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Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

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Module 1 – Lecture 4 MONEY TIME RELATIONSHIP. Prof. Dr. M.F. El-Refaie. Compound interest factors:. Based on the concept of time value of money and bound to the mechanism of compound interest, it is often required to interrelate some of the following quantities: - PowerPoint PPT Presentation
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Module 1 – Lecture 4 MONEY TIME RELATIONSHIP Prof. Dr. M.F. El- Refaie
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Page 1: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Module 1 – Lecture 4

MONEY TIME RELATIONSHIP

Prof. Dr. M.F. El-Refaie

Page 2: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Based on the concept of time value of money and bound to the mechanism of

compound interest, it is often required to interrelate some of the following quantities:

P is the value or sum of money at a time denoted as the present, m.u.

F is the value or sum of money at a future time, m.u.

A is the series of consecutive, equal, end-of-period amounts of money, m.u per unit

time.

n is the number of interest periods.

i is the interest rate.

The transformation multipliers (or time-value conversions) are called the “Compound

Interest Factors" or the “Engineering Economy Factors".

The names of these factors together with their standard abbreviations and notations are

given in Table 4.1.

Compound interest factors:

Page 3: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Table 4.1: Standard factor notations

Factor nameStandard notation

Single-payment present worth (SPPWF)

Single-payment compound amount (SPCAF)

Uniform series present worth (USPWF)

Capital recovery (CRF)

Sinking fund (SFF)

Uniform series compound amount (USCAF)

/ , %,P F i n

/ , %,F P i n

/ , %,P A i n

/ , %,A P i n

/ , %,A F i n

/ , %,F A i n

Page 4: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Table 4.2: Computations using standard notation

To findgivenfactorformula

pF

Fp

pA

AP

AF

FA

/ , %,P F P F i n

/ , %,F P F P i n

/ , %,P A P A i n

/ , %,A P A P i n

/ , %,A F A F i n

/ , %,F A F A i n

/ , %,P F i n

/ , %,F P i n

/ , %,P A i n

/ , %,A P i n

/ , %,A F i n

/ , %,F A i n

Page 5: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

SINGLE-PAYMENT COMPOUND AMOUNT FACTOR (SPCAF)

1 2 3 4 n-1n-20

P= given

F= ??

(SPCAF)

n

This factor was implicitly derived previously.

1n

F P i

/ / , , 1n

F P F P i n i

Page 6: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

If you have L.E.2000 now and invest it at 10%, how much will it be worth in 8 years?

1 2 3 40 5

i = 10 %

6 7

years

8

2000

F= ??

Example 4.1

Page 7: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Given: P = L.E. 2000, i = 10% per year, and n = 8 years Find: F

Using a calculator: Use a calculator to evaluate the (1+i)n term F = L.E.2000 (1 + 0.10)8

= L.E.4287.18

Solution

Using a spreadsheet package as Excel, the future worth calculation could be

determined from FV Function:

FV returns the future value of an investment based on periodic, constant

payments and a constant interest rate.

Syntax

FV (rate, nper, pmt, pv, type)

Page 8: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Where

Rate    is the interest rate per period.

nper    is the total number of payment periods in an annuity.

Pmt    is the payment made each period; it cannot change over the life of the annuity.

Typically, pmt contains principal and interest but no other fees or taxes .

Pv     is the present value, or the lump-sum amount that a series of future payments is

worth right now .

Type  is the number 0 or 1 and indicates when payments are due. 0 or omitted at the

end of the period, 1 at the beginning of the period

F = FV (10%, 8,, 2000, 0)

Page 9: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

The following sums are to be invested at 5 % interest rate:

6x105 m.u. now, 3x105 m.u. two years from now and 4x105

m.u. five years from now. How much will be accumulated ten

years from now ?

Example 4.2

Page 10: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

F = P1 (F/P, 5, 10) + P2 (F/P, 5, 8) + P3 (F/P, 5, 5) = 6 × 105 × (1.05)10 + 3×105 × (1.05)8 + 4×105 ×

(1.05)5

= 1.9311×106 m.u.

Solution

1 2 3 4

0

6x105 m.u.

5

i= 5 %

6 7

years

8

3x105 m.u. 4 x105

m.u.

F = ??

9 10

Page 11: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

P' = 6 × 105 (F/P, 5, 5) + 3 × 105 (F/P, 5, 3) + 4 × 105

= 6 × 105 × (1.05)5 + 3 × 105 × (1.05)3 + 4 × 105

= 15.1306 ×105 m.u.

F = P' (F/P, 5, 5) = 15.1306 × 105 × (1.05)5

= 19.311 × 105

Alternative solution

Page 12: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Solution of the Excel sheet is shown in figure where we can see in the equation bar that:

F = FV (5%, 10,, 600000, 0) + FV (5%, 8,, 300000, 0) + FV (5%, 5,, 400000, 0)

Solution

Page 13: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

1 2 3 4 n-1n-20

P= ??

F= given

(SPPWF)

n

SINGLE-PAYMENT PRESENT WORTH FACTOR (SPPWF)

/ 1n

P F i

1/ / , ,

1nP F P F i n

i

Page 14: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Suppose that L.E.1000 is to be received after 5 years. At an

annual interest rate of 12%,

What is the present worth of this amount?

Example 4.3

Page 15: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Solution

1 2 3 40 5

i = 12 %

years

P = ??

1000 L.E.

This example could be solved by one of three ways:

1-Using a calculator (Mathematical Formulas)P = 1000 × (1 + 0.12)-5 = 1000 × (0.5674) = L.E. 567.40 In order To have L.E.1000 in your savings account at the end of 5 years, you must deposit L.E.567.40 now.

Page 16: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

2 .Using interest tables

Interest Table For Discrete Compounding; i = 12%

PeriodsCompound

AmountFactor

PresentWorthFactor

CompoundAmountFactor

PresentWorthFactor

nF/PP/FF/AP/A

41.57350.63554.77933.0373

51.76230.56746.35283.6048

61.97380.50668.11524.1114

(0.5674)

P = 1,000 (P/F, 12%, 5) = L.E.567.40

Note: this method might be obsolete as it requires lots of data tables. These tables are formed using the simple formulae developed here in this lecture

Page 17: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

3. Using Spreadsheet as Excel, the present value calculation is determined from the

function PV:

PV functions returns the present value of an investment.

Syntax

PV (rate, nper ,pmt, fv, type)

Rate    is the interest rate per period

Nper    is the total number of payment periods in an annuity .

Pmt    is the payment made each period and cannot change over the life of the annuity .

Fv    is the future value, or a cash balance you want to attain after the last payment is

made. If fv is omitted, it is assumed to be 0

Type    is the number 0 or 1 and indicates when payments are due. 0 or omitted at the

end of the period, 1 at the beginning of the period.

Page 18: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Then the Solution of the above example is.P = PV (12%, 5, 0, 1000, 0).

Page 19: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Suppose you buy a share for L.E.10 and sell it for L.E.20, your profit is L.E.10.

If that happens within a year, your rate of return is an impressive 100% (L.E.10 /

L.E.10 = 1). If that takes 5 years, what would be the average annual rate of return

on your investment?

Example 4.4

1 2 3 40 5

i = ? %

years

L.E.10

L.E.20

Page 20: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Solution

This example could be solved by either of the following methods:

1. F = P (1 + i)n

20 = 10(1 + i) 5 Then i = 2(1/5) – 1 = 14.87 %

2. Using spreadsheet application such as Excel. A financial function such as RATE (n, 0, P, F) allows us to calculate an unknown interest rate. The precise command statement would be as follows:

i = RATE (5, 0, -10, 20) = 14.87%

Page 21: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

1 2 3 4 n-1n-20

F= ??(USCAF)

n

A =given

A series of equal installments paid at the ends of consecutive interest periods will accumulate to a certain future value after n periods.

UNIFORM-SERIES COMPOUND-AMOUNT FACTOR (USCAF)

Page 22: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Payment at end of period No.

Accumulate to

1

2

3

.

..

..

n-1.

n

11 ni

21 ni

31 ni

A

1A i

A series of equal payments

Dealing with each payment separately we can deduce the formula for determining

the future value F for any series of equal installments as shown in table.

Page 23: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

1 2 31 1 1 1 1

n n nF A i i i i

Multiplying by 1 i

1 21 1 1 1 1

n n ni F A i i i i

11 niAiF

1 1, ,

niF F i nAA i

The future compound amount is the summation of the right-hand column. Thus,

Subtracting

Page 24: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Suppose you make 15 equal annual deposits of 1000 L.E. each

into a bank account paying 5% interest per year. The first deposit

will be made one year from today. How much money can be

withdrawn from this bank account immediately after the 15th

deposit?

Example 4.5

Page 25: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

The value of A is L.E.1000, n = 15 years, and i = 5% per year.

Immediately after the 15th payment, the future equivalent amount is.

F = l000 (F/A, 5%, 15) = 1000 (1.0515 – 1) / 0.05

= 1000 × (21.5786)

= L.E. 21578.60

Solution:

Page 26: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Suppose you make an annual contribution of L.E.3000 to your saving

account at the end of each for 10 years. If your saving account earns 7%

interest annually how much can be withdrawn at the end of ten years

Example 4.6

1 2 3 40 5

i = 7 %

6 7

years

8

A = L.E.3000

9 10

F = ??

Page 27: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Given: A = L.E.3000, n = 10 years, and i = 7% per year.

Find: F

F = 3000(F/A, 7%, 10) = L.E.3000 (13.8164)

= L.E.41449.20.

To obtain the future value of the annuity on Excel, we may use the following

financial command:

= FV (7%, 10, 3000, 0, 0)

Solution

Page 28: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

You have an opportunity to purchase a piece of vacant land for m.u.30000 cash. If you bought it, you would plan to hold the property for 15 years and then sell it at a profit. During this period, you would have to pay annual property taxes of m.u.600. You would have no income from the property. Assuming that you want a 10 % rate of return from the investment, at what net price would you have to sell it 15 years hence?

Example 4.7

Page 29: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

F = 30000 (F/P,10%,15) + 600 (F/A,10%,15)

= 30000 (1.08)15 + 600 ((1.0815 – 1) / 0.08)

= m.u. 144380.93

Using Spreadsheets (Excel) the above example could be solved as,FV= (10%,15, 600, 30000, 0)

Solution

Page 30: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

11

ni

iFA

, ,1 1

n

A iA i nFF i

1 2 3 4 n-1n-20

A = ??

(SSF)

n

F = given

SINKING FUND FACTOR (SFF)

Page 31: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

An enterprising student is planning to have personal savings

totaling L.E.1000000 when she retires at age 65. She is now 20

years old. If the annual interest rate will average 7% over the next

45 years on her savings account, what equal end-of-year amount

must she save to accomplish her goal?

Example 4.8

Page 32: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

A = 1000000 (A/ F, 7%, 45)A= 1000000 (0.07/((1+0.07)45 –1)) A= 1000000(0.0035) = L.E.3500

The future amount, F, is L.E.1000000. The equal annual amount this student

must place in a sinking fund that grows to L.E.1000000 in 45 years at 7%

annual interest

Solution

Page 33: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Also the above example could be solved by spreadsheet as Excel using function called PMT,PMT function Calculates the payment for a loan based on constant payments and a constant interest rate.

Syntax

PMT (Rate, Nper, Pv, Fv, Type)

Description of the arguments in PMT function,

Rate    is the interest rate for the loan.

Nper    is the total number of payments for the loan.

Pv    is the present value, or the total amount that a series of future payments is worth

now; also known as the principal.

Fv    is the future value, or a cash balance you want to attain after the last payment is

made. If Fv is omitted, it is assumed to be 0 (zero), that is, the future value of a loan is 0.

Type    is the number 0 (zero) or 1 and indicates when payments are due.

Page 34: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

The Excel solution then is,PMT (7%, 45, 0, 1000000, 0)

Page 35: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

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Page 36: Module 1 – Lecture 4 MONEY TIME RELATIONSHIP

Engineering Engineering Economy Economy Module 1: Module 1:

Lectures 1-8Lectures 1-8 Dr. Mohamed F. El-RefaieDr. Mohamed F. El-Refaie

(Sunday and Tuesday, L3,L4,L5 and (Sunday and Tuesday, L3,L4,L5 and L6)L6)

Dr. Sayed KasebDr. Sayed Kaseb (Saturday and Wednesday, L1,L2,L7 (Saturday and Wednesday, L1,L2,L7

and L8)and L8)

Module 1 T

eaching T

eam


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