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.
Related searches for CHAPTER 3
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
PRINCIPLES OF MONEYTIME RELATIONSHIPS
RENTAL FEE PAID FOR THE USE OF SOMEONE ELSES MONEY
Interest
Rate
Money Supply
MS1
ie
Money Demand
Quantity of Money
Equality in terms of Economic Value
Given a: ownership
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
Finding Equivalent Values of Cash Flows Six ScenariosF ownershipn
………….
N
P0
Derivation by Recursion: F/P factor= 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 = ?? ownership
…………..
1 2 3 .. .. n1
n
0
Uniform Series Present Worth and Capital Recovery Factors$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 ownership
…………..
N
Sinking Fund and Series Compound amount factors (A/F and F/A)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) = $
A ownership1+n1G
A1+n2G
A1+2G
A1+G
0 1 2 3 n1 N
Linear Gradient ExampleThis represents a positive, increasing arithmetic gradient
(n1)G ownership
(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)
Present Worth: Gradient ComponentMultiply both sides by (1+i)
0 1 2 3 4 5 6 7
Gradient Example$700
$600
$500
$400
$300
$200
$100
0 1 2 3 4 …….. n1 n
Geometric Gradients: IncreasingA1
A1(1+g)
A1(1+g)2
A1(1+g)3
A1(1+g)n1
Now, factor out the A1 value and rewrite as..
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
Geometric Gradient Example (+g)$5,000
$3,000
F 5 n = $2000
0 1 2 . . . . . . ……. n
P = $1,000
Unknown Number of Yearsi = 5%/year; n is unknown!
F 5 n = $2000
0 1 2 . . . . . . ……. n
P = $1,000
Unknown Number of Yearsieff = (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%