INDR 202
ENGINEERING ECONOMICS
CHAPTER 5
PRESENT-WORTH ANALYSIS
SPRING 2015
INSTRUCTOR: BORA ÇEKYAY
COURSE CONTENT
2
Engineering Economic Decisions Time Value of Money
Money Management Inflation
Present-Worth Analysis
Annual-Equivalence Analysis Rate-of-Return Analysis
Benefit-Cost Analysis
Depreciation & Income Taxes Project Cash-Flow Analysis
Project Uncertainty
PRESENT-WORTH ANALYSIS
3
Project Cash Flows Payback Screening
Present-Worth Analysis
EXAMPLE 1: HYDROELECTRIC PLANT
4
Total undiscounted investment: $800K
Power generating capacity: 6 million kwhs Estimated annual after-tax sales: $120K
EXAMPLE 1: HYDROELECTRIC PLANT
5 Is the investment worthwhile?
Will the project ever generate profit?
BANK LOAN CASH FLOWS
6
Loan
PROJECT CASH FLOWS
7
Investment
Return
PAYBACK PERIOD
8
DISCOUNTED PAYBACK PERIOD
including time value of money
PAYBACK PERIOD
time to recover initial investment
CONVENTIONAL PAYBACK PERIOD
PAYBACK SCREENING
9
PAYBACK SCREENING
elimination of projects with payback periods longer than acceptable time limit (set by the management)
FORMAL PROJECT EVALUATION
CONVENTIONAL PAYBACK PERIOD
10
Goal: To recover initial investment
Method: Calculate cumulative cash flow until it becomes nonnegative.
Screening Guideline: Projects with conventional payback period less than or equal to the prescribed threshold proceed to formal evaluation.
EXAMPLE 2: CONVENTIONAL PAYBACK
11
𝑛 Cash Flow
0 -$105,000+$20,000 1 $15,000 2 $25,000 3 $35,000 4 $45,000 5 $45,000 6 $35,000
Cumulative Cash Flow
-$85,000 -$70,000 -$45,000 -$10,000 $35,000 $80,000 $115,000
EXAMPLE 2: CONVENTIONAL PAYBACK
12
$85,000
$15,000$25,000
$35,000 $45,000 $45,000 $35,000
0
1 2 3 4 5 6
EXAMPLE 2: CONVENTIONAL PAYBACK
13 -100,000 0 50,000 100,000 150,0000 1 2 3 4 5 6
Years
3.2 YEARS
EXAMPLE 3: CONVENTIONAL PAYBACK
14
𝑛 PROJECT 1 PROJECT 2
0 -$90,000 -$90,000
1 $30,000 $25,000
2 $30,000 $25,000
3 $30,000 $25,000
4 $1,000 $25,000
5 $1,000 $25,000
6 $1,000 $25,000
3 years vs. 3.6 years
DISCOUNTED PAYBACK PERIOD
15
Goal: To recover initial investment & interest
Method: Calculate cumulative discounted cash flow until it becomes nonnegative.
Screening Guideline: Projects with discounted
payback period less than or equal to the prescribed threshold proceed to formal evaluation.
EXAMPLE 2: DISCOUNTED PAYBACK
16
𝑛 Cash Flow
0 -$105,000+$20,000 1 $15,000 2 $25,000 3 $35,000 4 $45,000 5 $45,000 6 $35,000
Cost of Funds (15%)
$0 -$85,000(15%)=-$12,750 -$82,750(15%)=-$12,413 -$70,163(15%)=-$10,524 -$45,687(15%)=-$6,853 -$7,540(15%)=-$1,131 $36,329(15%)= $5,449
discounted payback period is 𝑛 = 4.172
Cumulative
Disc. Cash Flow
EXAMPLE 2: DISCOUNTED PAYBACK
17
𝑛 Cash Flow
0 -$105,000+$20,000 1 $15,000 2 $25,000 3 $35,000 4 $45,000 5 $45,000 6 $35,000
discounted payback period is 𝑛 = 4.172
Total Net Future Cost -$85,000 -$82,750 -$70,163 -$45,687 -$7,540 $36,329 $76,778
-‐85000(1.152) +15000(1.151)
PRESENT-WORTH ANALYSIS
18
Method: Compute equivalent net surplus at 𝑛 = 0 for given interest rate 𝑖 .
PRESENT-WORTH ANALYSIS
19
INFLOW
OUTFLOW 𝑃𝑊 𝑖 INFLOW
MINIMUM ATTRACTIVE RATE OF RETURN
20
Rate at which the firm can always invest the money in its investment pool
Possibly changes over project life Depends on:
Ø Risk-Free Real Return Ø Inflation Factor
MINIMUM ATTRACTIVE RATE OF RETURN
21
2% 4%
2% 4% 20%
RISK-FREE
INVESTMENT
RISKY
INVESTMENT
RISK-FREE RETURN INFLATION
EXAMPLE 5: NEW WAREHOUSE
22
𝑃𝑊 𝑖 = −$100,000 + $11,000 𝑃|𝐴, 𝑖, 35 + $25,000 𝑃|𝐹, 𝑖, 35
𝑃𝑊 5% = $84,624.5 > 0
MINIMUM ATTTRACTIVE RATE OF RETURN
23
INVESTMENT POOL
All funds in the firm’s treasury can be placed in investments that yield a return equal to MARR.
BORROWED FUNDS
EXAMPLE 4: PRESENT-WORTH ANALYSIS
24
INFLOW
OUTFLOW
$75,000
$24,400 $27,340
$55,760
0
1 2 3
𝑃𝑊(15%)>?@ABC = $24,400 𝑃|𝐹, 15%, 1 + $27,340 𝑃|𝐹, 15%, 2 +$55,760 𝑃|𝐹, 15%, 3 = $78,553
𝑃𝑊(15%)BDE@ABC = $75,000
𝑃𝑊 15% = $78,553 − $75,000 = $3,553 > 0
EXAMPLE 4: PRESENT-WORTH ANALYSIS
25
𝒊 (%) 𝑷𝑾(𝒊) 𝒊 (%) 𝑷𝑾(𝒊)
0 $32,500 20 -$3,412
2 $27,743 22 -$5,924
4 $23,309 24 -$8,296
6 $19,169 26 -$10,539
8 $15,296 28 -$12,662
10 $11,670 30 -$14,673
12 $8,270 32 -$16,580
14 $5,077 34 -$18,360
16 $2,076 36 -$20,110
17.45 $0 38 -$21,745
18 -$751 40 -$23,302
break-even
EXAMPLE 4: PRESENT-WORTH ANALYSIS
26
EXAMPLE 4: PRESENT-WORTH ANALYSIS
27
INFLOW
OUTFLOW
$75,000
$24,400 $27,340
$55,760
0
1 2 3
A firm that has $75,000 available for investment needs to decide between
investing in the project,
EXAMPLE 4: PRESENT-WORTH ANALYSIS
28
$75,000
PROJECT
0 1 2 3
INVESTMENT POOL
Return to
investme
nt pool
Earning from investing:
$24,400 𝐹|𝑃, 15%, 2
+$27,340 𝐹|𝑃, 15%, 1 +$55,760
= $119,470
Earning from not investing:
$75,000 𝐹|𝑃, 15%, 3 = $114,066
Net gain (future worth) of the project $5,404
EXAMPLE 4: PRESENT-WORTH ANALYSIS
29 𝑛 0 Beginning Balance Interest Payment -$75,000 Project Balance -$75,000net future worth of the project 𝐹𝑊 15%
𝑃𝑊 15% = $5,404 𝑃/𝐹, 15%, 3 = $3,553
EXAMPLE 4: PRESENT-WORTH ANALYSIS
30 60,000 40,000 20,000 0 -20,000 -40,000 -60,000 -75,000 -80,000 -100,000 -120,0000 1 2 3
Years
Terminal Project Balance
(Net Future Worth, Project Surplus)
EXAMPLE 5: NEW WAREHOUSE
31
Building cost: $100K
Revenue: $17K per year
Maintenance & administration cost: $4K per year Annual income taxes: $2K
Expected life: 35 years
EXAMPLE 5: NEW WAREHOUSE
32
𝑃𝑊 𝑖 = −$100,000 + $11,000 𝑃|𝐴, 𝑖, 35 + $25,000 𝑃|𝐹, 𝑖, 35
𝑃𝑊 5% = $84,624.5 > 0
𝑃𝑊 15% = −$27,029.9 < 0
𝑃𝑊 𝑖 = 0
FUTURE-WORTH ANALYSIS
33
INFLOW
OUTFLOW 𝐹𝑊 𝑖 INFLOW
EXAMPLE 4: FUTURE-WORTH ANALYSIS
34
INFLOW
OUTFLOW
$75,000
$24,400 $27,340
$55,760
0
1 2 3
𝐹𝑊(15%)>?@ABC = $24,400 𝐹|𝑃, 15%, 1 + $27,340 𝑃|𝐹, 15%, 2 +$55,760 = $119,470
𝐹𝑊(15%)BDE@ABC = $75,000 𝐹|𝑃, 15%, 3 = $114,066 𝐹𝑊 15% = $119,470 − $114,066 = $5,404 > 0
CAPITALIZED-EQUIVALENT WORTH
35
𝐶𝐸 𝑖 = 𝐴 𝑃|𝐴, 𝑖, ∞ =
𝐴
𝑖
𝐴
0
𝑃 = 𝐶𝐸(𝑖)
EXAMPLE 6: CAPITALIZED-EQUIVALENT
36
𝐶𝐸 10% = $1,000
0.1 +
$1,000
0.1 𝑃|𝐹, 10%, 10 = $13,855
10 $1,000
$2,000
𝑃 = 𝐶𝐸(10%)
EXAMPLE 7: BRIDGE CONSTRUCTION
37
Construction cost: $2M
Annual maintenance cost: $50K
Renovation cost: $500K every 15 years Planning horizon: Infinite
EXAMPLE 7: BRIDGE CONSTRUCTION
38
$500,000 $500,000 $500,000 $500,000
$2,000,000
$50,000
EXAMPLE 7: BRIDGE CONSTRUCTION
39 Construction
𝑃P = $2𝑀
Maintenance
𝑃R = $50,000
0.05 = $1𝑀
Renovation
𝑃S = $500,000 𝐴|𝐹, 5%, 15
0.05 = $463,423
Present Worth
EXAMPLE 7: BRIDGE CONSTRUCTION
40 RENOVATION
𝑃S = $500,000 𝐴|𝐹, 5%, 15
0.05 = $463,423
𝑃S = $500,000
INVESTMENT PROJECTS
41
Projects such that decision on any one project has no effect on decision on another project are
independent.
Mutually exclusive projects resolve the same problem or meet the same need. Acceptance of one project implies rejection of all others.
Capital budgeting seeks the best allocation of available funds to projects.
INVESTMENT PROJECTS
42
SERVICE PROJECTS
generate same revenue at different costs MINIMIZE PRESENT COST
REVENUE PROJECTS
ANALYSIS & REQUIRED SERVICE PERIODS
43
ANALYSIS PERIOD
time span over which economic effects of a project are evaluated
REQUIRED SERVICE PERIOD
time span over which service of a project is needed
If required service period is given, analysis period should be the same as required service period.
ANALYSIS PERIOD
44 Required Service Period Analysis Period = Required Service Period FINITE INFINITE Project Repeatability Likely Project Repeatability Unlikelyanalysis period = project lives
analysis period < project lives
analysis period > project lives
analysis period = max project life
analysis period =
lcm of project lives
analysis period =
one of project lives Case 1
Case 2
Case 3
EXAMPLE 8: ANALYSIS PERIOD = PROJECT LIVES
45
$450
$600
$500
0
$1,000
$1,400
$2,075 $2,110
$4,000
𝑃𝑊 10% U = $280.62 𝑃𝑊 10% V = $572.88
Calculate present worth for each project.
EXAMPLE 8: ANALYSIS PERIOD = PROJECT LIVES
46
𝑃𝑊 10% U = $280.62 𝑃𝑊 10% V = $572.88
Assume that unused funds are invested at MARR.
$450 $600 $500 0 $1,000 Project A $1,400 $2,075 $2,110 $4,000 Project B
$450 $600 $500 0
$1,000
Modified Project A
$3,000
$3,993
EXAMPLE 9: ANALYSIS PERIOD = PROJECT LIVES
47
Two mutually exclusive investment projects
MARR = 12%
𝑛 Project A Project B
0 -$4,500 -$2,900
1 $2,610 $1,210
2 $2,930 $1,720
3 $2,300 $1,500
𝑃𝑊 12% U = $1,803.23 𝑃𝑊 12% V = $619.2
Select A.
𝐹𝑊 12% U = $2,533.41 𝐹𝑊 12% V = $869.93
EXAMPLE 10: ANALYSIS PERIOD < PROJECT LIVES
48
Calculate present worth of each project over required service period.
Required service period = 2 years
Model A
$45,000
$60,000
0 1 2 3 4 5 6
$480,000 $80,000
$50,000
0 1 2 3
$300,000
EXAMPLE 10: ANALYSIS PERIOD < PROJECT LIVES
49
Estimate salvage value at the end of service period.
Model A Model B
$80,000
$50,000
0 1 2 3
$300,000
$90,000
$45,000
$60,000
0 1 2 3 4 5 6
$480,000
$250,000
𝑃𝑊 15% U = −300 − 80 𝑃/𝐴, 0.15,2 + 90 𝑃/𝐹, 0.15,2 = −$362
𝑃𝑊 15% V = −480 − 45 𝑃/𝐴, 0.15,2 + 250 𝑃/𝐹, 0.15,2 = −$364
EXAMPLE 11: ANALYSIS PERIOD > PROJECT LIVES
50
Calculate present worth of each project over required service period.
𝑛 Model A Model B
0 -$12,500 -$15,000
1 -$5,000 -$4,000
2 -$5,000 -$4,000
3 -$5,000+$2,000 -$4,000
4 -$4,000+$1,500
5
EXAMPLE 11: ANALYSIS PERIOD > PROJECT LIVES
51
𝑛 Model A Model B
0 -$12,500 -$15,000
1 -$5,000 -$4,000
2 -$5,000 -$4,000
3 -$3,000 -$4,000
4 -$11,000 -$2,500 5 -$11,000 -$11,000
Replacement projects
Same alternative or new technology? Purchase, lease, or subcontract?
Leasing Model A
Annual lease payment: $6,000 Annual operating cost: $5,000
𝑃𝑊 15% U = −$34,359 𝑃𝑊 15% V = −$31,031
EXAMPLE 12: ANALYSIS PERIOD = MAX PROJECT LIFE
52
Mutually exclusive alternatives for natural resource extraction: Process 1 that extracts all the coal in 10 years & Process 2 in 8 years
EXAMPLE 13: ANALYSIS PERIOD = MAX PROJECT LIFE
53
𝑃𝑊 15% WX>AA = $2,208,470
𝑃𝑊 15% YZ[\Z = $2,180,210
INFINITE ANALYSIS PERIOD
54
Project Repeatability Likely
Use the lowest common multiple of project lives.
Project Repeatability Unlikely
PROJECT REPEATIBILITY LIKELY
55
𝑃𝑊 15% U = −$53,657 𝑃𝑊 15% V = −$48, 534
Select B.
Model A: 3 Years Model B: 4 years
PROJECT REPEATIBILITY LIKELY
56
Model A Model B
0 -‐12500 -‐15000 1 -‐5000 -‐4000
2 -‐5000 -‐4000 3 -‐5000+2000-‐12500 -‐4000
4 -‐5000 -‐4000+1500-‐15000 5 -‐5000 -‐4000
6 -‐5000+2000-‐12500 -‐4000 7 -‐5000 -‐4000
8 -‐5000 -‐4000+1500-‐15000 9 -‐5000+2000-‐12500 -‐4000
10 -‐5000 -‐4000 11 -‐5000 -‐4000
12 -‐5000+2000 -‐4000+1500
PROJECT REPEATIBILITY LIKELY
57
PWA=-‐12500-‐5000(P/A,0.15,12)-‐10500(P/F,0.15,3) -‐10500(P/F,0.15,6) -‐ 10500(P/F,0.15,9)+2000(P/F,0.15,12)=-‐$53,657
PWB=-‐15000-‐4000(P/A,0.15,12)-‐13500(P/F,0.15,4) -‐ 13500(P/F,0.15,8)+1500(P/F,0.15,12)=-‐$48,534
𝑖S = 1 + 0.15 S − 1 = 0.521 𝑖] = 1 + 0.15 ] − 1 = 0.75
PWA=-‐12500-‐5000(P/A,0.15,12)-‐12500(P/A,0.52,3)+2000(P/A,0.52,4)=-‐$53,657
PWB=-‐15000-‐4000(P/A,0.15,12)-‐15000(P/A,0.75,2)+1500(P/A,0.75,3)=-‐$48,534
Cycle A – PW=-‐12500-‐5000(P/A,0.15,3)+2000(P/F,0.15,3)=-‐$22,601.09 Cycle B – PW=-‐15000-‐4000(P/A,0.15,4)+1500(P/F,0.15,4)=-‐$25,562.28 PWA=-‐22601.09-‐22601.09(P/A,0.521,3)=-‐$53,657
SUMMARY
58
Mutually exclusive alternatives are intended to meet the same need, so only one is accepted.
In revenue projects, income generated depends on the project choice while in service projects, it is the same regardless of which project is selected.
Analysis Period is the time span over which economic effects of an investment are evaluated.
Required Service Period is the time span over which the service of a project is needed.
SUMMARY
59
Present-Worth Analysis is an equivalence method of analysis in which projects’ cash flows are discounted to a lump sum amount at present time.
Minimum Attractive Rate of Return (MARR) is the interest rate at which a firm can always earn or borrow money.
Present-worth analysis should be conducted with respect to MARR, which is generally determined by management.
