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Chapter 2Interest Factors and Their Use

-One of the most important concept in Engineering Economy.-The study : How to use the interest factors and interest table in economycomputation.*This chapter : n = Number of year, i = Interest Rate per year2.1 Single-payment factors (F/P and P/F) / P.50 (n39)F, P, n and i RelationshipsGiven : P = Present Worth (Invest money at n=0)

Fn = Future Worth (Accumulated money after n year)(At the end of 1st year) F1 = P+P•i = P(1+i)(At the end of 2nd year) F2 = F1+F1i = P(1+i)+P(1+i)i

= P(1+i+i+i2) = P(1+i)2

(At the end of 3rd year) F3 = F2+F2i = P(1+i)3

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Then the general formula for n yearF = P(1+i)n = P(F/P, i%, n) (2.2)

The factor (1+i)n = Single Payment Compound Amount Factor (SPCAF)From Equation (2.2)

(2.3)Single Payment Present Worth Factor (SPPWF)

P=given P=?i=given i=given

0 1 2 n-2 n-1 n 0 1 2 n-2 n-1 n

(a) F=? (b) F=givenFig 2-1 / P.51 Cash flow diagrams for single-payment factor (a) find F and (b) find P.

1 ( / , %, )(1 )n

P F F P F i ni

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3

Note P, A, n, and i RelationshipsF, A, n, and i Relationships

Standard and Factor NotationX = Y(X/Y, i%, n)To find X when give Y at i% per year and n years

Example. F = P(F/P, 15%, 10) To find F when give P at 15% per year and 10 yearsSolution. F = P(F/P, 15%,10) = P(4.0456)

= P(1+0.15)10 = P(4.04555774)

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Ex. 2.3 / P.55An independent engineering consultant reviewed records and found that the costof office supplies varied as show in the pie chart of Figure 2-3. If the engineerwants to know the equivalent value in year 10 of only the three largest amount,what is it at an interest rate of 5% per year?

175$0Yr

600$300$

135$250$ 400$

1Yr2Yr

3Yr

4Yr

2 3 cos , 2.3.Figure Pie chart of ts Example

5Yr

3

5

F=?i=5%

0 1 2 3 4 5 6 7 8 9 10$300

$400$600Figure 2-4 Diagram for a future worth in year 10, Example 2.3.Solution. Draw the cash flow diagram for the values $600, $300 and $400 from theEngineer’s perspective (Figure 2-4). Use F/P factors to find F in year 10.

F = $600(F/P, 5%, 10)+$300(F/P, 5%, 8)+$400(F/P, 5%, 5)= $600(1.6289)+$300(1.4775)+$400(1.2763)= $1931.11

The problem could also be solved by finding the present worth in year 0 of the$300 and $400 costs using the P/F factors and then finding the future worth of thetotal in year 10.

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P = $600+$300(P/F, 5%, 2)+$400(P/F, 5%, 5)= $600+$300(0.9070)+$400(0.7835)= $1,185.50

F = $1,185.50(F/P, 5%, 10) = $1,185.50(1.6289)= $1,931.06

CommentIt should be obvious that there are a number of ways the problem could be worked,since any year could be used to find the equivalent total the costs before findingthe future value in year 10. As an exercise, work the problem using year 5 for theequivalent total and then determine the final amount in year 10. All answers shouldbe the same except for round-off error.

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Example.If deposit $600 now, $300 two year from now, and $400 five years from now. Howmuch account 10 years from now if i=5% per year.

i = 5% per year0 1 2 3 4 5 6 7 8 9 10

Solution. FT = F1 + F2 + F3

FT = P1(F/P, 5%, 10)+P2(F/P, 5%, 8)+P3(F/P, 5%, 5)= $600(1.6289)+$300(1.4775)+$400(1.2763)= $1,931.11

1 $600P 2 $300P

3 $400P

?TF

year

TP

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Alternative : Using FT = PT(F/P, 5%, 10)where PT = P1+P2(P/F, 5%, 2) + P3(P/F, 5%, 5)

= $600+$300(0.9070)+$400(0.7835)= $1,185.5

Then FT = $1,185.5(F/P, 5%, 10)= $1,185.5(1.6289)= $1,931.06

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2.2 Uniform-series present worth factor and capital recovery factor (P/A and A/P) / P.56(n43)The present worth of the uniform series show in Fig. 2-5 can determined by

considering each A value as a future worth F in the single-payment present-worthformula in equation (2.3) and then summing the present-worth values. The generalformula is

P=? i=given P=given i=given0 1 2 n-2 n-1 n 0 1 2 n-2 n-1 n

A=given A=?(a) (b)

Figure 2-5/P.57 Cash flow diagram used to determine (a) P of a uniform series and (b) A for a present worth.

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Where the terms in brackets represent the SPPWF for years 1 throughn, respectively. Factoring out A.

(2.4)Equation (2.4) may be simplified by multiplying both sides by 1/(1+i) toyield

(2.5)Subtracting Equation (2.4) from Equation (2.5) yields

Factoring out P and rearranging, we have

1 2 3 1

1 1 1 1 1...(1 ) (1 ) (1 ) (1 ) (1 )n nP Ai i i i i

2 3 4 1

1 1 1 1 1...1 (1 ) (1 ) (1 ) (1 ) (1 )n nP Ai i i i i i

1 1

1 11 (1 ) (1 )nP P Ai i i

1

1 1 111 (1 ) 1nP Ai i i

1 2 3 1

1 1 1 1 1...(1 ) (1 ) (1 ) (1 ) (1 )n nP A A A A Ai i i i i

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So, (2.6)

Diagram used to determine the present worth of a uniform series.P=?

i=given0 1 2 3 4 n-4 n-3 n-2 n-1 n

F1 F2 F3 F4 . . . A=given . . . . Fn-1 Fn

P = F1(P/F, i% ,1)+F2(P/F, i%, 2)+….+Fn(P/F, i%, n) ;For F1=F2=…=Fn=AP = A(P/F, i%,1)+A(P/F, i%, 2)+…+A(P/F, i%, n)

1 1; 0

(1 )

n

n

iP A i

i i

12

1 1 1( / ) ;( / ) ; ( / )( / ) ( / ) ( / )

P F F A P AF P A F A P

(P/A) = (F/A)(P/F)(A/F) = (P/F)(A/P)(A/P) = (F/P)(A/F)

F

P A

หมายเหตุ : n และ i เดียวกนั

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2.3 Sinking fund factor and uniform-series compound amount factor (A/F and F/A) P.60(n46)F=given F=?

i=given i=given0 1 2 n-2 n-1 n 0 1 2 n-2 n-1 n

A=? A=given(a) (b)

Figure 2-7/P.60 Cash flow diagrams to (a) find A, given F, and (b) find F, given A.

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Table 2.1 Computations using standard notionTo find Give Factor Equation Formula

P F (P/F, i%, n) P=F(P/F, i%, n) P=F[1/(1+i)n]F P (F/P, i%, n) F=P(F/P, i%, n) F=P(1+i)n

P A (P/A, i%, n) P=A(P/A, i%, n)A P (A/P, i%, n) A=P(A/P, i%, n)A F (A/F, i%, n) A=F(A/F, i%, n)F A (F/A, i%, n) F=A(F/A, i%, n)

Derived by multiplying the P/F and A/P factor formulas. In equation form this is,A=F(P/F, i%, n) (A/P, i%, n) = F(A/F, i%, n)

(1 ) 1(1 )

n

niP A

i i

(1 )(1 ) 1

n

ni iA Pi

(1 ) 1niA Fi

(1 ) 1niF Ai

8

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Table 2.2 Use of interest tableStandard notation i% n Table Factor value(F/A, 10%, 3) 10 3 Table 15 3.310(A/P, 7%, 20) 7 20 Table 12 0.09439(P/F, 25%, 35) 25 35 Table 25 0.0004

●In order to simplify the routine engineering economy calculations involvingthe interest factors, tables of factor values have been prepared for interest rate(i%) from 0.5 to 50% and time period (n) from 1 to 100.●This Interest Table, found in Appendix of any Engineering Economy Texts.

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Ex 2.4 / P.59 (Ex2.3 n44)How much money should you willing to pay now that will yield $600 per yearfor 9 years starting next year if i=16% per year.

A=$600

0 1 2 3 4 5 6 7 8 9 yeari= 16%

Solution. (1) P=A(P/A, 16%, 9) = $600(4.6065) = $2,763.90(2)

P=$600(P/F, 16%, 1)+$600(P/F, 16%, 2)+…+$600(P/F, 16%, 9)=$2,673.96

(3) P = A(F/A, 16%, 9) (P/F, 16%, 9)= $600(17.5185)(0.2630) = $2,764.42

?P

9

1( / ,16%, ) $600

nP F P F n where F

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Ex. 2.5 / P.61 (Resemble Ex2.5 n47)The president wants to know the equivalent future worth of a $1x103

million capital investment each year for 8 years, starting 1 year fromnow. At a rate of 14% per year.Solution. F=?

i=14% per year0 1 2 3 4 5 6 7 8 year

A=$1,000F=$1,000(F/A, 14%, 8) = $1,000(13.2328) = $13,232.80

The actual future worth is $13,232.80 million

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Ex. 2.6 / P.62How much money must Carol deposit every year starting 1 year from now 5.5%per year in order to accumulate $6,000 seven years from now?Solution. F=$6,000

i=5.5% per year0 1 2 3 4 5 6 7 year

A=?A=F(A/F, 5.5%, 7) = $6,000(0.12096) = $725.76 per year

ฝากเงินปีละ $725.76 เป็นเวลา 7 ปี จะไดรั้บเงินรวม $6,000 7 ถา้i=5.5% per year

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2.5 Arithmetic gradient factors (P/G and A/G) / P.65 (n50)-Cash flow series Same amount in each interest period- interest periodG = Uniform arithmetic change in magnitude of receipts or disbursements from one

time period to the next.Assumption : Payment that occurs at the end of period 1 (base payment) is not a part

of the gradient series.

increasedecrease

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Ex. 2.9 / P.66 (Ex2.8 n51)A sports apparel company expects to realize a revenue of $80,000 next

year from the sale of its products. However, sales expected to increaseuniformly to a level of $200,000 in 9 years. Determine the gradient andconstruct the cash-flow diagram.Solution. Total amount = $200,000

Revenue gain by year 9 = $200,000-$80,000 =$120,000

G = $15,000 per year

1GainGradientn

$120,000 $15,0009 1

per year

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Fig. 2-12 / P.66 Diagram for gradient series, Example 2.9 (Fig 2-13 n52)G=$15,000

0 1 2 3 4 5 6 7 8 9 year

$80,000 $95,000 $110,000$125,000 $140,000 $155,000 $170,000 $185,000

$200,000

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0 1 2 3 4 5 n-1 n

G 2G 3G 4G( 2)n G ( 1)n G

P=G(P/F, i%, 2)+2G(P/F, i%, 3)+3G(P/F, i%, 4)+….+[(n-2)G](P/F, i%, n-1)+[(n-1)G](P/F, i%, n)

Factoring out G yieldsP=G[(P/F, i%, 2)+2(P/F, i%, 3)+3(P/F, i%, 4)+….+(n-2)(P/F, i%, n-1)+

(n-1)(P/F, i%, n)]Replacing the symbols with the appropriate single-payment present-worth factorexpression of equation (2.3) yields.

(2.11)2 3 4 2 1

1 2 3 2 1...(1 ) (1 ) (1 ) (1 ) (1 )n n

n nP Gi i i i i

.2 11/ .66 .Fig P Conventional arithmetic gradient series without the base amount

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Multiplying both sides of equation 2.11 by (1+i)1 to simplify yields(2.12)

Subtracting Eq.(2.11) from Eq.(2.12), noting that the first term ofEq.(2.12) and the last term of Eq.(2.11) have no matching terms, yields thefollowing relations.

If we write the left side of this equation as P+P•i-P, factor out the n in the lastterm, and divide by I, we have

(2.13)Since the expression in the brackets is the present worth of a uniform series of1 for n years, we can substitute the expression for the P/A factor from Eq.(2.6)

(2.14)

11 2 3 2 1

1 2 3 2 1(1 ) ...(1 ) (1 ) (1 ) (1 ) (1 )n n

n nP i Gi i i i i

11 2 3 1

1 (2 1) (3 2) ( 1) ( 2) 1(1 ) ...(1 ) (1 ) (1 ) (1 ) (1 )n n

n n nP i P Gi i i i i

1 2 3 1

1 1 1 1 1...(1 ) (1 ) (1 ) (1 ) (1 )n n

nGi i i i i

1 2 3 1

1 1 1 1 1...(1 ) (1 ) (1 ) (1 ) (1 ) (1 )n n n

G n GnPi i i i i i i i

(1 ) 1 (1 ) 1(1 ) (1 ) (1 ) (1 )

n n

n n n nG i Gn G i nPi i i i i i i i i

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The arithmetic-gradient present worth factor, or P/G factor, may be expressedIn two forms:

or (2.15)

Remember: The gradient starts in year 2 and P is located in year 0. Equation(2.14) expressed as an engineering economy relation is

(2.16)

1 (1 ) 1( / , , )(1 ) (1 )

n

n ni nP G i n

i i i i

( / , , )P G P G i n

2

(1 ) 1( / , , )(1 )

n

ni inP G i ni i

13

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Gradient Factors●Present Worth at year 0 of the gradient G.

P=G(P/G, i%, n)Uniform Gradient Present Worth Factor (UGPWF) ~Present Worth Gradient Factor●The Equivalent Uniform Annual Worth of the Gradient; G

A = G(P/G, i%, n) (A/P, i%, n) = G(A/G, i%, n)Uniform Gradient Annual Worth Factor (UGAWF) ~Annual Worth Gradient Factor●Future Worth at year n of the Gradient; G

F = G(P/G, i%, n) (F/P, i%, n) = G(F/G, i%, n)Uniform Gradient Future Worth Factor (UGFWF) ~Future Worth Gradient Factor

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2.4 Interpolation in Interest Tables / P.63i or n FactorTabulated (Y1) Value 1 (X1)Desired (Y) Unlisted (X)Tabulated (Y2) Value 2 (X2)

ab

cd

X

Y

1y

2yy

x 2x1x cd

a b

a c ac db d b

a = y - y1 ; c = x - x1b =y2- y1 ; d = x2-x1

1 2Y Y Y

1 2X X X

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Ex. 2.7 / P.64 (A/P, 7.3%, 10) = ?Solution. From Interest Table (A/P, 7%, 10) = 0.14238

(A/P, 8%, 10) = 0.14903Setting : y1 = 7 x1 = 0.14238

y = 7.3 x = ?y2 = 8 x2 = 0.14903

a = y-y1 = 7.3-7 = 0.3b = y2-y1 = 8-7 = 1c = x-x1 = x- 0.14238d = x2-x1 = 0.14903-0.14238 = 0.00665So,

x – 0.14238 = 0.001995x = 0.14238 + 0.001995

(A/P, 7.3%, 10) = 0.144375Note. (A/P, 7.5%, 10) = =0.14571

0.3 (0.00665)1

ac db

1 ( / ,7%,10) ( / ,8%,10)2A P A P

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Linear InterpolationFrom Linear Equation : y(x) = Ax1+B ---------------(L)At x=x1 y1=Ax1+B -------------(1)At x=x2 y2=Ax2+B -------------(2)(2)-(1) : y2 – y1 = Ax2 – Ax1 = A(x2-x1)

----------------(3) Constant (Slop)From (1) y1 = Ax1+B

B = y1-Ax1Substituing B into (L)

y = Ax + y1 –Ax1Ax = y - y1 + Ax1

where A from Eq.(3)

2 1

2 1

y yAx x

11

y yx xA

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From Ex. 2.7 (A/P, 7.3 %, 10) = ?

y-y1 = 7.3 – 7 = 0.3 and x1 = 0.14238x = (0.3) (0.00665) + 0.14238 = 0.144375

From (A/P, i%, n) =Then (A/P, 7.3%, 10) =

2 1

2 1

8 7 10.00665 0.00665

y yAx x

(1 )(1 ) 1

n

ni ii

10

10

(0.073)(1 0.073) 0.14436(1 0.073) 1

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Ex. 2.8 / P.64 Find the value of (P/F, 4%, 48)From Interest Table : (P/F, 4%, 45) = 0.1712

(P/F, 4%, 48) = ?(P/F, 4%, 50) = 0.1407

Setting : y1 = 45 x1 = 0.1712y = 48 x = ?y2 = 50 x2 = 0.1407a = y – y1 = 48 – 45 = 3b = y2- y1 = 50 – 45 = 5c = x – x1 = x – 0.1712d = x2 – x1=0.1407 – 0.1712 = -0.0305

x – 0.1712 = -0.0183x = 0.1712 – 0.0183 = 0.1529

So, (P/F, 4%, 48) = 0.1529

3 ( 0.0305) 0.01835

ac db

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Linear Element Interpolation

At X=X1; ---------(1)At X=X2; ---------(2)From Eq.(1) & (2) ;

Substituing into Eq.(A)Riamanging;

1X 2X

( )y x

2Y1Y

L

2 1X X X

1 2( ) ( )y x x A

2dy slopdx

X2 1X X L

1 1 2 1Y X

2 1 2 2Y X

1 2 1 1 21 ( )X Y X YL

2 2 11 ( )Y YL

2 1 1 2 2 11 1( ) ( ) ( )y x X Y X Y Y Y XL L

2 1 1 1 21 1( ) ( ) ( )y x X X Y X X YL L

1 1 2 2( )y x N Y N Y

1 2 2 11 1( ) ; ( )N X X N X XL L

( )B

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Form Ex.2.8 (P/F, 4%, 48) =? X=48, Y=?(P/F, 4%, 45) =0.1712 X1=45, Y1=0.1712(P/F, 4%, 48) =0.1407 X2=50, Y2=0.1407

From (B) L=X2 – X1;Y=N1Y1+N2Y2

At X=48; Y=0.1529(P/F,4%, X) = (P/F, 4%, 48) = 0.1529

1 21 2 1 3(50 48) ; (48 45)5 5 5 5

N N

2 3(0.1712) (0.1407)5 5

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P, A and EUAW of Conventional G consideration / P.69(1) The base amount (A) begin in year 1 through year n.(2) For increasing G (+)

PT = PA + PG; AT = A + AG

(3) For decreasing G (-)PT = PA - PG; AT = A – AG

(PG at 2 year before G start; AG start at year at year 1 year n)

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Ex. 2.10 / P.69Depositing $500,000 1 year from now and increasing $100,000 each year for 9year thereafter if interest at 5% per year.Solution. From PT = PA + PG

PA = Present worth of the base amount (A=$500) = $500(P/A, 5%, 10)PG = Present worth of the gradient (G=$100) = $100(P/G, 5% ,10)Then PT = PA + PG

= $500(P/A, 5%, 10) + $100(P/G, 5% ,10)= $7,026.05 ($7,026,050)

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PT=?

0 1 2 3 4 5 6 7 8 9 10 year

PA=?

0 1 2 3 4 5 6 7 8 9 10 yearPG=?

0 1 2 3 4 5 6 7 8 9 10 year

$500 $600 $700 $800 $900 $1,000 $1,100 $1,200 $1,300 $1,400

$500A

$100G

$100 $200 $300 $400 $500 $600 $700 $800 $900

2 16 ( $1,000 )Figure Partitioned cash flow diagram in units

5%i per year

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Example. From Ex. 2.10Solving for Equivalent Uniform Annual Worth Series (EUAW = AT =?)From cash flow diagram AT = A + AG (For increasing G)

A = EUAW of the base amount = $500AG = EUAW of the gradient G = G(A/G, 5%, 10)Then AT = $500+$100(4.0991) = $909.91 per year ($909,910)

Alternative : From Ex. 2.10; PT = $7026.05AT = PT(A/P, 5%, 10) = $7026.05(0.12950) = $909.87 ($909,870)

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2.7 Determination of an unknown interest rate (For single payment, singlereceipt, A and G) / P.74Ex. 2.12 / P.75

If Laurel can make an investment in a friend’s business of $3,000 now in order toreceive $5,000 five years from now, what would be the rate of return on theinvestment?

i=? F=$5,0000 1 2 3 4 5

P=$3,000Solution. 1) Using (P/F) Factor

P = F(P/F, i%, n)(P/F, i%, 5) = (P/F) = (3,000/5,000) = 0.6

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From Interest Table; (P/F) factor of 0.6 for n=5 lies between 10 and 11%(P/F, 10%, 5) = 0.6209(P/F, 11%, 5) = 0.5935(P/F, i%, 5) = 0.6 i=?

By using Linear interpolating i=10.76% per year3,000 = 5,000(P/F ,10.76%, 5)

= 5,000(0.6)To verify the correctiveness of the answer

F(P/F, 10.76%, 5) = P =

= 5,000(0.5999) = 3,000 = P OK

1(1 )n

Fi

5

15,000(1 0.1076)

20

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Alternative : 2) Using (F/P) FactorF = P(F/P, i%, n)

5,000 = 3,000(F/P, i%, 5)(F/P, i%, n) = (5,000/3,000) = 1.6667

From Interest Table (F/P, 10%, 5) = 1.6105(F/P, 11%, 5) = 1.6851

By Interpolating; Setting Y1=1.6105 X1=10Y =1.6667 X =?Y2=1.6851 X2=11

Then a=Y-Y1 = 0.0562 c=X-X1=X-10b=Y2-Y1=0.0746 d=X2-X1=11-10=1

40

0.0562 (1) 0.7550.0746

ac db

5,000log(1 ) log log3,000

Fn iP

10-X = 0.755X = 10 + 0.755 =10.775% OKi = 10.76%

5,000 = 3,000(P/F, 10.76%, 5)Alternative : 3) Using (F/P) Formula

F=P(F/P, i%, n) = P(1+i)n

(1+i)n = F/PTaking log;

(1+i) = 10(0.04437) = 1.1076i = 1.1076-1 = 0.1076 = 10.76%

1,6667log(1 ) log 0.044375

i

21

41

Calculate of Unknown Years (Periods)Good for Pay off PeriodEx 2.14 / P.78How long will it take for $1,000 to double if the interest rate is 5% per year?Solution. For P=$1,000, F=$2,000, i=5%, n=?

1) Using (F/P) or (P/F) Factor Interest TableF=P(F/P, i%, n) or P=F(P/F, i%, n)

(F/P, 5%, n) = 2 or (P/F, 5%, n) = 0.5By interpolating n=14.2 years

2) Using (F/P) or (P/F) FormulaF=P(1+i)n = 2P = P(1+i)n

(1+i)n = 2 nlog(1+i) = log 2log 2 14.207

log(1 0.05)n years

FP

PF

42

From the cash flow diagram show below. Show all possible solution to compute P or F for i=10% per year$100 $100 $100 $100

0 1 2 3 4 5P=? (a)

A=$5500 1 2 3 4 5

(b) F=?A=$1,000

0 1 2 3 4 5P=? (c)

0 1 2 3(d) F=?

$100$150$150

Example.2.16 / P.80

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43

Fig a.P=100(P/F,10%,1)+100(P/A,10%,3)(P/F,10%,2) a1P=100(P/A,10%,5)-100(P/F,10%,2) a2P=[100(F/A,10%,5)-100(F/P,10%,3)](P/F,10%,5) a3P=100(P/F,10%,1)+ 100(P/F,10%,3)+ 100(P/F,10%,4)+ 100(P/F,10%,5) a4Fig b.F=550(F/A,10%,4)(F/P,10%,1) b1F=550(P/A,10%,4)(F/P,10%,5) b2F=550(F/A,10%,5)-550 b3F=550(F/P,10%,4)+550(F/P,10%,3)+ 550(F/P,10%,2)+ 550(F/P,10%,1) b4

44

Fig c.P=1,000(P/A,10%,4)(P/F,10%,1) c1P=1,000(F/A,10%,4)(P/F,10%,5) c2P=1,000(P/A,10%5)-1,000(P/F,10%,1) c3P=1,000(P/F,10%,2)+1,000(P/F,10%,3)+1,000(P/F,10%,4)+1,000(P/F,10%,5) c4Fig d.F=150(F/A,10%,3)-50(F/P,10%,2) d1F=100(F/P,10%,2)+150(F/A,10%,2) d2F=100(F/P,10%,2)+150(F/P,10%,1)+150 d3F=[100+50(A/G,10%,3)](F/A,10%,3)-50 d4F=[100+50(A/G,10%,2)](F/A,10%,2)(F/P,10%,1)-150 d5