> `!tbr̘.pՌ)kjkfq¸t0@[\M١$u2]A1T$"G}~:NR A!C4?u8;y,^qB_815CbźzA ͋pxωuZslHmqIuXO(ԘX3e~՛nZAYd%2btfc #^kG'hqlX[wE2Uēczܲ8=>^iyNCR=A:
^7;Z"Aq^Se&OG0!&K+YY_71ef6_*t8QsyFepƛh.@oQBU;y@d15yo"sv]#tq@)=o
=;٬%a Ʃ8ܲ
333J
ap7(i즥Gxl^Qr@YP6Pҗ2K]"cD*]űa2^g^nQ8ϲiXMM^Bљ59b%mh/QEW9$(e'}_Yv&?@X]GEg%pY*jxC%ÿl#4߷rZHq/4'Ew~n4uk]O=vv0iwm:/6E1,/6$}w)ǹk;c~fǼPw h? H!;D
ܬnv:]i.):?+}J?
<}g}^/*QZn/n:~R:_Nd&2#̠E4MfCR>k:s9}Q
9_̖ά1YTx\VQ%Ls1#6a[~kο
ju5Rlv+#]w'2t8}uGeVLW;*y+jڋilIkLWT)\I~4I(:?[%Ԅ+]r1;oluZS#f[J\n[kege[;15ɷ=cKߍ%'
$%&'()*+,./0124568:;<2,2$t!
Compound Interest(Interest that is computed on the original unpaid debt and the unpaid interest
Compound interest is most commonly used in practice
Total interest earned = In = P (1+i)n  P
Where,
P present sum of money
i interest rate
n number of periods (years)^I
IFuture Value of a Loan With Compound Interest..(Amount of money due at the end of a loan
F = P(1+i)1(1+i)2& ..(1+i)n or F = P (1 + i)n
Where,
F = future value and P = present value
Referring to slide #10, i = 9%, P = $100 and say n= 2. Determine the value of F.)4x)
x)*'Notation forCalculating a Future Value(((Formula:
F=P(1+i)n is the
single payment compound amount factor.
Functional notation:
F=P(F/P,i,n) F=5000(F/P,6%,10)
F =P(F/P) which is dimensionally correct.
 ="+
&c*+(Notation forCalculating a Present Value))(
P=F(1/(1+i))n=F(1+i)n is the
single payment present worth factor.
Functional notation:
P=F(P/F,i,n) P=5000(P/F,6%,10)
Interpretation of (P/F, i, n): a present sum P, given a future sum, F, n interest periods hence at an interest rate i per interest period
'
#c+,Spreadsheet FunctionQP = PV(i,N,A,F,Type)
F = FV(i,N,A,P,Type)
i = RATE(N,A,P,F,Type,guess)
Where, i = interest rate, N = number of interest periods, A = uniform amount, P = present sum of money, F = future sum of money, Type = 0 means endofperiod cash payments, Type = 1 means beginningofperiod payments, guess is a guess value of the interest rate
,QZZQP
mEquivalence QRelative attractiveness of different alternatives can be judged by using the technique of equivalence
We use comparable equivalent values of alternatives to judge the relative attractiveness of the given alternatives
Equivalence is dependent on interest rate
Compound Interest formulas can be used to facilitate equivalence computations&RZ4Technique of Equivalence(Determine a single equivalent value at a point in time for plan 1.
Determine a single equivalent value at a point in time for plan 2.Z)Engineering Economic Analysis Calculation**$Generally involves compound interest formulas (factors)
Compound interest formulas (factors) can be evaluated by using one of the three methods
Interest factor tables
Calculator
Spreadsheet
6..02;Given the choice of these two plans which would you choose?<<$
P1Resolving Cash Flows to Equivalent Present Values224P = $1,000(PA,10%,5)
P = $1,000(3.791) = $3,791
P = $5,000
Alternative 2 is better than alternative 1 since alternative 2 has a greater present value<2ZZfZAn Example of Future Value([Example: If $500 were deposited in a bank savings account, how much would be in the account three years hence if the bank paid 6% interest compounded annually?
Given P = 500, i = 6%, n = 3, use F = FV(6%,3,,500,0) = 595.91
Note that the spreadsheet gives a negative number to find equivalent of P. If we find P using F = $595.91, we get P = 500.An Example of Present Value(Example 35: If you wished to have $800 in a savings account at the end of four years, and 5% interest we paid annually, how much should you put into the savings account?
n = 4, F = $800, i = 5%, P = ?
P = PV(5%,4,,800,0) = $658.16
You should use P = $658.16Economic Analysis Methods$wThree commonly used economic analysis methods are
Present Worth Analysis
Annual Worth Analysis
Rate of Return Analysis
#"Present Worth Analysis$xSteps to do present worth analysis for a single alternative (investment)
Select a desired value of the return on investment (i)
Using the compound interest formulas bring all benefits and costs to present worth
Select the alternative if its net present worth (Present worth of benefits Present worth of costs) e" 0<II $$Present Worth Analysis$4Steps to do present worth analysis for selecting a single alternative (investment) from among multiple alternatives
Step 1: Select a desired value of the return on investment (i)
Step 2: Using the compound interest formulas bring all benefits and costs to present worth for each alternative
Step 3: Select the alternative with the largest net present worth (Present worth of benefits Present worth of costs)&t't'Present Worth Analysis$A construction enterprise is investigating the purchase of a new dump truck. Interest rate is 9%. The cash flow for the dump truck are as follows:
First cost = $50,000, annual operating cost = $2000, annual income = $9,000, salvage value is $10,000, life = 10 years. Is this investment worth undertaking?
P = $50,000, A = annual net income = $9,000  $2,000 = $7,000, S = 10,000, n = 10.
Evaluate net present worth = present worth of benefits present worth of costs Present Worth Analysis$Present worth of benefits = $9,000(PA,9%,10) = $9,000(6.418) = $57,762
Present worth of costs = $50,000 + $2,000(PA,9%,10)  $10,000(PF,9%,10)= $50,000 + $2,000(6..418)  $10,000(.4224) = $58,612
Net present worth = $57,762  $58,612 < 0 do not invest
What should be the minimum annual benefit for making it a worthy of investment at 9% rate of return?Lg$H"!Present Worth Analysis$Present worth of benefits = A(PA,9%,10) = A(6.418)
Present worth of costs = $50,000 + $2,000(PA,9%,10)  $10,000(PF,9%,10)= $50,000 + $2,000(6..418)  $10,000(.4224) = $58,612
A(6.418) = $58,612 A = $58,612/6.418 = $9,312.44@:s.Cost and Benefit Estimates$Present and future benefits (income) and costs need to be estimated to determine the attractiveness (worthiness) of a new product investment alternative! %Annual costs and Income for a Product&&$Annual product total cost is the sum of annual material, labor, and overhead (salaries, taxes, marketing expenses, office costs, and related costs), annual operating costs (power, maintenance, repairs, space costs, and related expenses), and annual first cost minus the annual salvage value.
Annual income generated through the sales of a product = number of units sold annuallyxunit priceZs
%%Rate of Return Analysis$Single alternative case
In this method all revenues and costs of the alternative are reduced to a single percentage number
This percentage number can be compared to other investment returns and interest rates inside and outside the organization''Rate of Return Analysis$ Steps to determine rate of return for a single standalone investment
Step 1: Take the dollar amounts to the same point in time using the compound interest formulas
Step 2: Equate the sum of the revenues to the sum of the costs at that point in time and solve for i&FF&&Rate of Return Analysis$sAn initial investment of $500 is being considered. The revenues from this investment are $300 at the end of the first year, $300 at the end of the second, and $200 at the end of the third. If the desired return on investment is 15%, is the project acceptable?
In this example we will take benefits and costs to the present time and their present values are then equated
((Rate of Return Analysis$$500 = $300(PF, n=1, i) + 300(PF, n=2, i) + $200(PF, n=3, i)
Now solve for i using trial and error method
Try 10%: $500 = ? $272 + $247 + $156 = $669 (not equal)
Try 20%: $500 = ? $250 + $208 + $116 = $574 (not equal)
Try 30%: $500 = ? $231 + $178 + $91 = $500 (equal) i = 30%
The desired return on investment is 15%, the project returns 30%, so it should be implementedNyZ
E 0` 33` Sf3f` 33g` f` www3PP` ZXdbmo` \ғ3y`Ӣ` 3f3ff` 3f3FKf` hk]wwwfܹ` ff>>\`Y{ff` R>& {p_/̴>?" dd@,?" dd@ " @ ` n?" dd@ @@``PR @ ` `p>>f(
6 `}
T Click to edit Master title style!
!
0. `
RClick to edit Master text styles
Second level
Third level
Fourth level
Fifth level!
S
0<6 ^ `
>*
0H; ^
@*
0@@ ^ `
@*H
0h ? 3380___PPT10.=? Default Design
0zrp
(
0Œ P
P*
0V
R*
d
c$ ?
0Ē
0
RClick to edit Master text styles
Second level
Third level
Fourth level
Fifth level!
S
6t _P
P*
6b _
R*
H
0h ? 3380___PPT10.Fu$0 $(
r
Sj>j
r
Shj `
j
H
0h ? 3380___PPT10.=?$
00$(
r
SP_ `
_
r
S_ `_
H
0h ? 3380___PPT10.=$
0@$(
r
S, j `@
j
r
STjp `j
H
0h ? 3380___PPT10.CP$
0P$(
r
S8^ `
^
r
Sj `j
H
0h ? 3380___PPT10.E`GaJ$
0`$(
r
Stj `
j
r
SLj@ `j
H
0h ? 3380___PPT10.E6n6
0,6(
,~
, s*Hj `
j
x
, c$j@j
H
,0h ? ̙3380___PPT10.Ns6
0<6(
<~
< s*j `
j
x
< c$j0j
H
<0h ? ̙3380___PPT10.NG(>6
06(
~
s*`j `
j
x
c$j0j
H
0h ? ̙3380___PPT10.NN6
0@6(
@~
@ s*j `
j
x
@ c$jj
H
@0h ? ̙3380___PPT10.ETp:K'
0
'(
r
SP `p
j
r
SpQ `
LB
c$D0 0 RB
s*D0 RB
s*D0 RB
s*D0 P
P
@
0Pp7
6$1,000 2
0d w
10 2
0,0
11 2
0
12 2
0! 0
4$580 2
0%p 0W
4$540 2H
0h ? 3380___PPT10.df
0C;D(
D~
D s* `
~
D s* `u
D
<p
00
D
< 0p6
OOriginal amount to be returned = $100
Interest to be returned = $100 x .09 = $9P0P
D
<W:
00L
D
C$AclockEoH
D0h ? ̙33
0 H4(
H~
H s*`c `
x
H c$d
H
6<2f
(
$I2 = $100 x (1+.09)2  $100 = $18.81P%0H
H0h ? ̙33
00L*(
L~
L s* `}
~
L s*p `
L
s*f
F = $100 (1 + .09)2 = $118.81R0"H
L0h ? ̙33
0ZR@(
~
s*P `}
x
c$t@
RB
s*DԔ RB
s*DԔH
0h ? ̙33
0P<(
~
s* `}
~
s*D `
H
0h ? ̙33<
0`<(
~
s* `}
~
s* `
H
0h ? ̙3380___PPT10.wZd)$
0`X$(
Xr
X SH `@
r
X S @ `
H
X0h ? 3380___PPT10.L0l
0~vPT(
T~
T s*$e `}
e
x
T c$e 0 e
T
s*He @
t:Both at the same interest rate and at the same time point.;0;
T
0eZ 
`Judge the relative attractiveness of the two alternatives from the comparable equivalent values.a1a H
T0h ? ̙33$
0$(
r
Sphe `
e
r
S@ke `e
H
0h ? 3380___PPT10.^[M
0*1M(
x
c$qe `}
`
r
Sre
e
 @
1#"2&@
e
<xue?
z
N$5,000 @`
<e?z
N$5,000 @`
<e?@z
MTotal @`
<De?
4z
@ @`
<e?4
z
N$1,000 @`
<Re?@4z
I5 @`
<:e?
4
@ @`
<l5e?
4
N$1,000 @`
<ğe?@ 4
I4 @`
<4e?
@ @`
<e?
N$1,000 @`
<e?@
I3 @`
<e?
b
@ @`
<e?b
N$1,000 @`
<e?@b
I2 @`
<`e?
b
@ @`
<(d?
b
N$1,000 @`
<?@b
I1 @`
<?
N$5,000 @`
<$!?
@ @`
<)?@
I0 @`
<"?
NPlan 2 @`
<??
NPlan 1 @`
<G?@
LYear @``B
0o
?@ZB
s*1
?@ZB
s*1
?@ZB
!
s*1
?@bbZB
"
s*1
?@ZB
#
s*1
?@ ZB
$
s*1
?@44ZB
%
s*1
?@zz`B
&
0o
?@
`B
'
0o
?@@
ZB
(
s*1
?
ZB
)
s*1
?
`B
*
0o
?

0L
LTo make a choice the cash flows must be altered so a comparison may be made.M0MH
0h ? 3380___PPT10. W%o6
0t6(
tr
t S$e `
e
r
t Spe e
r
t Sep`Q e
r
t SXe p`"e
LB
t
c$D`RB
t
s*D``PRB
t
s*DP
P
RB
t
s*DppRB
t
s*DRB
t
s*DRB
t
s*DLB
t
c$D``RB
t
s*D0 `H
t0h ? 3380___PPT10.N@6
0@6(
~
s* `
x
c$
H
0h ? ̙3380___PPT10.wZ`
6
0P6(
~
s*L `
x
c$$
H
0h ? ̙3380___PPT10.zZP/$
0`$(
r
S\ `
r
S4p `
H
0h ? 3380___PPT10.\3$
0$(
r
S `
r
S `
H
0h ? 3380___PPT10.f0
00(
x
c$` `
x
c$$ `
H
0h ? 3380___PPT10.f$
0p$(
r
S0 `
r
S`1 `
H
0h ? 3380___PPT10.\ ܮ@$
0$(
r
SM `
r
SN@ `
H
0h ? 3380___PPT10.^pR$
0$(
r
S% `
r
S `
H
0h ? 3380___PPT10.e0v$
0$(
r
SD `
r
Sp `
H
0h ? 3380___PPT10.
$
0$(
r
SЋj `@
j
r
Sxj `j
H
0h ? 3380___PPT10.`Po$
0$(
r
Sl `
r
S@ `
H
0h ? 3380___PPT10.gXs$
0$(
r
S趒 `
r
S@ `
H
0h ? 3380___PPT10.i.$
0$(
r
SI `
r
SlJ `
H
0h ? 3380___PPT10.h$
0 $(
r
SP `
r
SQ `
H
0h ? 3380___PPT10.jPq_roUeIgui_km
op7rv ϓ&dp 8Lx[_@utxљ,01uU(
/0DArialNew RomanTTܖ0ܖDSymbolew RomanTTܖ0ܖ DTimes New RomanTTering EconomyBasic Concepts
Cash FlowCategories of Cash FlowsCash Flow diagramsDrawing a Cash Flow Diagram An Example of Cash Flow DiagramCash Flow DiagramTime Value of MoneyCompound Interest.Future Value of a Loan With Compound Interest(Notation for Calculating a Future Value)Notation for Calculating a Present ValueSpreadsheet Function
Equivalence Technique of Equivalence*Engineering Economic Analysis Calculation<Given the choice of these two plans which would you choose?2Resolving Cash Flows to Equivalent Present ValuesAn Example of Future ValueAn Example of Present ValueEconomic Analysis MethodsPresent Worth AnalysisPresent Worth AnalysisPresent Worth AnalysisPresent Worth AnalysisPresent Worth AnalysisCost and Benefit Estimates&Annual costs and Income for a ProductRate of Return AnalysisRate of Return AnalysisRate of Return AnalysisRate of Return AnalysisFonts UsedDesign Template
Slide Titles#_0iseise@.
@n?" dd@ @@``>%'
$%&'()*+,./0124568:;<2,2$t!
Compound Interest(Interest that is computed on the original unpaid debt and the unpaid interest
Compound interest is most commonly used in practice
Total interest earned = In = P (1+i)n  P
Where,
P present sum of money
i interest rate
n number of periods (years)^I
IFuture Value of a Loan With Compound Interest..(Amount of money due at the end of a loan
F = P(1+i)1(1+i)2& ..(1+i)n or F = P (1 + i)n
Where,
F = future value and P = present value
Referring to slide #10, i = 9%, P = $100 and say n= 2. Determine the value of F.)4x)
x)*'Notation forCalculating a Future Value(((Formula:
F=P(1+i)n is the
single payment compound amount factor.
Functional notation:
F=P(F/P,i,n) F=5000(F/P,6%,10)
F =P(F/P) which is dimensionally correct.
 ="+
&c*+(Notation forCalculating a Present Value))(
P=F(1/(1+i))n=F(1+i)n is the
single payment present worth factor.
Functional notation:
P=F(P/F,i,n) P=5000(P/F,6%,10)
Interpretation of (P/F, i, n): a present sum P, given a future sum, F, n interest periods hence at an interest rate i per interest period
'
#c+,Spreadsheet FunctionQP = PV(i,N,A,F,Type)
F = FV(i,N,A,P,Type)
i = RATE(N,A,P,F,Type,guess)
Where, i = interest rate, N = number of interest periods, A = uniform amount, P = present sum of money
!"#$%&'()*+,./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuwxz{}~yvRoot EntrydO),
PicturesCurrent User*)SummaryInformation(PowerPoint Document(5DocumentSummaryInformation8 ܖ0ܖ@.
@n?" dd@ @@``>%'
$%&'()*+,./0124568:;<2,2$t!
Compound Interest(Interest that is computed on the original unpaid debt and the unpaid interest
Compound interest is most commonly used in practice
Total interest earned = In = P (1+i)n  P
Where,
P present sum of money
i interest rate
n number of periods (years)^I
IFuture Value of a Loan With Compound Interest..(Amount of money due at the end of a loan
F = P(1+i)1(1+i)2& ..(1+i)n or F = P (1 + i)n
Where,
F = future value and P = present value
Referring to slide #10, i = 9%, P = $100 and say n= 2. Determine the value of F.)4x)
x)*'Notation forCalculating a Future Value(((Formula:
F=P(1+i)n is the
single payment compound amount factor.
Functional notation:
F=P(F/P,i,n) F=5000(F/P,6%,10)
F =P(F/P) which is dimensionally correct.
 ="+
&c*+(Notation forCalculating a Present Value))(
P=F(1/(1+i))n=F(1+i)n is the
single payment present worth factor.
Functional notation:
P=F(P/F,i,n) P=5000(P/F,6%,10)
Interpretation of (P/F, i, n): a present sum P, given a future sum, F, n interest periods hence at an interest rate i per interest period
'
#c+,Spreadsheet FunctionQP = PV(i,N,A,F,Type)
F = FV(i,N,A,P,Type)
i = RATE(N,A,P,F,Type,guess)
Where, i = interest rate, N = number of interest periods, A = uniform amount, P = present sum of money, F = future sum of money, Type = 0 means endofperiod cash payments, Type = 1 means beginningofperiod payments, guess is a guess value of the interest rate
,QZZQ>Equivalence QRelative attractiveness of different alternatives can be judged by using the technique of equivalence
We use comparable equivalent values of alternatives to judge the relative attractiveness of the given alternatives
Equivalence is dependent on interest rate
Compound Interest formulas can be used to facilitate equivalence computations&RZ4Technique of Equivalence(Determine a single equivalent value at a point in time for plan 1.
Determine a single equivalent value at a point in time for plan 2.Z)Engineering Economic Analysis Calculation**$Generally involves compound interest formulas (factors)
Compound interest formulas (factors) can be evaluated by using one of the three methods
Interest factor tables
Calculator
Spreadsheet
6..02;Given the choice of these two plans which would you choose?<<$
P1Resolving Cash Flows to Equivalent Present Values224P = $1,000(PA,10%,5)
P = $1,000(3.791) = $3,791
P = $5,000
Alternative 2 is better than alternative 1 since alternative 2 has a greater present value<2ZZfZAn Example of Future Value([Example: If $500 were deposited in a bank savings account, how much would be in the account three years hence if the bank paid 6% interest compounded annually?
Given P = 500, i = 6%, n = 3, use F = FV(6%,3,,500,0) = 595.91
Note that the spreadsheet gives a negative number to find equivalent of P. If we find P using F = $595.91, we get P = 500.An Example of Present Value(Example 35: If you wished to have $800 in a savings account at the end of four years, and 5% interest we paid annually, how much should you put into the savings account?
n = 4, F = $800, i = 5%, P = ?
P = PV(5%,4,,800,0) = $658.16
You should use P = $658.16Economic Analysis Methods$wThree commonly used economic analysis methods are
Present Worth Analysis
Annual Worth Analysis
Rate of Return Analysis
#"Present Worth Analysis$xSteps to do present worth analysis for a single alternative (investment)
Select a desired value of the return on investment (i)
Using the compound interest formulas bring all benefits and costs to present worth
Select the alternative if its net present worth (Present worth of benefits Present worth of costs) e" 0<II $$Present Worth Analysis$4Steps to do present worth analysis for selecting a single alternative (investment) from among multiple alternatives
Step 1: Select a desired value of the return on investment (i)
Step 2: Using the compound interest formulas bring all benefits and costs to present worth for each alternative
Step 3: Select the alternative with the largest net present worth (Present worth of benefits Present worth of costs)&t't'Present Worth Analysis$A construction enterprise is investigating the purchase of a new dump truck. Interest rate is 9%. The cash flow for the dump truck are as follows:
First cost = $50,000, annual operating cost = $2000, annual income = $9,000, salvage value is $10,000, life = 10 years. Is this investment worth undertaking?
P = $50,000, A = annual net income = $9,000  $2,000 = $7,000, S = 10,000, n = 10.
Evaluate net present worth = present worth of benefits present worth of costs Present Worth Analysis$Present worth of benefits = $9,000(PA,9%,10) = $9,000(6.418) = $57,762
Present worth of costs = $50,000 + $2,000(PA,9%,10)  $10,000(PF,9%,10)= $50,000 + $2,000(6..418)  $10,000(.4224) = $58,612
Net present worth = $57,762  $58,612 < 0 do not invest
What should be the minimum annual benefit for making it a worthy of investment at 9% rate of return?Lg$H"!Present Worth Analysis$Present worth of benefits = A(PA,9%,10) = A(6.418)
Present worth of costs = $50,000 + $2,000(PA,9%,10)  $10,000(PF,9%,10)= $50,000 + $2,000(6..418)  $10,000(.4224) = $58,612
A(6.418) = $58,612 A = $58,612/6.418 = $9,312.44@:s.Cost and Benefit Estimates$Present and future benefits (income) and costs need to be estimated to determine the attractiveness (worthiness) of a new product investment alternative! %Annual costs and Income for a Product&&$Annual product total cost is the sum of annual material, labor, and overhead (salaries, taxes, marketing expenses, office costs, and related costs), annual operating costs (power, maintenance, repairs, space costs, and related expenses), and annual first cost minus the annual salvage value.
Annual income generated through the sales of a product = number of units sold annuallyxunit priceZs
%%Rate of Return Analysis$Single alternative case
In this method all revenues and costs of the alternative are reduced to a single percentage number
This percentage number can be compared to other investment returns and interest rates inside and outside the organization''Rate of Return Analysis$ Steps to determine rate of return for a single standalone investment
Step 1: Take the dollar amounts to the same point in time using the compound interest formulas
Step 2: Equate the sum of the revenues to the sum of the costs at that point in time and solve for i&FF&&Rate of Return Analysis$sAn initial investment of $500 is being considered. The revenues from this investment are $300 at the end of the first year, $300 at the end of the second, and $200 at the end of the third. If the desired return on investment is 15%, is the project acceptable?
In this example we will take benefits and costs to the present time and their present values are then equated
((Rate of Return Analysis$$500 = $300(PF, i, n=1) + 300(PF, i, n=2) + $200(PF, , n=3)
Now solve for i using trial and error method
Try 10%: $500 = ? $272 + $247 + $156 = $669 (not equal)
Try 20%: $500 = ? $250 + $208 + $116 = $574 (not equal)
Try 30%: $500 = ? $231 + $178 + $91 = $500 (equal) i = 30%
The desired return on investment is 15%, the project returns 30%, so it should be implementedNxZ
D8?$
0 $(
r
SDt `
r
Su `
H
0h ? 3380___PPT10.jPq_r(U0(201wU(
/0DArialNew RomanTTܖ0ܖDSymbolew RomanTTܖ0ܖ DTimes New RomanTTܖ0ܖ
!"#$%&'()Oh+'0`h
Engineering Economicsiseise53Microsoft Office PowerPoint@\:@0Ӎ=@K,"Gg l'y$xx'@Arial. 2
q17."Systemx7@Arial. '2
6Engineering Economics.@Arial. 2
K3November 3, 2004.՜.+,0
Onscreen Showsjsu5#'ArialSymbolTimes New RomanDefault DesignEngineering EconomicsEngineering EconomyEngine, F = future sum of money, Type = 0 means endofperiod cash payments, Type = 1 means beginningofperiod payments, guess is a guess value of the interest rate
,QZZQ>Equivalence QRelative attractiveness of different alternatives can be judged by using the technique of equivalence
We use comparable equivalent values of alternatives to judge the relative attractiveness of the given alternatives
Equivalence is dependent on interest rate
Compound Interest formulas can be used to facilitate equivalence computations&RZ4Technique of Equivalence(Determine a single equivalent value at a point in time for plan 1.
Determine a single equivalent value at a point in time for plan 2.Z)Engineering Economic Analysis Calculation**$Generally involves compound interest formulas (factors)
Compound interest formulas (factors) can be evaluated by using one of the three methods
Interest factor tables
Calculator
Spreadsheet
6..02;Given the choice of these two plans which would you choose?<<$
P1Resolving Cash Flows to Equivalent Present Values224P = $1,000(PA,10%,5)
P = $1,000(3.791) = $3,791
P = $5,000
Alternative 2 is better than alternative 1 since alternative 2 has a greater present value<2ZZfZAn Example of Future Value([Example: If $500 were deposited in a bank savings account, how much would be in the account three years hence if the bank paid 6% interest compounded annually?
Given P = 500, i = 6%, n = 3, use F = FV(6%,3,,500,0) = 595.91
Note that the spreadsheet gives a negative number to find equivalent of P. If we find P using F = $595.91, we get P = 500.An Example of Present Value(Example 35: If you wished to have $800 in a savings account at the end of four years, and 5% interest we paid annually, how much should you put into the savings account?
n = 4, F = $800, i = 5%, P = ?
P = PV(5%,4,,800,0) = $658.16
You should use P = $658.16Economic Analysis Methods$wThree commonly used economic analysis methods are
Present Worth Analysis
Annual Worth Analysis
Rate of Return Analysis
#"Present Worth Analysis$xSteps to do present worth analysis for a single alternative (investment)
Select a desired value of the return on investment (i)
Using the compound interest formulas bring all benefits and costs to present worth
Select the alternative if its net present worth (Present worth of benefits Present worth of costs) e" 0<II $$Present Worth Analysis$4Steps to do present worth analysis for selecting a single alternative (investment) from among multiple alternatives
Step 1: Select a desired value of the return on investment (i)
Step 2: Using the compound interest formulas bring all benefits and costs to present worth for each alternative
Step 3: Select the alternative with the largest net present worth (Present worth of benefits Present worth of costs)&t't'Present Worth Analysis$A construction enterprise is investigating the purchase of a new dump truck. Interest rate is 9%. The cash flow for the dump truck are as follows:
First cost = $50,000, annual operating cost = $2000, annual income = $9,000, salvage value is $10,000, life = 10 years. Is this investment worth undertaking?
P = $50,000, A = annual net income = $9,000  $2,000 = $7,000, S = 10,000, n = 10.
Evaluate net present worth = present worth of benefits present worth of costs Present Worth Analysis$Present worth of benefits = $9,000(PA,9%,10) = $9,000(6.418) = $57,762
Present worth of costs = $50,000 + $2,000(PA,9%,10)  $10,000(PF,9%,10)= $50,000 + $2,000(6..418)  $10,000(.4224) = $58,612
Net present worth = $57,762  $58,612 < 0 do not invest
What should be the minimum annual benefit for making it a worthy of investment at 9% rate of return?Lg$H"!Present Worth Analysis$Present worth of benefits = A(PA,9%,10) = A(6.418)
Present worth of costs = $50,000 + $2,000(PA,9%,10)  $10,000(PF,9%,10)= $50,000 + $2,000(6..418)  $10,000(.4224) = $58,612
A(6.418) = $58,612 A = $58,612/6.418 = $9,312.44@:s.Cost and Benefit Estimates$Present and future benefits (income) and costs need to be estimated to determine the attractiveness (worthiness) of a new product investment alternative! %Annual costs and Income for a Product&&$Annual product total cost is the sum of annual material, labor, and overhead (salaries, taxes, marketing expenses, office costs, and related costs), annual operating costs (power, maintenance, repairs, space costs, and related expenses), and annual first cost minus the annual salvage value.
Annual income generated through the sales of a product = number of units sold annuallyxunit priceZs
%%Rate of Return Analysis$Single alternative case
In this method all revenues and costs of the alternative are reduced to a single percentage number
This percentage number can be compared to other investment returns and interest rates inside and outside the organization''Rate of Return Analysis$ Steps to determine rate of return for a single standalone investment
Step 1: Take the dollar amounts to the same point in time using the compound interest formulas
Step 2: Equate the sum of the revenues to the sum of the costs at that point in time and solve for i&FF&&Rate of Return Analysis$sAn initial investment of $500 is being considered. The revenues from this investment are $300 at the end of the first year, $300 at the end of the second, and $200 at the end of the third. If the desired return on investment is 15%, is the project acceptable?
In this example we will take benefits and costs to the present time and their present values are then equated
((Rate of Return Analysis$$500 = $300(PF, i, n=1) + 300(PF, i, n=2) + $200(PF, i, n=3)
Now solve for i using trial and error method
Try 10%: $500 = ? $272 + $247 + $156 = $669 (not equal)
Try 20%: $500 = ? $250 + $208 + $116 = $574 (not equal)
Try 30%: $500 = ? $231 + $178 + $91 = $500 (equal) i = 30%
The desired return on investment is 15%, the project returns 30%, so it should be implementedNyZ
E8?$
0 $(
r
SDt `
r
Su `
H
0h ? 3380___PPT10.jPq_r2(<(2h01UU(
/0DArialNew RomanTTܖ0ܖDSymbolew RomanTTܖ0ܖ DTimes New RomanTTܖ0ܖ@.
@n?" dd@ @@``>%'
$%&'()*+,./0124568:;<2,2$t!
Compound Interest(Interest that is computed on the original unpaid debt and the unpaid interest
Compound interest is most commonly used in practice
Total interest earned = In = P (1+i)n  P
Where,
P present sum of money
i interest rate
n number of periods (years)^I
IFuture Value of a Loan With Compound Interest..(Amount of money due at the end of a loan
F = P(1+i)1(1+i)2& ..(1+i)n or F = P (1 + i)n
Where,
F = future value and P = present value
Referring to slide #10, i = 9%, P = $100 and say n= 2. Determine the value of F.)4x)
x)*'Notation forCalculating a Future Value(((Formula:
F=P(1+i)n is the
single payment compound amount factor.
Functional notation:
F=P(F/P,i,n) F=5000(F/P,6%,10)
F =P(F/P) which is dimensionally correct.
 ="+
&c*+(Notation forCalculating a Present Value))(
P=F(1/(1+i))n=F(1+i)n is the
single payment present worth factor.
Functional notation:
P=F(P/F,i,n) P=5000(P/F,6%,10)
Interpretation of (P/F, i, n): a present sum P, given a future sum, F, n interest periods hence at an interest rate i per interest period
'
#c+,Spreadsheet FunctionQP = PV(i,N,A,F,Type)
F = FV(i,N,A,P,Type)
i = RATE(N,A,P,F,Type,guess)
Where, i = interest rate, N = number of interest periods, A = uniform amount, P = present sum of money, F = future sum of money, Type = 0 means endofperiod cash payments, Type = 1 means beginningofperiod payments, guess is a guess value of the interest rate
,QZZQ>Equivalence QRelative attractiveness of different alternatives can be judged by using the technique of equivalence
We use comparable equivalent values of alternatives to judge the relative attractiveness of the given alternatives
Equivalence is dependent on interest rate
Compound Interest formulas can be used to facilitate equivalence computations&RZ4Technique of Equivalence(Determine a single equivalent value at a point in time for plan 1.
Determine a single equivalent value at a point in time for plan 2.Z)Engineering Economic Analysis Calculation**$Generally involves compound interest formulas (factors)
Compound interest formulas (factors) can be evaluated by using one of the three methods
Interest factor tables
Calculator
Spreadsheet
6..02;Given the choice of these two plans which would you choose?<<$
P1Resolving Cash Flows to Equivalent Present Values224P = $1,000(PA,10%,5)
P = $1,000(3.791) = $3,791
P = $5,000
Alternative 2 is better than alternative 1 since alternative 2 has a greater present value<2ZZfZAn Example of Future Value([Example: If $500 were deposited in a bank savings account, how much would be in the account three years hence if the bank paid 6% interest compounded annually?
Given P = 500, i = 6%, n = 3, use F = FV(6%,3,,500,0) = 595.91
Note that the spreadsheet gives a negative number to find equivalent of P. If we find P using F = $595.91, we get P = 500.An Example of Present Value(Example 35: If you wished to have $800 in a savings account at the end of four years, and 5% interest we paid annually, how much should you put into the savings account?
n = 4, F = $800, i = 5%, P = ?
P = PV(5%,4,,800,0) = $658.16
You should use P = $658.16Economic Analysis Methods$wThree commonly used economic analysis methods are
Present Worth Analysis
Annual Worth Analysis
Rate of Return Analysis
#"Present Worth Analysis$xSteps to do present worth analysis for a single alternative (investment)
Select a desired value of the return on investment (i)
Using the compound interest formulas bring all benefits and costs to present worth
Select the alternative if its net present worth (Present worth of benefits Present worth of costs) e" 0<II $$Present Worth Analysis$4Steps to do present worth analysis for selecting a single alternative (investment) from among multiple alternatives
Step 1: Select a desired value of the return on investment (i)
Step 2: Using the compound interest formulas bring all benefits and costs to present worth for each alternative
Step 3: Select the alternative with the largest net present worth (Present worth of benefits Present worth of costs)&t't'Present Worth Analysis$A construction enterprise is investigating the purchase of a new dump truck. Interest rate is 9%. The cash flow for the dump truck are as follows:
First cost = $50,000, annual operating cost = $2000, annual income = $9,000, salvage value is $10,000, life = 10 years. Is this investment worth undertaking?
P = $50,000, A = annual net income = $9,000  $2,000 = $7,000, S = 10,000, n = 10.
Evaluate net present worth = present worth of benefits present worth of costs Present Worth Analysis$Present worth of benefits = $9,000(PA,9%,10) = $9,000(6.418) = $57,762
Present worth of costs = $50,000 + $2,000(PA,9%,10)  $10,000(PF,9%,10)= $50,000 + $2,000(6..418)  $10,000(.4224) = $58,612
Net present worth = $57,762  $58,612 < 0 do not invest
What should be the minimum annual benefit for making it a worthy of investment at 9% rate of return?Lg$H"!Present Worth Analysis$Present worth of benefits = A(PA,9%,10) = A(6.418)
Present worth of costs = $50,000 + $2,000(PA,9%,10)  $10,000(PF,9%,10)= $50,000 + $2,000(6..418)  $10,000(.4224) = $58,612
A(6.418) = $58,612 A = $58,612/6.418 = $9,312.44@:s.Cost and Benefit Estimates$Present and future benefits (income) and costs need to be estimated to determine the attractiveness (worthiness) of a new product investment alternative! %Annual costs and Income for a Product&&$Annual product total cost is the sum of annual material, labor, and overhead (salaries, taxes, marketing expenses, office costs, and related costs), annual operating costs (power, maintenance, repairs, space costs, and related expenses), and annual first cost minus the annual salvage value.
Annual income generated through the sales of a product = number of units sold annuallyxunit priceZs
%%Rate of Return Analysis$Single alternative case
In this method all revenues and costs of the alternative are reduced to a single percentage number
This percentage number can be compared to other investment returns and interest rates inside and outside the organization''Rate of Return Analysis$ Steps to determine rate of return for a single standalone investment
Step 1: Take the dollar amounts to the same point in time using the compound interest formulas
Step 2: Equate the sum of the revenues to the sum of the costs at that point in time and solve for i&FF&&Rate of Return Analysis$sAn initial investment of $500 is being considered. The revenues from this investment are $300 at the end of the first year, $300 at the end of the second, and $200 at the end of the third. If the desired return on investment is 15%, is the project acceptable?
In this example we will take benefits and costs to the present time and their present values are then equated
((Rate of Return Analysis$$500 = $300(PF, i, n=1) + 300(PF, i, n=2) + $200(PF, i, n=3)
Now solve for i using trial and error method
Try 10%: $500 = ? $272 + $247 + $156 = $669 (not equal)
Try 20%: $500 = ? $250 + $208 + $116 = $574 (not equal)
Try 30%: $500 = ? $231 + $178 + $91 = $500 (equal) i = 30%
The desired return on investment is 15%, the project returns 30%, so it should be implementedNyZ
Er
01