The Tim e V alue of M oney
February 9, 2004
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Solar Hot Water Heating System (contd.)
• Assume : We use the solar heater to provide 50% of annual heat load
(with the remainder being supplied by auxiliary heater).
• So, if the collector efficiency is 50%, we need:
– 90 sq. ft. of collector area in Boston
– 45 sq. ft. of collector area in Tucson
and the collectors will deliver 45,000 BTU/day, on average.
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Solar Hot Water Heating System
• Requirement: 100 gpd @ 150 F
T h = 150 F
Solar
flux, q
Hot
water storage tank
Auxiliary heating
pump
T c = 40F
BTU required/day = 100 ( gpd ) x 8.33 (lb/gal) x 1 (BTU/lb- F) x (150-40)( F)
= ~ 90,000 BTU/day
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Sola r BT U d elivere d per sq . ft . pe r day |
January |
June |
Average |
Boston,MA |
500 |
2000 |
1000 |
Tucson, AZ |
1000 |
2500 |
2000 |
Note: We don’t include the cost of the hot water storage tank,
since the gas heater also requires such a tank.
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Thermal energy
Solar flux, q(t) x collector area x 0.5
Design heat load requirement
9 0,000 BTU/day
0
t 1
t 2
1
Time, t (yrs)
General problem: Choose collector area such that
capital cost of collector + cost of auxiliary energy is minimized.
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Syste m Cost |
Boston @ 90 sq. ft |
Tucson @ 45 sq ft |
Panels ($17/sq.ft) |
1530 |
765 |
Piping |
500 |
500 |
Pump & controls |
100 |
100 |
Installation |
500 |
500 |
Total |
$2630 |
$1865 |
Economic Comparison
Solar hot water
Gas heater
time
$2630 (Boston)
time
• A simplistic comparison: Assume equipment lifetime of 10 years.
Hence, “ annual cost” = $263. Avoided cost of natural gas = 45,000 BTU/day x 10 -6 (MCF/BTU) x 14 ($/MCF) x 365 days/yr x (1/0.8) =
$287/yr
• But this calculation understates the true cost of the solar investment
• Suppose the homeowner borrows the funds for the solar heating system.
• Assume the loan is for 10 years, at 10%/yr interest rate, and that the bank requires 10 equal annual payments.
• If the homeowner simply paid the bank $263/yr for 10 years, the bank wouldn’t be happy!
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Econom ic com parison (contd.)
P
A
• What is the value of A, the uniform annual payment, such that the loan P (at interest rate r) will be fully repaid after 10 years, with interest? (Assume payments are made at the end of each year.)
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After the end of t he f irst year, the homeowner pays interest at rat e r on the
principal, rP, and retires a portion of the principal, D 1 , where
A = Pr + D 1
(1)
After the end of t he second year, the homeowner pays interest on t he
residual principal of (P-D 1 ) and retires a further portion of the principal, D 2 , where
A = (P-D 1 )r + D 2
(2)
And substituting for D 1 in (2) and solving for D 2 we have:
D 2 = (A – Pr)(1+r)
(3)
After the end of t he th ird year, the homeowner pays interest on the
residual principal, P – D 1 – D 2 , and retires a further portion of the principal D 3 , where
A = (P-D 1 -D 2 )r + D 3
(4)
And substituting for D 1 and D 2 in (4) a nd solving for D 3 , we have:
D 3 = (A – Pr) (1+r) 2
(5)
And, by induction,
D n = (A – Pr) (1 + r) n-1
(6)
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A = P
r 1+ r
N
˘
( 1 )
1
r 1
N
And since
N
D n P
n 1
we can write
P (A Pr)(1 (1 r) (1 r) 2 ... (1 r) N 1 )
(1 (1 r) N ˘
(A Pr)
r
and solving for A , w e have
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N
˘
A = P
r 1+ r
N
1 r
1
Loa n paymen t schedule
Constant annual payments Loan duration: 10 yrs Interest rate: 10%/yr
180
160
140
120
100
80
Interest payment
Principal repaid
60
40
20
0
1
2 3
4
5
6
7
8
9
10
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Dollars
What is the price of natural gas above which the
solar heating system will be economic?
We solve for p * in:
A = I
r 1+ r N ˘
o
N
p * Q
1 r 1
where:
I o = investment cost of solar hot water heating system
P * = breakeven price of natural gas ($/thousand cu. ft., ($/MCF)) Q = annual gas requirement (in MCF)
= 45,000 (BTU/day) x 365 (day/yr) x 10 -6 (MCF/BTU) x 1/0.8
= 20. 5 MCF/yr
(where we have again assumed an 80% efficiency for the gas heater)
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Minimu m delivere d pric e o f natura l ga s abov e whic h residentia l sola r ho t wate r heatin g is
economica l ($/MCF)
10- yr loan w/ uniform annual payments
@ interest rate r (%/yr)
Threshold price of gas, p* ($/MCF)
Boston
Tucson
(I o = $2630 ) (I o = $1865 )
1 0 20. 9 14. 8
Note :
Average price of residential natural gas in Massachusetts during 2002 = ~$15/MCF Average price of residential natural gas in Arizona during 2002 = $12.36/MCF (Source: DOE Energy Information Administration web site)
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Tim e Valu e Equivalenc e Factors
(Discrete compounding, discrete payments)
compound amount factor (aka future- worth-of-an- annuity factor)
F A
i
Sinking fund fa ctor (A/F, i, N)
A F i ˘
(1 i) N 1
A A A A A A
Present worth of (P/A, i, N) an annuity factor
P A i(1 i) N
( 1 i) N 1 ˘
A A A A A
C apital recovery (A/P, i, N) factor
A P (1 i) N 1
i( 1 i) N ˘
P
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3 |
15 |
10.5 |
6 |
17.4 |
12.5 |
Facto r Name |
Facto r Notation |
Formula |
Cas h Flo w Diagram |
Future worth factor (compound amount f actor) |
(F/P, i, N) |
F=P(1+i) N |
F |
Present worth factor |
(P/F, i, N) |
P=F(1+i) -N |
|
P |
|||
U niform series |
(F/A, i, N) |
(1 i) N 1 ˘ |
F |
Effective interest rates, i a , for various nominal rates,
r, and compounding frequencies,m
Compounding frequency
Compounding periods per year,m
Effective rate i a for nominal rate of
6%
8%
10%
12%
15%
24%
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Frequency of compounding
• Compounding of interest occurs over different intervals -- annually, quarterly, monthly, daily, etc.
• Example: A loan offered at “ an annual interest rate of 12%, compounded quarterly”:
– Interest rate per quarter = 12/4 = 3%
– Effective annual interest rate = (1+0.03) 4 - 1 = 1.1255-1 = 0.1255 (i.e., 12.55%)
• Differentiate between nominal and effective interest rates :
– If i is the interest rate per period, and m is the number of compounding periods per year:
• Effective annual interest rate, i a = (1+i) m - 1
• Nominal interest rate, r = i.m
• And, i a = (1 + r/m) m - 1
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Annually |
1 |
6.00 |
8.00 |
10.00 |
12.00 |
15.00 |
24.00 |
Semiannually |
2 |
6.09 |
8.16 |
10.25 |
12.36 |
15.56 |
25.44 |
Quarterly |
4 |
6.14 |
8.24 |
10.38 |
12.55 |
15.87 |
26.25 |
Bimonthly |
6 |
6.15 |
8.27 |
10.43 |
12.62 |
15.97 |
26.53 |
Monthly |
12 |
6.17 |
8.30 |
10.47 |
12.68 |
16.08 |
26.82 |
Daily |
365 |
6.18 |
8.33 |
10.52 |
12.75 |
16.18 |
27.11 |
Continuous |
∞ |
6.18 |
8.33 |
10.52 |
12.75 |
16.18 |
27.12 |
Exampl e - - Valuatio n o f Bonds
• Bonds are sold by organizations to raise money
• The bond represents a debt that the organization owes to the bondholder (not a share of ownership)
• Bonds typically bear interest semi-annually or quarterly, and are redeemable for a specified maturit y value (also known as the fac e value) at a given maturit y date .
• Interest is paid in the form of regular ‘ premiums’. The flow of premiums constitutes an annuity, A, where
A = (face value) x (bond rate)
• The value of a bond at a given point in time is equal to the present worth of the remaining premium payments plus the present worth of the redemption payment (i.e., the face value)
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Exam ple -- Valuation of B onds (contd.)
• Consider a 10-year U.S. treasury bond with a face value of
$5000 and a bond rate of 8 percent, payable quarterly:
– Premium payments of $5000 x (0.08/4) = $100 occur four times per year 5000
100
0
10
Present worth at time zero
P A (P/A, r/4, 40) F(P/F, r/4,40)
where A = 100, F = 5000, and r = 8% and using our fo rmulae, we have
P
P = A
(1 + i ) N 1 ˘ 1 ˘
and since A Fi
P = F
i ( 1
i ) (1+ i)
N
F
N
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Effective interest rates, i a , for various nominal rates,
r, and com pounding frequencies, m
Compounding frequency
Compounding periods per year,m
Effective rate i a for nominal rate of
6%
8%
10%
12%
15%
24%
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Continuous compounding
• For the case of m compounding periods per year and nominal annual interest rate, r, the effective annual interest rate i a is given by:
i a = (1 + r/m) m - 1
• In the limiting case of continuou s compounding
i lim
a m
( 1 r ) m 1
m
W riting
i r m
r
i lim ( 1 i ) i 1
a i 0
e r 1
or
r ln( 1 i a )
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Annually |
1 |
6.00 |
8.00 |
10.00 |
12.00 |
15.00 |
24.00 |
Semiannually |
2 |
6.09 |
8.16 |
10.25 |
12.36 |
15.56 |
25.44 |
Quarterly |
4 |
6.14 |
8.24 |
10.38 |
12.55 |
15.87 |
26.25 |
Bimonthly |
6 |
6.15 |
8.27 |
10.43 |
12.62 |
15.97 |
26.53 |
Monthly |
12 |
6.17 |
8.30 |
10.47 |
12.68 |
16.08 |
26.82 |
Daily |
365 |
6.18 |
8.33 |
10.52 |
12.75 |
16.18 |
27.11 |
Continuous |
∞ |
6.18 |
8.33 |
10.52 |
12.75 |
16.18 |
27.12 |
Continuous Compounding, Discrete Cash Flows
(nominal annual interest rate r, continuously compounded, N periods)
annuity factor
e r 1
A F Sinking Fund Factor (A/F, r%, N) e r 1
P
A
Present Worth of an annuity Factor
(P/A, r %, N)
P A
A P e rN 1
e rN 1 ˘
e ( e 1)
rN r
A
P Capital Rec overy Factor (A/P, r %, N)
A P e rN 1
e rN ( e r 1) ˘
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To Find |
Given |
Factor Name |
Factor Symbol |
Factor formula |
F |
P |
Future Worth Factor* |
(F/P, r%, N) |
F = P ( e rN ) |
P |
F |
Present Worth Factor |
(P/F, r%, N) |
P F ( e rN ) |
F |
A |
Future Worth of an |
(F/A, r%, N) |
e rN 1 F A |
Example :
You need $25,000 immediately in order to make a down paym ent on a new hom e. Suppose that you can borrow the money from your insurance company. You will be required to repay the loan in equal payments, made every 6 months over the next 8 years. The nominal interest rate being charged is 7% com pounded continuously. What is the am ount of each payment?
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Exampl e - - paymen t an d compoundin g period s don ’ t coincide
Find the present worth of a series of quarterly payments of
$1000 extending over 5 years, if the nominal interest rate is 8%, compounded monthly.
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Continuou s cas h flows , continuou s compounding
In many applications, cash flows are a lso essentially continuous (or it is convenient to treat them a s such). We need to develop time value factors equivalent to those we have obtained for discrete cash flows.
Let X = continuous rate of flow of cash over a period (in units of , e.g., $/yr) Assume also: continuous compounding at a nominal rate of r %/yr.
First, derive a future (end of year) worth X equivalent to 1 year of continuous cash flow X
($/yr).
To do this, represent X as a uniform series of m discrete cash flows of magnitude X /m in the limit as m --> ∞
X X X X X
m m m m m
FW lim X (F/A, i%, m) where i r
m m
m
X lim
X ( 1 i ) m 1 ˘
m m i 0
lim X
( 1 i ) r i 1 ˘
i
r
X
e r 1 ˘ i
r ln(1 i a )
X a
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“ Funds flow” time value factors for continuous compounding
A
0
N
To Find: Given:
P A (P/ A , r%, N) Present worth of a continuous
annuity factor
A P ( A /P, r%, N) Continuous annuity from a present
e r N 1
re rN
re rN
future amount factor
re rN
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Exampl e o f continuou s cas h flow s & continuou s compounding
An oil refinery is considering an investment in upgrading a main pump. The upgrading is expected to result in a reduction in m aintenance labor and m aterials costs of about $3000 per year.
If the expected lifetime of the pump is three years, what is the
largest investment in the project that would be justified by the expected savings?
(Assume that the required rate of return on investments (before
taxes) is a nominal rate of 20%/ year, continuously compounded.)
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amount f actor |
e rN 1 |
|||
F |
P |
( F / P , r %, N ) |
Future worth of a continuous present amount factor |
e r N ( e r 1 ) re r |
P |
F |
( P / F , r %, N ) |
Present worth of a continuous |
e r 1 |
Example (from PSB):
A county government is c onsidering building a road from downtown to the airport to relieve congested traffic on the existing two-lane divided highway. Before allowing the sale of a bond to finance the road project, the county court has requested an estimate of future toll revenues over the bond life. The toll revenues are directly proportional to the growth of traffic over the years, so the following g rowth cash flow function is assumed to be reasonable:
F(t) = 5 (1 – e -0.1t ) (in millions of dollars)
The bond is to be a 25-year instrument, and will pay interest at an annual rate of 6%, continuously compounded.
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The Method of Laplace Transforms
Note the similarity between the expression for finding the pr esent value of a continuous cash flow stream, f(t) and the Laplace transform of the function
N
P
f ( t ) e d t
rt
0
and
L f(t) e st f ( t ) d t
0
Tables of Laplace transforms for a range o f functions are available in many mathematical handbooks, and the Laplace Transform method c an be used as an alternative to the traditional approach to evaluate equivalence values for continuous cash flows. For simple cash flow problems, there is not much computational advantage in using this method, but for complex cash flow situations the Laplace transform method may offer significant savings in computation.
For m ore d etails of the Laplace transform technique fo r cash flow modeling, see Park and Sharp-Bette, Chapter 3.
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Example
Find the present worth of a seasonal cash flow given by:
f ( t ) A sin 2 t
where f(t ) is given in $/yr, t is in years, and the cash flow is discounted at an annual rate r, continuously compounded
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