CHAPTER 3. PRINCIPLES OF MONEYTIME RELATIONSHIPS. Objectives Of This Chapter. Describe the return to capital in the form of interest Illustrate how basic equivalence calculation are made with respect to the time value of capital in Engineering Economy. Capital.
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CHAPTER 3
PRINCIPLES OF MONEYTIME RELATIONSHIPS
Exchange money for shares of stock as proof of partial ownership
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Interest
Rate
Money Supply
MS1
ie
Money Demand
Quantity of Money
Equality in terms of Economic Value
Given a:
Present sum of money
Future sum of money
Uniform endofperiod series
Present sum of money
Uniform endofperiod series
Future sum of money
Find its:
Equivalent future value
Equivalent present value
Equivalent present value
Equivalent uniform endofperiod series
Equivalent future value
Equivalent uniform endofperiod series
Fn
………….
N
P0
= P(1+i)3
In general:
FN = P(1+i)n
FN = P(F/P,i%,n)
P = F(P/F,i%,n) where
(P/F,i%,n) = (1+i)n
Solution:
P= 1000,000 (P/F, 10%, 40)=...
P = ??
…………..
1 2 3 .. .. n1
n
0
$A per period
[1]
[2]
[2]

[1]
=
[3]
The present worth point of an annuity cash flow is always one period to the left of the first A amount
A/P,i%,n factor
P = $250,000 + $250,000(P  A, 3%, 19)
P = $250,000 + $3,580,950 = $3,830,950
$F
…………..
N
Find $A given the Future amt.  $F
$A per period
0
A = $10,000 (A  P, 1%, 60) = $222 per month
A = $1,000,000 (A  F, 10%, 40) = $
A1+n1G
A1+n2G
A1+2G
A1+G
0 1 2 3 n1 N
This represents a positive, increasing arithmetic gradient
(n1)G
(n2)G
3G
2G
1G
0G
0 1 2 3 4 ……….. n1 n
We want the PW at time t = 0 (2 periods to the left of 1G)
Multiply both sides by (1+i)

2
1
A/G,i,n =
0 1 2 3 4 5 6 7
$700
$600
$500
$400
$300
$200
$100
0 1 2 3 4 …….. n1 n
A1
A1(1+g)
A1(1+g)2
A1(1+g)3
A1(1+g)n1
Now, factor out the A1 value and rewrite as..
(1)
(2)
Subtract (1) from (2) and the result is…..
Solve for Pg and simplify to yield….
For the case i = g
0 1 2 3 4 5 6 7
$1700
$1700(1.11)1
$1700(1.11)2
$1700(1.11)3
PW(8%) = ??
$1700(1.11)5
0 1 2 3 4 5
$5,000
$3,000
Fn = $2000
0 1 2 . . . . . . ……. n
P = $1,000
i = 5%/year; n is unknown!
Fn = $2000
0 1 2 . . . . . . ……. n
P = $1,000
ieff = (1+0.169/365)365 1=0.184 or 18.4% per year
12% nominal for various compounding periods
r/M = 3% per quarter and year 3.75 = 15th Quarter
P @yr. 3.75 = P qtr. 15
= 3000(P/A, 3%, 6)  500(P/G, 3%, 6) = $9713.60
F yr. 10 = F qtr. 40
= 9713.60(F/P, 3%, 25) + 1000(F/P, 3%, 40) =
= $23,600.34
Now, examine the impact of letting “m” approach infinity.
ieff.= er – 1
Solve e0.18 – 1 = 1.1972 – 1 = 19.72%/year
The 19.72% represents the MAXIMUM effective interest rate for 18% compounded anyway you choose!
er – 1 = 0.15
er = 1.15
ln(er) = ln(1.15)
r = ln(1.15) = 0.1398 = 13.98%