Chapter 4 : Time Value of money part 1 & 2

Engineer economics chp4 time value of money

1. CHAPTER 4:
TIME VALUE OF MONEY
D R . N O R H A Z R E N A B D U L H A M I D
F A K U L T I T E K N O L O G I K E J U R U T E R A A N
ENGINEERING ECONOMY
(BNP 30402)

2. CONTENTS:
•Introduction
•Simple interest
•Compound interest
•The concept Equivalence
•Cash flow-diagram table
•Applications of Future and Present values

3. INTRODUCTION
•Capitalrefers to wealth in the form of money or
property that can be used to produce more wealth.
•Engineering economy studies involve the
commitment of capital for extended periods of time.
•A dollar today is worth more than a dollar one or
more years from now.
Money has a time value. If money remains
uninvested, value is lost

4. INTRODUCTION (CONT.)
Interest: The Cost Of Money
•Interest is the difference between an ending amount of money and the
beginning amount
Interest = amount owed now –original amount
Example: ending amount owed is RM110,000 & Beginning amount owed is
RM100,000, therefore; Interest = RM10,000 (RM110,000 –RM100,000)
When money is borrowed the interest paid
is the charge to the borrower for the use of
the lender’s property
interest paid
Whenmoneyislentorinvested,the
interestearnedisthelender’sgain
fromprovidingagoodtoanotheruse
interest earned

5. SIMPLE INTEREST
•Whenthetotalinterestearnedorchargedislinearlyproportionaltotheinitial
amountoftheloan(principal),theinterestrate,andthenumberofinterest
periods,theinterestandinterestratearesaidtobesimple.
•Is not used frequently in modern commercial practice.

6. SIMPLE INTEREST (CONT.)
The total amount repaid at the end of Ninterest periods is P+ I.
P= principal amount lent or borrowed
N= number of interest periods (e.g., years)
i= interest rate per interest period

7. SIMPLE INTEREST (CONT.)
Example 1:
You borrow RM10,000 for 3 years at 5% simple annual interest. Calculate the interest earned
by the bank.
Interest = (P )(N)(i)
= (10,000)(0 .05)(3)
= RM1,500
Example 2
You borrow RM10,000 for 60 days at 5% simple interest per year (assume a 365 day year).
What is the total amount borrowed by end of 60 days?

8. COMPOUND INTEREST
•Compound interest means interest on top of interest.
•Compound interest is calculated each period on the original principaland allinterest
accumulated during past periods
•Compound interest is commonly used in personal and professional financial
transactions.

9. COMPOUND INTEREST (CONT.)
Compound interest reflects both the remaining principal and any
accumulated interest. For RM1,000 at 10%…
Period
(1)
Amount owed at
beginning of period
(2)
Interest amount for
period
[(1)x10%]
(3)
Amount owed at
end of period
[(1)+(2)]
1 RM1,000 RM100 RM1,100
2 RM1,100 RM110 RM1,210
3 RM1,210 RM121 RM1,331

10. COMPOUND INTEREST (CONT.)
Another shorter way to calculate the total amount due is
by using this formula:
Total due after a
number of years Principal (1 + interest rate)
number of years
=

11. COMPOUND INTEREST (CONT.)
Therefore,
•Compound interest reflects both the remaining principal and any
accumulated interest. For RM1,000 at 10%…
For year 1 : RM1,000 (1 + 0.1)
1
= RM1,100
For year 2 : RM1,000 (1 + 0.1)
2
= RM1,210
For year 3 : RM1,000 (1 + 0.1)
3
= RM1,331

12. COMPOUND INTEREST (CONT.)
Example:
If RM100 is placed in an account that earns 8% compounded
quarterly, what will its worth in 8 years?
Period
(1)
Amount owed at
beginning of period
(2)
Interest amount for
period
[(1)x8%]
(3)
Amount owed at end
of period
[(1)+(2)]
1 100 8.00 108
2 108 8.64 116.64
3 116.64 9.33 125.97
4 125.97 10.08 136.05
5 136.05 10.89 146.94
6 146.94 11.75 158.69
7 158.69 12.70 171.39
8 171.39 13.71 185.11

13. COMPOUND INTEREST (CONT.)
Example:
OR
100 (1 + 0.08)
8
= RM 185.09

14. NOMINAL AND EFFECTIVE INTEREST RATES
•The concepts of nominal and effective are used when interest is
compounded more than once each year.
•Nominal rates:
1.Often states as the annual interest rate for credit cards, loans, and
house mortgages.
2.Also known as Annual Percentage Rate (APR).
3.E.g. an APR of 15% is the same as nominal 15% per year or a
nominal 1.25% per month.

15. NOMINAL AND EFFECTIVE INTEREST RATES
(CONT.)
•Effective rates:
1.Commonly stated as annual rate of return for investments, certificates of deposit,
and savings account.
2.Also known as Annual Percentage Yield (APY).
3.It is used to compare the annual interest between loans with different
compounding terms (daily, monthly, annually, or other).
•The nominal rates will never exceeds the effective rate, and similarly APR < APY

16. NOMINAL AND EFFECTIVE INTEREST RATES
(CONT.)
Examples of interest statements and interpretations
(1)
Interest
Rate statement
(2)
Nominal or
Effective interest
(3)
Compounding
period
15% per year compoundedmonthly Nominal Monthly
15% peryear Effective Yearly
Effective 15% per year compounded monthly Effective Monthly
20% per year compounded quarterly Nominal Quarterly
Nominal 1%per month compounded weekly Nominal Weekly
2%per month Effective Monthly
Effective 6%per quarter Effective Quarterly
1% per weekcompounded continuously Nominal Continuously

17. NOMINAL AND EFFECTIVE INTEREST RATES
(CONT.)
We can find the effective interest, iby using the formula below:
r = nominal interest rate
M = number of compounding periods per year
(a.k.a. compounding frequency)

18. NOMINAL AND EFFECTIVE INTEREST RATES
(CONT.)
Example:
A.A visa credit card carries an interest rate of 1% per month on the unpaid
balance. Calculate the effective rate per semiannual and annual periods.

19. NOMINAL AND EFFECTIVE INTEREST RATES
(CONT.)
Solution:

20. NOMINAL AND EFFECTIVE INTEREST RATES
(CONT.)
Example:
B.If the card’s interest rate is stated at 3.5% per quarter, find the
effective semiannual and annual rates.

21. NOMINAL AND EFFECTIVE INTEREST RATES
(CONT.)
Solution:

22. THE CONCEPT OF EQUIVALENCE
Economic equivalence allows us to compare alternatives on a common basis
Each alternative can be reduced to an equivalent basis dependent on:
•Interest rate
•Amounts of money involved
•Timing of the monetary receipts or expenses
Using these elements we can “move”cash flows so that we can compare them at
particular points in time.

23. THE CONCEPT OF EQUIVALENCE (CONT.)
Different sums of money at different times may be equal in
economic value
-1 0 1
Interest rate = 10% per year
RM90 last year
RM100 now
RM110 one
year from now
Interpretation: RM90 last year, RM100 now, and RM110 one year from
now are equivalent only at an interest rate of 10% per year

24. THE CONCEPT OF EQUIVALENCE (CONT.)
Example:
CompanyXmakesautobatteriesavailabletoGeneralMotorsdealers.Ingeneralthebatteries
arestoredthroughouttheyear,anda5%costincreaseisaddedeachyeartocoverthe
inventorycarryingchargeforthedistributorshipowner.Makecalculationstoshowwhich
ofthefollowingstatementsaretrueandwhicharefalseaboutbatterycosts.
1.The amount RM98 now is equivalent to a cost of RM105.60 one year from now.
2.A truck battery cost of RM200 one year ago is equivalent to RM205 now.
3.A RM38 cost now is equivalent to RM39.90 one year from now.
4.A RM3,000 cost now is equivalent to RM2,887.14 one year ago.
5.The carrying charge accumulated one year on an investment of RM2,000 worth of
batteries is RM100.

25. THE CONCEPT OF EQUIVALENCE (CONT.)
Solution:
1.The amount RM98 now is equivalent to a cost of RM105.60 one year
from now.
2.A truck battery cost of RM200 one year ago is equivalent to RM205
now.

26. THE CONCEPT OF EQUIVALENCE (CONT.)
Solution:
3.A RM38 cost now is equivalent to RM39.90 one year from now.
4.A RM3,000 cost now is equivalent to RM2,887.14 one year ago.
5.The carrying charge accumulated one year on an investment of RM2,000
worth of batteries is RM100.

29. CASH FLOW
Cash flows are inflows and outflows of money.
Cash inflows -receipt, revenue, income, saving
•Samples of cash inflow estimates:
1.Revenues (from sales and contracts)
2.Operating cost reductions (resulting from an alternative)
3.Salvage value
4.Construction and facility cost savings
5.Receipt of loan principal
6.Income tax savings
7.Receipts from stock and bond sales

30. CASH FLOW (CONT.)
Cash outflows -cost, expense, disbursement, loss
Samples of cash outflow estimates:
1.First cost of assets
2.Engineering design costs
3.Operating costs (annual and incremental)
4.Periodic maintenance and rebuild costs
5.Loan interest and principal payments
6.Major expected/unexpected upgrade costs
7.Income taxes

31. CASH FLOW (CONT.)
Once the cash inflow and outflow estimates are developed, the net cash
flow can be determined.
Net cash flow= receipts -disbursements
= cash inflows -cash outflows
End-of-period convention: all cash flows and NCF
occur at the end of an interest period

32. CONSTRUCTING ACASHFLOWDIAGRAM

33. CONSTRUCTING A CASH FLOW DIAGRAM
•Cash flow diagram: a graphical representation of cash flows drawn on a time
scale.
•The diagram includes what is known, what is estimated and what is needed.
•Once the cash flow diagram is complete, another person should be able to
work the problem by looking at the diagram.
•The time line is a horizontal line divided into equal periods such as days,
months, or years.

34. CONSTRUCTING A CASH FLOW DIAGRAM
(CONT.)
•Eachcashflow,suchasapaymentorreceipt,isplottedalongthislineatthe
beginningorendoftheperiodinwhichitoccurs.
•Fundsthatyoupayoutsuchassavingsdepositsorleasepaymentsarenegative
cashflowsthatarerepresentedbyarrowswhichextenddownwardfromthe
timelinewiththeirbasesattheappropriatepositionsalongtheline.
•Fundsthatyoureceivesuchasproceedsfromamortgageorwithdrawalsfroma
savingaccountarepositivecashflowsrepresentedbyarrowsextendingupward
fromtheline.

35. CONSTRUCTING A CASH FLOW DIAGRAM
(CONT.)
+ Cash flow
-Cash flow

36. CONSTRUCTING ACASHFLOWDIAGRAM:
LENDERVIEWPOINT

37. CONSTRUCTING A CASH FLOW DIAGRAM
(CONT.)
Example:
Construct a cash flow diagram for the following cash flows: RM10,000
outflow at time zero, RM3,000 per year inflow in years 1 through 5 at an
interest rate of 10% per year, and an unknown future amount in year 5.

38. CONSTRUCTING A CASH FLOW DIAGRAM
(CONT.)

39. TheXYZCorporationinsiststhatitsengineersdevelopacash-flowdiagramofthe
proposal.
Aninvestmentof$10,000canbemadethatwillproduceuniformannualrevenueof
$5,310forfiveyearsandthenhaveamarket(recovery)valueof$2,000attheendof
year(EOY)five.
Annualexpenseswillbe$3,000attheendofeachyearforoperatingandmaintaining
theproject.
Drawacash-flowdiagramforthefive-yearlifeoftheproject.Usethecorporation’s
viewpoint.
CONSTRUCTING ACASHFLOWDIAGRAMExample 4-1

40. CONSTRUCTING ACASHFLOWDIAGRAMExample 4-1 (Solution)

41. CONSTRUCTING ACASHFLOWTABLEExample 4-2
Twofeasiblealternativesforupgradingtheheating,ventilation,andairconditioning
(HVAC)systemhavebeenidentified.EitherAlternativeAorAlternativeBmustbe
implemented.
Attheendofeightyears,theestimatedmarketvalueforAlternativeAis$2,000and
forAlternativeBitis$8,000.
Assumethatbothalternativeswillprovidecomparableservice(comfort)overan
eight-yearperiod,andassumethatthemajorcomponentreplacedinAlternativeB
willhavenomarketvalueatEOYeight.
(1)Useacash-flowtableandend-of-yearconventiontotabulatethenetcashflows
forbothalternatives.
(2)Determinetheannualnetcash-flowdifferencebetweenthealternatives(B−A).

42. The costs are as follows:
Alternative A Rebuild (overhaul) the existing HVAC system
• Equipment, labor, and materials to rebuild . . . . . . . . . . . $18,000
• Annual cost of electricity . . . . . . . . . . . . . . . . . . . . . . . . . . . 32,000
• Annual maintenance expenses . . . . . . . . . . . . . . . . . . . . . . . 2,400
Alternative B Install a new HVAC system that utilizes existing ductwork
• Equipment, labor, and materials to install . . . . . . . . . . . . $60,000
• Annual cost of electricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9,000
• Annual maintenance expenses . . . . . . . . . . . . . . . . . . . . . . . 16,000
• Replacement of a major component four years hence . . 9,400
CONSTRUCTING ACASHFLOWTABLEExample 4-2

43. CONSTRUCTING ACASHFLOWTABLEExample 4-2 (Solution)

45. TERMINOLOGY AND SYMBOLS
i= effective interest rate per interest period;% per year
N = number of compounding (interest) periods; years, months
P = present sum of money; equivalentvalue of one or more cash flows at a
reference point in time; the present; $ (a.k.a. present worth)
F = future sum of money; equivalentvalue of one or more cash flows at a
reference point in time; the future; $ (a.k.a. future worth)
A = end-of-period cash flows in a uniform series continuing for a certain
number of periods, starting at the end of the first period and continuing
through the last;currency per period, $ per year (a.k.a. annual worth)

46. SINGLE-PAYMENT FACTORS

47. SINGLE-PAYMENT FACTORS (CONT.)
1.The quantity (1 + i)
-N
is called the single payment present worth factor.
2.The term on is read “P given F at i% interest per period for Ninterest
periods.”
Functional
symbol

48. SINGLE-PAYMENT FACTORS (CONT.)
Description: Formula:
Findingthe interest rate, igiven P, F
and N
FindingN given P,F, and i

49. SINGLE-PAYMENT FACTORS (CONT.)
Example 1:
If RM100 is placed in an account that earns 8% interest per year, what will its worth in 8
years?
P = RM100, i= 8%, N = 8, F = ?
F = P (1 + i)
N
= 100 (1 + 0.08)
8
= 100 (1.08)
8
= 100 (1.8509)
= RM185.09
F = P (F/P, i%, N)
= 100 (F/P, 8%, 8)
= 100 (1.8509)
= RM185.09

50. SINGLE-PAYMENT FACTORS (CONT.)
Example 2:
What is the present worth of RM30,000 in year 8 at an interest rate of
10% per year?
F = RM30,000, N = 8, i= 10%, P = ?
P = F (1 + i)
-N
= 30,000 (1 + 0.10)
-8
= 30,000 (1.10)
-8
= 30,000 (0.4665)
= RM13,995
P = F (P/F, i%, N)
= 30,000 (P/F, 8%, 10)
= 30, 000 (0.4665)
= RM13,995

51. SINGLE-PAYMENT FACTORS (CONT.)
Example 3:
If an engineer invested RM15,000 on 1.1.1991 into a retirement account
that earned 6% per year interest, how much money will be in the account
on 1.1.2016?

52. Example 4:
The average price of gasoline in 2005 was RM2.31 per gallon. In
1993, the average price was RM1.07. what was the average annual
rate of increase in the price of gasoline over this 12-year period?

53. Example 4 (cont.):
The average price of gasoline in 2005 was RM2.31 per gallon. We computed
the average annual rate of increase in the price of gasoline to be 6.62%. If we
assume that the price of gasoline will continue to inflate at this rate, how long
will it be before we are paying RM5.00 per gallon?