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• To evaluate the economic profitability and liquidity of a single project
• Equivalent measures of a project’s profitability
– Present Worth (PW)
– Future Worth (FW)
– Annual Worth (AW)
– Internal Rate of Return
• Measures of liquidity
– Simple Payback Method
– Discounted Payback Method
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M
A
R
R
• Firms will set a minimum interest rate that all projects should earn it in order to be considered for funding
• Once established by the firm is termed the Minimum
Attractive Rate of Return (MARR)
• Numerous models exist to aid engineering managers in estimating what this rate should be in a given time period
• Personal Example:
• Assume you want to purchase a new computer
• Assume you have a charge card that carries a 18% per year interest rate.
• If you charge the purchase, YOUR cost of capital is the 18%
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• Firm’s raise capital from the following sources
• Equity – using the owner’s funds (retained earnings, cash on hand) belongs to the owners
• Debt – the firm borrows from outside the firm and pays an interest rate on the borrowed funds
• Once this “cost” is approximated, then, new project MUST return at least the cost of the funds used in the project PLUS some additional percent return
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• First, start with a “safe” investment possibility
• A firm could always invest in a short term CD paying around 4-5% (But investors will expect more than that!)
• The firm should compute it’s current weighted average cost of capital (Assume the weighted average cost of capital (WACC) is 10.25%)
•
Certainly, the MARR must be greater than the firms cost of capital in order to earn a “profit” or “return” that satisfies the owners!
• Thus, some additional “buffer” must be provided to account for risk and uncertainty
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• Assume a firm’s MARR = 12% with 2 projects, A and B
• A costs $400,000 with an estimated return of 13%/year
• B cost $100,000 with an estimated return of 14.5%/year
• What if the firm has a budget of say $150,000
• A cannot be funded – not sufficient funds!
• B is funded and earns 14.5% return or more
• A is not funded, hence, the firm looses the OPPORTUNITY to earn 13%
• Changes as the amount of investment capital changes over time
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1. Compute the present equivalent of the estimated cash flows using the MARR as the interest rate
2. If PW(MARR)

0, then the project is profitable
If PW(MARR) < 0, then the project is not profitable
• Example: Cost/Revenue Estimates
– Initial Investment: $50,000
– Annual Revenues: 20,000
– Annual Operating Costs: 2,500
– Salvage Value @ EOY 5: 10,000
– Study Period: 5 years
– MARR 20% per year
• Draw CFD
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• PW(20%) =-50,000 + (20,000 –2,500)(P|A,20%,5) +
10,000(P|F,20%,5)= $6,354.50
• PW = $6,354.50 tells us
– We have recovered our entire $50,000 investment,
– We have earned our desired 20% on this investment,
– We have made a lump sum equivalent profit of
$6,354.50 beyond what was expected (required)
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• Treasury bonds
– Issued by Federal Government
– Full backing of the Government
– 1 year or less; 2-10 year issues; and 10-30 year issues
• State and Municipal Bonds
– Issued by states and local governments
– Generally tax exempt by the Federal Government
– Used to finance state and local projects
• Mortgage Bonds
– Issued by Corporations
– Secured by the firm’s assets
– Money received by the firm is used to fund projects
– Buyers of these bonds are not owners – they are lenders to the
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• For a bond, let
Z = face, or par value
C = redemption or disposal price (usually Z ) r = bond rate (nominal interest) per interest period
N = number of periods before redemption i = bond yield (redemption ) rate per period
V
N
= value (price) of the bond N interest periods prior to redemption -- PW measure of merit
VN = C ( P / F, i%, N ) + rZ ( P / A, i %, N )
• Periodic interest payments to owner = rZ for N periods
• When bond is sold, receive single payment (C), based on the price and the bond yield rate ( i )
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• Example
Z = $5,000 (face value) r = 4.5% per year paid semiannually
N = 10 years
• The interest the firm would pay to the current bond holder is rZ=$5000(0.045/2)=$ 112.5 per …
• The bond holder, buys the bond and will receive
$112.50 every … months for the life of the bond
• How much will you consider this bond if you can earn
8%/yr c.q.?
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• What is fixed?
– The future interest payments are fixed (rZ)
– The future face value of the bond in fixed (Z)
• What can vary?
– The purchase price such that you earn at least the 8%/yr c.q.
• Draw CFD
• PW (2%) =$112.50 (P/A,4.04%,20) +$5,000
(P/F,2%,40) =$3788
• IF the buyer can buy this bond for $3,788 or less, he/she will earn at least the 8% c.q. rate.
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• Compute the future equivalent of the estimated cash flows using the MARR as the interest rate
• If FW(MARR) >0, then the project is profitable
• If FW(MARR)<0, then the project is not profitable
• Previous Example – FW Method
• FW(20%) =-50,000(F|P,20%,5)+(20,000-
2,500)(F|A,20%,5)+10,000= $15,813
• Since FW(20%)>0, the project is profitable
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• Popular with some managers who tend to thing in terms of
“$/year, $/months, etc.
• Easily understood-results are reported in $/time period
• AW(i%) = R - E - CR(i%)
(Eqn. 4.4)
R = annual equivalent revenues
E = annual equivalent expenses
CR = annual equivalent capital recovery cost
• CR is the equivalent uniform annual cost of capital invested
• CR(i%) = I (A/P,i%,N) - S (A/F,i%,N) (Eqn. 4-5)
– CR(20%) = $50,000(A|P,20%,5) - $10,000(A|F,20%,5)= $15,376
– AW(20%) = R – E – CR(20%), or
– AW(20%) = $20,000 - $2,500 - $15,376 = $2,124
• Since AW(20%) 
0, project is profitable 14

• Note: PW, FW, and AW are equivalent measures of profitability
• If PW>0, then FW>0 and AW>0
• From our example,
PW = $6,354.50 therefore,
FW = 6,354.50(F|P, 20%,5) = $15,812 and
AW = 6,354.50(A|P, 20%,5) = $2125
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• Benjamin Franklin, according to the American Bankers
Association, left $5,000 to the residents of Boston in 1791, with the understanding that it should be allowed to accumulate for a hundred years. By 1891 the $5,000 had grown to $322,000. A school was built, and $92,000 was set aside for a second hundred years of growth. In 1960, this second century fund had reached
$1,400,000. As Franklin put it, in anticipation: "Money makes money and the money that money makes, makes more money."
• Question: What average interest rate per year was earned from
1791 to 1891?
• Given: P = $5,000, n = 100, F = $322,000 , Find: i'% per year
• F = P(F|P, i'%, 100) or $322,000 = $5000(F|P, i'%, 100)
• F = P(1+i') n or 64.4 = (1+i') 100 or i' = 4.25% per year
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• The Internal Rate of Return (IRR) method solves for the interest rate that equates the equivalent worth of a project's cash outflows (expenditures) to the equivalent worth of cash inflows (receipts or savings)
• In other words, the IRR is the interest rate that makes the
PW, AW, and FW of a project's estimated cash flows equal to zero, or
PW(i') of cash inflow = PW(i') of cash outflow
• We commonly denote the IRR by i’, and use PW(i' %) =
0, AW(i' %) = 0, or FW(i' %) = 0
• In general, we must solve for i' by trial and error (unless we use an equation solver)
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• Once we know the value of the IRR for a project, we compare it to the MARR to determine whether or not the project is acceptable with respect to profitability
IRR = i'

MARR project is acceptable
IRR = i' < MARR project is unacceptable (reject)
• Difficulties with the IRR Method:
– The IRR Method assumes that recovered funds are reinvested at the IRR rather than the MARR
– Possible multiple IRRs
• Why should you learn the IRR Method?
• The majority of U.S. companies favor the IRR method for evaluating capital investment projects
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• Find i'% such that the PW(i'%) = 0
• 0 = -$50,000 + $17,500(P|A, i'%,5) + $10,000(P|F, i'%,5)
PW (20%) = 6354.50 tells us that i' > 20%
PW (25%) = 339.75 > 0, tells us that i'% > 25%
PW (30%) = -4,684.24 < 0, tells us that i'% < 30%
25% < i' < 30%
• Use linear interpolation to estimate i'%
• i' =25.3%>MARR
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• An individual approaches the Loan Shark Agency for $1,000 to be repaid in 24 monthly installments. The agency advertises an interest rate of 1.5% per month. They proceed to calculate a monthly payment in the following manner:
Amount you leave with
Credit investigation
$1,000
25
Credit risk insurance
Total
5
$1,030
Interest:
Total owed:
($1,030)(24)(0.015) = $371
$1,030 + ... = ...
Payment: $1,401/24 = $58.50 per month
• What effective annual interest rate is the individual paying?
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• Step 2 – Solve for i'
• PW(inflows @ i' %) - PW(outflows @ i' %) = 0
• $58.50 (P/A, i' %,24) - $1,000 = 0 or (P/A, i' %,24) =
17.0940
• From table … < i' < ...
• Using linear interpolation, we find that i' = … per month
• From equation 3-3 with r/M = 0.0292, the effective annual interest rate being charged is (1.0292) 12 - 1 =
0.4125 or 41.25% per year
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• In 1555 King Henry VIII borrowed money from his bankers on the condition that he pay 5% of the loan every 3 months, until a total of 40 payments were made.
Then the loan would be considered repaid. What effective annual interest did King Henry pay?
• Solution:
• … = … (P/A,i'%,40); therefore, (P/A,i'%,40) = 20, or i'
= 3.94% per quarter
• i/yr. = (1.0394) 4 - 1

0.1672 (or 16.72%)
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• $7,500 loan repaid in 24 monthly payments of $350 each
• Step 1: Draw a CFD
• Step 2: Find i'/month that establishes equivalence between cash outflows and cash inflows
0 =-$7,500 + $350 (P/A,i',24 months), i'

0.93% /month
• Step 3: Find i' per year, i'/year = (1.0093) 12 - 1 12%=…
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• Simple Payback Period - how many years it takes to recover the investment (ignoring the time value of money)
• Discounted Payback Period - how many years it takes to recover the investment (including the time value of money)
• Previous Example
EOY Simple Payback Cumulative PW(0%) Discounted Payback Cumulative PW(20%)
-----------------------------------------------------------------------------------
0 -$50,000 -$50,000
1
2
3
4
5
-32,500
-15,000
+2,500
+20,000
+47,500
PBP= 3 years
-35,417
-23,264
-13,137
-4,697
+6,354.50
PBP= 5 years
Discounted Payback:
@k=1, PW = -50,000 + 17,500(P|F, 20%, 1) = -35,417

Investment $40,000
Annual Expenses
Annual Revenues (next year)
$3,000 decreasing by $500/year thereafter to Yr.10 $10,000
Study Period 5 years
Market Value @ Yr. 5 $15,000
MARR (per Year) 12%
•
Use the PW, FW, AW, and PBP method to evaluate this investment
PW(12%)= -$40,000 + $7,000 (P/A,12%,5) - $500(P/G,12%,5) - 15,000
(P/F,12%,5)= -$9455 < 0 Reject
FW(12%)= -$40,000 (F/P,12%,5) + $7,000 (F/A,12%,5) - $500
(A/G,12%,5)(F/A,12%,5)+ $15,000= -$16,660 < 0 Reject
AW(i%) = R - E - CR(i%) where CR(i%) = I (A/P,i%,N) - S (A/F,i%,N)
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• PW(12%)= -$40,000 + $7,000 (P/A,12%,5) -
$500(P/G,12%,5) - 15,000 (P/F,12%,5)= -$9455 < 0
FW(12%)= -$40,000 (F/P,12%,5) + $7,000 (F/A,12%,5)
- $500 (A/G,12%,5)(F/A,12%,5)+ $15,000= -$16,660 <0
•
AW(i%) = R - E - CR(i%) where
CR(i%) = I (A/P,i%,N) - S (A/F,i%,N) or
CR(12%) = $40,000 (A/P,12%,5) - $15,000 (A/F,12%,5)
=$8,735
•
AW(12%) = $10,000 - $500(A/G,12%,5) - $3,000 -
$8,735= -$2,620 < 0 Reject
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EOY Cumulative PW(i=0%) Cumulative PW(i=12%)
0
1
2
3
4
5
-$40,000
-33,000
-26,500
-20,500
-15,000
+5,000
Simple PBP=5
-$40,000
-33,750
-28,568
-24,297
-20,802
-9,454
Discounted PBP>5
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• Avoid using this method as a primary analysis technique for selection projects
• Does not employ the time value of money (simple)
• If used, can lead to conflicting selections when compared to more technically correct methods like present worth!
• Disregards all cash flows past the payback time period
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