lec 4-eng economy - sinking fund factor and uniform- mme3109

7/27/2019 Lec 4-Eng Economy - Sinking Fund Factor and Uniform- mme3109

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Sinking fund factor and Uniform-seriescompound amount factor (A/F and F/A)

2

The simplest way to derive the A/F factoris to substitute into factors alreadydeveloped.We already know

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

1)1()1(

n

n

i

ii P A

Now if P from the first equation issubstituted in the second equation, we getthe equation as follows: (next slide)


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1)1( nii

F A

1)1()1(

)1(1

n

n

n i

ii

i F A

After simplification, the following equation is developed

The expression in brackets in the above equation isthe A/F or sinking fund factor . It determines theuniform annual series that is equivalent to agiven future worth F.

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1)1(ni

i F A

i

i A F

n 1)1(

Refer to the following equation (already developed)

The term in brackets is called uniform-seriescompound amount factor (USCAF) or F/A factor.

This equation can be rewritten as

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The cash flow diagram can be shown as


0 n-1n-221

i = given

A = ?

n

F = Given

When F/A factor is multiplied by A, it yieldsthe future worth of the uniform series, F .

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Remember that the future amount Foccurs in the same period as the lastA.

Table 2-3 summarizes the notations andequations.

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Sinking fund factor and Uniform-seriescompound amount factor (A/F and F/A )

Table 2-3: F/A and A/F Factors: Notationsand Equations

Factor Find/Given

Standard

NotationEquation

Equation with

factor formula

Excel

FunctionNotation Name

(F/A,i,n)

Uniform-seriescompoundamount

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

(A/F,i,n) Sinkingfund

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

i

i n 1)1(

1)1( niii

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Example 2.5Formasa Plastics has major fabrication plants

in Texas and Hong Kong. The president wantsto know the equivalent future worth of $1million capital investment cash for 8 years,starting 1 year from now. Formasa capitalearns at a rate of 14% per year.

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Example 2.5 (continued)Solution: The cash flow diagram shows the

annual payments starting at the end of

year 1 and ending in the year the futureworth is desired.

1 5432

i = 14%

A = $1,000,000

6

F = ?

0 7 8

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Example 2.5 (continued)Solution: The F value in 8 years is

1 5432

i = 14%

A = $1,000,000

6

F = ?

0 7 8

F = $1,000,000 (F/A,14%,8) = $1,000,000 (13.23281)

= $ 13,232,810

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Example 2.6How much money must Carol deposit every year starting

1 year from now at 5% per year in order toaccumulate $6000 seven years from now?

Solution: The cash flow diagram

1 20 543

i = 5%

A = ?

6

F = $6000

7

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A = $6000(A/F,5%,7) = $6000(0.12096)

= $725.76 per year.

The A/F factor value of 0.12096 was

computed using the factor formula

1)1(n

i

i F A

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Interpolation in interest tables

When it is necessary to locate a factorvalue for an i or n in the interest tables,the desired value can be obtained in one ofthe two ways:

(1) by using the formulas derived or(2) by linearly interpolating between thetabulated values.

However, the value obtained through linearinterpolation is not exactly correct, sincethe equations are nonlinear .

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Interpolation: calculation of the value of a function between the values already known


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Interpolation in interest tables15

Nonetheless, interpolation is sufficient in mostcases as long as the values of i and n are not toodistant from one another.

In linear interpretation it is necessary to set upthe known (values 1 and 2) and unknown factors asshown in Table 2-4:

Table 2-4: Linear Interpretation setup

tabulated value 1

desired unlistedtabulated value 2

ab d

c

Interpolation: calculation of the value of a function between the values already known


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Interpolation in interest tables16

A ratio equation is then set up and solved forthe value of unknown quantity, C

Or

Where a, b, c, and d represent the differencesbetween the numbers shown in the interesttables.

d

c

b

a

d b

ac


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Example 2.717

Determine the value of the A/P factor for an interest rate of 7.3%and n of 10 years, that is (A/P,7.3%,10)

Solution

The values of the A/P factor for interest rate of 7 and 8% and n=10are listed in interest tables (Tables 12 and 13)

7%

7.3%

8%

b

a

x

0.14903

0.14238c

d


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Example 2.718

The unknown X is the desired factor value.For the ratio equation

= 0.00199

Since the factor is increasing in value as the interest

rate increases from 7 to 8%, the value of c must beadded to the value of the 7% factor. ThusX = 0.14238+0.00199 = 0.14437.

Compare this with the exact factor value (0.144358)

14238.014903.078

73.7

d

b

ac


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Example 2.8

Find the value of the (P/F,7.3%,10)factor.

From the interest table, the values of theP/F factor for 45 and 50 years are found.

a

45

48

50

b x

0.1407

0.1712c

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Example 2.8

0183.01407.01712.045504548

d b

ac

From the equation,

Since the value of the factor decreases as n increases, c isSubtracted from the factor value for n = 45

X= 0.1712 0.0183 = 0.1529

Comment: Though it is possible to perform two-way linearInterpolation, it is much easier and more accurate to use theFactor formula or a spreadsheet function.

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Table 12 (Partial): Discrete cashflows: Compound interest factors 7%)

n

Single Payment Uniform series Payments Arithmetic Gradients

CompoundAmount F/P

PresentworthP/F

SinkingfundA/F

CompoundAmountF/A

CapitalRecoveryA/P

PresentworthP/A

GradientPres. worthP/G

Gradientuni. seriesA/G

91.8385 0.5439 0.08349 11.9780 0.15349 6.5152 23.1404 3.5517

10 1.9672 0.5083 0.07238 13.8164 0.14238 7.0236 27.7156 3.9461

11 2.1049 0.4751 0.06338 15.7836 0.13336 7.4987 32.4665 4.3296


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Table 13 (Partial): Discrete cashflows: Compound interest factors (8%)

n

Single Payment Uniform series Payments Arithmetic Gradients

CompoundAmount

F/P

PresentworthP/F

SinkingfundA/F

CompoundAmountF/A

CapitalRecoveryA/P

PresentworthP/A

GradientPres.worthP/G

Gradientuni. seriesA/G

91.990 0.5002 0.08008 12.4876 0.16008 6.2469 21.8081 3.4910

10 2.1589 0.4632 0.06903 14.4866 0.14903 6.7101 25.9768 3.8713

11 2.3316 0.4289 0.06008 16.6455 0.14008 7.1390 30.2657 4.2395

lec 4-eng economy - sinking fund factor and uniform- mme3109

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