Transcript Methods of Calculating Interest

Part II: Evaluating business & engineering
assets
Ch 5: Present worth analysis
Ch 6: Annual equivalence analysis
Ch 7: Rate-of-return analysis
(Conventional) Payback period
Useful for initial screening
Principle: How fast can I recover my initial investment?
Method: Based on cumulative cash flow (or accounting
profit)
Screening guideline: If the payback period is less than or
equal to some specified payback period, the project
would be considered for further analysis.
Weakness: Does not consider the time value of money
Conventional-payback method is based on actual dollars
Example 5.1: Payback period
N
0
1
2
3
4
5
6
Cash flow
-$105,000+$20,000
$35,000
$45,000
$50,000
$50,000
$45,000
$35,000
Cum. flow
-$85,000
-$50,000
-$5,000
$45,000
$95,000
$140,000
$175,000
Payback period should occur somewhere
between N = 2 and N = 3.
$45,000
$45,00
0
$35,000
4
5
6
4
5
Annual cash flow
$35,000
$25,000
$15,000
0
1
2
Years
3
Cumulative cash flow ($)
$85,000
150,000
3.2 years
payback period
100,000
50,000
0
-50,000
-100,000
0
1
2
3
6
Years (n)
Discounted payback period
Also used for initial screening
Principle: How fast can I recover my initial investment plus
interest?
Method: Based on cumulative discounted cash flow
Screening guideline: If the discounted payback period
(DPP) is less than or equal to some specified payback
period, the project would be considered for further
analysis.
Weakness: Cash flows occurring after DPP are ignored
Discounted-payback method considers time value of money
Example 5.2 Discounted payback period calculation
Period
Cash flow
Cost of funds
(15%)*
0
Cumulative
cash flow
0
-$85,000
-$85,000
1
15,000
-$85,000(0.15) = -$12,750
-82,750
2
25,000
-$82,750(0.15) = -12,413
-70,163
3
35,000
-$70,163(0.15) = -10,524
-45,687
4
45,000
-$45,687(0.15) =-6,853
-7,540
5
45,000
-$7,540(0.15) = -1,131
36,329
6
35,000
$36,329(0.15) = 5,449
76,778
* Cost of funds = (unrecovered beginning balance) X (interest rate)
Summary: payback period
Payback periods can be used as a screening tool for
liquidity, but we need a measure of investment worth for
profitability.
Net present worth measure
Principle: Compute the equivalent net surplus at n = 0 for a
given interest rate of i, or minimum attractive rate of
return (MARR)
Decision rule: Accept the project if the net surplus is
positive.
inflow
0
1
2
3
outflow
4
5
net surplus
PW(i)inflow
0
PW(i)outflow
PW(i) > 0
Guideline for selecting a MARR
real return
2%
inflation
4%
risk premium
0%
total expected return
6%
real return
2%
inflation
4%
risk premium
20%
total expected return
26%
U.S. Treasury bills
Risk-free
real return
Inflation
very safe
Risk
premium
very risky
Amazon.com
Example 5.3 - Tiger Machine Tool Company
inflow
$24,400
$27,340
0
1
$55,760
2
3
outflow
$75,000
PW (15%)
inflow
= $24 ,400 ( P / F ,15%, 1) + $27 ,340 ( P / F ,15%, 2 )
+ $55 ,760 ( P / F ,15%, 3) = $78,553
PW (15%)
outflow
= $75,000
PW (15%) = $78,553 - $75,000 = $3,553 > 0, Accept
Present worth profile
40
Reject
Accept
30
PW (i) ($ thousands)
20
Break even interest rate
(or rate of return)
10
$3553
17.46%
5
20
0
-10
-20
-30
0
10 15
25
i = MARR (%)
30
35
40
Capitalized equivalent worth
Principle: PW for a project with an annual receipt of A over
infinite service life
Equation: CE(i) = A(P/A, i, ) = A/i
A
0
P = CE(i)
Ex 5.4: A = $2 million, i = 8%, N = 
A/i = CE(i) = $2 million/0.08 = $25 million
Part II: Evaluating business & engineering
assets
Ch 5: Present worth analysis
Ch 6: Annual equivalence analysis
Ch 7: Rate-of-return analysis
Annual worth analysis
Principle: Measure an investment worth on annual basis
Benefit: By knowing the annual equivalent worth, we
can:
– seek consistency of report formats
– determine unit costs (or unit profits)
– facilitate unequal project life comparisons
Ex. 6.1: Computing equivalent annual worth
$9.0
$12
$10
$5
PW(15%) = $6.946
0
$8
1
2
3
4
5
6
$3.5
$6.946
$15
A = $1.835
0
1
2
3
4
5
6
0
AE(12%) = $6.946(A/P, 15%, 6) = $1.835
Annual equivalent worth – repeating cash flow cycles
$700
$500
$800
$800
$700
$400 $400 $500
$400 $400
Repeating cycle
$1,000
$1,000
First cycle: PW(10%) = -$1,000 + $500 (P/F, 10%, 1)
+ . . . + $400 (P/F, 10%, 5) = $1,155.68
AE(10%) = $1,155.68 (A/P, 10%, 5) = $304.87
Both cycles: PW(10%) = $1,155.68 + $1,155.68 (P/F, 10%, 5)
AE(10%) = $1,873.27 (A/P, 10%,10) = $304.87
When only costs are
involved, the AE method
is called the annual
equivalent cost.
Revenues must cover two
kinds of costs: operating
costs and capital costs.
annual equivalent costs
Annual equivalent cost
capital
costs
+
operating
costs
Capital (ownership) costs
S
Two transactions are
associated with owning
equipment: initial cost (I) &
salvage value (S).
Capital costs: taking these
items into consideration,
we calculate the capital
recovery costs as:
0
N
I
0
1
CR(i) = I(A/P,i,N) – S(A/F,i,N)
= (I – S)(A/P,i,N) + iS
Recall that (A/F,i,N) = (A/P,i,N) – i
2
3
CR(i)
N
Applying annual worth analysis
Ex. 6.3 – Equivalent worth per unit of time
$24,400
$55,760
$27,340
0
1
Recall ex 5.3
2
3
$75,000
Operating hours per year
2,000 hr
2,000 hr
2,000 hr
PW (15%) = $3553
AE (15%) = $3,553 (A/P, 15%, 3) = $1,556
Savings per machine hour = $1,556/2,000 = $0.78/hr
Part II: Evaluating business & engineering
assets
Ch 5: Present worth analysis
Ch 6: Annual equivalence analysis
Ch 7: Rate-of-return analysis
Why RoR measure is so popular?
Which statement is easier to understand?
This project will bring in a 15% rate of return on
investment.
This project will result in a net surplus of $10,000
in NPW.
Rate of return (RoR): a relative percentage method that measures
the yield as a percent of investment over the life of a project
Return on investment
Definition 1: Rate of return (RoR) is defined as the interest
rate earned on the unpaid balance of an installment loan.
Example: A bank lends $10,000 & receives annual
payments of $4,021 over 3 years. The bank is said to
earn a return of 10% on its loan of $10,000.
Rate of return
Definition 2: Rate of return (RoR) is the break-even interest
rate, i*, which equates the present worth of a project’s
cash outflows to the present worth of its cash inflows.
Mathematical relation:
PW(i*) = PW(i*)cash inflows – PW(i*)cash outflows = 0
Methods for finding rate of return
–
–
–
–
Using Excel’s financial command
Direct solution
Trial-and-error
“Cash Flow Analyzer” – online financial calculator
Excel command to find the rate of return:
=IRR(cell range, guess)
e.g., =IRR(C0:C7, 10%)
Direct solution methods
Project A
$1,000  $1,500( P / F , i ,4)
$1,000  $1,500(1  i ) 4
0.6667  (1  i ) 4
ln 0.6667
 ln(1  i )
4
0101365
.
 ln(1  i )
e
0.101365
PW (i )  $2,000 
$1,300 $1,500

0
2
(1  i ) (1  i )
1
, then
1 i
PW (i )  2,000  1,300 x  1,500 x 2
Let x 
Solve for x:
x  0.8 or -1.667
Solving for i yields
 1 i
ie
1
 10.67%
0.101365
Project B
1
1
 i  25%,  1667
.

 i  160%
1 i
1 i
Since  100%  i  , the project's i *  25%.
0.8 
See example 7.2
Trial and error method – Project C
Step 1: Guess an interest
rate, say, i = 15%
Step 2: Compute PW(i) at the
guessed i value
PW (15%) = $3,553
Step 3: If PW(i) > 0, then increase i.
If PW(i) < 0, then decrease i.
PW(18%) = -$749
Step 4: If you bracket the solution,
use a linear interpolation to
approximate the solution
3,553
0
-749
15%
i
18%
 3,553 
i  15%  3%

 3,553 749
Based on example 6.3
 17.45%
Basic decision rule
If ROR > MARR, accept
This rule does not work for a situation where
an investment has multiple rates of return
Return on invested capital
Definition 3: Return on invested capital is defined as the
interest rate earned on the unrecovered project balance
of an investment project. It is commonly known as
internal rate of return (IRR).
Example: A company invests $10,000 in a computer and
results in equivalent annual labor savings of $4,021 over
3 years. The company is said to earn a return of 10% on
its investment of $10,000.
Project balance calculation
0
1
2
3
Beginning
project balance
-$10,000
-$6,979
-$3,656
Return on
invested capital
-$1,000
-$697
-$365
+$4,021
Payment
received
-$10,000
+$4,021
+$4,021
Ending project
balance
-$10,000
-$6,979
-$3,656
0
The firm earns a 10% rate of return on funds that remain internally
invested in the project. Since the return is internal to the project, we call
it internal rate of return.
Multiple rates of return
If a project has more than one rate of return, how would
you make an accept/reject decision?
Simple Investment
Def: Initial cash flows are
negative, and only one
sign change occurs in the
net cash flows series.
Example: -$100, $250, $300
(-, +, +)
RoR: A unique RoR
Non-simple Investment
Def: Initial cash flows are
negative, but more than
one sign changes in the
remaining cash flow
series.
Example: -$100, $300, -$120
(-, +, -)
RoR: A possibility of multiple
RoR’s
Example 7.1 – Investment classification
Period
(N)
Project A
Project B
Project C
0
-$1,000
-$1,000
+$1,000
1
-500
3,900
-450
2
800
-5,030
-450
3
1,500
2,145
-450
4
2,000
Project A is a simple investment.
Project B is a nonsimple investment.
Project C is a simple borrowing.
How to proceed: if you encounter multiple
rates of return
Abandon the IRR analysis and use the PW criterion
For MARR = 15%
PW(15%) = –$1,000 + $2,300 (P/F,15%,1) – $1,320(P/F,15%,2 )
= $1.89 > 0
Accept the investment
To find the true rate of return (or return on invested capital)
for the project, see appendix 7A,
Incremental analysis (procedure)
Step 1: Compute the cash flow for the difference between
the projects (A,B) by subtracting the cash flow of the
lower investment cost project (A) from that of the higher
investment cost project (B).
Step 2: Compute the IRR on this incremental investment
(IRRB-A).
Step 3:Accept the investment B if and only if
IRRB-A > MARR
NOTE: Make sure that both IRRA and IRRB are greater than
MARR.
Ex 7.7 - Incremental rate of return
n
B1
B2
B2 - B1
0
1
2
3
-$3,000
1,350
1,800
1,500
-$12,000
4,200
6,225
6,330
-$9,000
2,850
4,425
4,830
IRR
25%
17.43%
15%
Given MARR = 10%, which project is a better choice?
Since IRRB2-B1=15% > 10%, and also IRRB2 > 10%, select B2.
Part III: Development of project cash flows
Ch 8: Accounting for depreciation & income taxes
– Accounting depreciation
– Book depreciation methods
– Tax depreciation methods
– How to determine “accounting profit”
– Corporate taxes
Asset depreciation: two concepts
Economic depreciation
the gradual decrease in
utility in an asset with
use & time
Physical
depreciation
Functional
depreciation
Depreciation
Accounting depreciation
the systematic allocation
of an asset’s value in
portions over its
depreciable life—often
used in engineering
economic analysis
Book
depreciation
Tax
depreciation
What can be depreciated?
 Assets used in business or held for production of income
 Assets having a definite useful life and a life longer than
one year
 Assets that must wear out, become obsolete or lose
value
A qualifying asset for depreciation must satisfy all three
conditions.
Examples?
Ex 8.1 – Cost basis
Cost of a new hole-punching machine
(Invoice price)
+ Freight
$62,500
725
+ Installation labor
2,150
+ Site preparation
3,500
Cost basis to use in depreciation
calculation
$68,875
Types of depreciation
Book depreciation
– In reporting net income to investors/stockholders
– In pricing decisions
Tax depreciation
– In calculating income taxes for the IRS
– In engineering economics, we use depreciation in the
context of tax depreciation
Methods of calculating depreciation
– straight line
– declining balance
– unit production
Straight-line (SL) method
Principle: A fixed asset provides its service in a uniform
fashion over its life
Annual depreciation:
Dn = (I – S) / N, and constant for all n.
Book value
Bn = I – N (D)
where I = cost basis
S = salvage value
N = depreciable life
Declining balance method
Principle: A fixed asset provides its service in a
decreasing fashion
Annual depreciation:
Dn = αBn-1 = αI(1 – α)n-1
Book value:
Bn = I(1 – α)n
where 0 < α < 2(1/N)
Note: if  is chosen to be the upper bound,  = 2(1/N),
we call it a 200% DB or double declining balance method.
Ex. 8.3 – Declining balance method
I = $10,000
n
Dn
Bn
0
$10,000
1 $4,000
6,000
2 2,400
3,600
3 1,440
2,160
4
864
1,296
5
518
778
Bn must be > S
N = 5 yr
S = $2,000
Dn = αBn-1 = αI(1 – α)n-1
Bn = i(1 – α)n
n
Dn
Bn
0
$10,000
1 $4,000
6,000
2 2,400
3,600
3 1,440
2,160
4
160
2,000
5
0
2,000
Units-of-production method
Principle: service units will be consumed in a non timephased fashion
Annual depreciation
Dn = Service units consumed for year(I - S)
total service units
Ex 8.5 – Units-of-production depreciation
Given:
I = $55,000
S = $5,000
total service units = 250,000 mi
usage for this year = 30,000 mi
Solution:
30, 000
Dep 
($55, 000  $5, 000)
250, 000
 3 

 ($50, 000)
 25 
 $6, 000
Modified accelerated cost recovery systems (MACRS)
Personal property
– definition: movable property; property of any kind
except real property
– depreciation based on DB method switching to SL
– half-year convention (all assets placed in service
mid year)
– zero salvage value
Real property
– permanent fixtures
– SL method
– mid-month convention
– zero salvage value
Year (n)
Calculation in %
MACRS (%)
1
(0.5)(0.40)(100%)
DDB
20%
2
(0.4)(100%-20%)
SL = (1/4.5)(80%)
DDB
32%
17.78%
3
(0.4)(100%-52%)
SL = (1/3.5)(48%)
DDB
19.20%
13.71%
4
(0.4)(100%-71.20%) switch
SL = (1/2.5)(29.80%) to SL
11.52%
11.52%
5
SL = (1/1.5)(17.28%) SL
11.52%
6
SL = (0.5)(11.52%)
5.76%
SL
MACRS for real property
Types of real property
27.5-year (residential)
39-year (commercial)
• SL method
• zero salvage value
• mid-month convention
Example: Placed a residential property in service in March.
Find the depreciation allowance in year 1.
D1 = (9.5/12)(100%/27.5) = 2.879%
Ex. 8.9 – Cash flow vs. net income
Item
Income
Cash flow
Gross income (revenue)
$50,000
$50,000
Expenses
Cost of goods sold
Depreciation
Operating expenses
20,000
4,000
6,000
-20,000
Taxable income
20,000
Taxes (40%)
Net income
Net cash flow
8,000
-6,000
-8,000
$12,000
$16,000
Ex 8.9 (cont.) – Net income versus net cash flow
net cash flows = net income + non-cash expense (depreciation)
$50,000
$40,000
$30,000
$20,000
$10,000
$0
net
cash flow
net income
$12,000
depreciation
$4,000
income taxes
$8,000
operating expenses
cost of goods sold
$6,000
$20,000
gross
revenue
Marginal & effective (average) tax rate for a taxable
income of $16,000,000
Taxable income
Marginal tax
rate
Amount of taxes
Cumulative taxes
First $50,000
15%
$7,500
$7,500
Next $25,000
25%
6,250
13,750
Next $25,000
34%
8,500
22,250
Next $235,000
39%
91,650
113,900
Next $9,665,000
34%
3,286,100
3,400,000
Next $5,000,000
35%
1,750,000
5,150,000
Remaining
$1,000,000
38%
380,000
$5,530,000
$5,530,000
Average tax rate =
 34.56%
$16,000,000
Capital gains & ordinary gains
Capital gains
Total gains
Ordinary gains
or
depreciation recapture
Cost basis
Book value
Salvage value