Lecture 11 - Sensitivity

Lecture 11 - Sensitivity - 540:311 DETERMINISTIC MODELS IN...

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Unformatted text preview: 540:311 DETERMINISTIC MODELS IN OPERATIONS RESEARCH Lecture 11: Chapter 5 Class [email protected]: Mon Mar 27th 10:20‐11:40am Prof. W. Art Chaovalitwongse Recall from Chapter 5 ‐> 6.1 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. •  if b1 > 120, x1 will be greater than 40 and will violate the demand constraint. •  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. Recall from Chapter 5 ‐> 6.1 Shadow Prices - It is important to determine how a constraint’s rhs changes the optimal z-value. We define: The shadow price for the i th 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 i th constraint is increased by one. This definition applies only if the change in the rhs of constraint i leaves the current basis optimal. Finishing constraint: 100 + Δ finishing hours are available (assuming the current basis remains optimal). The LP s optimal solution is then x1 = 20 + Δ and x2 = 60 – Δ with z = 3x1 + 2x2 = 3(20 + Δ) + 2(60 - Δ) = 180 + Δ. Thus, as long as the current basis remains optimal, a one-unit increase in the number of finishing hours will increase the optimal zvalue by $1. So, the shadow price for the first (finishing hours) constraint is $1. 6.3 – SensiGvity 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 (maximization problem). 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 – SensiGvity 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 – SensiGvity Analysis There can be two reasons why a change in an LP s parameters cause BV to no longer be optimal: 1.  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. 2.  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 – SensiGvity Analysis Six types of changes in an LP s parameters change the optimal solution: 1.  Changing the objective function coefficient of a nonbasic variable. 2.  Changing the objective function coefficient of a basic variable. 3.  Changing the right-hand side of a constraint. 4.  Changing the column of a nonbasic variable. 5.  Adding a new variable or activity. 6.  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. Diet Problem Define: BR = # of Brownies IC = # of Scoops of Ice Cream COLA = # of Cans of Cola PC = # of Pieces of Pineapple MIN 50 BR + 20 IC + 30 COLA + 80 PC SUBJECT TO 400 BR + 200 IC + 150 COLA + 500 PC >= 500 (Calorie constraint) 3 BR + 2 IC >= 6 (Chocolate constraint) 2 BR + 2 IC + 4 COLA + 4 PC >= 10 (Sugar constraint) 2 BR + 4 IC + COLA + 5 PC >= 8 (Fat constraint) END OBJECTIVE FUNCTION VALUE 1) 90.00000 VARIABLE VALUE REDUCED COST BR 0.000000 27.500000 IC 3.000000 0.000000 COLA 1.000000 0.000000 PC 0.000000 50.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 250.000000 0.000000 3) 0.000000 ‐2.500000 4) 0.000000 ‐7.500000 5) 5.000000 0.000000 OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE BR 50.000000 INFINITY 27.500000 IC 20.000000 18.333334 5.000000 COLA 30.000000 10.000000 30.000000 PC 80.000000 INFINITY 50.000000 RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 2 500.000000 250.000000 INFINITY 3 6.000000 4.000000 2.857143 4 10.000000 INFINITY 4.000000 5 8.000000 5.000000 INFINITY Case 1 •  The price of a brownie increases to 60 cents •  A piece of pineapple cheesecake decreases to 50 cents 22.5 = 50‐27.5 ≤ cost of a brownie(60) ≤ 50+∞= ∞ 30 = 80‐50 ≤ cost of pineapple cheesecake(50) ≤ 80+∞= ∞ Case 2 •  A scoop of ice cream increases to 30 cents •  A can of cola decreases to 25 cents ∑r j ≤ 1? j rj = rj = Δc j Ij −Δc j Dj if Δc j ≥ 0 where Δc j = change in obj coefficient if Δc j ≤ 0 where I j , D j = max allowable increase/decrease 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 righthand 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). Case 1 •  The calorie requirement is decreased to 400 calories •  The fat requirement is increased to 15 oz. ‐∞ = 500‐∞ ≤ calorie requirement(400) ≤ 500+250= 750 ‐∞ = 8‐∞ ≤ fat requirement(15) ≤ 8+5= 13 Case 2 •  The chocolate requirement is increased to 10 oz. •  The sugar requirement is decreased to 5 oz. ∑r j ≤ 1? j rj = rj = Δb j Ij −Δb j Dj if Δb j ≥ 0 where Δb j = change in RHS if Δb j ≤ 0 where I j , D j = max allowable increase/decrease ...
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This note was uploaded on 09/20/2011 for the course ENG 300 taught by Professor Albin during the Fall '11 term at Rutgers.

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