Chapter 4. Money – Time Relationships and Equivalence
A)Cash Flows and Cash Flow Diagrams
A cash flow is a plot of Cash Flows versus Time. The abscissa (x-axis) represents time and the ordinate(y-axis) represents the amount of cash flow.
There are three types of cash flow diagrams.
1)Basic Cash Flow – shows all cash flows for each period
2)Net Cash Flow – shows net cash flow for each period (most common)
3)Cumulative Cash Flow – shows cumulative net cash flow for each period
Basic Assumption – cash flows are assumed to occur at the end of the period; payments are assumed to be at the end of the period(instead of the beginning of the period)
B)Primary Mathematical Relationships
1)Geometric Series
S = a + ar + ar2 + ar3 + ar4 + + ar(n-1) = a[(rn -1) / (r – 1) ]
Examples
S = 1 + 2 + 4 + 8 = ? a=1, r=2, n=4, S=1[24 – 1]/ [2 – 1 ] = 15
S = 0.5 + 1 + 2 + 4 + 8 = ? Solve
a=
r=
n=
S=
S = 1 + 1.08 + 1.082 + 1.083 = ?Solve
a=
r=
n=
S=
2)Infinite Limit (continuous interest problems)
Lim [ 1 + 1/k]k = e = 2.71828
k→∞
if k = 1, Lim = 2if k = 2, Lim = 2.25
if k = 3, Lim = ?if k = 5, Lim = ?
if k = 10, Lim = ?if k = 100, Lim = ?
if k = 1000, Lim = ?
C) Interest Basics
1) Simple Interest (interest only on the principal)
I = niP where I = total interest
n = number of periods
i = interest rate(amount of interest per unit time)
P = principal
F = future worth = total amount at the end of period = P + I
F = P + I = P + niP = P (1 + in )
If P = $ 1,000, n = 2 years and I = 10%, then
F = 1,000 ( 1 + 2 x 0.10) = 1,200 = 1,000 + 100 + 100
Principalinterest interest
Year 1 Year 2
2) Compound Interest(interest on the interest as well as interest on the principal)
YearAmount +Interest =Total = F
1PiPP(1 + i)
2P(1 + i)iP(1 + i)P(1 + i)2
3P( 1 + i)2ip(1 + i)2P(1 + i)3
......
nP(1 + i)(n-1)iP(1 + i)(n-1)P(1 + i)n
If P = 1,000 and i = 10% and n=2
YearAmount +Interest=Total = F
11,0001001,100
21,1001101,210 vs 1,200 for simple interest
D) Interest Rates (Page 168+ )
1)Actual (effective) Interest
i = interest rate (amount per compounding period)
3%/quarter 1%/month12%/year0.03288%/day
2)Nominal Interest
r = interest rate/year = APR(annual percentage rate)
r = nominal interest
M = number of compounding periods per year
i = interest rate/compounding period
r = Mi
3%/quarter = 4 x 3% = 12%
1%/month = 12 x 1% = 12%
12%/year = 1 x 12% = 12%
0.03288%/day= 365 x 0.03288 = 12.0012%
3)Effective Annual Interest( ieff )
ieff = ( 1 + r/M)M -1 where r = nominal interest and m = number of compounding periods per year
for 1%/month, r = 12%, and ieff = (1 + 0.12/12)12 – 1 = 0.1268 = 12.68% instead of 12%(r)
for 0.04%/day, r = 14.6% and ieff = (1 + 0.146/365)365 -1 = 0.15716 = 15.72% versus 14.6%(r)
4)Continuous compounding
ieff = limit [( 1 + r/M)M – 1 ]= limit [ {1 + 1/(M/r)}M/r ]r -1 = er -1
M→∞ M→∞
if i = 1% /month and n = 12 months, then r = 12%(APR)
ieff = 12.68% compounded monthly
ieff = 12.75% compounded continuously
5) Types of Problems
Types of Interest
DiscreteContinuous
Discrete95%2-4%
Types of Payments Continuous0<1%
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IENG 377 Fall 2006
E) Compounding Factors
Notation: P = Present WorthA = uniform end-of-period payment i = ieff=effective annual interest
F = Future WorthG = uniform gradient amountn = number of periods
r = nominal annual interest
Compounding Factors - Discrete Payments & Discrete interest(See Table 15.1, page 591)
Payment TypeFactor NameFindGivenSymbolFormula
Single PaymentPresent WorthPF(P/F, i, n) (1 + i)- n
Future WorthFP(F/P, i, n)(1 + i)n
(compound amount)
Uniform PaymentSinking Fund AF(A/F,i,n)i / [(1 + i)n -1 ]
(Uniform Series)Capital Recovery AP(A/P,i,n)[i(1 + i)n] / [(1 + i)n -1]
Compound Amount FA(F/A,i,n)[(1 + i)n -1] / i
Present WorthPA(P/A,i,n)[(1 + i)n – 1] / [i(1 + i)n]
Uniform Gradient
Uniform Gradient Present WorthPG(P/G,i,n) {[((1+i)n-1)/(i2(1+i)n)]-[n/(i(1+i)n)]}
Uniform Gradient Future WorthFG(F/G,i,n){[((1+i)n -1)/i2 ) – [n/i] }
Uniform Gradient Uniform SeriesAG(A/G,i,n){ (1/i) - n/[(1+i)n -1] }
Payment TypeFactor NameFindGiven SymbolFormula
Geometric Gradient
Geometric Gradient Present WorthPA1,g (P/A1,g,i,n){[1- ((1+g)n/(1+i)n)]/[i-g]}
If i = g{n/(1+i)}
Geometric Gradient Future WorthFA1,g(F/A1,g,i,n){[(1+i)n- (1+g)n]/[i-g]}
If i = g{n(1+i)(n-1)}
Geometric Gradient Uniform SeriesAA1,g(A/A1,g,i,n) {[i(1+i)n - (1+g)n] / [(i-g)(1+i)n – 1]}
If i = g{[ni(1+i)n] / [(1+i)n -1]}
Notes
(g = f in text)
G is zero in years 0 and 1, the G in year 2, and 2G in year 3, ,,,, and (n-1)G in year n !!!!
To convert discrete expressions to continuous expressions, use er – 1 = i and ern = (1 + i)n
1)(F/A, i,n) = (P/A,i,n) x (F/P,i,n)
2)(P/G,i,n) = (F/G,i,n) x (P/F, i, n)
For continuous interest, use er = (1+i) and i = er -1 in expressions (similarly for g in geometric expressions (1+g) = eg and g = eg -1
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IENG 377 Fall 2006
G) Inflation, Dollars, Escalation
(1 + i) = (1 + f) (1 + x)
i = interest rate(including inflation)f = inflation ratex = inflation free interest rate
(market interest rate)
(real interest rate)
Current Dollars (future dollars, inflated dollars, nominal dollars, dollars received/expended)
Constant Dollars (inflation free dollars, real dollars, today’s dollars, deflated dollars, base year dollars)
If current dollars are used, then the interest rate used must include inflation. This is generally used.
If constant dollars are used, then the inflation free interest rate must be used. This is often done by civil engineers and government agencies; however care must be used in evaluating depreciation amounts.
If the interest rate is 10% and the inflation rate is 4%, what is the inflation free interest rate?
x = (1+i)/(1+f) –1 = (1.10)/(1.04) –1 = 1.0577 – 1 = 0.0577 = 5.77%
If the interest rate at the bank on savings is 2% and the inflation rate is 4%, what is the inflation free interest rate?
X = 1.02/1.04 – 1 = 0.981 = - .019= -1.9%
(why people don’t save money when rates are low!!!!)
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IENG 377 Fall 2006
Escalation – the same as a geometric gradient, except that the gradient occurs also in period 1.
Payment TypeFactor NameFindGiven SymbolFormula
Escalation Gradient
Escalation Gradient Present WorthPA1,g (P/A1,g,i,n){[(1+g)/(i-g)] * [1-((1+g)n/(1+i)n)]}
If i = g{n}
Escalation Gradient Future WorthFA1,g(F/A1,g,i,n){[(1+g)/(i-g)][(1+i)n- (1+g)n]}
If i = g{n(1+i)n}
Escalation Gradient Uniform SeriesAA1,g(A/A1,g,i,n) {[i(1+g)[(1+i)n-(1+g)n]]/[(i-g)[(1+i)n–1]]}
If i = g{[ni(1+i)(n-1)] / [(1+i)n -1]}
Formula Derivations
Compound Amount Factor F = A x ?
(sketch)
F = A + A(1+i) + A(1+i)2 + + A(1+i)(n-1)
= A (rn -1)/(r-1) = A [(1+i)n -1]/[(1+i) -1] = A[(1+i)n -1]/[i]
Present Worth FactorP = A x?
(sketch)
P = F/(1+i)n = { A[(1+i)n -1]/[i] } / {(1+i)n} = A[(1+i)n -1]/[i(1+i)n]
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IENG 377 Fall 2006