Download presentation

Presentation is loading. Please wait.

Published byKeith Parrick Modified about 1 year ago

1
INFO 631 Prof. Glenn Booker Week 7 – Chapters INFO631 Week 7

2
Break-Even Analysis Ch. 19 INFO631 Week 7 Slides adapted from Steve Tockey – Return on Software 2

3
Present Economy “Present Economy”: decision techniques not involving time-value of money –Decision made based on 1 or more decision variables and 1 or more objective functions –Techniques: Break-even (Ch 19) –Finds value of the decision variable where performance is identical between alternatives Optimization (Ch 20) –Finds value of the decision variable(s) with the best performance. 3INFO631 Week 7

4
Break-Even Analysis Outline The situation Decision variables and objective functions Break-even with two alternatives Break-even with three alternatives General case break-even analysis 4INFO631 Week 7

5
The Situation – Chicago Company Chicago-based software project team needs.Net training but hasn’t decided how many people need it Team finds reputable Los Angeles-based training company Chicago project manager has two options –Send people to LA for training Cost is $1620 per person for tuition, travel, expenses, … –Hire instructor to come to Chicago Cost is $17,975 including fee, instructor travel, & expenses At what point is it better to have the instructor come to Chicago instead of sending team members to LA? Essence of break-even analysis: knowing the break-even point, deciding is real easy Question to answer: You’re the manager of a Chicago-based SW team that needs this training. How to best use your project $? 5INFO631 Week 7

6
Decision Variables and Objective Functions Decision variable –Set of possible values for some choice in a decision analysis E.g., the number of people that get.Net training 6INFO631 Week 7

7
Decision Variables and Objective Functions Objective function –Equation relating values of the decision variable to performance of an alternative CostInLA = $1620 * #People CostInChicago = $17,975 –Types Income function: –Relates values of decision variables to income –Income function income Cost Function –Relates values of decision variables to cost –Cost function cost 7INFO631 Week 7

8
Break-Even With Two Alternatives Find value of decision variable where objective functions are equal Algebraic solution –Set the objective functions equal to each other and solve for the decision variable 8INFO631 Week 7

9
Break-Even With Two Alternatives –Example Know CostInLA = $1620 * #People and CostInChicago = $17,975 Set CostInLA = CostInChicago $1620 * #People = $17,975 #People = $17,975 / $1620 = 11.1 So what does this mean: If <= 11 people need training, send them to LA If > 11 people need training, have instructor come to Chicago 9INFO631 Week 7

10
Break-Even With Two Alternatives Graphical solution –Plot the objective functions and find the intersection #People Cost Chicago LA $0 $10K $20K Break-even point 10INFO631 Week 7

11
Break-Even With Three Alternatives Adding another alternative makes the analysis more complicated But basically still straight forward –Graphical –Algebraic Solution 11INFO631 Week 7

12
Example - Add option to Chicago Company New Option for Chicago Company: –LA training company makes an offer involving a session in Denver CostInDenver = $ $925 * #People –To Solve Easiest to see graphically –See next slide –Always look for the solution with the lowest cost! 12INFO631 Week 7

13
Break-Even With Three Alternatives Graphical Solution – New Option Added #People Cost Chicago LA $0 $10K $20K Denver CostInDenver = $ $925 * #People Break-even Denver-LA Break-even Denver- Chicago 13INFO631 Week 7

14
Break-Even With Three Alternatives How interpret? New option is viable for a middle case –If <= 7 people need training, send them to LA –If 8 to 13 people need training, send them to Denver –If > 14 people need training, have instructor come to Chicago 14INFO631 Week 7

15
Break-Even With Three Alternatives If the Denver option were –Unlikely, but possible--same break-even as Chicago:LA, Denver only viable at exactly 11 people CostInDenver = $10,000 + $725 * #People #People Cost Chicago LA $0 $10K $20K Denver 15INFO631 Week 7

16
Break-Even With Three Alternatives If the Denver option were –CostInDenver = $7,500 + $1300 * #People –Also a possibility: Denver always worse than LA or Chicago, it’s never viable #People Cost Chicago LA $0 $10K $20K Denver 16INFO631 Week 7

17
Break-Even With Three Alternatives, Algebraic Solution Solve LA:Chicago –CostInLA = $1620 * #People CostInChicago = $17,975 –CostInLA = CostInChicago –$1620 * #People = $17,975 –#People = $17,975 / $1620 = 11 Is Denver better or worse than LA:Chicago? –Cost at LA:Chicago break-even (11 people) is $17,975 –CostInDenver for 11 people = $ $925 * 11 = $15,175 –Denver is better than LA:Chicago, discard LA:Chicago 17INFO631 Week 7

18
Break-Even With Three Alternatives, Algebraic Solution Solve LA:Denver Is Chicago better or worse than LA:Denver? CostInLA = $1620 * #People CostInDenver = $5,000 + $925* #People CostInLA = CostInDenver $1620 * #People = $5,000 + $925 * #People ($1620 * #People) – ($925 * #People) = $5,000 $695 * #People = $5000 #People = $5000 / $695 = 7 Cost at LA:Denver break-even (7 people) is $11,475 CostInChicago for 11 people = $17,975 LA:Denver is better than Chicago, keep LA:Denver 18INFO631 Week 7

19
Break-Even With Three Alternatives, Algebraic Solution Solve Chicago:Denver Is LA better or worse than Chicago:Denver? CostInChicago = $17,975 CostInDenver = $5,000 + $925* #People CostInChicago = CostInDenver $17,975 = $5,000 + $925 * #People $12,975 = $925 * #People #People = $12,975 / $925 = 14 Cost at Chicago:Denver break-even (14 people) is $17,975 CostInLA for 14 people = $22,680 Chicago:Denver is better than LA, keep Chicago:Denver 19INFO631 Week 7

20
Break-Even With Three Alternatives, Algebraic Solution Sort the valid break-even points by increasing decision variable value Reason about the segments At 7 people, LA:Denver break-even At 14 people, Chicago:Denver break-even Denver must be best between 7 and 14 people, so LA must be best below 7, and Chicago must be best above 14 20INFO631 Week 7

21
General Case Break-Even 1.Calculate the break-even point for each pair of objective functions 1.With n functions, there will be (n*(n-1))/2 candidates 2.Discard all break-even points that are: 1.Dominated by any other objective function at that value of the decision variable 2.Outside the reasonable range of the decision variable (e.g., too-low values and too-high values) 3.Sort the remaining, valid break-even points in order of increasing decision variable value 4.Reason about the segments 1.When the same objective function is in consecutive break-even points, it’s the best between those points 21INFO631 Week 7

22
Key Points Break-even analysis chooses between two or more alternatives by figuring out which points, if any, would be indifferent between those alternatives A decision variable represents a set of possible values for some choice An objective function is an equation relating values of decision variables to performance of an alternative To find break-even points with two alternatives, set the objectives functions equal to each other and solve for the value of the decision variable where that happens –The graphical approach finds the intersection on a graph of the objective functions With three or more alternatives, break-even points between each pair need to be considered –Some points may be dominated by other alternatives and will need to be discarded 22INFO631 Week 7

23
Optimization Analysis Ch. 20 INFO631 Week 7 Slides adapted from Steve Tockey – Return on Software 23

24
Optimization Analysis Outline Introducing optimization One alternative, one decision variable Multiple alternatives, one decision variable One alternative, multiple decision variables Multiple alternatives, multiple decision variables 24INFO631 Week 7

25
Introducing Optimization Find the point where overall performance is most favorable Useful when an objective function has 2 or more competing components –One component increases with the decision variable –Other component decreases See also Ch 11 – Economic Life - Graph 25INFO631 Week 7

26
Introducing Optimization Can be applied to maximizing an income function –Finding max point on an income function rather then min point on a cost function –Use same techniques Just look for min point rather than max point 26INFO631 Week 7

27
Introducing Optimization Two ways to solve –Algebraic (elegant) and Uses differential calculus –Min/Max when 1 st derivative = 0 –Graphical (brute-force) Run multiple sample values for decision variable through function and narrow in on best result Methods –One Alternative, One Decision Variable Simplest –Multiple Alternatives, Single Decision Variable –Single Alternative, Multiple Decision Variables –Multiple Alternatives, Multiple Decision Variables 27INFO631 Week 7

28
One Alternative, One Decision Variable Performance = F( Decision variable ) Two ways to solve –Algebraic (elegant) and Uses differential calculus –Min/Max when 1 st derivative = 0 –Graphical (brute-force) Run multiple sample values for decision variable through function and narrow in on best result 28INFO631 Week 7

29
One Alternative, One Decision Variable Example: distributed application needs queuing buffer –More messages/packet reduces network overhead but increases average queue time per message –Overall queuing delay described by the cost function: TD is total delay in milliseconds r is number of messages queued per packet 18r component is average queuing delay 648/r component is variable network overhead 21 is fixed network overhead –What is optimum messages/packet? 29INFO631 Week 7

30
One Alternative, One Decision Variable: Algebraic Solution Find the first derivative Set the first derivative equal to zero and solve for the decision variable (on next slide) The optimum messages/packet = 6 30INFO631 Week 7

31
One Alternative, One Decision Variable: Algebraic Solution solve for the decision variable The optimum messages/packet = 6 and TD = 237 ms 31INFO631 Week 7

32
One Alternative, One Decision Variable: Graphical Solution Graphical solution (brute force) –Plot the objective function, then find maximum/minimum point –Note: In general, there can be several local min/max points on function Overall: Min = P Max = Q R & S = local min/max 32INFO631 Week 7

33
One Alternative, One Decision Variable: Graphical Solution r TD R = 6 33INFO631 Week 7

34
Multiple Alternatives, Single Decision Variable Multiple Alternatives and their functions are driven by the same, single decision –Cost 1 = F 1 ( Decision variable ) –Cost 2 = F 2 ( Decision variable ) – ….. 34INFO631 Week 7

35
Multiple Alternatives, Single Decision Variable Find where each alternative has optimum performance –Use single alternative, single decision variable techniques Use Graphical (brute force) or Algebraic (elegant) General Approach –Find the performance of each alternative at its optimum point –Select the alternative with the best performance at its optimum point –Choose alternate with best value at optimal point 35INFO631 Week 7

36
Multiple Alternatives, Single Decision Variable Graphical solution Alternative A1 has optimum point at P Alternative A2 has optimum point at Q A2 at Q is cheaper than A1 at P, so choose A2 and run it at Q 36INFO631 Week 7

37
Single Alternative, Multiple Decision Variables Performance = F( v1, v2, … ) –V1 = variable 1 –V2 = variable 2, Etc. Algebraic (elegant) solution –Use multiple differentiation Differential calculus Graphical (brute force) solution –Plot the surface and look for extremes –Systematic search algorithm –Monte Carlo analysis (Ch 24) 37INFO631 Week 7

38
Systematic Search of Decision Variable Space LowestDV1 := some selected value of DecisionVariable1 LowestDV2 := some selected value of DecisionVariable2 LowestDV3 := some selected value of DecisionVariable3 LowestCost = CostFunction ( LowestDV1, LowestDV2, LowestDV3 ) while DV1 runs over the range of DecisionVariable1 while DV2 runs over the range of DecisionVariable2 while DV3 runs over the range of DecisionVariable3 if CostFunction ( DV1, DV2, DV3 ) < LowestCost then LowestCost = CostFunction ( DV1, DV2, DV3 ) LowestDV1 := DV1 LowestDV2 := DV2 LowestDV3 := DV3 38INFO631 Week 7

39
Systematic Search of Decision Variable Space When systematic search completed –Optimum point will be (close to) LowestDV1, LowestDV2, LowestDV3 and Have the LowestCost Not much help when search space is very big –Too many combinations –Use Monte Carlo analysis (Ch 24) 39INFO631 Week 7

40
Multiple Alternatives, Multiple Decision Variables Multiple Alternatives and their functions are driven by the multiple decision variables –Performance 1 = F 1 ( v1, v2, … ) –Performance 2 = F 2 ( v1, v2, … ) –…. V1 = variable 1 V2 = variable 2, Etc. 40INFO631 Week 7

41
Multiple Alternatives, Multiple Decision Variables Same approach as multiple alternatives, single decision variable –Find where each alternative has optimum performance Use single alternative, multiple decision variable techniques –Find the performance of each alternative at its optimum point –Select the alternative with the best performance at its optimum point 41INFO631 Week 7

42
Key Points Optimization analysis is useful when objective functions have competing components: balance components to find where overall performance is best Two ways of finding optimum point on a single alternative with a single decision variable –Algebraic uses differential calculus –Graphical uses computed values To optimize more than one performance function with a single decision variable, first find optimum for each then select the one with best performance at its optimum point Two ways of finding optimum point on a single alternative with a multiple decision variables –Algebraic uses differential calculus –Graphical uses computed values Optimizing multiple performance functions with multiple decision variables is just like optimizing multiple alternatives with a single decision variable 42INFO631 Week 7

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google