This document provides an overview of engineering economics and the time value of money. It defines key terms like capital, simple interest, compound interest, and cash flows. It presents examples of calculating future and present value using formulas. It also discusses rules for relating cash flows that occur at different points in time by multiplying or dividing by (1+i) to move the cash flows forward or backward in time. Sample problems are worked through to find future value, present value, interest rate, and number of periods given various inputs.

1. CHAPTER 4
Part ….1
The Time Value of Money
Created By
Eng. Maysa Faroon Gharaybeh

2. • 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

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

4. 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.

5. Simple Interest & Compound
Interest

6. 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.

7. 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

8. 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

9. 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.

10. Example
period Amount Interest Amount
owed at amount per owed at
the period the end of
beginning period
of period
1 $1,000 $100 $1,100
2 $1,100 $110 $1,210
3 $1,210 $121 $1,331

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

12. 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;

13. 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

14. 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

15. Cash Flow Diagram Notation
1
1 2 3 4 5=N
1 Time scale with progression of time moving from left to
right; the numbers represent time periods (e.g., years,
months, quarters, etc...).

16. Cash Flow Diagram Notation
1
1 2 3 4 5=N
P =$8,000 2
1 Time scale with progression of time moving from left to
right; the numbers represent time periods (e.g., years,
months, quarters, etc...) and may be presented within a
time interval or at the end of a time interval.
2 Present expense (cash outflow) of $8,000 for lender.

17. Cash Flow Diagram Notation
A = $2,524 3
1
1 2 3 4 5=N
P =$8,000 2
1 Time scale with progression of time moving from left to
right; the numbers represent time periods (e.g., years,
months, quarters, etc...) and may be presented within a
time interval or at the end of a time interval.
2 Present expense (cash outflow) of $8,000 for lender.
3 Annual income (cash inflow) of $2,524 for lender.

18. Cash Flow Diagram Notation
A = $2,524 3
1
1 2 3 4 5=N
P =$8,000 2 4 i = 10% per year
1 Time scale with progression of time moving from left to
right; the numbers represent time periods (e.g., years,
months, quarters, etc...) and may be presented within a
time interval or at the end of a time interval.
2 Present expense (cash outflow) of $8,000 for lender.
3 Annual income (cash inflow) of $2,524 for lender.
4
4 Interest rate of loan.

19. Cash Flow Diagram Notation
A = $2,524 3 5
1
1 2 3 4 5=N
P =$8,000 2 4 i = 10% per year
1 Time scale with progression of time moving from left to
right; the numbers represent time periods (e.g., years,
months, quarters, etc...) and may be presented within a
time interval or at the end of a time interval.
2 Present expense (cash outflow) of $8,000 for lender.
3 Annual income (cash inflow) of $2,524 for lender.
Dashed-arrow line indicates
4 Interest rate of loan. 5 amount to be determined.

20. ‫اإلعالن واألعمى‬
‫جلس رجل أعمى على إحدى عتبات عمارة واضعا ً قبعته‬
‫بٌن قدمٌه وبجانبه لوحة مكتوب علٌها: أنا أعمى أرجوكم‬
‫ساعدونً . فمر رجل إعالنات باألعمى ووقف لٌرى أن‬
‫قبعته ال تحوي سوى قروش قلٌلة فوضع المزٌد فٌها و‬
‫دون أن ٌستأذن األعمى أخذ لوحته وكتب علٌها عبارة‬
‫أخرى وأعادها مكانها ومضى فً طرٌقه الحظ األعمى أن‬
‫قبعته قد امتألت بالقروش واألوراق النقدٌة فعرف أن‬
‫شٌئا ً 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 '‬

21. example 4-1 Page 141

22. Investment of $10,000
Initial
investment at
the beginning of
year 1

23. Uniform annual revenue of
$5,310 for five years(at the end
of each year )

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

25. Annual expenses of $3,000 for
the five years

26. Relating Present and Future
Equivalent Values of Single Cash
Flows

27. 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 N
0 1 2 3 4
F=?

28. 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 N
0 1 2 3 4
F=?

30. Example4-3

31. 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
F
0 N=
P=?

32. Example 4-4

33. 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)

34. 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)

35. Example 4-5

36. 4. Finding N when given P, F and i:
• Finding #of periods when given present &
future value at i% interest rate.
• N = log(FP) / log(1+i) (4-7)

37. Example 4-6

38. 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

40. 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(FP) / log(1+i) (4-7)
N = log(4,000/1,000)/log(1+0,08) = 18.01
Answer is 18 years…

41. 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
(F/P, 8%, 18) 3.9960
(F/P, 8%, N) 4
(F/P, 8%, 19) 4.3157
Then the closest one is N= 18 years .

43. • End of Chapter 4 PART 1
• See you with Chapter4 PART 2