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## INFO 631 Prof. Glenn Booker

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**INFO 631Prof. Glenn Booker**Week 7 – Chapters 19-20 INFO631 Week 7**Break-Even Analysis**Ch. 19 Slides adapted from Steve Tockey – Return on Software INFO631 Week 7**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. INFO631 Week 7**Break-Even AnalysisOutline**• The situation • Decision variables and objective functions • Break-even with two alternatives • Break-even with three alternatives • General case break-even analysis INFO631 Week 7**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 $? INFO631 Week 7**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 INFO631 Week 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 INFO631 Week 7**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 INFO631 Week 7**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 INFO631 Week 7**Break-Even With Two Alternatives**• Graphical solution • Plot the objective functions and find the intersection LA $20K Chicago Cost $10K Break-even point $0 #People INFO631 Week 7**Break-Even With Three Alternatives**• Adding another alternative makes the analysis more complicated • But basically still straight forward • Graphical • Algebraic Solution INFO631 Week 7**Example - Add option to Chicago Company**• New Option for Chicago Company: • LA training company makes an offer involving a session in Denver • CostInDenver = $5000 + $925 * #People • To Solve • Easiest to see graphically • See next slide • Always look for the solution with the lowest cost! INFO631 Week 7**Break-Even With Three Alternatives**• Graphical Solution – New Option Added CostInDenver = $5000 + $925 * #People Break-even Denver-LA LA $20K Chicago Cost Break-even Denver-Chicago $10K Denver $0 #People INFO631 Week 7**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 INFO631 Week 7**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 LA $20K Chicago Cost $10K Denver $0 #People INFO631 Week 7**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 $20K Chicago Cost Denver $10K LA $0 #People INFO631 Week 7**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 = $5000 + $925 * 11 = $15,175 • Denver is better than LA:Chicago, discard LA:Chicago INFO631 Week 7**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 INFO631 Week 7**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 INFO631 Week 7**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 INFO631 Week 7**General Case Break-Even**• Calculate the break-even point for each pair of objective functions • With n functions, there will be (n*(n-1))/2 candidates • Discard all break-even points that are: • Dominated by any other objective function at that value of the decision variable • Outside the reasonable range of the decision variable (e.g., too-low values and too-high values) • Sort the remaining, valid break-even points in order of increasing decision variable value • Reason about the segments • When the same objective function is in consecutive break-even points, it’s the best between those points INFO631 Week 7**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 INFO631 Week 7**Optimization Analysis**Ch. 20 Slides adapted from Steve Tockey – Return on Software INFO631 Week 7**Optimization AnalysisOutline**• Introducing optimization • One alternative, one decision variable • Multiple alternatives, one decision variable • One alternative, multiple decision variables • Multiple alternatives, multiple decision variables INFO631 Week 7**Introducing Optimization**See also Ch 11 – Economic Life - Graph • 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 INFO631 Week 7**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 INFO631 Week 7**Introducing Optimization**• Two ways to solve • Algebraic (elegant) and • Uses differential calculus • Min/Max when 1st 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 INFO631 Week 7**One Alternative, One Decision Variable**• Performance = F( Decision variable ) • Two ways to solve • Algebraic (elegant) and • Uses differential calculus • Min/Max when 1st derivative = 0 • Graphical (brute-force) • Run multiple sample values for decision variable through function and narrow in on best result INFO631 Week 7**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? INFO631 Week 7**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 INFO631 Week 7**One Alternative, One Decision Variable: Algebraic Solution**• solve for the decision variable • The optimum messages/packet = 6 and TD = 237 ms INFO631 Week 7**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 INFO631 Week 7**One Alternative, One Decision Variable: Graphical Solution**r TD 1 687 2 381 3 291 4 255 5 241 6 237 7 240 8 246 9 255 10 266 11 278 12 291 13 305 14 319 15 334 R = 6 INFO631 Week 7**Multiple Alternatives, Single Decision Variable**• Multiple Alternatives and their functions are driven by the same, single decision • Cost1 = F1( Decision variable ) • Cost2 = F2( Decision variable ) • ….. INFO631 Week 7**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 INFO631 Week 7**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 INFO631 Week 7**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) INFO631 Week 7**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 INFO631 Week 7**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) INFO631 Week 7**Multiple Alternatives, Multiple Decision Variables**• Multiple Alternatives and their functions are driven by the multiple decision variables • Performance1 = F1( v1, v2, … ) • Performance2 = F2( v1, v2, … ) • …. • V1 = variable 1 • V2 = variable 2, Etc. INFO631 Week 7**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 INFO631 Week 7**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 INFO631 Week 7