# Sensitivity Analysis 3.1 AN EXAMPLE FOR ANALYSIS

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2 3. An Example for Analysis 77 per case, which is higher than either of the other products. There is no limit on the demand for champagne glasses. Now what is the optimal product mix among the three alternatives? The formulation of the custom-molding example, including the new activity of producing champagne glasses, is straightforward. We have exactly the same capacity limitations hours of production capacity, cubic feet of warehouse capacity, and limit on six-ounce juice-glass demand and one additional decision variable for the production of champagne glasses. Letting x = Number of cases of six-ounce juice glasses, in hundreds; x = Number of cases of ten-ounce cocktail glasses, in hundreds; x 3 = Number of cases of champagne glasses, in hundreds; and measuring the contribution in hundreds of dollars, we have the following formulation of our custommolder example: subject to: Maximize z = 5x + 4.5x + 6x 3, (hundreds of dollars) 6x + 5x + 8x 3 60, (production capacity; hours) 0x + 0x + 0x 3 50, (warehouse capacity; hundreds of sq. ft.) x 8, (demand for 6 oz. glasses; hundreds of cases) x 0, x 0, x 3 0. If we add one slack variable in each of the less-than-or-equal-to constraints, the problem will be in the following canonical form for performing the simplex method: 6x + 5x + 8x 3 + x 4 = 60, () 0x + 0x + 0x 3 + x 5 = 50, (3) x + + x 6 = 8, (4) 5x + 4.5x + 6x 3 z = 0. (5) The corresponding initial tableau is shown in Tableau. After applying the simplex method as described Tableau Basic Current variables values x x x 3 x 4 x 5 x 6 x x x ( z) in Chapter, we obtain the final tableau shown in Tableau. Since the final tableau is in canonical form and all objective-function coefficients of the nonbasic variables are currently nonpositive, we know from Chapter that we have the optimal solution, consisting of x = 6 3 7, x = 4 7, x 6 = 4 7, and z = In this chapter we present results that depend only on the initial and final tableaus of the problem. Specifically, we wish to analyze the effect on the optimal solution of changing various elements of the problem data without re-solving the linear program or having to remember any of the intermediate tableaus Excel spreadsheet available at ()

5 80 Sensitivity Analysis 3. of producing champagne glasses. Had the new activity been priced out at the outset, using the shadow prices determined in Chapter, we would have immediately discovered that the opportunity cost of diverting resources from the current solution to the new activity exceeded its potential contribution. There would have been no need to consider the new activity further. This is an important observation, since it implies that the shadow prices provide a mechanism for screening new activities that were not included in the initial model formulation. In a maximization problem, if any new activity prices out negatively using the shadow prices associated with an optimal solution, it may be immediately dropped from consideration. If, however, a new activity prices out positively with these shadow prices, it must be included in the problem formulation and the new optimal solution determined by pivoting. General Discussion The concepts presented in the context of the custom-molder example can be applied to any linear program. Consider a problem in initial canonical form: Shadow price a x + a x + + a n x n +x n+ = b y a x + a x + + a n x n + x n+ = b y... a m x +a m x + + a mn x n + + x n+m = b m y m ( z) + c x +c x + + c n x n + 0x n+ +0x n+ + +0x n+m = 0 The variables x n+, x n+,..., x n+m are either slack variables or artificial variables that have been introduced in order to transform the problem into canonical form. Assume that the optimal solution to this problem has been foundand the corresponding final form of the objective function is: ( z) + c x + c x + + c n x n + c n+ x n+ + c n+ x n+ + + c n+m x n+m = z 0. (9) As we have indicated before, c j is the reduced cost associated with variable x j. Since (9) is in canonical form, c j = 0 if x j is a basic variable. Let y i denote the shadow price for the ith constraint. The arguments from the example problem show that the negative of the final objective coefficient of the variable x n+i corresponds to the shadow price associated with the ith constraint. Therefore: c n+ = y, c n+ = y,..., c n+m = y m. (0) Note that this result applies whether the variable x n+i is a slack variable (i.e., the ith constraint is a less-thanor-equal-to constraint), or whether x n+i is an artificial variable (i.e., the ith constraint is either an equality or a greater-than-or-equal-to constraint). We now shall establish a fundamental relationship between shadow prices, reduced costs, and the problem data. Recall that, at each iteration of the simplex method, the objective function is transformed by subtracting from it a multiple of the row in which the pivot was performed. Consequently, the final form of the objective function could be obtained by subtracting multiples of the original constraints from the original objective function. Consider first the final objective coefficients associated withthe original basic variables x n+, x n+,..., x n+m. Let π, π,..., π n be the multiples of each row that are subtracted from the original objective function to obtain its final form (9). Since x n+i appears only in the ith constraint and has a + coefficient, we should have: c n+i = 0 π i. Combining this expression with (0), we obtain: c n+i = π i = y i.

6 3.3 Variations in the Objective Coefficients 8 Thus the shadow prices y i are the multiples π i. Since these multiples can be used to obtain every objective coefficient in the final form (9), the reduced cost c j of variable x j is given by: c j = c j m a i j y i ( j =,,..., n), () i= and the current value of the objective function is: z 0 = m b i y i i= or, equivalently, m z 0 = b i y i () i= Expression () links the shadow prices to the reduced cost of each variable, while () establishes the relationship between the shadow prices and the optimal value of the objectivefunction. Expression () also can be viewed as a mathematical definition of the shadow prices. Since c j = 0 for the m basic variables of the optimal solution, we have: m 0 = c j a i j y i i= for j basic. This is a system of m equations in m unknowns that uniquely determines the values of the shadow prices y i. 3.3 VARIATIONS IN THE OBJECTIVE COEFFICIENTS Now let us consider the question of how much the objective-function coefficients can vary without changing the values of the decision variables in the optimal solution. We will make the changes one at a time, holding all other coefficients and righthand-side values constant. The reason for this is twofold: first, the calculation of the range on one coefficient is fairly simple and therefore not expensive; and second, describing ranges when more than two coefficients are simultaneously varied would require a system of equations instead of a simple interval. We return to consideration of the objective coefficient of x 3, a nonbasic variable in our example. Clearly, if the contribution is reduced from \$6 per case to something less it would certainly not become attractive to produce champagne glasses. If it is now not attractive to produce champagne glasses, then reducing the contribution from their production only makes it less attractive. However, if the contribution from production of champagne glasses is increased, presumably there is some level of contribution such that it becomes attractive to produce them. In fact, it was argued above that the opportunity cost associated with diverting resources from the optimal production schedule was merely the shadow price associated with a resource multiplied by the amount of the resource consumed by the activity. For this activity, the opportunity cost is \$6 4 7 per case compared to a contribution of \$6 per case. If the contribution were increased above the break-even opportunity cost, then it would become attractive to produce champagne glasses. Let us relate this to the procedures of the simplex method. Suppose that we increase the objective function coefficient of x 3 in the original problem formulation (5) by c 3, giving us: 5x + 4.5x + (6 + c 3 ) x } {{ } 3 z = 0. = c3 new

7 8 Sensitivity Analysis 3.3 Tableau 3 Basic Current variables values x x x 3 x 4 x 5 x 6 x x x ( z) c In applying the simplex method, multiples of the rows were subtracted from the objective function to yield the final system of equations. Therefore, the objective function in the final tableau will remain unchanged except for the addition of c 3 x 3. The modified final tableau is given in Tableau 3. Now x 3 will become a candidate to enter the optimal solution at a positive level, i.e., to enter the basis, only when its objective-function coefficient is positive. The optimal solution remains unchanged so long as: c 3 0 or c Equivalently, we know that the range on the original objective-function coefficient of x 3, say c new 3, must satisfy < c new if the optimal solution is to remain unchanged. Next, let us consider what happens when the objective-function coefficient of a basic variable is varied. Consider the range of the objective-function coefficient of variable x. It should be obvious that if the contribution associated with the production of six-ounce juice glasses is reduced sufficiently, we will stop producing them. Also, a little thought tells us that if the contribution were increased sufficiently, we might end up producing only six-ounce juice glasses. To understand this mathematically, let us start out as we did before by adding c to the objective-function coefficient of x in the original problem formulation (5) to yield the following modified objective function: (5 + c )x + 4.5x + 6x 3 z = 0. If we apply the same logic as in the case of the nonbasic variable, the result is Tableau 4. Tableau 4 Basic Current variables values x x x 3 x 4 x 5 x 6 x x x ( z) c However, the simplex method requires that the final system of equations be in canonical form with respect to the basic variables. Since the basis is to be unchanged, in order to make the coefficient of x zero in the final tableau we must subtract c times row 3 from row 4 in Tableau 4. The result is Tableau 5. By the simplex optimality criterion, all the objective-function coefficients in the final tableau must be

8 3.3 Variations in the Objective Coefficients 83 Tableau 5 Basic Current variables values x x x 3 x 4 x 5 x 6 x x x ( z) c c 4 7 c c nonpositive in order to have the current solution remain unchanged. Hence, we must have: ( ) c 0 that is, c 4, ( ) 4 7 c 0 that is, c 4, ( ) c 0 that is, c + 5 ; and, taking the most limiting inequalities, the bounds on c are: 4 c 5. If we let c new = c + c be the objective-function coefficient of x in the initial tableau, then: 4 7 cnew 5 5, where the current value of c = 5. It is easy to determine which variables will enter and leave the basis when the new cost coefficient reaches either of the extreme values of the range. When c new = 5 5, the objective coefficient of x 5 in the final tableau becomes 0; thus x 5 enters the basis for any further increase of c new. By the usual ratio test of the simplex method, { } { b i 4 Min i ā āis > 0 = Min 7, } 4 7 = Min {50, }, 3 is 35 4 and the variable x 6, which is basic in row, leaves the basis when x 5 is introduced. Similarly, when c new = 4 7, the objective coefficient of x 3 in the final tableau becomes 0, and x 3 is the entering variable. In this case, the ratio test shows that x leaves the basis. General Discussion To determine the ranges of the cost coefficients in the optimalsolution of any linear program, it is useful to distinguish between nonbasic variables and basic variables. If x j is a nonbasic variable and we let its objective-function coefficient c j be changed by an amount c j with all other data held fixed, then the current solution remains unchanged so long as the new reduced cost c new j remains nonnegative, that is, c new j = c j + c j m a i j y i = c j + c j 0. i= The range on the variation of the objective-function coefficient of a nonbasic variable is then given by: < c j c j, (3)

9 84 Sensitivity Analysis 3.4 so that the range on the objective-function coefficient c new j < c new j c j c j. = c j + c j is: If x r is a basic variable in row k and we let its original objective-function coefficient c r be changed by an amount c r with all other data held fixed, then the coefficient of the variable x r in the final tableau changes to: c new r = c r + c r m a ir y i = c r + c r. i= Since x r is a basic variable, c r = 0; so, to recover a canonical form with c r new = 0, we subtract c r times the kth constraint in the final tableau from the final form of the objective function, to give new reduced costs for all nonbasic variables, c new j = c j c r ā kj. (4) Here ā kj is the coefficient of variable x j in the kth constraint of the final tableau. Note that c new j will be zero for all basic variables. The current basis remains optimal if c new j 0. Using this condition and (4), we obtain the range on the variation of the objective-function coefficient: { } { } c j Max c j j ā ākj > 0 c r Min kj j ā ākj < 0. (5) kj The range on the objective-function coefficient cr new = c r + c r of the basic variable x r is determined by adding c r to each bound in (5). The variable transitions that occur at the limits of the cost ranges are easy to determine. For nonbasic variables, the entering variable is the one whose cost is being varied. For basic variables, the entering variable is the one giving the limiting value in (5). The variable that leaves the basis is then determined by the minimum-ratio rule of the simplex method. If x s is the entering variable, then the basic variable in row r, determined by: { } b r b i = Min ā rs i āis > 0, ā is is dropped from the basis. Since the calculation of these ranges and the determination of the variables that will enter and leave the basis if a range is exceeded are computationally inexpensive to perform, this information is invariably reported on any commercially available computer package. These computations are easy since no iterations of the simplex method need be performed. It is necessary only () to check the entering variable conditions of the simplex method to determine the ranges, as well as the variable that enters the basis, and () to check the leaving variable condition (i.e., the minimum-ratio rule) of the simplex method to compute the variable that leaves the basis. 3.4 VARIATIONS IN THE RIGHTHAND-SIDE VALUES Now let us turn to the questions related to the righthand-side ranges. We already have noted that a righthandside range is the interval over which an individual righthand-side value can be varied, all the other problem data being held constant, such that variables that constitute the basis remain the same. Over these ranges, the values of the decision variables are clearly modified. Of what use are these righthand-side ranges? Any change in the righthand-side values that keep the current basis, and therefore the canonical form, unchanged has no effect upon the objective-function coefficients. Consequently, the righthand-side ranges are such that the shadow prices (which are the negative of the coefficients of the slack or artificial variables in the final tableau) and the reduced costs remain unchanged for variations of a single value within the stated range.

10 3.4 Variations in the Righthand-Side Values 85 We first consider the righthand-side value for the demand limit on six-ounce juice glasses in our example. Since this constraint is not binding, the shadow price associated with it is zero and it is simple to determine the appropriate range. If we add an amount b 3 to the righthand side of this constraint (4), the constraint changes to: x + x 6 = 8 + b 3. In the original problem formulation, it should be clear that, since x 6, the slack in this constraint, is a basic variable in the final system of equations, x 6 is merely increased or decreased by b 3. In order to keep the current solution feasible, x 6 must remain greater than or equal to zero. From the final tableau, we see that the current value of x 6 = 4 7 ; therefore x 6 remains in the basis if the following condition is satisfied: x 6 = b 3 0. This implies that: b or, equivalently, that: b3 new = 8 + b Now let us consider changing the righthand-side value associated with the storage-capacity constraint of our example. If we add b to the righthand-side value of constraint (3), this constraint changes to: 0x + 0x + 0x 3 + x 5 = 50 + b. In the original problem formulation, as was previously remarked, changing the righthand-side value is essentially equivalent to decreasing the value of the slack variable x 5 of the corresponding constraint by b ; that is, substituting x 5 b for x 5 in the original problem formulation. In this case, x 5, which is zero in the final solution, is changed to x 5 = b. We can analyze the implications of this increase in the righthand-side value by using the relationships among the variables represented by the final tableau. Since we are allowing only one value to change at a time, we will maintain the remaining nonbasic variables, x 3 and x 4, at zero level, and we let x 5 = b. Making these substitutions in the final tableau provides the following relationships: x 3 35 b = 4 7, x 6 4 b = 4 7, x + 4 b = In order for the current basis to remain optimal, it need only remain feasible, since the reduced costs will be unchanged by any such variation in the righthand-side value. Thus, which implies: or, equivalently, x = b 0 (that is, b 50), x 6 = (that is, b ), x = (that is, b 90), b 90, 8 b new 40, where the current value of b = 50 and b new = 50 + b. Observe that these computations can be carried out directly in terms of the final tableau. When changing the ith righthand side by b i, we simply substitute b i for the slack variable in the corresponding constraint and update the current values of the basic variables accordingly. For instance, the change of storage capacity just considered is accomplished by setting x 5 = b in the final tableau to produce Tableau 6 with modified current values. Note that the change in the optimal value of the objective function is merely the shadow price on the storage-capacity constraint multiplied by the increased number of units of storage capacity.

11 86 Sensitivity Analysis 3.4 Variable Transitions Tableau 6 Basic Current variables values x x x 3 x 4 x 5 x 6 x b 7 7 x x ( z) b We have just seen how to compute righthand-side ranges so that the current basis remains optimal. For example, when changing demand on six-ounce juice glasses, we found that the basis remains optimal if b new , and that for b3 new < 6 7 3, the basic variable x 6 becomes negative. If b3 new were reduced below 6 3 7, what change would take place in the basis? First, x 6 would have to leave the basis, since otherwise it would become negative. What variable would then enter the basis to take its place? In order to have the new basis be an optimal solution, the entering variable must be chosen so that the reduced costs are not allowed to become positive. Regardless of which variable enters the basis, the entering variable will be isolated in row of the final tableau to replace x 6, which leaves the basis. To isolate the entering variable, we must perform a pivot operation, and a multiple, say t, of row in the final tableau will be subtracted from the objective-function row. Assuming that b3 new were set equal to in the initial tableau, the results are given in Tableau 7. Tableau 7 Basic Current variables values x x x 3 x 4 x 5 x 6 x x x ( z) t t 35 4 t t In order that the new solution be an optimal solution, the coefficients of the variables in the objective function of the final tableau must be nonpositive; hence, ) (that t 0 is, t 4, ( ) t 0 that is, t 4, ) 35 4 (that t 0 is, t 5, t 0 (that is, t 0), which implies: 0 t 4. Since the coefficient of x 3 is most constraining on t, x 3 will enter the basis. Note that the range on the righthandside value and the variable transitions that would occur if that range were exceeded by a small amount are

12 3.4 Variations in the Righthand-Side Values 87 easily computed. However, the pivot operation actually introducing x 3 into the basis and eliminating x 6 need not be performed. As another example, when the change b in storage capacity reaches in Tableau 6, then x 6, the slack on six-ounce juice glasses, reaches zero and will drop from the basis, and again x 3 enters the basis. When b = 90, though, then x, the production of six-ounce glasses, reaches zero and will drop from the basis. Since x is a basic variable in the third constraint of Tableau 6, we must pivot in the third row, subtracting t times this row from the objective function. The result is given in Tableau 8. Tableau 8 Basic Current variables values x x x 3 x 4 x 5 x 6 x 7 7 x x ( z) 54 t t 7 7 t t The new objective-function coefficients must be nonpositive in order for the new basis to be optimal; hence, t 0 (that is, t 0), ) (that t 0 is, t 4, ( ) 7 7 t 0 that is, t, t 0 (that is, t 5 which implies 0 t 5. Consequently, x 5 enters the basis. The implication is that, if the storage capacity were increased from 50 to 40 cubic feet, then we would produce only the ten-ounce cocktail glasses, which have the highest contribution per hour of production time. ), General Discussion In the process of solving linear programs by the simplex method, the initial canonical form: is transformed into a canonical from: a x + a x + + a n x n + x n+ = b, a x + a x + + a n x n + x n+ = b, a m x + a m x + + a mn x n } {{ + x n+m = b } m Original basic variables ā x + ā x + + ā n x n + β x n+ + β x n+ + + β m x n+m = b, ā x + ā x + + ā n x n + β x n+ + β x n+ + + β m x n+m = b,.. ā m x + ā m x + + ā mn x n + β m x n+ + β m x n+ + + β mn x n+m = b m. Of course, because this is a canonical form, the updated data ā i j and β i j will be structured so that one basic variable is isolated in each constraint. Since the updated coefficients of the initial basic (slack or artificial)

13 88 Sensitivity Analysis 3.5 variables x n+, x n+,..., x n+m in the final tableau play such an important role in sensitivity analysis, we specifically denote these values as β i j. We can change the coefficient b k of the kth righthand side in the initial tableau by b k with all the other data held fixed, simply by substituting x n+k b k for x n+k in the original tableau. To see how this change affects the updated righthand-side coefficients, we make the same substitution in the final tableau. Only the terms β ik x n+k for i =,,..., m change in the final tableau. They become β ik (x n+k b k ) = β ik x n+k β ik b k. Since β ik b k is a constant, we move it to the righthand side to give modified righthand-side values: b i + β ik b k (i =,,..., m). As long as all of these values are nonnegative, the basis specified by the final tableau remains optimal, since the reduced costs have not been changed. Consequently, the current basis is optimal whenever b i +β ik b k 0 for i =,,..., m or, equivalently, Max i { } b i β ik > 0 β ik { bi b k Min i β ik } β ik < 0. (6) The lower bound disappears if all β ik 0, and the upper bound disappears if all β ik 0. When b k reaches either its upper or lower bound in Eq. (6), any further increase (or decrease) in its value makes one of the updated righthand sides, say b r + β rk b k, negative. At this point, the basic variable in row r leaves the basis, and we must pivot in row r in the final tableau to find the variable to be introduced in its place. Since pivoting subtracts a multiple t of the pivot row from the objective equation, the new objective equation has coefficients: c j tā r j ( j =,,..., n). (7) For the new basis to be optimal, each of these coefficients must be nonpositive. Since c j = 0 for the basic variable being dropped and its coefficient in constraint r is ā r j =, we must have t 0. For any nonnegative t, the updated coefficient c j tā r j for any other variable remains nonpositive if ā r j 0. Consequently we need only consider ā r j < 0, and t is given by { } c j t = Min j ār j < 0. (8) ā r j The index s giving this minimum has c s tā rs = 0, and the corresponding variable x s can become the new basic variable in row r by pivoting on ā rs. Note that this pivot is made on a negative coefficient. Since the calculation of the righthand-side ranges and the determination of the variables that will enter and leave the basis when a range is exceeded are computationally easy to perform, this information is reported by commercially available computer packages. For righthand-side ranges, it is necessary to check only the feasibility conditions to determine the ranges as well as the variable that leaves, and it is necessary to check only the entering variable condition (8) to complete the variable transitions. This condition will be used again in Chapter 4, since it is the minimum-ratio rule of the so-called dual simplex method. 3.5 ALTERNATIVE OPTIMAL SOLUTIONS AND SHADOW PRICES In many applied problems it is important to identify alternative optimal solutions when they exist. When there is more than one optimal solution, there are often good external reasons for preferring one to the other; therefore it is useful to be able to easily determine alternative optimal solutions. As in the case of the objective function and righthand-side ranges, the final tableau of the linear program tells us something conservative about the possibility of alternative optimal solutions. First, if all reduced costs of the nonbasic variables are strictly negative (positive) in a maximization (minimization) problem, then there is no alternative optimal solution, because introducing any variable into the basis at a positive level would reduce (increase) the value of the objective function. On the other hand, if one or more of the reduced costs are zero, there may exist alternative optimal solutions. Suppose that, in our custom-molder example,

14 3.5 Alternative Optimal Solutions and Shadow Prices 89 the contribution from champagne glasses, x 3, had been From Section 3. we know that the reduced cost associated with this activity would be zero. The final tableau would look like Tableau 9. Tableau 9 Basic Current Ratio variables values x x x 3 x 4 x 5 x 6 test x x x / 7 ( z) This would imply that x 3, production of champagne glasses, could be introduced into the basis without changing the value of the objective function. The variable that would be dropped from the basis would be determined by the usual minimum-ratio rule: { b i Min i ā i3 } āi3 > 0 = Min { 4 7, } { = Min 4, 45 In this case, the minimum ratio occurs for row 3 so that x leaves the basis. The alternative optimal solution is then found by completing the pivot that introduces x 3 and eliminates x. The values of the new basic variables can be determined from the final tableau as follows: x 3 = 45, x = x 3 = x 6 = x 3 = x = x 3 = z = ( 45 ( 45 7 ( 45 ) = , ) = 8, ) = 0, Under the assumption that the contribution from champagne-glass production is 6 4 7, we have found an alternative optimal solution. It should be pointed out that any weighted combination of these two solutions is then also an alternative optimal solution. In general, we can say that there may exist an alternative optimal solution to a linear program if one or more of the reduced costs of the nonbasic variables are zero. There does exist an alternative optimal solution if one of the nonbasic variables with zero reduced cost can be introduced into the basis at a positive level. In this case, any weighted combination of these solutions is also an alternative optimal solution. However, if it is not possible to introduce a new variable at a positive level, then no such alternative optimal solution exists even though some of the nonbasic variables have zero reduced costs. Further, the problem of finding all alternative optimal solutions cannot be solved by simply considering the reduced costs of the final tableau, since there can in general exist alternative optimal solutions that cannot be reached by a single pivot operation. Independent of the question of whether or not alternative optimal solutions exist in the sense that different values of the decision variables yield the same optimal value of the objective function, there may exist alternative optimal shadow prices. The problem is completely analogous to that of identifying alternative optimal solutions. First, if all righthand-side values in the final tableau are positive, then there do not exist alternative optimal shadow prices. Alternatively, if one or more of these values are zero, then there may exist }.

16 3.7 The Computer Output An Example 9 illustrate the typical computer output for the variation of the custom-molder example presented in the first few sections of this chapter. The results shown in Fig. 3. were obtained on a commercial time-sharing system. Note that the output includes a tabular display of the problem as stated in Eq. (). The first four numbered sections of this output should be compared to Tableau for this example, given in Section 3.. The optimal value of the objective function is z = hundred dollars, while the associated values of the decision variables are as follows: production of six-ounce juice glasses x = hundred cases, and production of ten-ounce cocktail glasses x = 4 7 hundred cases. Note that there is slack in the constraint on demand for six-ounce juice glasses of 4 7 hundred cases, which corresponds to variable x 6. Finally, champagne glasses are not produced, and hence x 3 = 0. Note that the reduced cost associated with production of champagne glasses is 4 7, which is the amount the objective function would decrease per hundred cases if champagne glasses were in fact produced. In our discussion of shadow prices in Section 3., it is pointed out that the shadow prices are the negative of the reduced costs associated with the slack variables. Thus the shadow price on production capacity is 4 hundred dollars per hour of production time, and the shadow price on storage capacity is 35 hundred dollars per hundred square feet of storage space. The shadow price on six-ounce juice glasses demand is zero, since there remains unfulfilled demand in the optimal solution. It is intuitive that, in general, either the shadow price associated with a constraint is nonzero or the slack (surplus) in that constraint is nonzero, but both will not simultaneously be nonzero. This property is referred to as complementary slackness, and is discussed in Chapter 4. Sections 5 and 7 of the computer output give the ranges on the coefficients of the objective function and the variable transitions that take place. (This material is discussed in Section 3.3.) Note that the range on the nonbasic variable x 3, production of champagne glasses, is one-sided. Champagne glasses are not currently produced when their contribution is \$6 per case, so that, if their contribution were reduced, we would certainly not expect this to change. However, if the contribution from the production of champagne glasses is increased to \$6 4 7 per case, their production becomes attractive, so that variable x 3 would enter the basis and variable x, production of six-ounce juice glasses, would leave the basis. Consider now the range on the coefficient of x, production of six-ounce juice glasses, where the current contribution is \$5 per case. If this contribution were raised to \$5 5 per case, the slack in storage capacity would enter the basis, and the slack in juice glass, demand would leave. This means we would meet all of the juice-glass demand and storage capacity would no longer be binding. On the other hand, if this contribution were reduced to \$4 4 7 per case, variable x 3, production of champagne glasses, would enter the basis and variable x, production of six-ounce juice-glasses, would leave. In this instance, juice glasses would no longer be produced. Sections 6 and 8 of the computer output give the ranges on the righthand-side values and the variable transitions that result. (This material is discussed in Section 3.4.) Consider, for example, the range on the righthand-side value of the constraint on storage capacity, where the current storage capacity is 50 hundred square feet. If this value were increased to 40 hundred square feet, the slack in storage capacity would enter the basis and variable x, production of six-ounce glasses, would leave. This means we would no longer produce six-ounce juice glasses, but would devote all our production capacity to ten-ounce cocktail glasses. If the storage capacity were reduced to 8 hundred square feet, we would begin producing champagne glasses and the slack in the demand for six-ounce juice glasses would leave the basis. The other ranges have similar interpretations. Thus far we have covered information that is available routinely when a linear program is solved. The format of the information varies from one computer system to another, but the information available is always the same. The role of computers in solving linear-programming problems is covered more extensively in Chapter 5. Excel spreadsheets available at and

17 9 Sensitivity Analysis 3.7 Figure 3. Solution of the custom-molder example.

18 3.7 Simultaneous Variations Within the Ranges SIMULTANEOUS VARIATIONS WITHIN THE RANGES Until now we have described the sensitivity of an optimal solution in the form of ranges on the objectivefunction coefficients and righthand-side values. These ranges were shown to be valid for changes in one objective-function coefficient or righthand-side value, while the remaining problem data are held fixed. It is then natural to ask what can be said about simultaneously changing more than one coefficient or value within the ranges. In the event that simultaneous changes are made only in objective-function coefficients of nonbasic variables and righthand-side values of nonbinding constraints within their appropriate ranges, the basis will remain unchanged. Unfortunately, it is not true, in general, that when the simultaneous variations within the ranges involve basic variables or binding constraints the basis will remain unchanged. However, for both the ranges on the objective-function coefficients when basic variables are involved, and the righthand-side values when binding constraints are involved, there is a conservative bound on these simultaneous changes that we refer to as the 00 percent rule. Let us consider first, simultaneous changes in the righthand-side values involving binding constraints for which the basis, and therefore the shadow prices and reduced costs, remain unchanged. The righthand-side ranges, as discussed in Section 3.4, give the range over which one particular righthand-side value may be varied, with all other problem data being held fixed, such that the basis remains unchanged. As we have indicated, it is not true that simultaneous changes of more than one righthand-side value within these ranges will leave the optimal basis unchanged. However, it turns out that if these simultaneous changes are made in such a way that the sum of the fractions of allowable range utilized by these changes is less than or equal to one, the optimal basis will be unchanged. Let us consider our custom-molder example, and make simultaneous changes in the binding production and storage capacity constraints. The righthand-side range on production capacity is 37.5 to 65.5 hundred hours, with the current value being 60. The righthand-side range on storage capacity is 8 to 40 hundred cubic feet, with the current value being 50. Although it is not true that the optimal basis remains unchanged for simultaneous changes in the current righthand-side values anywhere within these ranges, it is true that the optimal basis remains unchanged for any simultaneous change that is a weighted combination of values within these ranges. Figure 3. illustrates this situation. The horizontal and vertical lines in the figure are the ranges for production and storage capacity, respectively. The four-sided figure includes all weighted combinations of these ranges and is the space over which simultaneous variations in the values of production and storage capacity can be made while still ensuring that the basis remains unchanged. If we consider moving from the current righthand-side values of 60 and 50 to b new and b new respectively, where b new 60 and b new 50, we can ensure that the basis remains unchanged if 60 b new bnew As long as the sum of the fractions formed by the ratio of the change to the maximum possible change in that direction is less than or equal to one, the basis remains unchanged. Hence, we have the 00 percent rule. Since the basis remains unchanged, the shadow prices and reduced costs also are unchanged. A similar situation exists in the case of simultaneous variations in the objective-function coefficients when basic variables are involved. It is not true that the basis remains unchanged for simultaneous changes in these coefficients anywhere within their individual ranges. However, it is true that the optimal basis remains unchanged for any simultaneous change in the objective-function coefficients that is a weighted combination of the values within these ranges. If, for example, we were to increase all cost coefficients simultaneously, the new optimal basis would remain unchanged so long as the new values c new, c new, and c3 new satisfy: c new cnew cnew Again, the sum of the fractions formed by the ratio of the change in the coefficient to the maximum possible change in that direction must be less than or equal to one.

19 94 Sensitivity Analysis 3.7 Figure 3. Simultaneous variation of righthand-side values. General Discussion Let us analyze the 00 percent rule more formally. If we first consider simultaneous variations in the righthandside values, the 00 percent rule states that the basis remains unchanged so long as the sum of the fractions, corresponding to the percent of maximum change in each direction, is less than or equal to one. To see that this must indeed be true, we look at weighted combinations of the solution for the current values and the solutions corresponding to particular boundaries of the ranges. In Fig. 3., the shaded area contains all weighted combinations of the current values, the lower bound on b, and the upper bound on b. Let x 0 j, j =,,..., n, be an optimal solution to the given linear program, and let a i j be the coefficients of the initial tableau corresponding to the basic variables. Then, j B a i j x 0 j = b i (i =,,..., m), (9) where j B indicates that the sum is taken only over basic variables. Further, let x k j, j =,,..., n, be the optimal solution at either the upper or lower limit of the range when changing b k alone, depending on the direction of the variation being considered. Since the basis remains unchanged for these variations, these solutions satisfy: { a i j x k j = b i = bk + bk max for i = k, (0) b i for i = k. j B It is now easy to show that any nonnegative, weighted combination of these solutions must be an optimal feasible solution to the problem with simultaneous variations. Let λ 0 be the weight associated with the

20 3.7 Simultaneous Variations Within the Ranges 95 optimal solution corresponding to the current values, Eq. (9), and let λ k be the weight associated with the optimal solution corresponding to the variation in b k in Eq. (0). Consider all solutions that are nonnegative weighted combinations of these solutions, such that: m λ k =. () k=0 The corresponding weighted solution must be nonnegative; that is, m x j = λ k x k j 0 ( j =,,..., n), () k=0 since both λ k and x k j are nonnegative. By multiplying the ith constraint of Eq. (9) by λ 0 and the ith constraint of Eq. (0) by λ k and adding, we have m λ k m a i j x k j = λ k b i. jb Since the righthand-side reduces to: m k=0 k=0 λ k b i = k =i k=0 λ k b k + λ i (b i + bi max ) = b i + λ i bi max, we can rewrite this expression by changing the order of summation as: ( m ) a i j λ k x k j = b i + λ i bi max (i =,,..., m). (3) j B k=0 Expressions () and (3) together show that the weighted solution x j, j =,,..., n, is a feasible solution to the righthand-side variations indicated in Eq. (0) and has the same basis as the optimal solution corresponding to the current values in Eq. (9). This solution must also be optimal, since the operations carried out do not change the objective-function coefficients. Hence, the basis remains optimal so long as the sum of the weights is one, as in Eq. (). However, this is equivalent to requiring that the weights λ k, k =,,..., m, corresponding to the solutions associated with the ranges, while excluding the solution associated with the current values, satisfy: m λ k. (4) k= The weight λ 0 on the solution corresponding to the current values is then determined from Eq. (). Expression (4) can be seen to be the 00 percent rule by defining: or, equivalently, which, when substituted in Eq. (4) yields: b k λ k b max k λ k = m k= b k b max k b k b max k. (5)

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