economy Ch4 part.1 _by louy Alhami

CHAPTER 4 Part ….1

The Time Value of Money

Created By

Eng. Maysa Faroon Gharaybeh

• Engineering is :

The profession in which a knowledge of mathematical and natural science is applied to develop ways to utilize Economically , the materials and the forces of the nature for the benefits of the mankind .

• Engineering Economy

Involves the evaluation of the economic merits of the proposed solutions of the engineering problems.

• Benefits VS Costs

CAPITAL

• Wealth in the form of money or property that can be used to produce more wealth.

why Consider Return to Capital?

1. Interest (profit) pay the providers of capital for forgoing its use during the time the capital is being used .

2. Interest (profit) is a payment for the risk.

Simple Interest & Compound Interest

Simple Interest

• The total interest earned or charged is

linearly proportional to the initial amount of

the loan (principal), the interest rate and the

number of interest periods for which the

principal is committed.

When applied, total interest “I” may be found by

I = ( P ) ( N ) ( i ), where

•P = principal amount lent or borrowed

•N = number of interest periods ( e.g., years )

•i = interest rate per interest period

Example:

If someone borrowed $1,000 for 3 years with a simple interest of 10%........... then the total interest is

I = P*N*i

I = 1,000 * 3 * 0.1 = $300

The total amount owed = P+I = 1,000+300 =

$1,300

Compound Interest

• Whenever the interest charge for any interest period is based on the remaining principal amount plus any accumulated interest charges up to the beginning of that period.

Example

period Amount owed at the beginning of period

Interest amount per period

Amount owed at the end of period

1 $1,000 $100 $1,100

2 $1,100 $110 $1,210

3 $1,210 $121 $1,331

Economic Equivalence

• Established when we are indifferent between a future payment, or a series of future payments, and a present sum of money .

Considers the comparison of alternative options, or proposals, by reducing them to an equivalent basis, depending on:

1. interest rate;

2. amounts of money involved;

3. timing of the affected monetary receipts and/or expenditures;

Cash Flow Diagrams / Table Notation

• i = effective interest rate per interest period

• N = number of compounding periods (e.g., years)

• P = present sum of money; the equivalent value of one or more cash flows at the present time reference point

Cash Flow Diagrams / Table Notation

• F = future sum of money; the equivalent value of one or more cash flows at a future time reference point

• A = end-of-period cash flows (or equivalent end-of-period values ) in a uniform series continuing for a specified number of periods, starting at the end of the first period and continuing through the last period

Cash Flow Diagram Notation

1 2 3 4 5 = N 1

1 Time scale with progression of time moving from left to

right; the numbers represent time periods (e.g., years,

months, quarters, etc...).


1 2 3 4 5 = N 1



months, quarters, etc...) and may be presented within a

time interval or at the end of a time interval.

P =$8,000 2

2 Present expense (cash outflow) of $8,000 for lender.


1 2 3 4 5 = N 1





P =$8,000 2


A = $2,524 3

3 Annual income (cash inflow) of $2,524 for lender.


1 2 3 4 5 = N 1





P =$8,000 2


A = $2,524 3


4

Interest rate of loan. 4 4

i = 10% per year


1 2 3 4 5 = N 1





P =$8,000 2


A = $2,524 3


i = 10% per year 4

4 Interest rate of loan.

5

5 Dashed-arrow line indicates

amount to be determined.

اإلعالن واألعمى جلس رجل أعمى على إحدى عتبات عمارة واضعا ً قبعته

أنا أعمى أرجوكم : بٌن قدمٌه وبجانبه لوحة مكتوب علٌها

فمر رجل إعالنات باألعمى ووقف لٌرى أن . ساعدونً

قبعته ال تحوي سوى قروش قلٌلة فوضع المزٌد فٌها و

دون أن ٌستأذن األعمى أخذ لوحته وكتب علٌها عبارة

األعمى أن أخرى وأعادها مكانها ومضى فً طرٌقه الحظ

قبعته قد امتألت بالقروش واألوراق النقدٌة فعرف أن

شٌئاً قد تغٌر وأدرك أن ما سمعه من الكتابة هو ذلك

التغٌٌر فسأل أحد المارة عما هو مكتوب علٌها فكانت

اآلتً

نحن فً فصل الربٌع لكننً ال أستطٌع رؤٌة جماله

غير وسائلك عندما ال تسير األمور كما يجب '

example 4-1 Page 141

Investment of $10,000

Initial investment at the beginning of year 1

Uniform annual revenue of $5,310 for five years(at the end

of each year )

Market recovery value of $2,000 at the end of year five

Annual expenses of $3,000 for the five years

Relating Present and Future Equivalent Values of Single Cash

Flows

If an amount of P dollars is invested at a point of time and i% is the interest rate (growth rate ) per period the amount will grow to a future amount F

P

0

N

F = ?

1 2 3 4

1. Finding F when given P:

• Finding future value when given present value

• F = P ( 1+i ) N (4-2)

– (1+i)N single payment compound amount factor

– functionally expressed as F =P ( F / P, i%, N )

– predetermined values of this are presented in column 2 of Appendix C of text.

P

0

N

F = ?

1 2 3 4

Example4-3

2. Finding P when given F:

• Finding present value when given future value

• P = F [ (1 + i ) ] - N (4-4)

– (1+i)-N single payment present worth factor

– functionally expressed as P = F ( P / F, i%, N )

– predetermined values of this are presented in column 3 of Appendix C of text

P = ?

0 N = F

Example 4-4

3-Important Rules

1. Cash flows cannot be added or subtracted unless they occur at the same point of time

2. To move a cash flow forward in time by one time unit multiply by (1+i)

3. To move a cash flow backward in time by one time unit divide by (1+i)

3. Finding i when given P, F and N:

• Finding interest rate when given present & future value for N period of time.

• i = (F/P)1/N -1 (4-6)

Example 4-5

4. Finding N when given P, F and i:

• Finding #of periods when given present & future value at i% interest rate.

• N = log(F\P) / log(1+i) (4-7)

Example 4-6

Q 4-9 page195 Q……Your cousin want to bay a fancy watch with

$425 , instead you suggest that she bay an inexpensive watch with $25 and invest the difference (i.e. $400) for 40 years with an interest of 9% per year how much she will get after 40 years ?

Solution…….

F = p(1 + i)N = $400(1+0.09)40 = $12,563.76

OR F = P(F/P, 9%,40) = $400*31.4094 =$12,563.76

Appendix C12

Q 4-17 page 195

Q …… How long dose it take for $1,000 to quadruple(1000*4) in value when the interest rate is 8%?

Solution …..

P = $1,000 , F= $ 4,000 and i = 8%

F = p(1 + i)N

N = log(F\P) / log(1+i) (4-7)

N = log(4,000/1,000)/log(1+0,08) = 18.01

Answer is 18 years…

Alternative solution:

$4,000 =$1,000 (F/P, 8%, N)

the value of (F/P, 8%, N) = 4 if you look at

Table C-11,you find that (F/P, 8%, N) = 3.9960 when N = 18 years

and (F/P, 8%, N) = 4.3157 when N = 19 years

Then the closest one is N= 18 years .

(F/P, 8%, 18) 3.9960

(F/P, 8%, N) 4

(F/P, 8%, 19) 4.3157

• End of Chapter 4 PART 1

• See you with Chapter4 PART 2

economy Ch4 part.1 _by louy Alhami

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Transcript of economy Ch4 part.1 _by louy Alhami