INTEREST FORMULASUNIT III

3.1 Simple and compound interest : The rental rate for a sum of money is usually expressed as the percent of the sum that is to be paid for its use for a period one year Interest rates are also quoted for periods other than one year ,known as interest periods.This section compares simple and compound interest approaches for determining the effect of the time value of money.3.1.1 Simple interest: Under simple interest ,the interest owed upon repayment of a loan is proportional to the length of time the principal sum has been borrowedThe interest earned may be found in the following manner.

Let I represent the interest earned , P the principal amount , n the interest period and i the interest rate. Then I= PniSuppose that $ 1000 is borrowed at a simple interest rate of 12% per annum. At the end of one year ,the interest owed would be I= $1000(1)(0.12)=$120 The principal plus interest would be $1120 and would be due at the end of the year.3.1.2 Compound interest : When a loan is made for several interest periods ,interest is calculated and payable at the end of each interest period.

There are a number of loan repayment plansThese range from paying the interest when it is due to accumulating the interest until the loan is due. For example ,the payments on a 4 year loan of $1000 at 16% interest per annum , payable when due ,would be calculated as shown in Table 3.1If the borrower does not pay the interest earned at the end of each period and is charged interest on the total amount owed ( principal plus interest ) the interest is said to be compounded.The interest owed in the previous year becomes part of the total amount owed for this year.

TABLE 3.1 CALCULATION OF COMPOUND INTEREST WHEN INTEREST IS PAID ANNUALLYTable 3.2 CALCULATION OF COMPOUND INTEREST WHEN INTEREST IS PERMITTED TO COMPOUND

YearAmount owed at beginning of yearInterest to be paid at end of yearAmount owed at end of yearAmount to be paid by borrower at end of year1$1,000.00$160.00$1,160.00$160.002$1,000.00$160.00$1,160.00$160.003$1,000.00$160.00$1,160.00$160.004$1,000.00$160.00$1,160.00$1,160.00

YearAmount owed at beginning of year (A)Interest to be added to loan at end of year (B)Amount owed at end of year(A+B)Amount paid by borrower at end of year1$1,000.00$1000X0.16=160$1000(1.16)=$1,160.00$00.002$1,160.00$1160.00X0.16=185.60$1000(1.16)2=1,345.60 00.003$1,345.60$1,345.60X0.16=215.30$1000 (1.16)3=1560.90 00.004$1560.90$1560.90X0.16=249.75$1000(1.16)4=1810.64$1,810.64

This years interest charge includes interest that has been earned on previous charges.For example, a loan of $1000 at 16% interest compounded annually for 4 year period will produce the result shown in Table 3.2Although the financial arrangements shown in Tables 3.1 and 3.2 require that the interest be calculated on the unpaid balance ,the two cases produce different results because of the way payments are made.In the first case, payment of interest at the time it is due avoids the payment of interest on interest.The reverse is true in the second payment scheme.Thus , the effect of compound interest depends upon the payment amounts and when they are made.

3.2 DESCRIBING CASH FLOWS OVER TIME:In most engineering economy studies ,only small elements of an enterprise are considered.For example, studies are often made to evaluate the consequences of the purchase of a single equipment item in a complex of many facilities.In such cases it would be desirable to isolate the individual item from the whole by some means analogous to the free-body diagram in mechanics.Thus it would be necessary to itemize all receipts and all disbursements that would arise from the acquisition and operation of the equipment being considered.

The cash flow is the actual inflow (receipts) and outflow (disbursements) at different points in time that occur over the life of an investment.To aid in identifying and recording the economic effects of investment alternatives, a graphical description of each alternative s cash transactions may be used.This graphical description, referred to as such a flow diagram, will provide the information necessary for analyzing an investment proposal.

A cash flow diagram represents receipts received during a period of time by an upward arrow ( an increase in cash) located at the period s end. The arrows height may be proportional to the magnitude of the receipts during that period.Similarly , disbursements during a period are represented by a downward arrow ( a decrease in cash)These arrows are then placed on a time scale that spans all time periods covered by the proposed investment.

$1000$1000 $1160$1160$160$160$160$160$160$160 BorrowerLenderFigure 3.1 Cash flow diagramAs an example ,consider the cash flow diagram in Figure 3.1,which pertain to the simple loan transaction described in Table 3.1.In this example the borrower receives $1000 ,and this amount appears as a positive cash flow on the borrowers cash flow diagram.Each year the borrower pays $160 in interest ; these amounts plus repayment of the $1000 borrowed ,appear as negative cash flows. Also shown in Figure 3.1 is the lenders cash flow diagramThe lender experiences a negative cash flow of $1000 ,followed by positive cash flows for interest received and also for repayment of the original amount loaned.

Since there are two parties to every transaction, it is important to note that the cash flow diagram directions in cash flow diagram depend upon the point of view taken.When an investment alternative has both receipts and disbursements occurring simultaneously, a net cash flow may be calculated.Net cash flow is the arithmetic sum of the receipts (+) and the disbursements (-) that occur at the same point in time.The utilization of net cash flow implies that the net dollars received or disbursed have the same effect on an investment decision as an investments total receipts and disbursements considered separately.

To facilitate describing investment cash flows ,the following notation will be adopted.Let Ft= net cash flow at time t where Ft 0 represents a net cash receiptIn engineering economy studies, disbursements made to implement an alternative are considered to take place at the beginning of the period embraced by the alternative.Receipts and disbursements occurring during the life of the alternative are usually assumed to occur at the end of the year or interest period in which they occur.This year end convention is adopted for describing cash flows over time and for developing applicable cash flow diagram

3.3 INTEREST FORMULAS (DISCRETE COMPOUNDING, DISCRETE PAYMENTS):The interest formulas derived in this section apply to the common situation of annual compounding interest and annual payments The following symbols will be used .Let i=annual interest raten=the number of annual interest periodsP= a present principal amount;A= a single payment ,in a series of n equal payments ,made at the end of each annual interest periodF= a future amount ,n annual interest periods hence.

Four important points apply in the derivation and use of interest factors for annual payments1.The end of one year is the beginning of the next year2. P is at the beginning of a year regarded as being the present3.F is at the end of the n th year from a time regarded as being the present.4.An A occurs at the end of each year of the period under considerationWhen P and A are involved, the first A of the series occurs one year after P.When F and A are involved ,the last A of the series occurs simultaneously with F.

3.3.1 Single payment compound amount factor ,(F/P,i,n)If an amount P is invested now and earns at the rate i per year ,how much principal and interest are accumulated after n years? The cash flow diagram for this situation is shown in Figure 3.2Since this transaction does not provide any payments until the investment is terminated ,interest is compounded as shown in Table 3.2There the interest earned is added to the principal at the end of each annual interest period.By substituting general terms in place of numerical values in Table 3.2 ,the results shown in Table 3.3 are obtained.The resulting factor , (1+i)n ,is known as the single payment compound amount factor and designated (F/P , i , n)

nn-1 1 2 3FP0Figure 3.2 Single present amount and single future amountThis factor may be used to find the future amount ,F, of a present principal amount ,P.The relationship is F= P (1+i)n or F= P (F/P , i , n)

TABLE 3.2 DERIVATION OF SINGLE PAYMENT COMPOUND AMOUNT FACTOR

YearAmount at beginning of yearInterest earned during yearCompound amount at end of year1PP iP +Pi = P(1+i)12P(1+i)P(1+i)iP(1+i)+P(1+i) i =P(1+i)23P(1+i)2P(1+i)2.iP(1+i)2+P(1+i)2.i =P(1+i)3nP(1+i)n-1P(1+i)n-1.iP(1+i)n-1+P (1+i)n-1.i=P(1+i)n =F

The designer used to identify the single payment compound -amount factor is (F/P, i, n)It appears over the parentheses where the value of the factor is to be entered.The first element in the designator ,F/P ,represents a ratio that identifies what the factor must be multiplied by P in order to find F.Thus , a P value must be multiplied by an F/P designated factor to find the F value that is equivalent F= P( F/P).The i represents the interest rate per period and the n the number of periods between the occurrence of P and F.

Problem: If $1000 is invested at 16% interest compounded annually at the beginning of year one , the compound amount at the end of the fourth year will be F=$1000(1+0.16)4=$1000(1.811)=$1,811 or by the use of the factor designation and its associated tabular value, (F/P,16,4) F=$1000( 1.811)=$1,811

3.3.2 Single payment present worth factor (P/F,i,n)The singlepayment compound amount relationship i.e F=P(1+i)n may be solved for P as follows : P=F{1/(1+i)n}The resulting factor ,1/(1+i)n is known as the singlepayment present worth factor and is designated (P/F , i, n) This factor may be used to find the present worth ,P ,of a future amount ,F.

Problem : How much must be invested now at 16% compounded annually so that $ 1811 can be received 4 years hence?This calculation is P=1811 {1/(1+0.16)4}=1811(0.5523)=$1000 or by using the factor designation and interest tables P/F,16,4 P= $1811(0.5523)= $10003.3.3 Equal PaymentSeries Compound Amount Factor(F/A,i,n) :In some engineering economy studies , it is necessary to find the single future value that would accumulate from a series of equal payments occurring at the end of succeeding interest periods. Such a series of cash flows is presented in Figure 3.3

The sum of the compound amounts of the several payments may be calculated by use of the single-payment compound amount factor.For example the calculation of the compound amount of a series of five $100 payments made at the end of each year at 12% interest compounded annually is shown in Table 3.4It is apparent that the tabular method is cumbersome for calculating the compound amount for an extensive series.Therefore it is desirable to derive a compact solution for this type of situation.If A represents a series of n equal payments , such as the $ 100 series in Table 3.4,then F= A(1)+A(1+i) +..+ A( 1+i) n-2 +A (1+i) n-1The total future amount ,F, is equal to the sum of individual future amounts calculated for each payment , A.

0 1 2 3 n-1 n A A A A AF Figure 3.3 Equal annual series and single future amount TABLE 3.4 THE COMPOUND AMOUNT OF A SERIES OF YEAR END PAYMENTS

End of yearYear end payment times compound amount factorCompound amount at end of 5 yearsTotal compound amount1$100 (1.12)0$100 2$100(1.12)1$112.00

3$100 (1.12)2$125.444S100 (1.12)3 $140.495S100 ( 1.12)4$157.35 $635.28

Multiplying this equation by (1+i) results in F(1+i) = A(1+i)+A(1+i)2+.+A(1+i)n-1+A(1+i)nSubtracting the first equation from the second gives F(1+i) = A(1+i)+A(1+i)2+.+A(1+i)n-1+A(1+i)n -F=-A-A(1+i)-A(1+i)2+.-A(1+i)n-1F(1+i)-F= - A + A (1+i)nSolving for F gives F={-A+A(1+i)n}/i=A{-1+(1+i)n}/i(3.4)The resulting factor {(1+i)n-1}/ i is called equal payment series- compound- amount- factor and designated ( F/A ,i, n)

This factor may be used to find the compound amount ,F, of an equal payment series ,A.For example ,the future amount of a $100 payment deposited at the end of the next 5 years and earning 12% per annum will be F=$100{(1+0.12)5 -1}/0.12=$100( 6.353)=$635 which agrees with the result found in Table 3.4Using the factor designation and the interest tables F/A,12,5 gives F=$100(6.353)=$635

3.3.4 Equal paymentseries sinking fund factor ( A/F, i, n )The equal-payment series compound amount relationship of equation (3.4) may be solved for A as follows : A= F{i/[(1+i)n)-1]}(3.5)The resulting factor i/[(1+i)n-1], is known as the equal payment-series sinking fund factor and is designated (A/F , i , n)This factor may be used to find the required end of year payments , A, to accumulate a future amount , F, as shown in Figure 3.3

If for example , it is desired to accumulate $635 by making a series of five equal annual payments at 12% interest compounded annually , the required amount of each payment will be A= $635 {0.12/[(1+0.12)5-1]}= $635 ( 0.1574)=$100 A/F,12,5or A=$63(0.1574)=$1003.3.5 Equal-Payment Series Capital-Recovery Factor,(A/P,i,n)A deposit of amount P is made now at an annual interest rate i.The depositor wishes to withdraw the principal plus earned interest , in a series of equal year-end amounts over the next n years. When the last withdrawal is made , there should be no funds left on depositThe cash flow diagram for this situation is illustrated in Figure 3.4

0 1 2 3 n-1 n P A A A A AFigure 3.4 . Equal annual series and single present amountIt has been shown previously that F is related to A by the equal-payment-series sinking-fund factor and that F and P are linked by the single-payment compound amount factorThe substitution of P(1+i)n for F in equation (3.5) gives A=P (1+i)n .i/[(1+i)n-1]=P.{i(1+i)n/[(1+i)n-1]} .(3.6)

The resulting factor , i(1+i)n/[(1+i)n-1], is known as the equal payment capital recovery factor and is designated (A/P , i , n) This factor may be used to find the end of period payments , A, that will be provided by a present amount ,P.For example,$1000 invested at 15% interest compounded annually will provide for eight equal year end payments of A=$1000 {0.15 (1+0.15)8/(1+0.15)8-1} A/P,15,8=$1000(0.2229)=$223 or A= $1000 ( 0.2229 )=$223

As each annual withdrawal is made , the amount remaining on deposit is smaller than the amount remaining after the previous withdrawal. Because the interest earned is based on the amount on deposit , the interest earned each year also diminishesThe equal payment series capital recovery factor accounts for these year-by-year changes in what happens to be a complicated relationship between interest earned and amount withdrawn.

3.3.6 Equal-payment series present worth Factor ,(P/A,i,n)To find what single amount must be deposited now so that equal end of period payments can be made, P must be found in terms of AThe equal payment-series capital-recovery factor [Equation (3.6)] may be solved for P as follows P= A{ [(1+i)n-1]/i(1+i)n}(3.7)The resulting factor ,{ (1+i)n-1}/i(1+i)n, is known as the equal payment series present worth factor and is designated ( P/A , i, n)

This factor may be used to find the present worth ,P, of a series of equal periodic payments, A, as depicted in Figure 3.4For example ,the present worth of series of eight equal annual payments of $223 at an interest rate of 15% compounded annually will be P= $223 { (1+ 0.15)8-1)/0.15 (1+0.15 )8} =$223 (4.4873)=$1000 or P/A,15,8P=$223 (4.4873)=$1,000

3.3.7 Uniform Gradient Series Factor,(A/G,i,n):In some cases , periodic payments do not occur in an equal series.They may increase or decrease by a constant amount.In general ,a uniformly increasing series of payments for n interest periods may be expressed as G,2G, (n-1)G , as shown in Fig 3.5,where G denotes the annual change in the magnitude of the payments.One way of evaluating such a series is to apply the interest formulas developed previously to each payment in the series.

0 1 2 3 n-1 nFigure 3.5 A uniformly increasing gradient series G2G(n-1 )G(n- 2)G 0 1 2 3 4 5 6Figure 3.6 A uniformly decreasing gradient series End of year$5000$4400$3800$3200$2600$2000

This method will yield good results but will be time consumingAnother approach is to reduce the uniformly increasing or decreasing series of payments to an equal payment series , A, shown in Column 4 of Table 3.5. Let G=annual change or gradient; n= the number of years; A=the equal annual payment. The gradient sereis,G,2G,,(n-1)G can be expressed as sum of identical components of size G as presented in Column 3 of Table 3.5.

TABLE 3.5 GRADIENT SERIES AND AN EQUIVALENT SET OF SERIES

(1) End of year(2) Gradient series(3)Set of series equivalent to Gradient series(4)Annual series0000100A2GGA32GG+GA43GG+G+GA...n-1(n-2}GG+G+G+.+GAn(n-1)GG+G+G+.+G+GA

To find the future amount, F, at time n that will result from the gradient series displayed in Column 3,take each column of G values in Column 3 and find its future amount . Add these future amounts for all the columns of G in column 3.The total future amount can be represented as follows: F=(G/i){(1+i)n-1)/i}-{n.G/i}. (3.8)From Equation (3.5)find A that equals F for the gradient series in Table 3.5:The resulting factor A=G[(1/i)-(n/(1+i)n-1)] is called the uniform-gradient series factor and is designated (A/G , i, n)

Compounding frequency considerations:The derivations to this point have involved interest periods of only one year . In practice, however, cash flows or loan agreements may require that interest be paid more frequently, such as half- year, each quarter ,or each month.Such agreements result in interest periods of one-half year , one quarter year, or one-twelfth year and the compounding of interest twice, four times, or twelve times a year respectively .Interest rates associated with this more frequent compounding are normally quoted on an annual basis according to the following convention.

The nominal rate of interest is expressed on an annual basis and is determined by multiplying the actual or effective interest rate per interest period by the number of compounding periods per year.Nominal and effective interest rates:It is possible to establish a relationship between the effective interest rate for any time interval and the nominal interest rate per year . Letr= nominal interest rate per year*i=effective interest rate in the time interval.l= length of the time interval ( in years)m=reciprocal of the length of the compounding period ( in years)*The nominal interest rate is commonly referred to in financial transactions as the annual percentage rate , APR

The effective interest rate for any time interval is given by i={1+(r/m)}l.m. -1 ..(3.14)If the interest is compounded only once in the time interval , then l.m=1 and i=r/m.(3.15)To find the applicable effective interest rate for any time interval , the following relationship may be used.i={1+(r/m)}c-1, c >1 .(3.16) where c is the number of compounding periods in the time interval (c=l . m).When c=1,Equation (3.16) reduces to Equation (3.15) and either may be used to find the effective rate per compounding period.

For example ,if the nominal interest rate is 9% compounded monthly , the effective rate per month is i=0.09/12=0.0075 or 0.75% If c=1,the effective rate is found from Equation (3.16) for any time interval. 1.Nominal rate of 12% compounded monthly with time interval of one year (c=12) i={1+(0.12/12)}12 -1=0.1268 or 12.68% per year.When using the interest formulas for the discrete payments/discrete compounding situation ,it is important to recognize that interest formulas require i, the effective interest rate , and n , the number of periods involved , and these two must always be time compatible.

Example: If $2000 is to be borrowed at present and repaid 3 years later , the single payment ,F, shown in Figure 3.10 can be correctly computed for a variety of assumptions . For an interest rate of 6% compounded monthly ,the following three calculations provide the same result. Assume the periods are as follows: F/P,6.167,3 Years i=6.167% per year , n=3 years , F=$2000(1.967)Quarters i=1.507% per year , n=12 quarters , F=$2000(1.967)Months i=6%/12=0.5%per month , n=36 months, F=$2000(1.967)Note that for each example , i and n are stated in terms of the same time period.As long as the cash flow is correctly represented by the interest formula , this consistency between i and n will always provide identical solutions.

Years 1 2 3F$2000Quarters48 12Months 12 24 36Figure 3.10 Borrowing cash flow

Continuous compounding : As a limit , interest may be considered to be compounded as infinite number of times per year-that is ,continuously.Under these conditions ,the effective annual interest for continuous compounding is derived from Equation (3.14) with l=1 asia= lim {1+(r/m)}m - 1 m But since {1+(r/m)}m= [ {1+r/m}m/r]r and lim {1+r/m}m/r=e=2.7182m

Then ia= lim [{1+(r/m)m/r}]r-1=er-1 m Therefore ,when interest is compounded continuously, ia=effective annual interest rate = er-1Comparing interest rates:The effective interest rates corresponding to a nominal annual interest rate of 18% compounded annually ,semiannually , quarterly , monthly , weekly ,daily and continuously are shown in Table3.7

TABLE 3.7EFFECTIVE ANNUAL INTEREST RATES FOR VARIOUS COMPOUNDING PERIODS AT A NOMINAL RATE OF 18%

Compounding frequencyNumber of periods per yearEffective interest rate per periodEffective annual interest rateAnnually118.0000%18.0000%Semiannually29.0000%18.8100Quarterly44.500019.2517Monthly121.500019.5618Weekly520.364219.6843Daily3650.049319.7217Continuously0.000019.7217

Since the effective interest rate represents actual interest earned ,this rate should be used to compare the benefits of various nominal rates of interest.For example ,one might be confronted with the problem of determining whether it is more desirable to receive 16% compounded annually or 15% compounded monthly.The effective rate of interest per year for16% compounded annually is ,of course,16%,while for 15% compounded monthly the effective annual interest rate is ia= (1+{0.15/12})12-1=16.08%.Thus ,15% compounded monthly yields an actual rate of interest that is higher than 16% compounded annually.

3.6 INTEREST FORMULAS(CONTINUOUS COMPOUNDING,DISCRETE PAYMENTS):In certain economic valuations , it is reasonable to assume that continuous compounding interest more nearly represents the situation than does discrete compounding.Also , the assumption of continuous compounding may be more convenient from a computational standpoint in some applications.Therefore this section presents interest formulas that may be used in those cases where discrete payments and continuous-compounding interest seem appropriate.The following symbols will be used.

Let r= the nominal annual interest raten = the number of annual periods;P= a present principal sum;A= a single payment ,in a series of n equal payments , made at the end of each annual periodF= a future sum , n annual periods hence3.6.1 Single payment Compound Amount Factor:The single payment compound amount factor may be expressed as a function of the number of compounding periods as follows:For annual compounding F=P(1+r)nFor semiannual compounding ;F=P(1+{r/2})2n

For monthly compounding ; F=P(1+{r/12})12nIn general , if there are m compounding periods per year ,F= P{1+(r/m)}mnWhen interest is assumed to compound continuously , the interest earned is instantaneously added to the principal at the end of each infinitesimal interest period.For continuous compounding , the number of compounding periods per year is considered to be infinite. Therefore,F=P[lim (1+{r/m})mn] m

By rearranging terms F=P{lim[(1+r/m)m/r]rn} m But lim (1+r/m)m/r=e=2.7182 m Therefore , F=P.ernThe resulting factor ,ern ,is the single payment compound amount factor for continuous compounding interest and is F/P,r,n designated [ ] Note that any continuous-compounding , discrete payment factor may be derived from its discrete compounding ,discrete-payment counterpart by substituting the effective continuous interest rate for i .

3.6.2 Single payment present worth factor : The single payment compound amount relationship in Equation (3.18) may be solved for P as follows:P=F{1/er.n}.The resulting factor ,e-r.n, is the single-payment present-worth factor for continuous compounding interest and is designated P/F ,r , n [ ]3.6.3 Equal- Payment Series Present Worth factor: By considering each payment in a series individually ,the total present worth of the series is a sum of the individual presentworth amounts as follows:

P= A(e-r)+A(e-r2)+..+A(e-r.n) =A . e-r{1+e-r+e-r2+..+e-r(n-1)} n-1 which is A.e-r times the geometric series {1/er}j. j=0Therefore ,P=A. e-r {1-er.n}/{1-e-r} =A [(1-e-r.n)/(er-1)] ..(3.20)The resulting factor,(er-1)/(1-e-r.n) is the equal-payment-series present worth factor for continuous-compounding interest and is designated P/A , r, n [ ]

3.6.4 Equal Payment Series Capital Recovery Factor:The equal payment series present worth relationship may be solved for A follows.A=P[(er-1)/(1- e-r.n)](3.21)The resulting factor ,( er-1)/(1-e-rn) is the equal payment series capital recovery factor for continuous-compounding interest and is designated A,P,r,n [ ]3.6.5 Equal payment series sinking fund factor:The substitution of F.e-r.n for P in the equal-payment series capital recovery relationship results in

A=F. e-r.n[(er-1)/(1-e-rn)]=F{(er-1)/(ern-1)}The resulting factor ,(er-1)/(ern-1),is the equal-payment series sinking fund factor for continuous-compounding interest and is designated A/F,r,n [ ]3.6.6 Equal Payment Series Compound Amount factor:The equal-payment series sinking fund relationship may be solved for F as follows:F=A[(ern-1)/(er-1)

The resulting factor ,( ern-1)/(er-1) ,is the equal-payment series compound-amount factor-for continuous -compounding interest and is designated F/A , r, n [ ]3.6.7 Uniform Gradient Series Factor:The equivalent annual payment ,A, corresponding to a linear gradient G, number of years , n, and interest rate ,r ,may be found in a similar manner as for annual compounding,It can be shown that A= G[{1/(er-1)}- {n/(ern-1)}]

The resulting factor [{1/(er-1)}- {n/(ern-1)}] is called the uniform-gradient -series factor for continuous compounding interest and is designated A/G , r , n [ ]3.7Summary of interest formulas and designations:

PaymentFactorFind GivenDiscrete paymentsUnder discrete compoundingDiscrete paymentsUnder continuous compoundingSingle paymentCompound amountFP F/P,i,nF=P(1+i)n=P( ) F/P,r,nF=P.er.n=P[ ]Single paymentPresent -worthPF P/F,i,nP=F.{1/(1+i)n}=F( ) P/F,r,nP=F.{1/ern}=F[ ]

Equal payment seriesCompound -AmountFA F/A,i,nF=A[{(1+i)n-1}/i]=A( )F=P.e[(ern-1)/(er-1)= F/A,r,nA[ ]Equal Payment seriesSinking fundAFA=F[i/{(1+i)n-1}] A/F,i,n F=( )A=F[(er-1)/(ern-1) A/F,r,n =F[ ]EqualPayment seriesPresent worthPAP=A[{ (1+i)n-1}/i(1+i)n] P/A,i,n =A( )P=A[(1-e-m)/(er-1)] P/A,r,n =A[ ]Equal payment seriesCapital recoveryAPA=P[{i(1+i)n}/{(1+i)n-1}] A/P,i,n =P( )A=P[(er-1)/(1-e-m)] A/P,r,n =P[ ]Gradient seriesUniform Gradient -SeriesAGA=G[(1/i)-n/{(1+i)n-1}] A/G,i,n =G ( )A=G[(1/(er-1)-n/(ern-1)] A/G,r,n=G[ ]