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# aaoczc2252

## by Perumal_Gopi42 on Nov 02, 2007

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Keep Smiling & Mailing ...

Keep Smiling & Mailing
With Warm Regards,
P.Gopi,
Inspection Engr,
CHEM P&S - QA&I,
L&T, E&C Div, Powai
Mumbai - 400072.
Extn - 4279 / 3049
Phone: +9122 6705 4279
Fax: +9122 6705 1383 / 1890

Keep Smiling & Mailing
With Warm Regards,
P.Gopi,
Inspection Engr,
CHEM P&S - QA&I,
L&T, E&C Div, Powai
Mumbai - 400072.
Extn - 4279 / 3049
Phone: +9122 6705 4279
Fax: +9122 6705 1383 / 1890

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## aaoczc2252Presentation Transcript

• Sensitivity Analysis The optimal solution of a LPP is based on the conditions that prevailed at the time the LP model was formulated and solved. In the real world, the decision environment rarely remains static and it is essential to determine how the optimal solution changes when the parameters of the model are changed. That is what sensitivity analysis does. It provides efficient computational techniques to study the dynamic behavior of the optimal solution resulting from making changes in the parameters of the model.
• In studying the sensitivity analysis, we should be familiar with the lingo that is being used in LPP situations. A general LPP is of the form Maximize (or Minimize) subject to the constraints
• The RHS constants of the constraints, are referred to resources or availabilities of the problem. The objective coefficients, are referred to as unit profits (or unit costs). The decision variables, are referred to as units of activities 1, 2, …, n .
• Dual Price of a constraint This measure actually represents the unit worth of a resource - that is it gives the contribution to the objective function resulting from a unit increase or decrease in the availability of a resource. In terms of duality theory, the dual price of a resource (=constraint) i, is precisely the value of the optimal dual variable y i associated with the constraint i . (Did you understand why it is called “dual price”?). Other non-suggestive names include shadow prices and simplex multipliers .
• Reduced cost of a variable x j (=activity j) is defined as Cost of consumed resources per unit of activity x j - profit per unit of activity x j = z j - c j
• Changes affecting feasibility
• The feasibility of the current optimum solution may be affected only if
• The RHS of the constraints are changed
• OR
• (2) A new constraint is added to the model.
• In both cases, infeasibility occurs when at least one element of the RHS of the optimal tableau becomes negative – that is, one or more of the current basic variables become negative.
• is changed to First we consider the changes in the optimal solution due to changes in the RHS b i . We note that the optimal solution is given by where b is the old RHS and B is the basic matrix. Remember B -1 is found in the optimal tableau below the entries which initially had an identity submatrix. The optimal value is given by When , the corresponding new solution and new objective value are got by replacing with .
• Example 4.3-2 (Pages 135-136) TOYCO assembles three types of toys: trains, trucks and cars using three operations. The daily limits on the available times for the three operations are 430, 460, and 420 minutes respectively. The profits per toy train, truck and car are \$3, \$2, and \$5 respectively. The assembly times per train at the three operations are 1, 3, and 1 minute respectively. The corresponding times per truck and per car are (2, 0, 4) and (1, 2, 0) minutes respectively. A zero indicates that the operation is not used.
• Letting x 1 , x 2 , and x 3 represent the daily number of units assembled of trains, trucks and cars, the LPP model is: Maximize subject to The optimal tableau is given in the next slide. ( Note: x 4 , x 5 , and x 6 are slack variables there.)
• x6 0 2 0 0 -2 1 1 20 x3 0 3/2 0 1 0 1/2 0 230 z 1 4 0 0 1 2 0 1350 Basic z x1 x2 x3 x4 x5 x6 Sol x2 0 -1/4 1 0 1/2 -1/4 0 100 This is the optimal tableau
• Suppose that TOYCO wants to change the capacities of the three operations according to the following cases:
• Use Sensitivity analysis to determine the optimal solution in each case. Solution: We note ( b ) New Solution is
• Since this is feasible, it is optimal and the new optimal value = ( c ) New Solution is This is not feasible. So we apply dual simplex method to restore feasibility. We note new z =
• x6 0 2 0 0 -2 1 1 400 x3 0 3/2 0 1 0 1/2 0 400 z 1 4 0 0 1 2 0 1900 Basic z x1 x2 x3 x4 x5 x6 Sol x2 0 -1/4 1 0 1/2 -1/4 0 -50 x6 0 1 4 0 0 0 1 200 x5 0 1 -4 0 -2 1 0 200 z 1 2 8 0 5 0 0 1500 x3 0 1 2 1 1 0 0 300 This is the new optimal tableau.
• Feasibility Range of the Elements of the RHS Another way of looking at the effect of changing the availabilities of the resources, b i , is to determine the range for which the current solution remains feasible. For example if, in the TOYCO model, b 2 is changed to b 2 + D 2 = 460+ D 2 , we want to find the range of D 2 so that the current solution remains optimal.
• When b 2 is changed to b 2 + D 2 = 460+ D 2 , the new solution is (current optimal solution + D 2 times the 2 nd column of B -1 .)
• is feasible if Or Thus current solution remains optimal if RHS of the 2 nd constraint lies between 440 and 860 (the other RHSs being the same).
• Problem 5 Problem Set 4.5B Page 151 HiDec produces two models of electronic gadgets that use resistors, capacitors, and chips. The following table summarizes the data of the situation: Unit Resources Requirements Resource Model 1 Model2 Maximum Availability (units) (units) (units) Resistor 2 3 1200 Capacitor 2 1 1000 Chips 0 4 800 Unit Profit(\$) 3 4
• Let x 1 , x 2 be the amounts produced of Models 1 and 2 respectively. Then the above model becomes the LPP Maximize subject to (Resistors) (Capacitors) (Chips) Taking the slack variables as s 1 , s 2 , s 3 , the optimal tableau is:
• x2 0 0 1 1/2 -1/2 0 100 s3 0 0 0 - 2 2 1 400 z 1 0 0 5/4 1/4 0 1750 Basic z x1 x2 s1 s2 s3 Sol x1 0 1 0 -1/4 3/4 0 450 This is the optimal tableau
• (a) Determine the status of each resource Answer: Since s 1 = 0 = s 2 , the resistor and the capacitor resources are scarce. Since s 3 > 0, the chips resource is abundant. (b) In terms of the optimal profit, determine the worth of one resistor, one capacitor and one chip. Answer: They are respectively y 1 , y 2 , y 3 the dual optimal solution and hence are 5/4, 1/4, 0 respectively.
• (c) Determine the range of applicability of the dual prices for each resource. Resistor: If D 1 is the increase in the resource 1, the new optimal solution is given by  0 gives
• Similar calculations show that, for a change D 2 in the capacitor, the range of feasibility is given by And for a change D 3 in the chips, the range of feasibility is given by (d) If the available number of resistors is increased to 1300 units, find the new optimum solution. The new solution is: This is feasible and hence optimal. And z = 1875. Yes, as D 1 = 100
• (g) A new contractor is offering to sell HiDec additional resistors at 40 cents each but only if HiDec would purchase at least 500 units. Should HiDec accept the offer?
• Addition of a new constraint
• The addition of a new constraint to an existing model can lead to one of two cases:
• The new constraint is redundant , meaning that it is satisfied by the current optimal solution and hence can be dropped altogether from the model altogether.
• The current solution violates the new constraint in which case the dual simplex method can be used to recover feasibility.
• We introduce this constraint in the final simplex tableau as an additional row where the usual additional (slack or surplus) variable is taken as the basic variable for this new row. Because this new row will probably have non-zero coefficients for some of the other basic variables, the conversion to proper form for Gaussian elimination is done and then reoptimization is done in the usual way. The following example illustrates this.
• Example: Consider the LPP Maximize subject to Applying the Simplex method, the optimal tableau is given in the next slide. Now a new constraint solution. is added. Find the new optimal
• x2 0 0 1 - 3/2 0 - 1/2 1/2 5 x1 0 1 0 1/2 0 1/2 1/2 15 z 1 0 0 3/2 0 3/2 1/2 25 Basic z x1 x2 x3 s1 s2 s3 Sol s1 0 0 0 1 1 - 1 - 2 10 s4 0 3 -2 1 0 0 0 1 28 s4 0 0 0 0 z 1 0 0 0 0 3/7 2/7 3/7 22 x2 0 0 1 0 0 4/7 5/7 -3/7 8 x1 0 1 0 0 0 1/7 3/7 1/7 14 s1 0 0 0 0 1 - 12/7 -15/7 2/7 8 x3 0 0 0 1 0 5/7 1/7 -2/7 2 0 0 0 -7/2 0 - 5/2 -1/2 1 -7
• Changes affecting Optimality
• Changes in the objective coefficients c j
• The changes in the objective coefficients, c j , affect &quot; z j - c j &quot; and hence the optimality. We recompute the z-Row by replacing c j with c' j . We note that C B will also change to C' B .
In the previous problem, we replace the objective function with . Find the new optimal solution.
• Thus c 1 is changed from 2 to 3; c 2 is changed from -1 to -2; and c 3 is changed from 1 to 3; We note that in z-Row, the coefficients of basic variables x 1 , x 2 , and x 4 will remain zero now also. We have only to calculate the new coefficients of the non-basic variables only and the new z. Coefficient of x 3 = z 3 – c 3
• Coefficient of x 5 = z 5 – c 5 Coefficient of x 6 = z 6 – c 6
• Since all z j – c j are  0, the current solution remains optimal. Of course the new z is If in the previous problem, we replace the objective function with You can verify that z 6 – c 6 = -1/2 < 0. Thus optimality is spoiled. By applying the regular Simplex method, we can show that the new optimum solution is x 1 = 10, x 2 = 0, x 3 = 0 with new z = 20.
• Addition of a new activity (= Changes in the coefficients of an existing activity) Suppose a new activity n+1 is added with coefficients
• To find the effect of this on the current optimal solution, we pretend this activity was present initially with all coefficients zero. Hence the new coefficients are calculated using the formulae: In z-row, the coefficient of x n +1 is And in the constraint matrix its coefficients are If z n+1 – c n+1 satisfies the optimality condition, the current solution is optimal. Else we apply regular simplex method to restore optimality.
• The same procedure is adopted when the coefficients of an existing variable are changed. If the variable ia basic variable, we should see that the new tableau is in proper form (i.e. coefficients of other basic variables in that column should be made zero).
• In the previous problem the coefficients of the (non-basic) variable x 3 are changed from to Using sensitivity analysis, find the new optimal solution and value. The only change in the optimal tableau will be the x 3 column. We calculate the new x 3 column and the coefficient of x 3 in the z -row.
• We first note that Hence the new x 3 column is
• New coefficient of x 3 in z -row = Thus optimality is disturbed. We replace the x 3 column in the original optimal tableau by the new values and then find the new optimal solution and optimal value by regular simplex method.
• x2 0 0 1 0 - 1/2 1/2 5 x1 0 1 0 0 1/2 1/2 15 z 1 0 0 0 3/2 1/2 25 Basic z x1 x2 x3 s1 s2 s3 Sol s1 0 0 0 1 - 1 - 2 10 z 1 0 0 0 1/4 5/4 0 55/2 x2 0 0 1 0 1/4 -3/4 0 15/2 x1 0 1 0 0 1/12 5/12 1/3 95/6 x3 0 0 0 1 1/6 - 1/6 -1/3 5/3 -3/2 6 -1/2 -3/2
• In the previous problem a new variable x 7 is introduced with coefficients Using sensitivity analysis, find the new optimal solution and value. This is like the previous case. We assume the variable x 7 was already present with all coefficients 0.
• We first note that Hence the new x 7 column is
• New coefficient of x 7 in z -row = Hence the original solution remains optimal with the same objective value and it does not help by introducing this new variable.
• In the previous problem the coefficients of the (basic) variable x 1 are changed from to Using sensitivity analysis, find the new optimal solution and value. The only change in the optimal tableau will be the x 1 column. We calculate the new x 1 column and the coefficient of x 1 in the z -row.
• We first note that Hence the new x 1 column is
• New coefficient of x 1 in z -row = We replace the x 1 column in the original optimal tableau by the new values.
• x2 0 1 -3/2 0 - 1/2 1/2 5 x1 0 0 1/2 0 1/2 1/2 15 z 1 0 3/2 0 3/2 1/2 25 Basic z x1 x2 x3 s1 s2 s3 Sol s1 0 0 1 1 - 1 - 2 10 z 1 0 1/2 0 0 1/2 0 5 s3 0 0 1 -1 0 0 1 20 x1 0 1 -1/2 1 0 1/2 0 5 s1 0 0 2 -1 1 - 1 0 50 2 0 1 -1 1 0 0 1/2 0 1/2 -1/2 -5 0 0 1 -1 0 0 1 20
• Problem 5 Problem Set 4.5B Page 151 HiDec produces two models of electronic gadgets that use resistors, capacitors, and chips. The following table summarizes the data of the situation: Unit Resources Requirements Resource Model 1 Model2 Maximum Availability (units) (units) (units) Resistor 2 3 1200 Capacitor 2 1 1000 Chips 0 4 800 Unit Profit(\$) 3 4
• Let x 1 , x 2 be the amounts produced of Models 1 and 2 respectively. Then the above model becomes the LPP Maximize subject to (Resistors) (Capacitors) (Chips) Taking the slack variables as x 3 , x 4 , x 5 , the optimal tableau is:
• (c) Determine the range of applicability of the dual prices for each resource. Resistors: If D 1 is the increase in the resource 1, the new optimal solution is given by  0 gives
• Similar calculations show that, for a change D 2 in the capacitor, the range of feasibility is given by And for a change D 3 in the chips, the range of feasibility is given by (d) If the available number of resistors is increased to 1300 units, find the new optimum solution. The new solution is: This is feasible and hence optimal. And z = 1875.
• (e) If the available number of chips is reduced to 350 units, will you be able to determine the new optimal solution directly from the given information? Explain. Answer: NO, as the feasibility is not spoilt only if the number of chips is (f) If the availability of capacitors is limited by the range of applicability computed in (c), determine the corresponding range of the optimal profit and the corresponding ranges for the number of units to be produced of Models 1 and 2.
• If the number of capacitors is changed from b 2 = 400 to b 2 = 400+ D 2 , where then Hence
• (g) A new contractor is offering to sell HiDec additional resistors at 40 cents each but only if HiDec would purchase at least 500 units. Should HiDec accept the offer? Answer: Since the new value of b 1 , the number of resistors is (at least) 1700, the feasibility is spoilt (as D 1 > 200). The new solution is (taking D 1 =500) And new z = 2375. We restore Optimality by Dual Simplex.
• Problem 9 Problem Set 4.5E Page 159
• Consider the HiDec model in Problem 5, Problem Set 4.5B
• Determine the conditions that will keep the current solution optimum if the unit profits of Models 1 and 2 are changed simultaneously.
• Suppose that the objective function is changed to
Determine the associated optimum solution.
• Solution. (a) Let c 1 = 3 be changed to c 1 =3+ d 1 . Let c 2 = 4 be changed to c 2 =4+ d 2 . Thus the new coefficient of the non-basic variable s 1 in the z -row is z 3 – c 3 for retaining optimality
• Thus the new coefficient of the non-basic variable s 2 in the z -row is z 4 – c 4 for retaining optimality These are the two conditions on d 1 and d 2 .
• Solution (b): Here new objective function is Thus d 1 = 5 - 3 =2 and d 2 = 2 - 4 = -2. New z 3 – c 3 = New z 4 – c 4 = Thus optimality is spoiled. And we reoptimize using regular Simplex method. Note also new z =2450
• Problem 3 Problem Set 4.5F Page 161 In the TOYCO model, suppose that a new toy (fire engine) requires 3, 2, and 4 minutes respectively, on operations 1, 2, and 3. Determine the new optimal solution when the profit per unit is given by (a) \$5 (b) \$10 (Optimal tableau in next page) Solution: This amounts to introducing a new variable x 7 with coefficients as follows.
• We first note that Hence the new x 7 column is
• New coefficient of x 7 in z -row = Thus in case (a) the current solution is optimal and it does not help to introduce the new toy. in case (a) And = -3 in case (b) In case (b), optimality is spoiled and we restore it using Regular simplex method. (see the next slide)
• Consider the LPP Maximize subject to The optimal tableau is given in the next slide.
• We first note that Hence the new x 1 column is
• New coefficient of x 1 in z -row = Thus optimality is spoiled. We try to restore it.