problem ofExample 6.3:

(a)subject to

(b)(c)(d)

Solution. Writing the problem in the Simplex tableau, we obtain the initial cal form as shown in Table 6-10 From the initial tableau, the basic feasible solutionis

canoni-Note that the cost function in the last row is in terms of only nonbasic variables x1

and x2 Thus, coefficients in the x1and x2columns and the last row are the reduced

cost function: f = 0 from the last row of the tableau

TABLE 6-10 Solution of Example 6.7 by the Simplex Method

Initial tableau: x3is identified to be replaced with x2 in the basic set.

Third tableau: Reduced cost coefficients in nonbasic columns are

nonnegative; the tableau gives optimum point.

Example 6.8 demonstrates the solution of the profit maximization problem by the Simplexmethod.

cost coefficients c¢ j Scanning the last row, we observe that there are negative

coeffi-cients Therefore, the current basic solution is not optimum In the last row, the mostnegative coefficient of -5 corresponds to the second column Therefore, we select x2

to become a basic variable, i.e., elimination should be performed in the x2column

This fixes the column index q to 2 in Eq (6.15) Now taking the ratios of the right side parameters with positive coefficients in the second column bi/ai2, we obtain a

minimum ratio for the first row as 4 This identifies the first row as the pivot rowaccording to Step 4 Therefore, the current basic variable associated with the first row,

x3 , should become nonbasic Now performing the pivot step on column 2 with a12asthe pivot element, we obtain the second canonical form (tableau) as shown in Table6-10 For this canonical form the basic feasible solution is

The cost function is f = -20 (0 = f + 20), which is an improvement from f = 0.

Thus, this pivot step results in a move from (0, 0) to (0, 4) on the convex polyhedron

of Fig 6-2

The reduced cost coefficient corresponding to the nonbasic column x1is negative.Therefore, the cost function can be reduced further Repeating the above-mentioned

process for the second tableau, we obtain a21= 2 as the pivot element, implying that

x1 should become basic and x4should become nonbasic The third canonical form is

shown in Table 6-10 For this tableau, all the reduced cost coefficients c¢j sponding to the nonbasic variables) in the last row are ≥0 Therefore, the tableau yields

(corre-the optimum solution as x1= 1, x2= 5, x3= 0, x4= 0, f = -29 (f + 29 = 0), which

cor-responds to the point C (1,5) in Fig 6-2

nonbasic variables: x1=x3=0

basic variables: x2=4,x4=2

EXAMPLE 6.8 Solution of Profit Maximization Problem by

the Simplex Method

Use the Simplex method to find the optimum solution for the profit maximizationproblem of Example 6.2

Solution. Introducing slack variables in the constraints of Eqs (c) through (e) inExample 6.2, we get the LP problem in the standard form:

(a)subject to

(b)(c)

128

(d)(e)Now writing the problem in the standard Simplex tableau, we obtain the initial

canonical form as shown in Table 6-11 Thus the initial basic feasible solution is x1=

0, x2= 0, x3= 16, x4= x5= 1, f = 0, which corresponds to point A in Fig 6-1 The initial cost function is zero, and x3, x4, and x5are the basic variables

Using the Simplex procedure, we note that a22= 1

–

14is the pivot element This implies

that x4should be replaced by x2in the basic set Carrying out the pivot operation usingthe second row as the pivot row, we obtain the second tableau (canonical form) shown

in Table 6-11 At this point the basic feasible solution is x1= 0, x2= 14, x3= 2, x4=

0, x5=5

–

12, which corresponds to point E in Fig 6-1 The cost function is reduced to

-8400 The pivot element for the next step is a11, implying that x3should be replaced

by x1in the basic set Carrying out the pivot operation, we obtain the third canonicalform shown in Table 6-11 At this point all reduced cost coefficients (corresponding

to nonbasic variables) are nonnegative, so according to Theorem 6.4, we have the

1

TABLE 6-11 Solution of Example 6.8 by the Simplex Method

Initial tableau: x4is identified to be replaced with x2 in the basic set.

— 1/14 = 14 ¨ smallest

–

14

1 –

5/12

— 17/336 = 140

Fig 6-1 The optimum value of the cost function is -8800 Note that c¢j, ing to the nonbasic variables x3and x4, are positive Therefore, the global optimum isunique, as may be observed in Fig 6-1 as well.

correspond-EXAMPLE 6.9 LP Problem with Multiple Solutions

Solve the following problem by the Simplex method: maximize z = x1+ 0.5x2subject

to 2x1+ 3x2£ 12, 2x1+ x2£ 8, x1, x2≥ 0

Solution. The problem was solved graphically in Section 3.4 of Chapter 3 It has

multiple solutions as may be seen in Fig 3-7 We will solve the problem using the

Simplex method and discuss how multiple solutions can be recognized for genera1

LP problems The problem is converted to standard LP form:

(a)subject to

(b)(c)(d)Table 6-12 contains iterations of the Simplex method The optimum point is reached

in just one iteration because all the reduced cost coefficients are nonnegative in thesecond canonical form (second tableau) The solution is given as

basic variables: x1 = 4, x3= 4

nonbasic variables: x2 = x4= 0

optimum cost function: f= -4

The solution corresponds to point B in Fig 3-7 In the second tableau, the reduced

cost coefficient for the nonbasic variable x2 is zero This means that it is possible to

make x2basic without any change in the optimum cost function value This suggests

existence of multiple optimum solutions Performing the pivot operation in column 2,

we find another solution given in the third tableau of Table 6-12 as:

This solution corresponds to point C on Fig 3-7 Note that any point on the line B–Calso gives an optimum solution Multiple solutions can occur when the cost function

is parallel to one of the constraints For the present example, the cost function is allel to the second constraint, which is active at the solution

par-TABLE 6-12 Solution by the Simplex Method for Example 6.9

Initial tableau: x4is identified to be replaced with x1 in the basic set.

Second tableau: First optimum point; reduced cost coefficients in

nonbasic columns are nonnegative; the tableau gives optimum

solution c ¢3= 0 indicates the possibility of multiple solutions x3 is

identified to be replaced with x2 in the basic set to obtain another

In general, if a reduced cost coefficient corresponding to a nonbasic variable is zero in

the final tableau, there is a possibility of multiple optimum solutions From a practical

stand-point, this is not a bad situation Actually, it may be desirable because it gives the designeroptions; any suitable point on the straight line joining the two optimum designs can be

selected to better suit the needs of the designer Note that all optimum design points are global

solutions as opposed to local solutions.

Example 6.10 demonstrates how to recognize an unbounded feasible set (solution) for aproblem

EXAMPLE 6.10 Identification of an Unbounded Problem

with the Simplex Method

Solve the LP problem: maximize z = x1- 2x2subject to 2x1- x2≥ 0, -2x1+ 3x2£ 6,

x1 , x2≥ 0

Solution. The problem has been solved graphically in Section 3.5 It can be seenfrom the graphical solution (Fig 3-8) that the problem is unbounded We will solvethe problem using the Simplex method and see how we can recognize unboundedproblems Writing the problem in the standard Simplex form, we obtain the initial

canonical form shown in Table 6-13 where x3and x4are the slack variables (note thatthe first constraint has been transformed as -2x1+ x2£ 0) The basic feasible solutionis

Scanning the last row, we find that the reduced cost coefficient for the nonbasic

vari-able x1is negative Therefore, x1should become a basic variable However, a pivotelement cannot be selected in the first column because there is no positive element.There is no other possibility of selecting another nonbasic variable to become basic;

the reduced cost coefficient for x2(the other nonbasic variable) is positive Therefore,

no pivot steps can be performed, and yet we are not at the optimum point Thus, thefeasible set for the problem is unbounded The foregoing observation will be true in

general For unbounded problems, there will be negative reduced cost coefficients for

nonbasic variables but no possibility of pivot steps

The basic Simplex method of Section 6.3 is extended to handle “≥ type” and equality straints A basic feasible solution is needed to initiate the Simplex solution process Such asolution is immediately available if only “£ type” constraints are present However, for the

con-“≥ type” and equality constraints, an initial basic feasible solution is not available To obtainsuch a solution, we must introduce artificial variables for the “≥ type” and equality con-straints, define an auxiliary minimization LP problem, and solve it The standard Simplexmethod can still be used to solve the auxiliary problem This is called Phase I of the Simplex

procedure At the end of Phase I, a basic feasible solution for the original problem becomesavailable Phase II then continues to find a solution to the original LP problem The method

is illustrated with examples

6.4.1 Artificial Variables

When there are “≥ type” constraints in the LP problem, surplus variables are subtracted fromthem to transform the problem to the standard form The equality constraints, if present, arenot changed because they are already in the standard form For such problems, an initial basicsolution cannot be obtained by selecting the original design variables as nonbasic (settingthem to zero), as is the case when there are only “£ type” constraints, e.g., for all the exam-ples in Section 6.3 To obtain an initial basic feasible solution, the Gauss-Jordan elimination

procedure can be used to convert the Ax = b in the canonical form However, an easier way

is to introduce nonnegative auxiliary variables for the “≥ type” and equality constraints, define

an auxiliary LP problem, and solve it using the Simplex method The auxiliary variables are

called artificial variables and are different from the surplus variables They have no

physi-cal meaning; however, with their addition we obtain an initial basic feasible solution for theauxiliary LP problem by treating them as basic along with any slack variables for “£ type”constraints All other variables are treated as nonbasic (i.e., set to zero)

For the sake of simplicity of discussion, let us assume that each constraint of the standard

LP problem requires an artificial variable We shall see later in examples that constraints that

do not require an artificial variable can also be treated routinely Recalling that the standard

LP problem has n variables and m constraints, the constraint equations augmented with the

artificial variables are written as

(6.18)

where xn +j , j = 1 to m are the artificial variables The constraints of Eq (6.18) can be written

in the summation notation as

(6.19)

Thus the initial basic feasible solution is obtained as xj = 0, j = 1 to n, and xn +i = bi, i= 1

to m Note that the artificial variables basically augment the convex polyhedron of the

original problem The initial basic feasible solution corresponds to an extreme point (vertex)located in the new expanded space The problem now is to traverse extreme points in theexpanded space until an extreme point is reached in the original space When the originalspace is reached, all artificial variables will be nonbasic (i.e., they will have zero value) Atthis point the augmented space is literally removed so that future movements are only amongthe extreme points of the original space until the optimum is reached In short, after creatingartificial variables, we eliminate them as quickly as possible The preceding procedure is

called the two-phase Simplex method of LP.

6.4.2 Artificial Cost Function

To eliminate the artificial variables from the problem, we define an auxiliary function called

the artificial cost function, and minimize it subject to the constraints of Eq (6.19) and

non-negativity of all the variables The artificial cost function is simply a sum of all the artificial

variables and will be designated as w:

6.4.3 Definition of Phase I Problem

Since the artificial variables are introduced to simply obtain an initial basic feasible solutionfor the original problem, they need to be eliminated eventually This elimination is done bydefining an LP problem called the Phase I problem The objective of this problem is to makeall the artificial variables nonbasic so that they have zero value In that case, the artificial costfunction in Eq (6.20) will be zero, indicating the end of Phase I However the Phase I problem

is not yet in a form suitable to initiate the Simplex method The reason is that the reduced

cost coefficients c¢jof the nonbasic variables in the artificial cost function are not yet able to determine the pivot element and perform the pivot step Currently, the artificial cost

avail-function in Eq (6.20) is in terms of the basic variables xn+1, , xn +m Therefore the reduced

cost coefficients c¢jcannot be identified They can be identified only if the artificial cost

func-tion w is in terms of the nonbasic variables x1, , xn To obtain w in terms of nonbasic

vari-ables, we use the constraint expressions to eliminate the basic variables from the artificial

cost function Calculating xn+1, , xn +mfrom Eqs (6.18) and substituting into Eq (6.20), we

obtain the artificial cost function w in terms of the nonbasic variables as

stan-the number of artificial variables is less than m—stan-the total number of constraints Accordingly,

the number of artificial variables required to obtain an initial basic feasible solution is also

less than m This implies that the sums in Eqs (6.21) and (6.22) are not for all the m

con-straints They are only over the constraints requiring an artificial variable

6.4.4 Phase I Algorithm

The standard Simplex procedure described in Section 6.3 can now be employed to solve theauxiliary optimization problem of Phase I During this phase, the artificial cost function isused to determine the pivot element The original cost function is treated as a constraint andthe elimination step is also executed for it This way, the real cost function is in terms of thenonbasic variables only at the end of Phase I, and the Simplex method can be continued

during Phase II All artificial variables become nonbasic at the end of Phase I Since w is the sum of all the artificial variables, its minimum value is clearly zero When w= 0, an extreme

point of the original feasible set is reached w is then discarded in favor of f and iterations continue in Phase II until the minimum of f is obtained Suppose, however, that w cannot

be driven to zero This will be apparent when none of the reduced cost coefficients for the

artificial cost function is negative and yet w is greater than zero Clearly, this means that we cannot reach the original feasible set and, therefore, no feasible solution exists for the

original design problem, i.e., it is an infeasible problem At this point the designer should

re-examine the formulation of the problem, which may be over-constrained or improperlyformulated

to

i m

j n

6.4.5 Phase II Algorithm

In the final tableau from Phase I, the artificial cost row is replaced by the actual cost tion equation and the Simplex iterations continue based on the algorithm explained in Section6.3 The basic variables, however, should not appear in the cost function Thus, pivot stepsneed to be performed on the cost function equation to eliminate the basic variables from it

func-A convenient way of accomplishing this is to treat the cost function as one of the equations

in the Phase I tableau, say the second equation from the bottom Elimination is performed

on this equation along with others In this way, the cost function is in the correct form tocontinue with Phase II The artificial variable columns can also be discarded for Phase II cal-culations However, they are kept in the tableau because they provide information that can

be useful for postoptimality analysis

For most LP problems, the Simplex method yields one of the following results as trated in the examples:

illus-1 If there is a solution to the problem, the method will find it (Example 6.11)

2 If the problem is infeasible, the method will indicate that (Example 6.12)

3 If the problem is unbounded, the method will indicate that (Example 6.10, Example6.13)

4 If there are multiple solutions, the method will indicate that, as seen in Examples 6.6and 6.9

EXAMPLE 6.11 Use of Artificial Variable for

“≥ Type” Constraints

Find the optimum solution for the following LP problem using the Simplex method:

maximize z = y1+ 2y2subject to 3y1+ 2y2£ 12, 2y1+ 3y2≥ 6, y1≥ 0, y2is unrestricted

(a)subject to

(b)(c)(d)

where x4is a slack variable for the first constraint and x5is a surplus variable for thesecond constraint It can be seen that if we select the real variables as nonbasic, i.e.,

x1 = 0, x2= 0, x3= 0, the resulting basic solution is infeasible because x5= -6 fore, we need to use the two-phase algorithm Accordingly, we introduce an artificial

There-variable x6in the second constraint as

x i≥0; i=1 to 5

2x1+3x2-3x3-x5=6

3x1+2x2-2x3+x4=12

minimize f = -x1-2x2+2x3

The artificial cost function is defined as w = x6 Since w should be in terms of basic variables (x6is basic), we substitute for x6from Eq (e) and obtain w as

non-(f )The initial tableau for Phase I is shown in Table 6-14 The initial basic variables

are x4= 12 and x6= 6 The nonbasic variables are x1= x2= x3= x5= 0 Also w = 6 and f= 0 This corresponds to the infeasible point D in Fig 6-4 According to the

Simplex algorithm, the pivot element is a22, which implies that x2should become basic

and x6 should become nonbasic Performing the pivot step, we obtain the second

tableau given in Table 6-14 For the second tableau, x4= 8 and x2= 2 are the basicvariables and all others are nonbasic This corresponds to the feasible point A in Fig.6-4 Since all the reduced cost coefficients of the artificial cost function are nonneg-ative and the artificial cost function is zero, an initial basic feasible solution for theoriginal problem is obtained Therefore, this is the end of Phase I

For Phase II, column x6should be ignored in determining pivots For the next step,

the pivot element is a15in the second tableau according to Steps 1 and 2 of the Simplex

method This implies that x4should be replaced by x5as a basic variable The thirdtableau is obtained as shown in Table 6-14 The last tableau yields an optimum solu-

tion for the problem, which is x5= 12, x2= 6, x1= x3= x4= 0, and f = -12 The tion for the original design problem is then y1= 0, y2= 6, and z = 12, which agrees with the graphical solution of Fig 6-4 Note that the artificial variable column (x6) in

solu-the final tableau is solu-the negative of solu-the surplus variable column (x5) This is true for all

C D

TABLE 6-14 Solution by the Two-Phase Simplex Method for Example 6.11

Initial tableau: x6is identified to be replaced with x2 in the basic set.

Second tableau: End of Phase I Begin Phase II x4 is identified to be

replaced with x5 in the basic set.

Third tableau: Reduced cost coefficients in nonbasic columns are

nonnegative; the third tableau gives the optimum solution End of

EXAMPLE 6.12 Use of Artificial Variables for Equality

Constraints (Infeasible Problem)

Solve the LP problem: maximize z = x1+ 4x2subject to x1+ 2x2£ 5, 2x1+ x2= 4,

x1 - x2≥ 3, x1, x2≥ 0

Solution. The constraints for the problem are plotted in Fig 6-5 It can be seen thatthe problem has no feasible solution We will solve the problem using the Simplexmethod to see how we can recognize an infeasible problem Writing the problem inthe standard LP form, we obtain

(a)subject to

(b)(c)(d)(e)

FIGURE 6-5 Constraints for Example 6.12 Infeasible problem.

TABLE 6-15 Solution for Example 6.12 (Infeasible Problem)

Initial tableau: x5is identified to be replaced with x1 in the basic set.

vari-cial cost function is not zero (w = 1) Therefore there is no feasible solution to theoriginal problem

EXAMPLE 6.13 Use of Artificial Variables (Unbounded

(b)(c)(d)

where x3is a slack variable, x4is a surplus variable, and x5is an artificial variable.Note that the right side of the first constraint is zero so it can be treated as either

“£ type” or “≥ type.” We will treat it as “£ type.” Note also that the second constraint

is “≥ type,” so we must use an artificial variable and an artificial cost function to findthe initial basic feasible solution The solution for the problem is given in Table 6-16

For the initial tableau x3= 0 and x5= 2 are basic variables and all others are

nonba-sic Note that this is a degenerate basic feasible solution The solution corresponds to

point A (the origin) in Fig 6-6 Scanning the artificial cost row, we observe that there

are two possibilities for pivot columns, x1or x2 If x2is selected as the pivot column,

A

B

C

D Feasible region unbounded

x2

x1

FIGURE 6-6 Constraints for Example 6.13 Unbounded problem.

then the first row must be the pivot row with a12= 1 as the pivot element This will

make x2basic and x3nonbasic However, x2will remain zero and the resulting tion will be degenerate, corresponding to point A One more iteration will be neces-

solu-sary to move from A to D If we choose x1as the pivot column, then a21= 1 will be

the pivot element making x1as basic and x5as nonbasic Carrying out the pivot step,

we obtain the second tableau as shown in Table 6-16 The basic feasible solution is

x1 = 2, x3= 2, and other variables are zero This solution corresponds to point D inFig 6-6 This is the basic feasible solution for the original problem because the arti-

ficial cost function is zero, i.e., w = 0 The original cost function has also reduced

from 0 to -6 This is the end of Phase I Scanning the cost function row, we find that

the reduced cost coefficient c¢4is negative, but the pivot element cannot be determined,

i.e., x4cannot be made basic (all elements in the x4column are negative in the secondtableau) This indicates the problem to be unbounded

TABLE 6-16 Solution for Example 6.13 (Unbounded Problem)

Initial tableau: x5is identified to be replaced with x1 in the basic set.

6.4.6 Degenerate Basic Feasible Solution

It is possible that during iterations of the Simplex method, a basic variable attains zero value,i.e., the basic feasible solution becomes degenerate What are the implications of this situa-tion? We shall discuss them in Example 6.14

EXAMPLE 6.14 Implications of Degenerate Basic

(a)subject to

(b)(c)(d)(e)(f)

where x3and x4are slack variables, x5and x6are surplus variables, and x7and x8areartificial variables The two-phase Simplex procedure takes three iterations to reachthe optimum point These iterations are given in Table 6-17 It can be seen that in the

third tableau, the basic variable x4has zero value so the basic feasible solution is

degenerate At this iteration, it is determined that x5 should become basic so x5is thepivot column We need to determine the pivot row We take ratios of the right sides

with the positive elements in the x5column This determines the second row as thepivot row because it has the lowest ratio (zero) In general, if the element in the pivotcolumn and the row that gives degenerate basic variable is positive, then that row mustalways be the pivot row; otherwise, the new solution cannot be feasible Also, in thiscase, the new basic feasible solution will be degenerate, as for the final tableau inTable 6-17 The only way the new feasible solution can be nondegenerate is when theelement in the pivot column and the degenerate variable row is negative In that case

TABLE 6-17 Solution for Example 6.14 (Degenerate Basic Feasible Solution)

Initial tableau: x8is identified to be replaced with x1 in the basic set.

–3 5 –3

the new basic feasible solution will be nondegenerate It is theoretically

possible for the Simplex method to fail by cycling between two degenerate basic sible solutions However, in practice this usually does not happen The final solution

fea-for this problem is

cal design problems Thus we are interested not only in the optimum solution but also in how

it changes when the parameters change The changes may be either discrete (e.g., when weare uncertain about which of several choices is the value of a particular parameter) or con-

tinuous The study of discrete parameter changes is often called sensitivity analysis, and that

of continuous changes is called parametric programming There are five basic parametric

changes affecting the solution:

1 Changes in the cost function coefficients, cj

2 Changes in the resource limits, bi

3 Changes in the constraint coefficients, aij

4 The effect of including additional constraints

5 The effect of including additional variables

A thorough discussion of these changes, while not necessarily difficult, is beyond ourscope In principle, we could imagine solving a new problem for every change Fortunately,for a small number of changes there are useful shortcuts Almost all computer programs for

LP problems provide some information about parameter variations We shall study the

para-metric changes defined in items 1 through 3 The final tableau for the LP problem contains

all the information needed to study these changes We shall describe the information

con-tained in the final tableau and its use to study the three parametric changes For other tions, full length texts on linear programming may be consulted

varia-It turns out that the optimum solution of the altered problem can be computed using theoptimum solution of the original problem and the information in the final tableau as long aschanges in the parameters are within certain limits This is especially beneficial for problems

that take a long time to solve In the following discussion we use a¢ ij, c¢ j, and b¢ ito represent

the corresponding values of the parameters aij, cj, and biin the final tableau

6.5.1 Changes in Resource Limits

First, we study how the optimum value of the cost function for the problem changes if we

change the right side parameters, bi’s (also known as resource limits), of the constraints The

constraint variation sensitivity Theorem 4.7 of Section 4.5 can be used to study the effect ofthese changes Use of that theorem requires knowledge of the Lagrange multipliers for theconstraints Theorem 6.5 gives a way of recovering the multipliers for the constraints of an

LP problem from the final tableau.

Theorem 6.5 Lagrange Multiplier Values Let the standard LP problem be solved using

the Simplex method (1) For “£ type” constraints, the Lagrange multiplier equals the reducedcost coefficient in the slack variable column associated with the constraint (2) For “=” and

“≥ type” constraints, the Lagrange multiplier equals the reduced cost coefficient in the ficial variable column associated with the constraint (3) The Lagrange multiplier is always

arti-≥ 0 for the “£ type” constraint, always £ 0 for the “arti-≥ type” constraint, and free in sign forthe “= type” constraint

In Section 4.5, the physical meaning of the Lagrange multipliers was described There, the

Lagrange multipliers were related to derivatives of the cost function with respect to the right

side parameters Equality and inequality constraints were treated separately with vi and uiastheir Lagrange multipliers, respectively However, in this section, we use a slightly different

notation We use ei as the right side parameter of any constraint and yias its Lagrange plier Using this notation and Theorem 4.7, we obtain the following derivative of the cost func-tion with respect to the right side parameters, and change in the optimum cost function:

multi-(6.23)

It is noted here that Theorem 6.5 and Eq (6.23) are applicable only if changes in the right

side parameters are within certain limits, i.e., there are upper and lower limits on changes

in the resource limits for which Eq (6.23) is valid The changes need not be small any more

as was stipulated for nonlinear problems in Section 4.5 Calculations for the limits are

dis-∂

∂f = - = - = - ( - )

cussed later in this section Note the calculation for Df remains valid for simultaneous changes

to multiple constraints; in that case all the changes are added

It is also noted that Theorem 4.7 and Eq (6.23) were discussed for the general problem

written as minimization of a cost function with “ £ type” and equality constraints However,

Eq (6.23) is applicable to “≥ type” constraints as well as long as care is exercised in using

appropriate signs for the Lagrange multiplier yiand the change Dei We shall demonstrate use

of Theorem 6.5 and Eq (6.23) with examples

It is also important to note that if an inequality is inactive at the optimum, then its slack

or surplus variable is greater than 0 Therefore its Lagrange multiplier is 0 to satisfy the

switching condition, y i s i= 0 (except for the abnormal case where both the Lagrange plier and the constraint function have zero value) This observation can help in verifying thecorrectness of the Lagrange multipliers recovered from the final LP tableau Example 6.15describes recovery of the Lagrange multipliers from the final tableau for the “£ type” constraints

multi-EXAMPLE 6.15 Recovery of Lagrange Multipliers for

“£ Type” Constraint

Consider the problem: maximize z = 5x1- 2x2subject to the constraints 2x1+ x2£ 9,

x1 - 2x2£ 2, -3x1+ 2x2£ 3, x1, x2≥ 0 Solve the problem by the Simplex method.Recover Lagrange multipliers for the constraints

Solution. Constraints for the problem and cost function contours are plotted in Fig

6-7 The optimum solution is at point C and is given as x1= 4, x2= 1, z = 18 The

problem is transformed to the standard form as

6 8

C

D

F E

x2

x1

G H

FIGURE 6-7 Graphical solution for Example 6.15.

(a)subject to

(b)(c)(d)(e)

where x3, x4, and x5are the slack variables Solving the problem using the Simplexmethod, we obtain the sequence of calculations given in Table 6-18 From the finaltableau,

TABLE 6-18 Solution for Example 6.15 by the Simplex Method

Initial tableau: x4is identified to be replaced with x1 in the basic set.

Third tableau: Reduced cost coefficients in nonbasic columns are

nonnegative; the tableau gives optimum point.

where f = -(5x1- 2x2); note that Eq (6.23) is valid for a minimization problem only.

If the right side of the first constraint changes from 9 to 10, the cost function f changes

by Df = -1.6(10 - 9) = -1.6, i.e., the new value of f will be -19.6 (z = 19.6) Point F

in Fig 6-7 gives the new optimum solution for this case If the right side of the second

constraint changes from 2 to 3, the cost function f changes by Df = -1.8(3 - 2) =

-1.8 to -19.8 Point G in Fig 6-7 gives the new optimum solution Note that anysmall change in the right side of the third constraint will have no effect on the cost

function When the right side of first and second constraints are changed to 10 and 3

simultaneously, the net change in the cost function is -(1.6 + 1.8), i.e., new f will be

-21.4 The new solution is at point H in Fig 6-7

f e

f e

It is noted (as in Section 4.5) that the Lagrange multipliers are very useful for practical

design problems Their values give the relative effect of changes in the right side parameters

of constraints (resource limits) Using their relative values, the designer can determine the

most profitable way to adjust the resource limits, if necessary and possible The Lagrange

multipliers are also called the dual variables (or, dual prices) The concept of duality in linear

programming is described in Chapter 7 Example 6.16 demonstrates recovery of Lagrangemultipliers for equality and “≥ type” constraints

EXAMPLE 6.16 Recovery of Lagrange Multipliers for “=”

and “≥ Type” Constraints

Solve the following LP problem and recover proper Lagrange multipliers for the

con-straints: maximize z = x1+ 4x2subject to x1+ 2x2£ 5, 2x1+ x2= 4, x1- x2≥ 1, x1,

x2≥ 0

Solution. Constraints for the problem are plotted in Fig 6-8 It can be seen that lineE–C is the feasible region for the problem and point E gives the optimum solution.Converting the problem to standard Simplex form, we obtain:

(a)subject to

(b)(c)

2x1+x2+x5=4

x1+2x2+x3=5

minimize f = -x1-4x2

where x3is a slack variable, x4is a surplus variable, and x5and x6are artificial ables The problem, solved in Table 6-19, takes just two iterations to reach theoptimum The solution from the final tableau is

Note that the artificial variable column (x6) is the negative of the surplus variable

column (x4) for the third constraint Using Theorem 6.5, the Lagrange multipliers forthe constraints are

1 For x1+ 2x2£ 5: y1= 0 (c¢3in the slack variable column x3)

2 For 2x1+ x2= 4: y2=5

–3(c¢5in the artificial variable column x5)

3 For x1- x2≥ 1: y3= -7

–3(c¢6in the artificial variable column x6)

When the right side of the third constraint is changed to 2 (i.e., x1- x2≥ 2), the cost

function f = (-x1- 4x2) changes by

(f)

That is, the cost function will increase by 7

–3, from -13

–3to -2 (z = 2) This can be also

observed from Fig 6-8 We shall demonstrate that same result is obtained when thethird constraint is written in the “£ form” (-x1+ x2£ -1) The Lagrange multiplier forthe constraint is 7

–3, which is the negative of the preceding value Note that it is also c¢4

F

D

E Optimum point

x2

x1

FIGURE 6-8 Constraints for Example 6.16 Feasible region: line E-C.

in the surplus variable column x4 When the right side of the third constraint is changed

to 2 (i.e., it becomes -x1+ x2£ -2), the cost function f = (-x1- 4x2) changes by

(g)

which is same as before

When the right side of the equality constraint is changed to 5 from 4, the cost

Df = -y3De3= - [- - -( )]=

7

73

TABLE 6-19 Solution for Example 6.16 with Equality Constraint

Initial tableau: x6is identified to be replaced with x1 in the basic set.

Third tableau: Reduced cost coefficients in nonbasic columns are

nonnegative; the tableau gives the optimum point End of Phase I End

6.5.2 Ranging Right Side Parameters

When the right side of a constraint is changed, the constraint boundary moves parallel to itself, changing the feasible region for the problem However, the isocost lines do not change.

Since the feasible region is changed, the optimum solution may change, i.e., the design ables as well as the cost function may change There are, however, certain limits on changes for which the set of active constraints at the optimum point is not altered That is,

vari-if the changes are within certain limits, the sets of basic and nonbasic variables do not change

In that case, the solution of the altered problem can be obtained from the information contained in the final tableau Otherwise, Eq (6.23) cannot be used, and more iterations ofthe Simplex method are needed to obtain solution for the altered problem Theorem 6.6describes determination of the limits and the new right sides when the changes are withinthe limits

Theorem 6.6 Limits on Changes in Resources Let D kbe the possible change in the right

side bk of the kth constraint If Dksatisfies the following inequalities, then no more iterations

of the Simplex method are required to obtain the solution for the altered problem and Eq.(6.23) can be used to determine changes to the cost function:

(6.24)

where

b i¢= right side parameter for the ith constraint in the final tableau

a ij¢= parameters in the jth column of the final tableau; the jth column corresponds to xj

which is the slack variable for a “£ type” constraint, or the artificial variable for anequality, or “≥ type” constraint

r i = negative of the ratios of the right sides with the parameters in the jth column

Dk= possible change in the right side of the kth constraint; the slack or the artificial variable for the kth constraint determines the index j of the column whose elements

are used in the Inequalities (6.24)

Furthermore, the new right side parameters b≤j due to a change of Dk in bk are given as

(6.25)

Using Eq (6.25) and the final tableau, new values for the basic variables in each row can

be obtained Equation (6.25) is applicable only if Dk is in the range determined by

Inequalities (6.24) To determine the range, we first determine the column index j according

to the rules given in Theorem 6.6 Then using the elements in the jth column, we determine the ratios ri = -b¢ i/a ij The largest negative ratio ri¢ gives the lower limit on change Dk

in bk If there is no a ij¢ > 0 (i.e., there is no negative ri), then the said ratio cannot be

found, and so there is no lower bound on Dk, i.e., the lower limit is -• The smallest

posi-tive ratio rigives the upper limit on change Dkin bk If there is no a¢ij< 0 (i.e., there is no

positive ri), then the said ratio cannot be found, and so there is no upper bound on Dk, i.e.,the upper limit is • Example 6.17 demonstrates calculations of the ranges for the right sideparameters and new values for the right side (i.e., the basic variables) for a problem with

<

¢ =

EXAMPLE 6.17 Ranges for Resource Limits—“£ Type”

Constraints

Find ranges for the right side parameters of constraints for the problem solved inExample 6.15

Solution. The graphical solution for the problem is shown in Fig 6-7 The final

tableau for the problem is given in Table 6-18 For the first constraint, x3is the slack

variable, and so j = 3 in Inequalities (6.24) for calculation of range for D1, the change

to the constraint’s right side The ratios of the right side parameters with the elements

in column 3, riof Eq (6.24) are calculated as

(a)

Since there is no positive ri, there is no upper limit on D1 The lower limit is mined as the largest element among the negative ratios according to the Inequality(6.24) as:

deter-(b)Thus, limits for D1are -5 £ D1£ • and the range on b1is obtained by adding the

current value of b1= 9 to both sides, as

(c)

For the second constraint (k = 2), x4is the slack variable Therefore, we will use

elements in column x4of the final tableau (a¢ i4, j= 4) in the inequalities of Eq (6.24)

The ratios of the right side parameters with the elements in column 4, riof Eq (6.24),are calculated as

(d)

According to the inequalities in Eq (6.24), lower and upper limits for D2are given as

(e)

Therefore, the allowed decrease in b2is 9.286 and the allowed increase is 2.5 Adding

2 to the above inequality (the current value of b2), the range on b2is given as

(f)Similarly, for the third constraint, the ranges for D3and b3are:

a i i i

= - ¢

¢ = -- -

-ÏÓ

¸

˛={ - - }4

a i i i

= - ¢

¢ = - -

-ÏÓ

¸

˛= -{ - - }3

10 - 9 = 1 Also, j = 3, so we use the third column from Table 6-18 in Eq (6.25) andobtain new values of the variables as

(h)(i)(j)The other variables remain nonbasic, so they have zero values The new solution cor-responds to point F in Fig 6-7 Similarly, if the right side of the second constraint is

changed from 2 to 3, the new values of the variables, using Eq (6.25) and the x4

column from Table 6-18, are calculated as x2= 0.6, x1= 4.2, and x5= 14.4 This tion corresponds to point G in Fig 6-7

solu-When the right sides of two or more constraints are changed simultaneously, wecan use Eq (6.25) to determine new values of the design variables However, we have

to make sure that the new right sides do not change the basic and nonbasic sets ofvariables, i.e., the vertex that gives the optimum solution is not changed Or, in otherwords, no new constraint becomes active As an example, let us calculate the newvalues of design variables using Eq (6.25) when the right sides of the first and thesecond constraints are changed to 10 and 3 from 9 and 2, respectively:

(k)(l)(m)

It can be verified that the new solution corresponds to point H in Fig 6-7

Find ranges for the right side parameters of the problem solved in Example 6.16

Solution. The final tableau for the problem is given in Table 6-19 The graphical

solution for the problem is given in Fig 6-8 In the tableau, x3is a slack variable for

the first constraint, x4is a surplus variable for the third constraint, x5is an artificial

variable for the second constraint, and x6 is an artificial variable for the third

con-straint For the first constraint, x3is the slack variable, and therefore, index j for use

in Inequalities (6.24) is determined as 3 Using the same procedure as for Example6.17, the ranges for D1and b1are calculated as -2 £ D1£ • and 3 £ b1£ •

Since the second constraint is an equality, the index j for use in Eq (6.24) is mined by the artificial variable x5for the constraint, i.e., j = 5 Accordingly, ratios ri

deter-in Eq (6.24) and the range for D2are calculated as

(a)(b)

The range for b2can be found by adding the current value of b2= 4 to both sides ofthe above inequality as 2 £ b2£ 6

The third constraint is a “≥ type”, so index j for use in Inequalities (6.24) is

deter-mined by its artificial variable x6; i.e., j = 6 Accordingly, ratios riin Eq (6.24) andthe range for D3are calculated as

(c)(d)

The limits on changes in b3are (add current value of b3= 1 to both sides of the aboveinequality) -1 £ b3£ 2

New values of design variables. We can use Eq (6.25) to calculate the new values

of the design variables for the right side changes that remain within the previouslydetermined ranges It can be seen that since the first constraint is not active, it doesnot affect the optimum solution as long as its right side remains within the range 3 £

(g)

To determine the new values of design variables when the right side of the third

con-straint is changed from 1 to 2, we use the x6column (j = 6) from Table 6-19 in

Eq (6.25) and obtain the new solution as

(h)(i)

= - ¢

¢ = - --

-ÏÌÓ

¸

˝

˛= -{ - }6

21

= - ¢

¢ = -- -

-ÏÌÓ

¸

˝

˛={ - - }5

21

6.5.3 Ranging Cost Coefficients

If a cost coefficient ck is changed to ck + Dck, we like to find an admissible range on Dcksuch

that the optimum design variables are not changed Note that when the cost coefficients are

changed, the feasible region for the problem does not change However, orientation of the

cost function hyperplane and value of the cost function change Limits on the change Dckfor

the coefficient ck depend on whether xkis a basic variable at the optimum Thus, we mustconsider the two cases separately Theorems 6.7 and 6.8 give ranges for the cost coefficientsfor the two cases

Theorem 6.7 Range for Cost Coefficient of Nonbasic Variables Let c k be such that x* kis

not a basic variable If this ck is replaced by any ck + Dck, where -c¢ k £ Dck £ •, then the

optimum solution (design variables and the cost function) does not change Here, c¢ k is the

reduced cost coefficient corresponding to x* kin the final tableau

Theorem 6.8 Range for Cost Coefficient of Basic Variables Let c k be such that x* k is

a basic variable, and let x* k = b¢ r(a superscript * is used to indicate optimum value) Then, the range for the change Dckin ckfor which the optimum design variables do not change isgiven as

(6.26)

where

a rj¢ = element in the rth row and the jth column of the final tableau The index r is determined by the row that determines x* k Index j corresponds to each of the

nonbasic columns excluding artificial columns (Note: if no a¢ ij> 0, then there is no

upper limit; if no a¢ ij< 0, then there is no lower limit.)

c¢j = reduced cost coefficient in the jth nonbasic column excluding artificial variable

columns

d j = ratios of the reduced cost coefficients with the elements in the rth row

corresponding to nonbasic columns excluding artificial columns

When Dcksatisfies Inequality (6.26), the optimum value of the cost function is f * + Dck x* k

To determine possible changes in the cost coefficient of a basic variable, the first step is

to determine the row index r for use in Inequalities (6.26) This represents the row mining the basic variable x* k After r has been determined, we determine ratios of the reduced

deter-cost coefficients and elements in the rth row according to the rules given in Theorem 6.8.

The lower limit on Dckis determined by the maximum of the negative ratios The upper limit

is determined by the minimum of the positive ratios Example 6.19 demonstrates the dure for the “£ type” constraints and Example 6.20 demonstrates it for the equality and

Determine ranges for the cost coefficients of the problem solved in Example 6.15

Solution. The final tableau for the problem is given in Table 6-18 The problem is

solved as a minimization of the cost function f = -5x1+ 2x2 Therefore, we will find

ranges for the cost coefficients c1= -5 and = c2= 2 Note that since both x1and x2arebasic variables, Theorem 6.8 will be used.

Since the second row determines the basic variable x1, r = 2 (the row number) for use in Inequalities (6.26) Columns 3 and 4 are nonbasic; therefore j = 3, 4 are the

column indices for use in Eq (6.26) After calculating the ratios dj, the range for Dc1

For the second cost coefficient, r = 1 (the row number) because the first row mines x2as a basic variable After calculating the ratios dj, the range for Dc2is calcu-lated as

= ¢

¢ =

-ÏÌÓ

, ; max . min ;

DD

= ¢

¢ =

ÏÌÓ

, ; D min , ; or D

EXAMPLE 6.20 Ranges for Cost Coefficients—Equality and

“≥ Type” Constraints

Find ranges for the cost coefficients of the problem solved in Example 6.16

Solution. The final tableau for the problem is given in Table 6-19 In the tableau,

x3 is a slack variable for the first constraint, x4 is a surplus variable for the third

constraint, and x5and x6are artificial variables for the second and third constraints,

respectively Since both x1and x2are basic variables, we will use Theorem 6.8 to find

ranges for the cost coefficients c1= -1 and c2= -4 Note that the problem is solved

as minimization of the cost function f = -x1- 4x2 Columns 4, 5, and 6 are nonbasic.However, since artificial columns 5 and 6 must be excluded, only column 4 can beused in Eq (6.26)

To find the range for Dc1, r = 3 is used because the third row determines x1as a

basic variable Using Inequalities (6.26) with r = 3 and j = 4, we have

For the second cost coefficient, r = 2 because the second row determines x2as a

basic variable Using Eq (6.26) with r = 2 and j = 4, the range for Dc2is obtained as

-• £ Dc2£ 3.5 Thus the range for c2with current value c2= -4 is given as -• £ c2

£ -0.5 If c2changes from -4 to -3, the new value of the cost function is given as

113

¸

˝

˛£Dc £ • or - £Dc £ •

*6.5.4 Changes in the Coefficient Matrix

Any change in the coefficient matrix A in Eq (6.10) changes the feasible region for the

problem This may change the optimum solution for the problem depending on whether the

change is associated with a basic variable Let aij be replaced by aij + Daij We shall

deter-mine limits for Daijso that with minor computations the optimum solution for the changedproblem can be obtained We must consider the two cases: (i) when the change is associatedwith a nonbasic variable and (ii) when the change is associated with a basic variable Resultsfor these two cases are summarized in Theorems 6.9 and 6.10, respectively

Theorem 6.9 Change Associated with a Nonbasic Variable Let j in a ij be such that xjis

not a basic variable and k be the column index for the slack or artificial variable associated with the constraint of the ith row Define a vector

With this notation, if Daijsatisfies one of the following sets of inequalities, then the optimum

solution (design variables and cost function) does not change when aijis replaced by any

a ij + Daij:

(6.29)or,

(6.30)

Also, if R = 0, the solution does not change for any value of Daij.

To use the theorem, a first step is to determine indices j and k Then we determine the

vector cBof Eq (6.27), and the scalar R of Eq (6.28) Conditions of Inequalities (6.29) and

(6.30) then determine whether the given Daijwill change the optimum solution If the ities are not satisfied, then we have to re-solve the problem to obtain the new solution

inequal-Theorem 6.10 Change Associated with a Basic Variable Let j in a ij be such that xjis a

basic variable and let x* j = b¢ t (i.e., t is the row index that determines optimum value of xj) Let the index k and the scalar R be defined as in Theorem 6.9 Let Daijsatisfy the followinginequalities:

(6.31)(6.32)and

(6.33)(6.34)and

(6.35)Note that the upper and lower limits on Daijdo not exist if the corresponding denominators

in Eqs (6.31) and (6.33) do not exist If Daijsatisfies the above inequalities, then the optimumsolution of the changed problem can be obtained without any further iterations of the Simplex

method If b¢r for r = 1 to m is replaced by

Da ij£ ¢c j R when R<0, and Da ij≥ - • when R=0

Da ij≥ ¢c j R when R>0, and Da ij £ • when R=0

cB =[c B1c B2 .c Bm]

in the final tableau, then the new optimum values for the basic variables can be obtained

when aij is replaced by aij + Daij In other words, if x* j = b¢ r, then x¢ j = b≤ r, where x¢ jrefers to theoptimum solution for the changed problem

To use the theorem, we need to determine indices j, t, and k Then we determine the stants Ar and Bqfrom Eqs (6.32) and (6.34) With these, ranges on Daijcan be determinedfrom Inequalities (6.31) and (6.33) If Daijsatisfy these inequalities, Eq (6.36) determinesthe new solution If the inequalities are not satisfied, the problem must be re-solved for thenew solution

Excel Solver can be used to solve linear programming problems The procedure for solvingthe problem is basically the same as explained for the solution of nonlinear equations inChapter 4 An Excel worksheet needs to be prepared to enter all the data and equations forthe problem Then, the Solver dialog box is activated under the Tools menu There, the objec-tive function, design variables, and the constraints are defined, and the problem is solved

We shall demonstrate this process by solving the problem given in Example 6.16

The Excel worksheet for the problem can be organized in many different ways Figure 6-9 shows one possible format for setting up the problem The original problem is entered

at the top of the sheet just as a reminder Other cells containing data about the problem areexplained as follows:

A10 to A15: row designations

C11, D11: starting values for the variables x1and x2, respectively; currently set to 0C12, D12: objective function coefficients for x1and x2

C13 to D15: constraint coefficient matrix

E12: formula to calculate the objective function value using the design

variable values in the cells C11 and D11E13: formula to calculate the left side of constraint 1

E14: formula to calculate the left side of constraint 2

FIGURE 6-9 Excel worksheet for the problem of Example 6.16.