engineering economy chapter 4: the time value of money the objective of chapter 4 is to explain time...
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Engineering Economy
Chapter 4: The Time Value of Money
The objective of Chapter 4 is to explain time value of money calculations and to illustrate economic equivalence.
Money has a time value.
• Capital refers to wealth in the form of money or property that can be used to produce more wealth.
• Engineering economy studies involve the commitment of capital for extended periods of time.
Simple Interest: infrequently used
When the total interest earned or charged is linearly proportional to the initial amount of the loan (principal), the interest rate, and the number of interest periods, the interest and interest rate are said to be simple.
Computation of simple interest
The total interest, I, earned or paid may be computed using the formula below.
P = principal amount lent or borrowed
N = number of interest periods (e.g., years)
i = interest rate per interest period
The total amount repaid at the end of N interest periods is P + I.
Example:
If $5,000 were loaned for five years at a simple interest rate of 7% per year, the interest earned would be
So, the total amount repaid at the end of five years would be the original amount ($5,000) plus the interest ($1,750), or $6,750.
Compound interest reflects both the remaining principal and any accumulated interest. For $1,000 at 10%…
Compound interest: frequently used
Period
(1)
Amount owed at beginning of
period
(2)=(1)x10%
Interest amount for period
(3)=(1)+(2)
Amount owed at end of
period
1 $1,000 $100 $1,100
2 $1,100 $110 $1,210
3 $1,210 $121 $1,331
Economic equivalence allows us to compare alternatives on a common basis.
• Each alternative can be reduced to an equivalent basis dependent on– interest rate,– amount of money involved, and– timing of monetary receipts or expenses.
• Using these elements we can “move” cash flows so that we can compare them at particular points in time.
Notation used in formulas for compound interest calculations.
i = effective interest rate per interest periodN = number of compounding (interest) periodsP = present sum of money; equivalent value of one or more cash flows at a reference point in time; the presentF = future sum of money; equivalent value of one or more cash flows at a reference point in time; the futureA = end-of-period cash flows in a uniform series continuing for a certain number of periods, starting at the end of the first period and continuing through the last
A cash flow diagram is a tool for clarifying and visualizing a series of cash flows.
Finding F when given P
),,/( % NiPFPF
NiPF )1(
Single payment compound amount factor Ni )1( ),,/( % NiPF
$8.712,114641.18000
)1.1(8000)1.01(8000 44
F
F
NiPF )1(
),,/( % NiPFPF
$8.712,11)4641.1(8000
)4,10,/(8000 %
F
PFF
Finding P when given F
),,/( % NiFPFP
NiPF )1( From
NiFP )1(
Single payment present worth factor Ni )1( ),,/( % NiFP
NiFP )1(
$302,66302.010000
)08.1(10000)08.01(10000 66
P
P
),,/( % NiFPFP
$302,6)6302.0(10000
)6,8,/(10000 %
P
FPP
Finding Interest rate (i) when given P, F, and N
1/ /1 NPFi
NN PFiiPF /1)/()1()1(
Finding N when given P, F, and i
)/log()1(
)/()1()1(
log PFi
PFiiPF
N
NN
)1(
)/log(
log i
PFN
1/ /1 NPFi
107.1/31.2 12/1 i
yearper%62.6
0662.010662.1
i
i
)1(
)/log(
log i
PFN
years05.12
)0662.1(
)1645.2log(
)0662.01(
)31.2/5log(
loglog
N
N
Interpolation Example:
suppose you wish to have 100,000$ after 27 years, what amount should be deposited now to provide for it at 7% interest rate per year.
),,/( % NiFPFP )27,7,/(000,100 %FPP
1314.0)30,7,/(
1842.0)25,7,/(
%
%
FP
FP
25 0.184227 x30 0.1314 3025
1314.01842.0
3027
1314.0
x
)27,7,/(1631.0 %FPx
$163101631.0000,100 P
NiFP )1(
$16093
1609.0000,100)07.01(000,100 27
P
P
Example:
suppose you wish to have 100,000$ after 27 years, what amount should be deposited now to provide for it at 7% interest rate per year.
Interpolation Example:
suppose you wish to have 100,000$ after 20 years, what amount should be deposited now to provide for it at 5.5% interest rate per year.
),,/( % NiFPFP )20,5.5,/(000,100 %FPP
3118.0)20,6,/(
3769.0)20,5,/(
%
%
FP
FP
5 0.37695.5 x6 0.3118
65
3118.03769.0
65.5
3118.0
x
)20,5.5,/(3444.0 %FPx
$270,343427.0000,100)055.1(000,100)1( 20 NiFP
$440,343444.0000,100 P
Finding F when given A
i
iAF
N 1)1(
),,/( % NiAFAF
Uniform series compound amount factor i
i N 1)1( ),,/( % NiAF
$526,559,3)762.154(23000
)40,6,/(23000
),,/(
%
%
F
AFF
NiAFAF
Finding P when given A
N
N
ii
iAP
)1(
1)1(
),,/( % NiAPAP
Uniform series present worth factor N
N
ii
i
)1(
1)1(
),,/( % NiAP
$54.490)3514.16(30
)20,2,/(30
),,/(
%
%
P
APP
NiAPAP
Finding A when given F
1)1( Ni
iFA
),,/( % NiFAFA
Uniform series sinking fund factor 1)1(
Ni
i),,/( % NiFA
Example:
What uniform annual amount should be deposited each year in order to accumulate 2143.6$ at the time of the eighth annual deposit given i = 10%
$45.187)0874.0(6.2143
)8,10,/(6.2143
),,/(
%
%
A
FAA
NiFAFA
Finding A when given P
1)1(
)1(N
N
i
iiPA
),,/( % NiPAPA
Uniform series capital recovery factor 1)1(
)1(
N
N
i
ii),,/( % NiPA
$5.436)0291.0(15000
)36,25.0,/(15000
),,/(
%
%
A
PAA
NiPAPA
Finding The number of cash flows in an annuity (N)
1. Given A, P, and i
A
PNiAP ),,/( %
2. Given A, F, and i
A
FNiAF ),,/( %
iLogPiA
A
N
1
log
iLogA
AFi
N
1
log
),,/( % NiAPAP ),8,/(000,10000,100 % NAP
10),8,/( % NAP
8181.9)20,8,/( % AP
0168.10)21,8,/( % AP
$98181000,108181.9 P
$100168000,100168.10 P
10,000$ for 20 years $98181P
21th payment equivalents to 1819$ (100,000-98,181) in present
$5.156,90338.51819)21,8,/(1819 % PFF
iLogPiA
A
N
1
log
years91.20)08.1log(
)5log(
08.108.0000,100000,10
000,10log
LogN
Finding The interest rate (i) given A, N, and F or P
N
N
ii
iAP
)1(
1)1(Can’t be solved for i
i
iAF
N 1)1(Can’t be solved for i
Finding The interest rate (i) given A, N, and F or P
),,/( % NiAFAF
10)8,,/( % iAF
)8,,/(000,6000,60 %iAF
8975.9)8,6,/( % AF
2598.10)8,7,/( % AF
i6% 9.8975
i 10
7% 10.2598
)8,,/( %iAF
Linear Interpolation
2598.108975.9
76
2598.1010
7
i
i = 6.28%
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