Graphical Sensitivity Analysis and Computer Solution

Learning Objectives Utilize graphical sensitivity analysis to answer what-if questions for the objective function coefficients and the right-Hand side values Use a spreadsheet Solver and TORA to find optimal solution for an LP model Understand the result of the spreadsheet Solver and TORA and interpret them

Sensitivity Analysis Assumed that parameters are known with certainty Parameters include the objective function coefficients, constraint quantity values, and constraint coefficients Parameters are simply estimates Effect of a parameter on the solution Parameter changes and their effects on the model solution is known as sensitivity analysis

Changes in Objective Function Coefficients General format Z=c1X1+c2X2 ;Z=40X1+50X2 Change c1 from 40 to 100 Will make the objective function more steeper; point C will become optimal Change c2 from $50 to $100 Will make the objective function more flatter; point A optimal Z=100X1+50X2=$3000 X1=0 X2=20 Z=2000 X1=30 X2=0 Z=3000 A B Z=40X1+100X2=$2000 C

Range of Optimality Changing in ci called range of optimality Interested in a range of values for a coefficient Graphical solution Max Z = 40 x1 + 50 x2 s.t. x1 + 2 x2 ≤ 40 4 x1 +3 x2 ≤ 120 Slope of Z function is -4/5 x2 =Z/50 – (4/5)x1 A X1=24 chairs X2=8 tables Z=$ 1360 B C

Upper Limit Determination for c1 If slope increases to -4/3 Point C becomes optimal Line BC is optimal; parallel to 4x1+3x2 What objective function coefficient for x1 will make the objective function slope=-4/3? Determined as (-c1 /50)=(-4/3); c1 =66.67 Steeper slope gives upper limit Slope=-4/3 A B Slope=-4/5 C

Lower Limit Determination for c1 If slope decreases to -1/2 Point A becomes optimal AB is optimal; parallel to x1 + 2x2= 40; Slope is -1/2 What objective function coefficient for x2 will make the objective function slope=-1/2? (-c1 /50)=(-1/2); c1 =$25 Flatter slope gives lower limit Slope=-4/5 A Slope=-1/2 B C

Complete Sensitivity Range Complete sensitivity range: 25 ≤ c1 ≤ 66.67 The profit can vary between $25.00 and $66.67, optimal solution point (x1=24 and x2=8) will not change (can you verify it?) Z will change depending on value of c1 Same analysis can be used for c2

Sensitivity Range for c2 If slope decreases (upper limit of C2) to -1/2 Points on AB are optimal (-40 /C2) = (-1/2); C2=$80 If slope increases (lower limit) to -4/3 Points on BC are optimal (-40/ C2) = (-4/3); C2 = 30 Range 30 ≤ c2 ≤ 80 Profit can vary between $30 and $80 and the optimal solution (x1=24 and x2=8) will not change Slope=-4/5 A x2 = -Z/c2 -40/c2 x1 B Slope=-1/2 C

Simultaneous Changes Sensitivity range applies only if one coefficient is varied and the other one held constant c2: $30 ≤ profit for table ≤ $80, this is true only if c1 remains constant Simultaneous changes is overly complex and time-consuming using graphical analysis Graphical analysis is a tedious way to perform sensitivity analysis

Feasibility Range Sensitivity ranges for the right-hand side of constraints called feasibility range Recall Max Z = $40 x1 + 50 x2 s.t. x1 + 2 x2 + s1 = 40 (labor, hr) 4 x1 + 3 x2 + s2 = 120 (wood, Ib) Consider an increase in the availability of labor hours (from 40 to 60 hrs) without changing solution mix

New Feasible Space Feasible region changes from OABC to OA'B'C Point B' is the new optimal solution point instead of B Solution mix did not change even though the values of x1 and x2 did change (from x1 =24, x2=8 to x1 =12, x2 =24) X1=12 X2=24 Z=1680 A’ B’ A X1=24 X2=8 Z=1360 B O C

Upper Limit of Labor Constraint Increase RHS value to 80 hours (upper limit) Solution space changes to OA'C (A‘ optimal point) If we go beyond 80 hours, s1 increases X1=0 X2=40 S1=0 S2=0 Z=2000 A’ X1+2X2=80 A B X1+2X2=40 O C

Lower Limit of Labor Constraint Decrease the RHS value to 30 hours (lower limit) Feasible region is OA'C with optimal point is at C Feasibility range 30≤Labor hours≤80 A’ X1+2X2=30 X1+2X2=40 O C

Upper Limit for Wood Constraint Increase RHS of wood constraint from 120 to 160 lb Solution space is OAC’ with optimal solution at C’ If we go beyond 160 lb, s2 increases 4X1+3X2=160 4X1+3X2=120 A X1=40 X2=0 S1,S2=0 Z=1600 B O C C’

Lower Limit for Wood Constraint Decrease it to 60 lbs New feasible region ... Optimal point … Feasibility range 60 ≤ Pounds ≤160 X1=0 X2=20 S1,S2=0 Z=800 A 4X1+3X2=120 4X1+3X2=60 B C’ O C

Unit Worth a Resource Feasibility range provides a single measure called unit worth of a shadow or resource price Shows the rate of change in the optimal value as a result of making changes in the availability of a resource Increased labor constraint from 40 to 80 hours, Z increased from $1360 to $2000 Unit worth of labor constraint is $16 (2000-1360)/(80-40)=16 Shows a1-hour change in labor constraint will change Z by $16

Dual Price Represents the unit worth of a resource Gives the contribution to the objective function resulting from a unit increase or decrease in the availability of a resource Dual prices for labor and wood resources are $16/hour and $6/lb

RHS Sensitivity Analysis Classify the constraint as either binding or nonbinding. A binding constraint must pass through the optimum solution point. If it is not, it is nonbinding. A binding constraint represents a scarce resource, whereas a nonbinding constraint represents an abundant resource. An increase in a scarce resource can improve the value of the objective function

Binding and Nonbinding Example In the previous example, only constraints (1) and (2) are binding because they pass the optimum solution point Line (1) and (2) corresponds to material A and B, respectively. 1 C 2

Increasing RHS of Resource A Moving line (1) upward parallel to itself (increasing the RHS), will incorporate gradually portions of the triangle CDK. Solution space is AFEKB When point K is reached, constraint (2) and (4) becomes …, with the optimum solution occurring at K. Constraint (1) will become redundant 1 D Material A=7 tons K E 4 F Material A=6 tons C A 2 B

Question How much do we increase the level of material A? Steps: Determine point K 2x1 +x2 = 8 line (2) x2=2 line (4) x1=3 and x2=2 2. Substitute point K in the left-hand side of constraint (1) x1 +2x2 =3+2*2 =7 tons Z value: Z=3x1 +2x2 =3*3 +2*2=13

Class Problem Consider constraint (2). Determine how much to increase the level of this resource ? Solution space is AFED... Point … is the new optimal point Point … is the intersection of line … and line … the value of point … x2 =0 , x1 =6 Maximum allowable level of material B is 2x1 +x2 =2 (6) + 2(0) =12 Z value is … E D C 1 F 2 A B J

Decreasing Nonbinding Resource Constraint (4) (maximum limit on the demand for interior paint) is a nonbinding resource Lower it down until passes point C. Point C has x1 = 3 1/3 and x2 =1 1/3 Maximum limit on constraint (4) is, therefore, x2 = 1 1/3 without changes the current solution mix 4 C

Class Problem What is the maximum limit on constraint (3)? Constraint (… ) is a … resource, because it is not passing point … Maximum limit on constraint (3) is … Z=… C 3

Summary of the Sensitivity Problem 1 Resource Type Maximum change in level of resources Maximum change in revenue z 1 Scarce 7-6=1 13-12 2/3=1/3 2 12-8=4 18-12 2/3=5 1/3 3 Abundant -2-1=-3 12 2/3 -12 2/3=0 4 1 1/3-2=-2/3 12 2/3 - 12 2/3=0

Sensitivity Analysis-Which Resources? Under limited budgeted consideration, which resources should receive higher priorities? Let yi be the worth unit of resource i yi=(maximum change in optimum z)/(maximum allowable increase in resource) For constraint (1) or material A y1=(13- 12 2/3)/(7-6)=1/3 per ton (in $1000) For constraint (2) y2=(18 – 12 2/3)/(12-8)=4/3 per ton (in $1000)

Summary of the Worth per Unit for All Resources Type Value of yi 1 2 3 4 Scarce Abundant y1 =1/3 y2 =4/3 y3 =0 y4 =0 Which resources should receive higher priority?

Sensitivity Analysis-Sensitivity Problem 3 How much change in the objective function coefficients? Will change the slope of the objective function line Will change the set of binding and nonbinding resources

Sensitivity Problem 3- Two Types of Situation By how much can a coefficient be changed (increased or decreased) without changing the optimal solution point? By how much can a coefficient be changed (increased or decreased) to reverse the status of a given resource?

Case #1-Without Changing the Optimal Solution An increase in c1 or a decrease in c2 causes Z to rotate in a clockwise direction A decrease in c1 or an increase in c2 causes Z to rotate in counterclockwise direction Point C will remain optimum as long as the slope of Z varies between the slopes of constraints (1) and (2) Points D and C are optimal when Z line coincides with constraint (1) This is also true for points C and B Decrease c1 Increase c2 D E C F A B Increase c1 Decrease c2

Determining Allowable Range for c1 c1 can be changed until Z coincides with line (2) or decreased until coincides with line (1) Slope of Z =c1x1+2x2 is c1/2 Slope of line 1 (x1+2x2=6) is ½ Slope of line 2 (2x1+x2=8) is 2 Minimum value of c1 c1/2=1/2;or c1=1 Maximum value of c1 c1/2=2;or c1=4 Range of c1; 1c1  4 When c1=1, either C or D is optimal D C 1 2

Class Problem What is the range for c2 Slope of z=3x1+c2x2 is … Slope of line 1 is … Slope of line 2 is … Minimum value of c2=… Maximum value of c2=…

Computer Solution of the LP Model Simplex method was used to solve LP Problems Then computer was used to solve LP models using steps of simplex method Computers popularized the LP technique More than a dozen software that solve LP problems Solve LP problems using Excel and TORA

Solution by Excel and TORA Excel is more time-consuming than TORA However the excel spreadsheet can be used as a template for programs TORA is a menu-driven software Not require user manual

Sensitivity Analysis with Excel (continued)