Presentation: H. Sarper 22 October 2013: go to page 26, section 6.5

Chapter 6Sensitivity Analysis and Dualityto accompanyIntroduction to Mathematical Programming: Operations Research, Volume 14th edition, by Wayne L. Winston and Munirpallam Venkataramanan Presentation: H. Sarper 22 October 2013: go to page 26, section 6.5

6.1 – A Graphical Introduction to Sensitivity Analysis • Sensitivity analysis is concerned with how changes in an LP’s parameters affect the optimal solution. max z = 3x1 + 2x2 2 x1 + x2 ≤ 100 (finishing constraint) x1 + x2 ≤ 80 (carpentry constraint) x1 ≤ 40 (demand constraint) x1,x2 ≥ 0 (sign restriction) Reconsider the Giapetto problem from Chapter 3 shown to the right: Where: x1 = number of soldiers produced each week x2 = number of trains produced each week.

6.1 – A Graphical Introduction to Sensitivity Analysis The optimal solution to the Giapetto problem is z = 180, x1 = 20, x2 = 60 (Point B in the figure to the right) and it has x1, x2, and s3 (the slack variable for the demand constraint) as basic variables. How would changes in the problem’s objective function coefficients or the constraint’s right-hand sides change this optimal solution?

6.1 – A Graphical Introduction to Sensitivity Analysis Graphical Analysis of the Effect of a Change in an Objective Function Coefficient • Point A is optimal if -c1/2 ≥ -1 or 0 ≤ c1 ≤ 2 ( -1 is the carpentry constraint slope). • Point B is optimal if -2 ≤ -c1/2 ≤ -1 or 2 ≤ c1 ≤ 4 (between the slopes of the carpentry and finishing constraint slopes). • Point C is optimal if -c1/2 ≤ -2 or c1 ≥ 4 ( -2 is the finishing constraint slope). Recall from the Giapetto problem, if the isoprofit line is flatter than the carpentry constraint, Point A(0,80) is optimal. Point B(20,60) is optimal if the isoprofit line is steeper than the carpentry constraint but flatter than the finishing constraint. Finally, Point C(40,20) is optimal if the slope of the isoprofit line is steeper than the slope of the finishing constraint. Since a typical isoprofit line is c1x1 + 2x2 = k, we know the slope of the isoprofit line is just -c1/2. In summary:

6.1 – A Graphical Introduction to Sensitivity Analysis A graphical analysis can also be used to determine whether a change in the rhs of a constraint will make the basis no longer optimal. Letting b1 = number of finishing hours in the Giapetto LP to the right, we see a change in b1shifts the finishing constraint parallel to its current position. The current optimal point (Point B) is where the carpentry and finishing constraints are binding.

6.1 – A Graphical Introduction to Sensitivity Analysis If we change the value of b1, then as long as the point where the finishing and carpentry constraints intersect are binding remains feasible, the optimal solution will still occur where these constraints intersect. We see that if b1 > 120, x1 will be greater than 40 and will violate the demand constraint. Also, if b1 < 80, x1 will be less than 0 and the nonnegativity constraint for x1 will be violated. Therefore: 80 ≤ b1 ≤ 120 The current basis remains optimal for 80 ≤ b1 ≤ 120, but the decision variable values and z-value will change. In a constraint with a positive slack (or positive excess) in an LPs optimal solution, if we change the rhs of the constraint to a value in the range where the basis remains optimal, the optimal solution to the LP remains the same.

6.1 – A Graphical Introduction to Sensitivity Analysis Shadow Prices - It is important to determine how a constraint’s rhs changes the optimal z-value. We define: The shadow price for the ith constraint of an LP is the amount by which the optimal z-value is improved (increased in a max problem or decreased in a min problem) if the rhs of the ith constraint is increased by one. This definition applies only if the change in the rhs of constraint i leaves the current basis optimal. Using the finishing constraint as an example, we know, 100 + D finishing hours are available (assuming the current basis remains optimal). The LP’s optimal solution is then x1 = 20 + D and x2 = 60 – D with z = 3x1 + 2x2 = 3(20 + D) + 2(60 - D) = 180 + D. Thus, as long as the current basis remains optimal, a one-unit increase in the number of finishing hours will increase the optimal z-value by $1. So, the shadow price for the first (finishing hours) constraint is $1.

6.1 – A Graphical Introduction to Sensitivity Analysis • The Importance of Sensitivity Analysis Sensitivity analysis is important because: • Values of LP parameters might change. If a parameter changes, sensitivity analysis shows it is unnecessary to solve the problem again. For example in the Giapetto problem, if the profit contribution of a soldier changes to $3.50, sensitivity analysis shows the current solution remains optimal. • Uncertainty about LP parameters. In the Giapetto problem for example, if the weekly demand for soldiers is at least 20, the optimal solution remains 20 soldiers and 60 trains. Thus, even if demand for soldiers is uncertain, the company can be fairly confident that it is still optimal to produce 20 soldiers and 60 trains.

6.2 – Some Important Formulas An LP’s optimal tableau can be expressed in terms of the LP’s parameters. The formulas developed in this section are used in the study of sensitivity analysis, duality, and advanced LP topics. Assume that we are solving a max problem that has been prepared for solution by the Big M method with the LP having m constraints and n variables. Although some of the variables may be slack, excess, or artificial, we choose to label them x1, x2, …,xn. The LP may then be written as shown. max z = c1x1 + c2x2 + … + cnxn s.t. a11x1 + a12x2 + … + a1nxn = b1 a21x1 + a22x2 + … + a2nxn = b2 …. …. … … … amx1 + am2x2 + … + amnxn = bm xi ≥ 0 (i = 1, 2, …, n)

6.2 – Some Important Formulas As an example, consider to the right the Dakota Furniture problem from Section 4.3 (without the x2 ≤ 5 constraint). Suppose we have found the optimal solution to the LP with the optimal tableau show to the right.

6.2 – Some Important Formulas • Define: BV = {BV1, BV2, …, BVn} to be the set of basic variables in the optimal tableau. NBV = {NBV1, NBV2, …, NBVn} the set of nonbasic variables in the optimal tableau. Dakota Problem xBV = vector listing the basic variables in the optimal tableau. xNBV = vector listing the nonbasic variables in the optimal tableau. cBV = row vector of the initial objective coefficients for the optimal tableau’s basic variables. cNBV = row vector of the initial objective coefficients for the optimal tableau’s nonbasic variables. Since BV = {s1,x3,x1}, cBV = {0 20 60} Since NBV = {x2,s2,s3}, cNBV = {0 20 60}

6.2 – Some Important Formulas Dakota Problem Define: B is an m x m matrix whose jth column is the column for BVj in the initial tableau. Aj is the column (in the constraints) for the variable xj. N is the m x (n-m) matrix whose columns are the columns for the nonbasic variables (in NBV order) in the initial tableau. NBV = {x2,s2,s3} for Dakota problem.

6.2 – Some Important Formulas Dakota Problem • Define: The m x 1 column vector b is the right-hand side of the constraints in the initial tableau. We can now use matrix algebra to determine how an LP’s optimal tableau (with the set of basic variables BV) is related to the original LP. z = cBVxBV + cNBVxNBV s.t. BxBV + NxNBV = b xBV, xNBV ≥ 0 We observe the Dakota LP may be written as:

6.2 – Some Important Formulas • Using the format on the previous slide, the Dakota problem is written: max z = s.t.

6.2 – Some Important Formulas • Multiplying the constraints through by B-1 yields: B-1BxBV + B-1NxNBV = B-1b orxBV + B-1NxNBV = B-1b Using the Gauss-Jordan method for the Dakota problem we know: B-1 = Substituting into xBV + B-1NxNBV = B-1b yields:

6.2 – Some Important Formulas • Conclusions: Column for xj in optimal tableau’s constraints = B-1aj Example: Column x2 in the Dakota optimal tableau = B-1a2 Right-hand side of optimal tableau’s constraints = B-1b Example:

6.2 – Some Important Formulas • Determining the Optimal Tableau’s Row 0 in Terms of BV and the Initial LP We multiple the constraints BxBV + NxNBV = b through by the vector cBVB-1. cBVxBV + cBVB-1NxNBV = cBVB-1b We know the original objective function: z - cBVxBV + cNBVxNBV = 0 Adding the two equations together and eliminating the optimal tableau’s basic variables we obtain: z + (cBVB-1N – cNBV) xNBV = cBVB-1b

6.2 – Some Important Formulas The coefficient of xj in row 0 is: cBVB-1(column of N for xj) – (coefficient of xj in cNBV) = cBVB-1aj - cj And the rhs of row 0 is cBVB-1b Letting be the coefficient of xj in the optimal tableau’s row 0 = cBVB-1aj - cj showing and the rhs of the optimal tableau’s row 0 = cBVB-1b

6.2 – Some Important Formulas • Formulas for computing the optimal tableau from the initial LP xj column in optimal tableau’s constraints = B-1aj Right-hand side of optimal tableau’s constraints = B-1b = cBVB-1aj - cj Coefficient of slack variable si in optimal row 0 = ith element of cBVB-1 Coefficient of excess variable ei in optimal row 0 = -(ith element of cBVB-1) Coefficient of artificial variable ai in optimal row 0 = (ith element of cBVB-1) + M (max problem) Right-hand side of optimal row 0 = cBVB-1b

6.3 – Sensitivity Analysis How do changes in an LP’s parameters (objective function coefficients, right-hand sides, and technological coefficients) change the optimal solution? Let BV be the set of basic variables in the optimal tableau. Given a change in an LP, determine if the BV remains optimal. From Chapter 4 we know the simplex tableau (for a max problem) for a set of basic variables is optimal if and only if each constraint has a nonnegative rhs and each variable has a nonnegative coefficient. Whether a tableau is feasible and optimal depends only upon the rhs of the constraints and the objective function coefficients of each variable in row 0 For example,, if an LP has variables x1, x2, …, x6 , the tableau to the right would be optimal. z + x2 + x4 + x6 = 6 = 1 = 2 = 3 This tableau’s optimality is not affected by parts of the tableau that are omitted.

6.3 – Sensitivity Analysis Suppose we have solved an LP and have found the BV is an optimal basis. Use the following procedure to determine if any change in the LP will cause the BV to no longer be optimal. Step 1 Using the formulas of Section 6.2 determine how changes in the LP’s parameters change the right hand side row 0 of the optimal tableau (the tableau having BV as the set of basic variables). Step 2 If each variable in row 0 has a nonnegative coefficient and each constraint has a nonnegative rhs, BV is still optimal. Otherwise, BV is no longer optimal. If BV is no longer optimal, find the new optimal solution by using the Section 6.2 formulas to recreate the entire tableau for BV and then continuing the simplex algorithm with the BV tableau as your starting tableau.

6.3 – Sensitivity Analysis • There can two reasons why a change in an LP’s parameters cause BV to no longer be optimal: • A variable (or variables) in row 0 may have a negative coefficient. In this case, a better (larger z-value) bfs can be obtained by pivoting in a nonbasic variable with a negative coefficient in row 0. If this occurs, the BV is now a suboptimal basis. • A constraint (or constraints) may now have a negative rhs. In this case, at least one member of BV will now be negative and BV will no longer yield a bfs. If this occurs, we say they BV is now an infeasible basis.

6.3 – Sensitivity Analysis Six types of changes in an LP’s parameters change the optimal solution: • Changing the objective function coefficient of a nonbasic variable. • Changing the objective function coefficient of a basic variable. • Changing the right-hand side of a constraint. • Changing the column of a nonbasic variable. • Adding a new variable or activity. • Adding a new constraint.

6.4 – The 100% Rule 100% Rule for Changing Objective Function Coefficients Depending on whether the objective function coefficient of any variable with a zero reduced cost in the optimal tableau is changed, there are two cases to consider: Case 1 – All variables whose objective function coefficients are changed have nonzero reduced costs in the optimal row 0. In Case 1, the current basis remains optimal if and only if the objective function coefficient for each variable remains within the allowable range given on the LINDO printout. If the current basis remains optimal, both the values of the decision variables and objective function remain unchanged. If the objective coefficient for any variable is outside the allowable range, the current basis is no longer optimal. Case 2 – at least one variable whose objective function coefficient is changed has a reduced cost of zero.

6.4 – The 100% Rule The 100% Rule for Changing Right-Hand Sides Case 1 – All constraints whose right-hand sides are being modified are nonbinding constraints. In case 1, the current basis remains optimal if and only if each right-hand side remains within its allowable range. Then the values of the decision variables and optimal objective function remain unchanged. If the right-hand side for any constraint is outside its allowable range, the current basis is no longer optimal. Case 2 – At least one of the constraints whose right-hand side is being modified is a binding constraint (that is, has zero slack or excess).

6.5 – Finding the Dual of an LP Associated with any LP is another LP called the dual. Knowledge of the dual provides interesting economic and sensitivity analysis insights. When taking the dual of any LP, the given LP is referred to as the primal. If the primal is a max problem, the dual will be a min problem and visa versa. Define the variables for a max problem to be z, x1, x2, …,xn and the variables for a min problem to be w, y1, y2, …, yn. Finding the dual to a max problem in which all the variables are required to be nonnegative and all the constraints are ≤ constraints (called normal max problem) is shown on the next slide.

6.5 – Finding the Dual of an LP max z = c1x1+ c2x2 +…+ cnxn s.t. a11x1 + a12x2 + … + a1nxn ≤ b1 a21x1 + a22x2 + … + a2nxn ≤ b2 … … … … am1x1 + am2x2 + … + amnxn ≤ bm xj ≥ 0 (j = 1, 2, …,n) Normal max problem It’s dual mim w = b1y1+ b2y2 +…+ bmym s.t. a11y1 + a12y2 + … + am1ym ≥ c1 a21y1 + a22y2 + … + am2ym ≥ c2 … … … … a1ny1 + a2ny2 + …+ amnym ≥ cn yi ≥ 0 (i = 1, 2, …,m) Normal min problem It’s dual

6.5 – Economic Interpretation of the Dual Problem • Interpreting the Dual of a the Dakota (Max) Problem The primal is: max z = 60x1 + 30x2 + 20x3 s.t. 8x1 + 6x2 + x3 ≤ 48 (Lumber constraint) 4x1 + 2x2 + 1.5x3 ≤ 20 (Finishing constraint) 2x1 + 1.5x2 + 0.5x3 ≤ 8 (Carpentry constraint) x1, x2, x3 ≥ 0 The dual is: min w = 48y1 + 20y2 + 8y3 s.t. 8y1 + 4y2 + 2y3 ≥ 60 (Desk constraint) 6y1 + 2y2 + 1.5y3 ≥ 30 (Table constraint) y1 + 1.5y2 + 0.5y3 ≥ 20 (Chair constraint) y1, y2, y3 ≥ 0

6.5 – Economic Interpretation of the Dual Problem The dual is: min w = 48 y1 + 20y2 +8y3 s.t. 8y1 + 4y2 + 2y3 ≥ 60 (Desk constraint) 6y1 + 2y2 + 1.5y3 ≥ 30 (Table constraint) y1 + 1.5y2 + 0.5y3 ≥ 20 (Chair constraint) y1, y2, y3 ≥ 0 Relevant information about the Dakota problem dual is shown below.

6.5 – Economic Interpretation of the Dual Problem The first dual constraint is associated with desks, the second with tables, and the third with chairs. Decision variable y1 is associated with lumber, y2 with finishing hours, and y3 with carpentry hours. Suppose an entrepreneur wants to purchase all of Dakota’s resources. The entrepreneur must determine the price he or she is willing to pay for a unit of each of Dakota’s resources. To determine these prices we define: y1 = price paid for 1 board ft of lumber y2 = price paid for 1 finishing hour y3 = price paid for 1 carpentry hour The resource prices y1, y2, and y3 should be determined by solving the Dakota dual.

6.5 – Economic Interpretation of the Dual Problem The total price that should be paid for these resources is 48 y1 + 20y2 + 8y3. Since the cost of purchasing the resources is to minimized: min w = 48y1 + 20y2 + 8y3 is the objective function for the Dakota dual. In setting resource prices, the prices must be high enough to induce Dakota to sell. For example, the entrepreneur must offer Dakota at least $60 for a combination of resources that includes 8 board feet of lumber, 4 finishing hours, and 2 carpentry hours because Dakota could, if it wished, use the resources to produce a desk that could be sold for $60. Since the entrepreneur is offering 8y1 + 4y2 + 2y3 for the resources used to produce a desk, he or she must chose y1, y2, and y3 to satisfy: 8y1 + 4y2 + 2y3 ≥ 60

6.5 – Economic Interpretation of the Dual Problem Similar reasoning shows that at least $30 must be paid for the resources used to produce a table. Thus y1, y2, and y3 must satisfy: 6y1 + 2y2 + 1.5y3 ≥ 30 Likewise, at least $20 must be paid for the combination of resources used to produce one chair. Thus y1, y2, and y3 must satisfy: y1 + 1.5y2 + 0.5y3 ≥ 20 The solution to the Dakota dual yields prices for lumber, finishing hours, and carpentry hours. In summary, when the primal is a normal max problem, the dual variables are related to the value of resources available to the decision maker. For this reason, dual variables are often referred to as resource shadow prices.

Presentation: H. Sarper 22 October 2013: go to page 26, section 6.5