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Sensitivity analysis continued… BSAD 30 Dave Novak Source: Anderson et al., 2013 Quantitative Methods for Business 12 th edition – some slides are directly.

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Presentation on theme: "Sensitivity analysis continued… BSAD 30 Dave Novak Source: Anderson et al., 2013 Quantitative Methods for Business 12 th edition – some slides are directly."— Presentation transcript:

1 Sensitivity analysis continued… BSAD 30 Dave Novak Source: Anderson et al., 2013 Quantitative Methods for Business 12 th edition – some slides are directly from J. Loucks © 2013 Cengage Learning

2 Overview Special case LP problems Alternative optimal solutions Infeasibility Unbounded problems Sensitivity analysis continued Non-intuitive shadow prices Example

3 Feasible region Recall – using a graphical solution procedure, the optimal solution to an LP is found at the extreme points to the feasible region The feasible region for a two-variable LP problem (x 1, x 2 ) can be nonexistent, a single point, a line, a polygon, or an unbounded area

4 Feasible region Any linear program falls in one of four categories: 1. has a unique optimal solution 2. has alternative optimal solutions 3. is infeasible 4. has an objective function that can be increased without bound (unbounded) A feasible region may be unbounded and yet still have optimal solutions This is common in minimization problems and is possible in maximization problems

5 Alternative optimal solution In some cases, an LP can have more than one optimal solution – this is referred to alternative optimal solutions There is more than one extreme point that minimizes or maximizes the OF value

6 Alternative optimal solution In the graphical method, if the objective function line is parallel to a boundary constraint in the direction of optimization, there are alternate optimal solutions All points on this line segment are optimal

7 Alternative optimal solution Consider the following LP Max 4x 1 + 6x 2 s.t. x 1 < 6 2x 1 + 3x 2 < 18 2x 1 + 3x 2 < 18 x 1 + x 2 < 7 x 1 + x 2 < 7 x 1 > 0 and x 2 > 0 x 1 > 0 and x 2 > 0

8 Alternative optimal solution example n Boundary constraint 2x 1 + 3x 2 < 18 (constraint #2) and OF line 4x 1 + 6x 2 are parallel. All points on line segment A – B are optimal solutions x1x1 x 2 76543217654321 1 2 3 4 5 6 7 8 9 10 Constraint #2: 2x 1 + 3x 2 < 18 Constraint #3: x 1 + x 2 < 7 Constraint #1: x 1 < 6 OF: Max 4x 1 + 6x 2 A B

9 Infeasibility No solution to the LP problem satisfies all the constraints, including the non-negativity conditions Graphically, this means a feasible region does not exist Causes include A formulation error has been made Management’s expectations are too high Too many restrictions have been placed on the problem (i.e. the problem is over-constrained)

10 Infeasibility Consider the following LP Max 2x 1 + 6x 2 s.t. 4x 1 + 3x 2 < 12 2x 1 + x 2 > 8 x 1, x 2 > 0

11 Infeasibility x1x1 Constraint #1: 4x 1 + 3x 2 < 12 Constraint #2: 2x 1 + x 2 > 8 2 4 6 8 10 4 8 2 6 10 x2x2 There are no points that satisfy both constraints, so there is no feasible region (and no feasible solution)

12 Unbounded The solution to a maximization LP problem is unbounded if the value of the solution may be made indefinitely large without violating any of the constraints This leads to an infinite production scenario For real problems, this is usually the result of improper formulation Quite likely, a constraint has been inadvertently omitted

13 Unbounded Consider the following LP Max 4x 1 + 5x 2 s.t. x 1 + x 2 > 5 3x 1 + x 2 > 8 x 1, x 2 > 0

14 Unbounded x1x1 Constraint #2: 3x 1 + x 2 > 8 Constraint #1: x 1 + x 2 > 5 Max 4x 1 + 5x 2 6 8 10 2 4 6 8 10 4 2 x2x2

15 Standard form of an LP Involves formulating each constraint as a strict equality and: ADDING a slack variable to ≤ (less than or equal to) constraints SUBTRACTING a surplus variable from ≥ (greater than or equal to) constraints Equality constraints do not have a slack or surplus variable

16 Standard form of an LP Max 5x 1 + 6x 2 s.t. x 1 < 6 2x 1 + 3x 2 ≥ 5 x 1 + x 2 < 10 x 1 + 5x 2 ≥ 2 x 1 > 0 and x 2 > 0

17 Standard form of an LP

18 Sensitivity analysis Recall that last class we discussed: Changes in objective function coefficients Changes in RHS values How to interpret shadow prices Sensitivity analysis involves changing only one coefficient at a time! All other coefficients are held constant

19 Changes in OF coefficients Change in OF coefficient does not impact the feasible region or the existing extreme points in any way, although it impacts the value associated with the optimal solution In a two-decision variable graphical problem, a change in an OF coefficient changes the slope of the OF line Regardless of how many decision variables are considered, changes in OF coefficients do not change the feasible region

20 Changes in OF coefficients Here we identified the range of optimality The current solution remains optimal for all changes to OF coefficients within this range The allowable increase and allowable decrease provide the range of values each OF coefficient can take on where the current optimal solution remains optimal Outside of this range, the slope of the OF line changes enough that a different extreme point in the feasible region becomes the new optimal solution

21 Changes in RHS constraint coefficients Change in RHS coefficients are a bit more tricky to interpret because changing the RHS of a constraint can impact both the feasible region and change the extreme points for the problem In a two-decision variable graphical problem, a change in RHS coefficient can increase or decrease the area of the feasible region

22 Changes in RHS constraint coefficients Here we identified the range of feasibility The allowable increase and allowable decrease associated with the shadow price for each constraint provides the range of values that each RHS coefficient can take on where the: Current binding constraints remain binding AND Interpretation of the shadow price is valid

23 Changes in RHS constraint coefficients Outside of the range of feasibility the interpretation of the shadow price does not hold, AND a different set of constraints become binding When the RHS of a constraint changes outside of the range of feasibility, you must re-formulate and re-solve the LP

24 Changes in the LHS constraint coefficients Classical sensitivity analysis provides no information about changes resulting from a change in the LHS coefficients in a constraint We must change the coefficient and rerun the model to learn the impact the change will have on the solution

25 Non-intuitive shadow prices Constraints with variables naturally on both the left-hand (LHS) and right-hand (RHS) sides often lead to shadow prices that have a non-intuitive explanation

26 Example of non-intuitive shadow price Recall from last class that Olympic Bike is introducing two new lightweight bicycle frames, the Deluxe (x 1 ) and the Professional (x 2 ), to be made from special aluminum and steel alloys The objective was to maximize total profit, subject to limits on the availability of aluminum and steel

27 Olympic bike example Consider the problem from Lecture 16 Max 10x 1 + 15x 2 s.t. 2x 1 + 4x 2 < 100 3x 1 + 2x 2 < 80 3x 1 + 2x 2 < 80 x 1 > 0 and x 2 > 0 x 1 > 0 and x 2 > 0 ObjectiveFunctionObjectiveFunction “Regular”Constraints“Regular”Constraints Non-negativity Constraints ConstraintsNon-negativity

28 Olympic bike example Let’s introduce an additional constraint The number of Deluxe frames produced (x 1 ) must be greater than or equal to the number of Professional frames produced (x 2 )

29 Olympic bike example Max 10x 1 + 15x 2 s.t. 2x 1 + 4x 2 < 100 3x 1 + 2x 2 < 80 3x 1 + 2x 2 < 80 x 1 > x 2 x 1 > x 2 x 1 > 0 and x 2 > 0 x 1 > 0 and x 2 > 0 ObjectiveFunctionObjectiveFunction “Regular”Constraints“Regular”Constraints Non-negativity Constraints ConstraintsNon-negativity

30 Olympic answer report

31 Olympic sensitivity report

32 Olympic bike example Interpret the shadow prices of: Constraint #1

33 Olympic bike example Interpret the shadow prices of: Constraint #2

34 Olympic bike example Interpret the shadow prices of: Constraint #3

35 Olympic bike example Interpret the range of optimality for OF coefficient: c 1

36 Olympic bike example Interpret the range of optimality for OF coefficient: c 2

37 Reduced costs The reduced cost of a variable is typically the shadow price of the corresponding non- negativity constraint So most, variables have a reduced cost = 0 x 1, x 2, etc. ≥ 0 We will discuss a situation when the reduced costs ≠ 0 in a four decision variable problem before the final exam

38 Summary Special case LP problems Alternative optimal solutions Infeasibility Unbounded problems Sensitivity analysis continued Non-intuitive shadow prices Example


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