Unformatted text preview: 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 behaviour 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) z = c1 x1 + c2 x2 + ... + cn xn subject to the constraints a11 x1 + a12 x2 + ... + a1n xn = b1 a21 x1 + a22 x2 + ... + a2 n xn = b2 . am1 x1 + am 2 x2 + ... + amn xn = bm x1 , x2 ,..., xn ≥ 0 The RHS constants of the constraints, b1 , b2 ,..., bm are referred to resources or availabilities of the problem. The objective coefficients, c1 , c2 ,..., cn are referred to as unit profits (or unit costs). The decision variables, x1 , x2 ,..., xn 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 yi associated with the constraint i. (Did you understand why it is called “dual price”?). Other nonsuggestive names include shadow prices and simplex multipliers. Reduced cost of a variable xj (=activity j) is defined as Cost of consumed resources per unit of activity xj  profit per unit of activity xj = zj  cj Changes affecting feasibility The feasibility of the current optimum solution may be affected only if (1) 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. First we consider the changes in the optimal solution due to changes in the RHS bi. We note that the optimal solution is given by xB = B b
where b is the old RHS and B is the basic matrix. Remember B1 is found in the optimal tableau below the entries which initially had an identity submatrix. −1 The optimal value is given by z = C B B b When b is changed to b′ , the corresponding new solution and new objective value are got by replacing b with b′ . −1 Example 4.32 (Pages 135136) 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 x1, x2, and x3 represent the daily number of units assembled of trains, trucks and cars, the LPP model is: Maximize subject to z = 3 x1 + 2 x2 + 5 x3
x1 + 2 x2 + x3 ≤ 430 3 x1 + 2 x3 ≤ 460 ≤ 420 x1 + 4 x2 x1 , x2 , x3 ≥ 0 The optimal tableau is given in the next slide. (Note: x4, x5, and x6 are slack variables there.) Basic z z 1 x2 x3 x6 0 0 x1 4 3/2 2 x2 0 1 0 0 x3 0 0 1 0 x4 1 0 2 x5 2 1/2 1 x6 0 0 0 1 Sol 1350 100 230 20 0 1/4 1/2 1/4 This is the optimal tableau Suppose that TOYCO wants to change the capacities of the three operations according to the following cases: 460 500 300 500 (b) 400 (c) 800 (a) 400 600 200 450 700 (d) 350 Use Sensitivity analysis to determine the optimal solution in each case. Solution: We note 1 / 2 − 1 / 4 0 0 −1 C B = [ 2 5 0] ; B = 1 / 2 0 − 2 1 1 (b) New Solution is x2 1 / 2 − 1 / 4 0 500 150 x = B −1b′ = 0 1 / 2 0 400 = 200 3 x6 − 2 1 1 600 0 Since this is feasible, it is optimal and the new optimal value = 3 × 0 + 2 ×150 + 5 × 200 = 1300
(c) New Solution is x2 1 / 2 − 1 / 4 0 300 − 50 x = B −1b′ = 0 1 / 2 0 800 = 400 3 x6 − 2 400 1 1 200 This is not feasible. So we apply dual simplex method to restore feasibility. We note new z = 3 ×0 + 2 × −50 + 5 × 400 =1900 Basic z z 1 x2 x3 x6
z x1 4 x2 0 1 0 0 8 4 2 4 x3 0 0 1 0 0 0 1 0 x4 1 x5 2 x6 0 0 0 1 0 0 0 1 Sol 1900 50 400 400 1500 200 300 200 0 1/4 0 3/2 0 1 0 0 0 2 2 1 1 1 1/2 1/4 0 1/2 2 5 2 1 0 1 0 1 0 0 x5 x3 x6 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, bi, is to determine the range for which the current solution remains feasible. For example if, in the TOYCO model, b2 is changed to b2+D2= 460+D2, we want to find the range of D2 so that the current solution remains optimal. When b2 is changed to b2+D2= 460+D2, the new solution is 1 / 2 − 1 / 4 0 430 x2 460 + D x = B −1b′ = 0 1 / 2 0 2 3 − 2 x6 1 1 420 100 − 1 / 4 D2 230 + 1 / 2 D = 2 20 D2 (current optimal solution + D2 times the 2nd column of B1.) 100 − 1 / 4 D2 230 + 1 / 2 D = 2 20 + D2 is feasible if 100 − 1 / 4 D2 ≥ 0 or D2 ≤ 400 230 + 1 / 2 D2 ≥ 0 or D2 ≥ −460 20 + D2 ≥ 0 or D2 ≥ −20 − 20 ≤ D2 ≤ 400 Or Thus current solution remains optimal if RHS of the 2nd 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:
Resource Resistor Capacitor Chips Unit Profit($) Unit Resources Requirements Model 1 Model2 Maximum Availability (units) (units) (units) 2 2 0 3 3 1 4 4 1200 1000 800 Let x1, x2 be the amounts produced of Models 1 and 2 respectively. Then the above model becomes the LPP Maximize z = 3 x1 + 4 x2 subject to 2 x1 + 3 x2 ≤ 1200 2 x1 + x2 ≤ 1000 4 x2 ≤ 800 x1 , x2 ≥ 0 (Resistors) (Capacitors) (Chips) Taking the slack variables as s1, s2, s3, the optimal tableau is: Basic z z 1 x1 s3 x2 0 0 0 x1 0 1 0 0 x2 0 0 0 1 s1 s2 5/4 1/4 1/4 3/4 2 2 1/2 1/2 s3 0 0 1 0 Sol 1750 450 400 100 This is the optimal tableau (a) Determine the status of each resource Answer: Since s1 = 0 = s2, the resistor and the capacitor resources are scarce. Since s3 > 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 y1, y2, y3 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 D1 is the increase in the resource 1, the new optimal solution is given by x1 −1 / 4 3 / 4 0 1200 + D1 s = −1 B b′ = − 2 2 1 1000 3 x2 1 / 2 −1 / 2 0 800 450 −1 / 4 D1 = 400 + − 2 D1 100 1 / 2 D1 ≥ 0 gives − 200 ≤ D1 ≤ 200 Similar calculations show that, for a change D2 in the capacitor, the range of feasibility is given by − 200 ≤ D2 ≤ 200 And for a change D3 in the chips, the range of feasibility is given by D3 ≥ −400 (d) If the available number of resistors is increased to 1300 units, find the new optimum solution. The new solution is: Yes, as D1 = 100 425 This is feasible 450 − 25 x1 s = 400 + − 200 = 200 and hence 3 x2 150 optimal. 100 50 And z = 1875. (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? ...
View
Full
Document
 Spring '11
 ritadubey
 Optimization, Resistor, BMW Sports Activity Series, optimal solution, optimal tableau

Click to edit the document details