MAN231 OPERATION RESEARCH

Analisis sensitivitas (Analisis postoptimal) Josef H. Nudu, josef@mail.uajy.ac.id ,josef_nudu@yahoo.com 0815 7882 19712009

Sensitivity AnalysisKondisi optimal yang diperoleh dari iterasi didasarkan pada model dan kondisi saat model dibuat.Untuk itu perlu dilakukan langkah lanjutan untuk mengetahui sensitivitas/ketangguhan solusi yang diperoleh terhadap perubahan koefisien variabel dan nilai kendala.Karena itu, analisis lanjutan yang dilakukan disebut analisis sensitivitas atau analisis postoptimal (setelah kondisi optimal diperoleh).

Akibat yang mungkin terjadiPerubahan koefisien fungsi objectifBisa merubah kondisi optimalitasPerubahan nilai kendalaPerubahan RHSPerubahan koefisien LHSBisa merubah fisibilitasPenambahan kendala baruBisa merubah fisibilitasPengurangan kendalaBisa merubah fisibilitas

Perhatikan bahwa untuk setiap tablo, bentuk yang diperoleh adalahZ 1cBB-1A ccBB-1 cBB-1bxB 0B-1AB-1B-1bA = matrik kendala awalB = matrik basisXn = matrik kolomXB = matrik kolom basiscB = matrik koefisien basis pada fungsi tujuanb = matrik nilai kendala

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 formMaximize (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 constraintThis 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 non-suggestive 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 feasibilityThe 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 bi. We note that the optimal solution is given bywhere 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 byWhen, the correspondingnew solution and new objective value are got by replacingwith .

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 x1, x2, and x3 represent the daily number of units assembled of trains, trucks and cars, the LPP model is:Maximizesubject toThe optimal tableau is given in the next slide. (Note: x4, x5, and x6 are slack variables there.)

x6 0 2 0 0 -2 1 1 20x3 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 Solx2 0 -1/4 1 0 1/2 -1/4 0 100This 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 isThis is not feasible. So we apply dual simplex method to restore feasibility. We notenew z =

x6 0 2 0 0 -2 1 1 400x3 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 Solx2 0 -1/4 1 0 1/2 -1/4 0 -50x6 0 1 4 0 0 0 1 200x5 0 1 -4 0 -2 1 0 200 z 1 2 8 0 5 0 0 1500x3 0 1 2 1 1 0 0 300This is the new optimal tableau.

Feasibility Range of the Elements of the RHSAnother 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 (current optimal solution + D2 times the 2nd column of B-1.)

is feasible if 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 151HiDec produces two models of electronic gadgets that use resistors, capacitors, and chips. The following table summarizes the data of the situation:Unit Resources RequirementsResource Model 1 Model2 Maximum Availability(units) (units) (units) Resistor 2 3 1200Capacitor 2 1 1000Chips 0 4 800Unit Profit($) 3 4

Let x1, x2 be the amounts produced of Models 1 and 2 respectively. Then the above model becomes the LPPMaximizesubject to(Resistors)(Capacitors)(Chips)Taking the slack variables as s1, s2, s3, the optimal tableau is:

x2 0 0 1 1/2 -1/2 0 100s3 0 0 0 - 2 2 1 400 z 1 0 0 5/4 1/4 0 1750 Basic z x1 x2 s1 s2 s3 Solx1 0 1 0 -1/4 3/4 0 450This is the optimal tableau

(a) Determine the status of each resourceAnswer: 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 0 gives

Similar calculations show that, for a change D2 in the capacitor, the range of feasibility is given byAnd for a change D3 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 D1 = 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 constraintThe 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 LPPMaximizesubject toApplying 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 5x1 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 Sols1 0 0 0 1 1 - 1 - 2 10s4 0 3 -2 1 0 0 0 1 28s4 0 0 0 0 z 1 0 0 0 0 3/7 2/7 3/7 22x2 0 0 1 0 0 4/7 5/7 -3/7 8x1 0 1 0 0 0 1/7 3/7 1/7 14s1 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 OptimalityChanges in the objective coefficients cj The changes in the objective coefficients, cj , affect "zj - cj" and hence the optimality. We recompute the z-Row by replacing cj with c'j . We note that CB will also change to C'B. In the previous problem, we replace the objective function with .Find the new optimal solution.

Thus c1 is changed from 2 to 3; c2 is changed from -1 to -2; and c3 is changed from 1 to 3; We note that in z-Row, the coefficients of basic variables x1 , x2 , and x4 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 x3 = z3 c3

Coefficient of x5 = z5 c5Coefficient of x6 = z6 c6

Since all zj cj 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 z6 c6 = -1/2 < 0. Thus optimality is spoiled. By applying the regular Simplex method, we can show that the new optimum solution is x1 = 10, x2 = 0, x3 = 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 xn+1 is And in the constraint matrix its coefficients are If zn+1 cn+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 x3 are changed fromtoUsing sensitivity analysis, find the new optimal solution and value.The only change in the optimal tableau will be the x3 column. We calculate the new x3 column and the coefficient of x3 in the z-row.

We first note thatHence the new x3 column is

New coefficient of x3 in z-row = Thus optimality is disturbed. We replace the x3 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 5x1 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 Sols1 0 0 0 1 - 1 - 2 10 z 1 0 0 0 1/4 5/4 0 55/2x2 0 0 1 0 1/4 -3/4 0 15/2x1 0 1 0 0 1/12 5/12 1/3 95/6x3 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 x7 is introduced with coefficientsUsing sensitivity analysis, find the new optimal solution and value.This is like the previous case. We assume the variable x7 was already present with all coefficients 0.

We first note thatHence the new x7 column is

New coefficient of x7 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 x1 are changed fromtoUsing sensitivity analysis, find the new optimal solution and value.The only change in the optimal tableau will be the x1 column. We calculate the new x1 column and the coefficient of x1 in the z-row.

We first note thatHence the new x1 column is

New coefficient of x1 in z-row = We replace the x1 column in the original optimal tableau by the new values.

x2 0 1 -3/2 0 - 1/2 1/2 5x1 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 Sols1 0 0 1 1 - 1 - 2 10 z 1 0 1/2 0 0 1/2 0 5s3 0 0 1 -1 0 0 1 20x1 0 1 -1/2 1 0 1/2 0 5s1 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

Terima Kasih

AKA Sensitivity AnalysisQ: How might an existing constraint be changed? - RHS coefficient changes; different machinery could be more or less productive; new design could use fewer parts; Q: Why might a new constraint be added? Preference for particular customers; minimum quantity of a particular product needed; Q: Why might the objective function change? Decreased profitability due to increased competition (or increased profitability due to demand outstripping supply); Next class well see how easy it is to deal with such changes!