3/31/2015

1

MS-291: Engineering Economy(3 Credit Hours)

So for ….1. Introduction What is Economics? Economics for Engineers ? What is Engineering Economy ? Performing Engineering Economy

Study ? Some Basic Concepts Utility & Various cost concept, Time value of

money (TVM), Interest rate and Rate of Returns, Cash Flow, EconomicEquivalence, Minimum Attractive Rate of Return, Cost of Capital andMARR, Simple and compound interest rates

2. Various Type of Factors

Factors Single payment Factors P/F, F/P

Uniform Series Factors P/A, A/P, F/A, A/F

Gradient Series Factors Arithmetic Gradient and Geometric Gradient

3. Dealing with Shifted Series Shifted uniform series Shifted series and single cash flows Shifted gradients

This is basicallythree“FoundationalPillars” we needfor using “variousengineeringeconomy criteria”for decisionmaking

3/31/2015

2

MS291: Engineering Economy

Chapter 4Nominal and Effective

Interest Rates

Content of the Chapter

• Interest Rate: important terminologies• Nominal and Effective Rate of Interest• Effective Annual Interest Rate• Converting Nominal rate into Effective Rate• Calculating Effective Interest rates• Equivalence Relations: PP and CP• Continuous Compounding• Varying Interest Rates

3/31/2015

3

Lets start with aSimple Example

15% per year

Compounded daily

Paid $1000 from credit cardHow muchyou going topay after 1year ?

1000+150 = $1150 ?

But ..is Due amountafter a year is really$1150 ? Lets docheck!!!

Rate is 15% per year but compounding is daily … so the rate at per day is 0.15/365 =0.000411 per day or 0.0411% per day

Days

1

Interest earned

Amount x r =0.411

Amount ($)

1000

Total due ($)

1000.411

2 0.4111691000.411 1000. 82269

3 0.4111691000. 82269 1001.233507

- -------------- ------

365 0.477311161.338553 1161.815863

1161.815863 …. Butthis is around 16.81%

rate … rather than15% stated

If compoundingperiod is less than ayear …. We facesuch situation!!!

3/31/2015

4

1161.815863 …. But this isaround 16.81% rate … rather

than 15% stated

Nominal Interest Rate (15%) Effective Interest Rate (16.81%)• denoted by (r)• does not include any consideration of

the compounding ofinterest(frequency)

• It is given as: r = interest rate perperiod x number of compoundingperiods

• Denoted by (i)• take accounts of the effect of the

compounding period• commonly express on an annual

basis (however any time maybeused)

Interest rate is same foreach period

But interest “due” isincreasing in everyperiod

Previous Learning

• Our learning so for is based “one” interest rate that’scompounded annually

• Interest rates on loans, mortgages, bonds & stocks arecommonly based upon interest rates compounded morefrequently than annually

• When amount is compounded more than once annually,distinction need to be made between nominal andeffective rate of interests

3/31/2015

5

Interest Rate:important terminologies

Interest period (t) – period of time over which interest is expressed.For example, 1% per month.

New time-based definitions to understand and remember

Compounding period (CP) – The time unit over which interest is charged or earned.For example,10% per year, here CP is a year.

Compounding frequency (m) – Number of times compounding occurs within theinterest period t.

For example, at i = 10% per year, compounded monthly, interest would becompounded 12 times during the one year interest period.

Examples of interest rateStatements

= 1 year= 1 month

= 12

Annual interest rate of 8% compounded monthly …interest period (t)

compounding period (CP)compounding frequency (m)

Annual interest rate of 6% compounded weekly …= 1 year= 1 Week

= 52

interest period (t)compounding period (CP)

compounding frequency (m)

3/31/2015

6

IMPORTANT: CompoundingPeriod and Interest Rate

• Some times, Compounding period is not mentioned inInterest statement

• For example, an interest rate of “1.5% per month”………..It means that interest is compounded eachmonth; i.e., Compounding Period is 1 month.

• REMEMBER: If the Compounding Period is notmentioned it is understood to be the same as the timeperiod mentioned with the interest rate.

Calculating EffectiveInterest Rate

• Effective interest rate per compoundingperiod can be calculated as follows:

==

3/31/2015

7

Example:

Three different bank loan rates for electricgeneration equipment are listed below.Determine the effective rate on the basis of thecompounding period for each rate

(a) 9% per year, compounded quarterly(b) 9% per year, compounded monthly(c) 4.5% per 6 months, compounded weekly

Example: CalculatingEffective Interest rates per CP

a. 9% per year, compounded quarterly.b. 9% per year, compounded monthly.c. 4.5% per 6 months, compounded weekly.

3/31/2015

8

Class Practice:2 minute

For nominal interest rate of 18% per yearcalculate the effective interest rate

i. If compounding period is yearlyii. If compounding period is semi-annuallyiii. If compounding period is quarterlyiv. If compounding period is monthlyv. If compounding period is weeklyvi. If compounding period is daily

18%

9%

4.5%

1.5%

0.346%

0.0493 %

Effective Annual InterestRates

• Nominal rates are converted into Effective Annual Interest Rates (EAIR)via the equation:

where ia = effective annual interest ratei = effective rate for one compounding period (r/m)

m = number times interest is compounded per year

• When we talk about “Annual” we consider year as theinterest period t , and the compounding period CP can beany time unit less than 1 year

= (1 + ) −1 = (1 + ) −1

3/31/2015

9

ExampleFor a nominal interest rate of 12% per year, determine the nominal and effectiverates per year for(a) quarterly, and(b) monthly compounding

Solution:

(a) Nominal r per year = 12% per year

Effective i per year = (1 + 0.03)4 – 1 = 12.55% per year(b) Nominal r per month = 12/12 = 1.0% per month

Effective I per year = (1 + 0.01)12 – 1 = 12.68% per year

Nominal r per quarter = 12/4 = 3.0% per quarter

ia = (1 + i)m – 1

where ia = effective annual interest ratei = effective rate for one compounding

period (r/m)m = number times interest is compounded

per year

15% per yearCompounded daily

Effective I per year = (1 + 0.15/365)365 – 1 = 16.81% per year

Class Practice:3 minutes

For nominal interest rate of 18% peryear determine, nominal andeffective interest rates per year ….

i. If compounding period is yearlyii. If compounding period is semi-

annuallyiii. If compounding period is

quarterlyiv. If compounding period is

monthlyv. If compounding period is

weekly

= (1 + ) −1= (1 + ) −1

3/31/2015

10

r = 18% per year,compounded CP-ly

Economic Equivalence:From Chapter 1

• Different sums of money at different times may be equalin economic value at a given rate

01

$100 now

$110Rate of return = 10% per year

$100 now is economically equivalent to $110 one year fromnow, if the $100 is invested at a rate of 10% per year

Year

Economic Equivalence: Combination of interest rate (rate of return) andtime value of money to determine different amounts of money at differentpoints in time that are economically equivalent …..Compounding/Discounting (F/P, P/F, F/A, P/G etc.)

1

3/31/2015

11

Equivalence Relations: PaymentPeriod(PP) & Compounding

Period(CP)

• The payment period (PP) is the length oftime between cash flows (inflows oroutflows)

Months0 1 2 3 4 5 6 7 8 9 10 11 12

│PP │1 month

E.g., r = nominal 8% per year, compounded semi-annuallyCP

6 monthsCP

6 months

PP = CP, PP >CP, or PP<CP

Equivalence Relations: PaymentPeriod(PP) and Compounding Period

• It is common that the lengths of the payment period and thecompounding period (CP) do not coincide

• To do correct calculation of Economic Equivalence…Interest rate must coincide with compounding period

• It is important to determine if PP = CP, PP >CP, or PP<CPLength of Time Involves Single

Amount(P and F Only)

Involves GradientSeries (A, G, or g)

PP = CP P/F , F/P P/A, P/G, P/gF/A etc.PP > CP

PP < CP P/F, F/P P/A, P/G, F/A etc.

3/31/2015

12

Determine the future value of $100 after 2 years at creditcard stated interest rate of 15% per year, compoundedmonthly.

P = $100, r = 15%, m = 12EIR /month = 15/12 = 1.25%

n = 2 years or 24 months

F = P(F/P, i, n)

Solution:

F = P(F/P, 0.0125, 24)F = 100(F/P, 0.0125, 24)F = 100(1.3474)F = 100(1.3474)F = $134.74

Alternative Method

i = (1 + r/m)m – 1= (1+0.15/12)12 – 1= 16.076%

F = P(F/P, 16.076%, 2)F = P(F/P, i, n)

F = 100(1.3456)F = $134.56The results are slightly different because of the rounding off16.076% to 16.0%

Case I:When PP>CP for Single Amount

for P/F or F/P

F = ?

Interpolationneeded

Case I:When PP>CP for Single

Amount for P/F or F/P Step 1: Identify the number of compounding

periods (M) per year Step 2: Compute the effective interest rate per

payment period (i) i = r/M Step 3: Determine the total number of payment

periods (n) Step 4: Use the SPPWF or SPCAF with i and N above

3/31/2015

13

Case II: When PP >CPfor Series for P/A or F/A

• For series cash flows, first step is to determine relationshipbetween PP and CP

• Determine if PP ≥ CP, or if PP < CP• When PP ≥ CP, the only procedure (2 steps) that can be

used is as follows:First, find effective i per PPExample: if PP is in quarters, must find effective i/quarter

Second, determine n, the number of A values involvedExample: quarterly payments for 6 years yields n = 4×6 = 24

• You can use then the standard P = A(P/A, i , n) or F = A(F/A, i, n)etc.

Example: PP >CP forSeries for P/A or F/A

For the 7 years, ExcelonEnergy has paid $500every 6 months for asoftware maintenancecontract. What is theequivalent total amountafter the last payment, ifthese funds are takenfrom a pool that has beenreturning 8% per year,compounded quarterly?

CP = Quarter

PP > CPEffective rate (i) per 6 months = (1+r/m)m -1

Solution:

i= (1+0.04/2)2 – 1 => 4.04%

r = 8 % per year or 4% per 6 months &m=2/ quarter

Since, total time is 7 years and PP is 6 monthswe have total 7x2=14 payments

F = 500(F/A, 0.0404, 14)F = 500(18.3422) => $9171.09

F = A(F/A, i, n)

PP = 6 months

Because ineach PPamount getcompoundedtwice

3/31/2015

14

Case III: EconomicEquivalence when PP< CP

• If a person deposits money each month into a savings account whereinterest is compounded quarterly, do all the monthly deposits earninterest before the next quarterly compounding time?

• If a person's credit card payment is due with interest on the 15th of themonth, and if the full payment is made on the 1st, does the financialinstitution reduce the interest owed, based on early payment? Anyone ?

• The Usual answers are NO!!!! … Some time possible for bigcooperation's

Months0 1 2 3 4 5 6 7 8 9 10 11 12

│PP │1 month

CP: 3 months = 1 quarter

Case III: EconomicEquivalence when PP< CP

1. Inter-period cash flowsearn no interest (mostcommon)

• positive cash flows aremoved to beginning of theinterest period (nointerest earned) in whichthey occur and negativecash flows are moved tothe end of the interestperiod (no interest paid)

2. inter-period cash flowsearn compound interest

cash flows are notmoved and equivalentP, F, and A values aredetermined using theeffective interest rateper payment period

Two policies:

3/31/2015

15

Example 4.11: Example: CleanAir Now (CAN) Company

Last year AllStar Venture Capital agreed to invest funds in CleanAir Now (CAN), a start-up company in Las Vegas that is anoutgrowth of research conducted in mechanical engineering atthe University of Nevada–Las Vegas. The product is a newfiltration system used in the process of carbon capture andsequestration (CCS) for coal-fired power plants. The venturefund manager generated the cash flow diagram in Figure in$1000 units from AllStar’s perspective. Included are payments(outflows) to CAN made over the first year and receipts (inflows)from CAN to AllStar. The receipts were unexpected this firstyear; however, the product has great promise, and advanceorders have come from eastern U.S. plants anxious to becomezero-emission coal-fueled plants. The interest rate is 12% peryear, compounded quarterly, and AllStar uses the no-interperiod-interest policy. How much is AllStar in the “red” at the endof the year?

Example 4.11: Example: Clean AirNow (CAN) Company

The venture fund manager generated the cash flow diagram in $1000units from AllStar’s perspective as given below. Included are payments(outflows) to CAN made over the first year and receipts (inflows)from CAN to AllStar. The receipts were unexpected this first year;however, the product has great promise, and advance orders havecome from eastern U.S. plants anxious to become zero-emission coal-fueled plants. The interest rate is 12% per year, compounded quarterly,and AllStar uses the no-inter period-interest policy. How much isAllStar in the “red” at the end of the year?

3/31/2015

16

Example: Clean Air Now(CAN) Company

Given cash flows

Negative Cash flows (outflows) at theend of CP period

Positive Cash flows (inflows) at the startof CP period

Example: Clean Air Now (CAN)Company

Solution:Effective rate per quarter = 12/4 = 3%

Now

F = 1000[-150(F/P, 3%, 4)-200(F/P, 3%, 3) +(180-175 )(F/P, 3%, 2)+ 165(F/P, 3%, 1)-50]

F = $ (-262111) Investment after one year

3/31/2015

17

Case III: EconomicEquivalence when PP< CP

• If a person deposits money each month into a savings account whereinterest is compounded quarterly, do all the monthly deposits earninterest before the next quarterly compounding time?

• If a person's credit card payment is due with interest on the 15th of themonth, and if the full payment is made on the 1st, does the financialinstitution reduce the interest owed, based on early payment? Anyone ?

• The Usual answers are NO!!!! … Some time possible for bigcooperation's

Months0 1 2 3 4 5 6 7 8 9 10 11 12

│PP │1 month

CP: 3 months = 1 quarter

Case III: EconomicEquivalence when PP< CP

1. Inter-period cash flowsearn no interest (mostcommon)

• positive cash flows aremoved to beginning of theinterest period (nointerest earned) in whichthey occur and negativecash flows are moved tothe end of the interestperiod (no interest paid)

2. inter-period cash flowsearn compound interest

cash flows are notmoved and equivalentP, F, and A values aredetermined using theeffective interest rateper payment period

Two policies:

3/31/2015

18

Example 4.11: Example: CleanAir Now (CAN) Company

Last year AllStar Venture Capital agreed to invest funds in CleanAir Now (CAN), a start-up company in Las Vegas that is anoutgrowth of research conducted in mechanical engineering atthe University of Nevada–Las Vegas. The product is a newfiltration system used in the process of carbon capture andsequestration (CCS) for coal-fired power plants. The venturefund manager generated the cash flow diagram in Figure in$1000 units from AllStar’s perspective. Included arepayments (outflows) to CAN made over the first year andreceipts (inflows) from CAN to AllStar. The receipts wereunexpected this first year; however, the product has greatpromise, and advance orders have come from eastern U.S.plants anxious to become zero-emission coal-fueled plants. Theinterest rate is 12% per year, compounded quarterly, and AllStaruses the no-inter period-interest policy. How much is AllStar inthe “red” at the end of the year?

Example 4.11: Example: Clean AirNow (CAN) Company

The venture fund manager generated the cash flow diagram in $1000units from AllStar’s perspective as given below. Included are payments(outflows) to CAN made over the first year and receipts (inflows)from CAN to AllStar. The receipts were unexpected this first year;however, the product has great promise, and advance orders havecome from eastern U.S. plants anxious to become zero-emission coal-fueled plants. The interest rate is 12% per year, compounded quarterly,and AllStar uses the no-inter period-interest policy. How much isAllStar in the “red” at the end of the year?

3/31/2015

19

Example: Clean Air Now(CAN) Company

Given cash flows

Negative Cash flows (outflows) at theend of CP period

Positive Cash flows (inflows) at the startof CP period

Example: Clean Air Now (CAN)Company

Solution:Effective rate per quarter = 12/4 = 3%

Now

F = 1000[-150(F/P, 3%, 4)-200(F/P, 3%, 3) +(180-175 )(F/P, 3%, 2)+ 165(F/P, 3%, 1)-50]

F = $ (-262111) Investment after one year

12% per year, compounded quarterly

3/31/2015

20

Continuous Compounding

• If we allow compounding to occur more and more frequently, thecompounding period becomes shorter and shorter and m , thenumber of compounding periods per payment period, increases.

• Continuous compounding is present when the duration of thecompounding period (CP), becomes infinitely small and, thenumber of times interest is compounded per period (m), becomesinfinite.

• Businesses with large numbers of cash flows each day considerthe interest to be continuously compounded for all transactions.

• As m approaches infinity, the effective interest ratei = (1 + r/m)m – 1 must be written and use as; i = er – 1

Example: ContinuousCompounding

Example: If a person deposits $500 into an account every 3 monthsat an interest rate of 6% per year, compounded continuously, howmuch will be in the account at the end of 5 years?

Solution: Payment Period: PP = 3 monthsNominal rate per three months: r = 6%/4 = 1.50%

Effective rate per 3 months: i = e0.015 – 1 = 1.51%F = 500(F/A,1.51%,20) = $11,573

Practice:

Example 4.12 & 4.13

3/31/2015

21

Varying Interest Rates• Interest rate does not remain constant full life time of a

project

• In order to do incorporate varying interest rates in ourcalculations, normally, engineering studies do consideraverage values that do care of these variations.

• But sometimes variation can be large and having significanteffects on Present or future values calculated via usingaverage values

• Mathematically, varying interest rates can beaccommodated in engineering studies

Example: Varying Interest RatesGiven below the cash flow calculate the Present value.

= 70,000 (1.8080) + 35,000 (0.9174)(0.8734) + 25,000(0.9091)(0.8734)(0.0.9174)= $172,816

$70,000

0 1 2 3 4

$35,000$25,000

P = ?

i=7%

i=9% i=10%

i=7%

Year

P =70,000(P/A, 7%, 2) + 35,000 (P/F, 9%, 1) (P/F, 7%, 2)+ 25000(P/F, 10%, 1) (P/F, 9%, 1) (P/F, 7%, 2)

3/31/2015

22

Varying Interest Rates

• When interest rates vary over time, use the interestrates associated with their respective time periods tofind P

• The general formula for varying interest rate is given as:

P = F1(P/F, i1, 1) + F2(P/F, i1, 1)(P/F, i2, 1) + …..+ Fn (P/F, i1, 1)(P/F, i2, 1) …(P/F, in, 1)

• For single F or P only the last term of the equation canbe used.

• For uniform series replace “F” with “A”

Example: Varying Interest Rates

Calculate (a) the Present value (b) the uniform Annualworth A of the following Cash flow series

P = 100(P/A, 10%, 5) + 160 (P/A, 14%, 3) (P/F, 10%, 5)= 100(3.7908) + 160(2.3216)(0.6209)

= $609.72

P = ?

P = F1(P/F, i1, 1) + F2(P/F, i1, 1)(P/F, i2, 1) + …..+ Fn (P/F, i1, 1)(P/F, i2, 1) …(P/F, in, 1)

$100

01 2 3 4 5

$100 $100 $100 $100

7 86

$160 $160 $160

i=14%i=10%Year

3/31/2015

23

Problem 4.57: Varying InterestRates

(b) Calculate the uniform Annual worthA of the following Cash flow series

609.72 = A(3.7908) + A(2.3216)(0.6209)A = 609.72 / 5.2323A = $ 116.53 per year

P = 609.72i=14%i=10%

Year

A = ?

01 2 3 4 5 7 86

P = 100(P/A, 10%, 5) + 150 (P/A, 14%, 3) (P/F, 10%, 5)= 100(3.7908) + 160(2.3216)(0.6209)

= $609.72

Year0 1 2 3 4 5 7 86

THANK YOU