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Engineering Economy
IPE - 3115
Course Teacher
Ridwan Mustofa
Lecturer
Dept of Industrial Engineering and Management
Khulna University of Engineering & Technology
Sensitivity
• Sensitivity is concerned with variability
• Variance associated with input parameters impact the output variable the
most
• The MARR as a parameter
• Interest rates and other interest factors tend to be more stable from project to project
• The analyst can limit the range over which these type of parameters vary
Determining Sensitivity to Parameter Variation
• A parameter is a variable or factor for which an estimate or stated value is
required to conduct the analysis at hand.
• Examples:
• P, F, A;
• i, n;
• Future costs, salvages, etc.
• Sensitivity analysis
• Seeks to determine what parameters matter most in an economic analysis
Visualizing the Impact of Parameters
• Plot the PW, AW, or ROR vs. input parameters
• Steps
• Pre-select the desired input parameters
• Select the probable range and increment of variation for each
parameter
• Select the measure of worth
• Compute the results for each parameter
• Graphically display the results by plotting the parameter vs. the
measure of worth
Problem 1
Wild Rice, Inc. expects to purchase a new asset for automated rice handling. Most
likely estimates are a first cost of $80,000, zero salvage value, and a cash flow
before taxes (CFBT) per year t that follows the relation $27,000 − 2000t. The
MARR for the company varies over a wide range from 10% to 25% per year for
different types of investments. The economic life of similar machinery varies from
8 to 12 years. Evaluate the sensitivity of PW by varying
(a) MARR, while assuming a constant n value of 10 years
Solution :
Taking MARR interval on 5 % that is 10%, 15%, 20%, 25%
Now,
PW10 = -80,000 + 25000 (P/A, 10%, 10) – 2000 (P/G, 10%, 10)
= -80,000 + 25,000 * 6.1446 – 2000 * 22.8913
= $ 27,832
PW15 = -80,000 + 25000 (P/A, 15%, 10) – 2000 (P/G, 15%, 10)
= -80,000 + 25,000 * 5.0188 – 2000 * 16.9795
= $ 11,511
PW20 = -80,000 + 25000 (P/A, 20%, 10) – 2000 (P/G, 20%, 10)
= -80,000 + 25,000 * 4.1925 – 2000 * 12.8871
= $ -962
PW25 = -80,000 + 25000 (P/A, 25%, 10) – 2000 (P/G, 25%, 10)
= -80,000 + 25,000 * 3.5705 – 2000 * 9.9870
= $ -10,711.5
Problem 2
A company planning to borrow $10.5 million for a plant expansion is
not sure what the interest rate will be when it applies for the loan. The
rate could be as low as 10% per year or as high as 12% per year for a 5-
year loan. The company will only move forward with the project if the
annual worth of the expansion is below $5.7 million. The M&O cost is
fixed at $3.1 million per year. The salvage could be $2 million if the
interest rate is 10% or $2.5 million if it is 12% per year. Is the decision
to move forward with the project sensitive to the interest rate and
salvage value estimates?
Solution
Required AW < $5.7 million
10%:
AW= -10,500,000(A/P,10%,5) – 3,100,000 + 2,000,000(A/F,10%,5)
= -10,500,000(0.26380) – 3,100,000 + 2,000,000(0.16380)
= $-5,542,300 (< $-5,700,000)
12%:
AW = -10,500,000(A/P,12%,5) – 3,100,000 + 2,500,000(A/F,12%,5)
= -10,500,000(0.27741) – 3,100,000 + 2,500,000(0.15741)
= $-5,619,280 (< $-5,700,000)
The decision is not sensitive since both AW values are below $5.7 million.
Problem 3
An engineer collected average cost and revenue data for Arenson’s FC1 handheld
financial calculator.
Fixed cost $300,000 per year
Cost per unit $40
Revenue per unit $70
(a) What is the range in breakeven quantity if there is possible variation in the fixed
cost from $200,000 to $400,000 per year? (Use $50,000 increments.)
(b) What is the incremental change in the breakeven quantity for each $50,000
change in fixed cost?
Solution
(a)
Q = FC/(70-40)
= FC/30
_ FC, $ QBE, units
200,000 6667
250,000 8333
300,000 10,000
350,000 11,667
400,000 13,333
(b) The change in QBE is 1667 units for each $50,000 increase in FC.
Formalized Sensitivity Analysis Using Three
Estimates
• Given an input parameter of interest
• Provide three estimates for that parameter
• A pessimistic estimate, P
• A most likely estimate, ML
• An optimistic estimate, O
• Note: This approach comes from PERT/CPM analysis and is based upon
the beta distribution
Three Estimates: Example 18.3
• Three alternatives (A, B, C) with 4 Parameters
• First cost, salvage value, AOC, and life
• For each parameter we formulate
Parameter
P pessimistic estimate
ML most likely estimate
O optimistic estimate
Problem 4
An engineer is evaluating three alternatives for new equipment at Emerson
Electronics. She has made three estimates for the salvage value, annual operating
cost, and life. The estimates are presented on an alternative-by-alternative basis in
Table. For example, alternative B has pessimistic estimates of S = $500,
AOC = $4000, and n = 2 years. The first costs are known, so they have the same
value. Perform a sensitivity analysis and determine the most economical alternative,
using AW analysis at a MARR of 12% per year.
Strategy First Cost SV AOC Life
Alt A.
P -20,000 0 -11,000 3
ML -20,000 0 -9,000 5
O -20,000 0 -5,000 8
Alt. B
P -15,000 500 -4,000 2
ML -15,000 1,000 -3,500 4
O -15,000 2,000 -2,000 7
Alt. C
P -30,000 3,000 -8,000 3
ML -30,000 3,000 -7,000 7
O -30,000 3,000 -3,500 9
AWB-ML = -15000 (A/P, 12%,4) + 1000 (A/F, 12%,4)
-3500
= -15000 * 0.32923 + 1000 * 0.20923
– 3500
= $-8229.22
Similarly, Find all the Annual worth of 3 alternatives
and 3 possibilities.
Ex. 18.3
$0
$5,000
$10,000
$15,000
$20,000
$25,000
1 2 3 4 5 6 7 8 9
Life - Years
A.
Cost
Alt A
Alt B
Alt C
Economic Variability and The Expected Value
• Expected Value
• Long-run average based upon occurrence and probability of
occurrence
• Definition of Expected Value
1
( ) ( )
m
i i
i
E X X P X
=
= 
Xi = value of the variable X for i from 1 to m different values
P(Xi) = probability that a specific value of X will occur
Subject to:
1
( ) 1.0
m
i
i
P X
=
=

ANA airlines plans to offer several new electronic services on flights between
Tokyo and selected European destinations. The marketing director estimates that
for a typical 24-hour period there is a 50% chance of having a net cash flow of
$5000 and a 35% chance of $10,000. He also estimates there is a small 5% chance
of no cash flow and a 10% chance of a loss of $1000, which is the estimated extra
personnel and utility costs to offer the services. Determine the expected net cash
flow.
Solution
Let NCF be the net cash flow in dollars, and let P (NCF) represent the associated
probabilities.
E (NCF) = 5000*(0.5) + 10,000*(0.35) + 0*(0.05) + 1000*(0.1)
= $5900
Problem:
Lite-Weight Wheelchair Company has a substantial investment in tubular steel bending equipment. A
new piece of equipment costs $5000 and has a life of 3 years. Estimated cash flows depend on
economic conditions classified as receding, stable, or expanding. A probability is estimated that each
of the economic conditions will prevail during the 3-year period. Apply expected value and PW
analysis to determine if the equipment should be purchased. Use a MARR of 15% per year.
Solution:
The PW values for the three scenarios are
PW R = -5000 + 2500( P/F ,15%,1) + 2000( P/F ,15%,2) + 1000( P/F ,15%,3)
= -5000 + 4344
= -$656
PWS = -5000 + 5708
= $708
PW E = -5000 + 6309
=$1309
Only in a receding economy will the cash flows not return the 15% to justify the investment.
The expected present worth is
E(PW) = 656*(0.4) + 708*(0.4) + 1309*(0.2)
=$283
At 15%, E (PW)> 0; the equipment is justified, using an expected value analysis.
For Practice from EXERCISE of the Book Engineering Economy
by Leeland and Blank (7th Edition) :
Sensitivity to Parameter Variation: 18.4,18.5,18.6,18.7,18.9,18.12,18.13
Three Estimates: 18.19,18.20,18.21,18.22
Expected Values: 18.24,18.26, 18.32,18.33,18.34
Decision Trees:

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Chapter 18 - Sensitivity Analysis.pdf

  • 1. Engineering Economy IPE - 3115 Course Teacher Ridwan Mustofa Lecturer Dept of Industrial Engineering and Management Khulna University of Engineering & Technology
  • 2. Sensitivity • Sensitivity is concerned with variability • Variance associated with input parameters impact the output variable the most • The MARR as a parameter • Interest rates and other interest factors tend to be more stable from project to project • The analyst can limit the range over which these type of parameters vary
  • 3. Determining Sensitivity to Parameter Variation • A parameter is a variable or factor for which an estimate or stated value is required to conduct the analysis at hand. • Examples: • P, F, A; • i, n; • Future costs, salvages, etc. • Sensitivity analysis • Seeks to determine what parameters matter most in an economic analysis
  • 4. Visualizing the Impact of Parameters • Plot the PW, AW, or ROR vs. input parameters • Steps • Pre-select the desired input parameters • Select the probable range and increment of variation for each parameter • Select the measure of worth • Compute the results for each parameter • Graphically display the results by plotting the parameter vs. the measure of worth
  • 5. Problem 1 Wild Rice, Inc. expects to purchase a new asset for automated rice handling. Most likely estimates are a first cost of $80,000, zero salvage value, and a cash flow before taxes (CFBT) per year t that follows the relation $27,000 − 2000t. The MARR for the company varies over a wide range from 10% to 25% per year for different types of investments. The economic life of similar machinery varies from 8 to 12 years. Evaluate the sensitivity of PW by varying (a) MARR, while assuming a constant n value of 10 years
  • 6. Solution : Taking MARR interval on 5 % that is 10%, 15%, 20%, 25% Now, PW10 = -80,000 + 25000 (P/A, 10%, 10) – 2000 (P/G, 10%, 10) = -80,000 + 25,000 * 6.1446 – 2000 * 22.8913 = $ 27,832 PW15 = -80,000 + 25000 (P/A, 15%, 10) – 2000 (P/G, 15%, 10) = -80,000 + 25,000 * 5.0188 – 2000 * 16.9795 = $ 11,511 PW20 = -80,000 + 25000 (P/A, 20%, 10) – 2000 (P/G, 20%, 10) = -80,000 + 25,000 * 4.1925 – 2000 * 12.8871 = $ -962 PW25 = -80,000 + 25000 (P/A, 25%, 10) – 2000 (P/G, 25%, 10) = -80,000 + 25,000 * 3.5705 – 2000 * 9.9870 = $ -10,711.5
  • 7. Problem 2 A company planning to borrow $10.5 million for a plant expansion is not sure what the interest rate will be when it applies for the loan. The rate could be as low as 10% per year or as high as 12% per year for a 5- year loan. The company will only move forward with the project if the annual worth of the expansion is below $5.7 million. The M&O cost is fixed at $3.1 million per year. The salvage could be $2 million if the interest rate is 10% or $2.5 million if it is 12% per year. Is the decision to move forward with the project sensitive to the interest rate and salvage value estimates?
  • 8. Solution Required AW < $5.7 million 10%: AW= -10,500,000(A/P,10%,5) – 3,100,000 + 2,000,000(A/F,10%,5) = -10,500,000(0.26380) – 3,100,000 + 2,000,000(0.16380) = $-5,542,300 (< $-5,700,000) 12%: AW = -10,500,000(A/P,12%,5) – 3,100,000 + 2,500,000(A/F,12%,5) = -10,500,000(0.27741) – 3,100,000 + 2,500,000(0.15741) = $-5,619,280 (< $-5,700,000) The decision is not sensitive since both AW values are below $5.7 million.
  • 9. Problem 3 An engineer collected average cost and revenue data for Arenson’s FC1 handheld financial calculator. Fixed cost $300,000 per year Cost per unit $40 Revenue per unit $70 (a) What is the range in breakeven quantity if there is possible variation in the fixed cost from $200,000 to $400,000 per year? (Use $50,000 increments.) (b) What is the incremental change in the breakeven quantity for each $50,000 change in fixed cost?
  • 10. Solution (a) Q = FC/(70-40) = FC/30 _ FC, $ QBE, units 200,000 6667 250,000 8333 300,000 10,000 350,000 11,667 400,000 13,333 (b) The change in QBE is 1667 units for each $50,000 increase in FC.
  • 11. Formalized Sensitivity Analysis Using Three Estimates • Given an input parameter of interest • Provide three estimates for that parameter • A pessimistic estimate, P • A most likely estimate, ML • An optimistic estimate, O • Note: This approach comes from PERT/CPM analysis and is based upon the beta distribution
  • 12. Three Estimates: Example 18.3 • Three alternatives (A, B, C) with 4 Parameters • First cost, salvage value, AOC, and life • For each parameter we formulate Parameter P pessimistic estimate ML most likely estimate O optimistic estimate
  • 13. Problem 4 An engineer is evaluating three alternatives for new equipment at Emerson Electronics. She has made three estimates for the salvage value, annual operating cost, and life. The estimates are presented on an alternative-by-alternative basis in Table. For example, alternative B has pessimistic estimates of S = $500, AOC = $4000, and n = 2 years. The first costs are known, so they have the same value. Perform a sensitivity analysis and determine the most economical alternative, using AW analysis at a MARR of 12% per year.
  • 14. Strategy First Cost SV AOC Life Alt A. P -20,000 0 -11,000 3 ML -20,000 0 -9,000 5 O -20,000 0 -5,000 8 Alt. B P -15,000 500 -4,000 2 ML -15,000 1,000 -3,500 4 O -15,000 2,000 -2,000 7 Alt. C P -30,000 3,000 -8,000 3 ML -30,000 3,000 -7,000 7 O -30,000 3,000 -3,500 9 AWB-ML = -15000 (A/P, 12%,4) + 1000 (A/F, 12%,4) -3500 = -15000 * 0.32923 + 1000 * 0.20923 – 3500 = $-8229.22 Similarly, Find all the Annual worth of 3 alternatives and 3 possibilities.
  • 15. Ex. 18.3 $0 $5,000 $10,000 $15,000 $20,000 $25,000 1 2 3 4 5 6 7 8 9 Life - Years A. Cost Alt A Alt B Alt C
  • 16. Economic Variability and The Expected Value • Expected Value • Long-run average based upon occurrence and probability of occurrence • Definition of Expected Value 1 ( ) ( ) m i i i E X X P X = =  Xi = value of the variable X for i from 1 to m different values P(Xi) = probability that a specific value of X will occur Subject to: 1 ( ) 1.0 m i i P X = = 
  • 17. ANA airlines plans to offer several new electronic services on flights between Tokyo and selected European destinations. The marketing director estimates that for a typical 24-hour period there is a 50% chance of having a net cash flow of $5000 and a 35% chance of $10,000. He also estimates there is a small 5% chance of no cash flow and a 10% chance of a loss of $1000, which is the estimated extra personnel and utility costs to offer the services. Determine the expected net cash flow. Solution Let NCF be the net cash flow in dollars, and let P (NCF) represent the associated probabilities. E (NCF) = 5000*(0.5) + 10,000*(0.35) + 0*(0.05) + 1000*(0.1) = $5900
  • 18. Problem: Lite-Weight Wheelchair Company has a substantial investment in tubular steel bending equipment. A new piece of equipment costs $5000 and has a life of 3 years. Estimated cash flows depend on economic conditions classified as receding, stable, or expanding. A probability is estimated that each of the economic conditions will prevail during the 3-year period. Apply expected value and PW analysis to determine if the equipment should be purchased. Use a MARR of 15% per year.
  • 19. Solution: The PW values for the three scenarios are PW R = -5000 + 2500( P/F ,15%,1) + 2000( P/F ,15%,2) + 1000( P/F ,15%,3) = -5000 + 4344 = -$656 PWS = -5000 + 5708 = $708 PW E = -5000 + 6309 =$1309 Only in a receding economy will the cash flows not return the 15% to justify the investment. The expected present worth is E(PW) = 656*(0.4) + 708*(0.4) + 1309*(0.2) =$283 At 15%, E (PW)> 0; the equipment is justified, using an expected value analysis.
  • 20. For Practice from EXERCISE of the Book Engineering Economy by Leeland and Blank (7th Edition) : Sensitivity to Parameter Variation: 18.4,18.5,18.6,18.7,18.9,18.12,18.13 Three Estimates: 18.19,18.20,18.21,18.22 Expected Values: 18.24,18.26, 18.32,18.33,18.34 Decision Trees: