Week 7 – Chapters 19-20
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Ch. 19
Slides adapted from Steve Tockey – Return on Software
INFO631 Week 7 2 www.ischool.drexel.edu

• “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 3 www.ischool.drexel.edu

Break-Even Analysis
• 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 4 www.ischool.drexel.edu

• 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 5 www.ischool.drexel.edu

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 6 www.ischool.drexel.edu

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 7 www.ischool.drexel.edu

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 8 www.ischool.drexel.edu

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 9 www.ischool.drexel.edu

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 10 www.ischool.drexel.edu

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

• 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 12 www.ischool.drexel.edu

Break-Even With Three Alternatives
• Graphical Solution – New Option Added
CostInDenver = $5000 + $925 * #People
Break-even
Denver-LA
LA
$20K
Chicago
Cost
$10K
Denver
Break-even
Denver-
Chicago
$0
#People
INFO631 Week 7 13 www.ischool.drexel.edu

• 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 14 www.ischool.drexel.edu

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 15 www.ischool.drexel.edu

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
$10K
Denver
LA
$0
#People
INFO631 Week 7 16 www.ischool.drexel.edu

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 17 www.ischool.drexel.edu

Break-Even With Three Alternatives,
Algebraic Solution
• Solve 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
• Is Chicago better or worse than
LA:Denver?
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 18 www.ischool.drexel.edu

Break-Even With Three Alternatives,
Algebraic Solution
• Solve 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
• Is LA better or worse than
Chicago:Denver?
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 19 www.ischool.drexel.edu

Break-Even With Three Alternatives,
Algebraic Solution
• Sort the valid break-even points by increasing decision variable value
At 7 people, LA:Denver break-even
At 14 people, Chicago:Denver break-even
• Reason about the segments
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 20 www.ischool.drexel.edu

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
INFO631 Week 7 21 www.ischool.drexel.edu

• 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 22 www.ischool.drexel.edu

Ch. 20
Slides adapted from Steve Tockey – Return on Software
INFO631 Week 7 23 www.ischool.drexel.edu

Optimization Analysis
• 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 24 www.ischool.drexel.edu

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
Economic life
AE(i)
Costs
Total cost
Operating and maintenance cost
Capital recovery with return, CR(i)
Number of years the asset is kept
INFO631 Week 7 25 www.ischool.drexel.edu

• 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 26 www.ischool.drexel.edu

• 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
INFO631 Week 7 27 www.ischool.drexel.edu

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
INFO631 Week 7 28 www.ischool.drexel.edu

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

18r

648

21 r
• 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 29 www.ischool.drexel.edu

One Alternative, One Decision Variable:
Algebraic Solution
• Find the first derivative dTD dr

18

648 r 2
• Set the first derivative equal to zero and solve for the decision variable
(on next slide) dTD dr

18

648 r
2

0
• The optimum messages/packet = 6
INFO631 Week 7 30 www.ischool.drexel.edu

• solve for the decision variable r

648

6
18
• The optimum messages/packet = 6 and
TD = 237 ms
TD

18 * 6

648

21

237
6
INFO631 Week 7 31 www.ischool.drexel.edu

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 o r m a
P e r f n c e
Overall minimum
P
Overall maximum
Q R
Decision variable
S
Overall:
Min = P
Max = Q
R & S = local min/max
INFO631 Week 7 32 www.ischool.drexel.edu

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
700
600
500
400
300
200
100
0 r
R = 6
TD
INFO631 Week 7 33 www.ischool.drexel.edu

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

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 35 www.ischool.drexel.edu

Multiple Alternatives, Single Decision
Variable
• Graphical solution
Optimum point for A1
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
$
Cost for A1
Optimum point for A2
Cost for A2
P Q
Decision variable
INFO631 Week 7 36 www.ischool.drexel.edu

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 37 www.ischool.drexel.edu

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 38 www.ischool.drexel.edu

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 39 www.ischool.drexel.edu

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

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 41 www.ischool.drexel.edu

• 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 42 www.ischool.drexel.edu