Engineering Economics Fundamentals: EIT Review Hugh Miller Colorado School of Mines Mining...

Engineering Economics

Fundamentals: EIT Review

Hugh MillerColorado School of MinesMining Engineering DepartmentFall 2008

Basics

Notation Never use scientific notation

Significant Digits Maximum of 4 significant figures unless the first digit is a “1”, in which case a

maximum of 5 sig figs can be used In general, omit cents (fractions of a dollar)

Year-End Convention Unless otherwise indicated, it is assumed that all receipts and disbursements

take place at the end of the year in which they occur.

Numerous Methodologies for Solving Problems Use the method most easy for you (visualize problem setup)

Concept of Interest

If you won the lotto, would you rather get $1 Million now or $50,000 for 25 years?

What about automobile and home financing? What type of financing makes more economic sense?

Interest: Money paid for the use of borrowed money.

Put simply, interest is the rental charge for using an asset over some period of time and then, returning the asset in the same conditions as we received it.

→ In project financing, the asset is usually money

Why Interest exist?

Taking the lender’s view of point:

Risk: Possibility that the borrower will be unable to pay Inflation: Money repaid in the future will “value” less Transaction Cost: Expenses incurred in preparing the loan

agreement Opportunity Cost: Committing limited funds, a lender will

be unable to take advantage of other opportunities. Postponement of Use: Lending money, postpones the

ability of the lender to use or purchase goods.

From the borrowers perspective …. Interest represents a

cost !

Simple InterestSimple Interest is also known as the Nominal Rate of Interest

Annualized percentage of the amount borrowed (principal) which is paid for the use of the money for some period of time.

Suppose you invested $1,000 for one year at 6% simple rate; at the end of one year the investment would yield:

$1,000 + $1,000(0.06) = $1,060

This means that each year interest gives $60

How much will you earn (including principal) after 3 years?

$1,000 + $1,000(0.06) + $1,000(0.06) + $1,000(0.06) = $1,180

Note that for each year, the interest earned is only calculated over $1,000.Does this mean that you could draw the $60 earned at the end of each year?

TermsIn most situations, the percentage is not paid at the end of the period, where the interestearned is instead added to the original amount (principal). In this case, interest earned fromprevious periods is part of the basis for calculating the new interest payment.This “adding up” defines the concept of Compounded Interest

Now assume you invested $1,000 for two years at 6% compounded annually;

At the end of one year the investment would yield:

$1,000 + $1,000 ( 0.06 ) = $1,060 or $1,000 ( 1 + 0.06 )

Since interest is compounded annually, at the end of the second year the investment would be worth:

[ $1,000 ( 1 + 0.06 ) ] + [ $1,000 ( 1 + 0.06 ) ( 0.06 ) ] = $1,124 Principal and Interest for First Year Interest for Second Year

Factorizing:

$1,000 ( 1 + 0.06 ) ( 1 + 0.06 ) = $1,000 ( 1 + 0.06 )2 = $1,124

How much this investment would yield at the end of year 3?

Solving Interest Problems

Step #1: Abstracting the Problem

Interest problems based upon 5 variables:

P, F, A, i, and n

Determine which are given (normally three) and what needs to be solved


Step #2: Draw a Cash Flow Diagram

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P

A1 A2 A3 A4 A5 A6

F

Time

Receipts

Disbursements

Interest FormulasThe compound interest relationship may generally be expressed as:

F = P (1+r)n (1)

Where F = Future sum of moneyP = Present sum of moneyr = Nominal rate of interestn = Number of interest periods

Other variables to be introduced later:A = Series of n equal payments made at the end of each periodi = Effective interest rate per period

Notation: (F/P,i,n) means “Find F, given P, at a rate i for n periods” This notation is often shortened to F/P

Interest Formulas

r = Nominal rate of interest

i = Effective interest rate per period

Nominal Interest is the periodic interest rate times the number of periods per year:

Nominal annual interest rate of 12% based upon monthly compounding means 1% interest rate per month compounded

When the compounding frequency is annually: r = i

When compounding is performed more than once per year, the effective rate (true annual rate) always exceeds the nominal annual rate: i > r

Future Value

Example: Find the amount which will accrue at the end of Year 6 if $1,500 is invested now at 6% compounded annually.

Method #1: Direct Calculation

(F/P,i,n)F = P (1+r)n

Given Find Fn = P = i =

Future Value


Method #1: Direct Calculation

(F/P,i,n)F = P (1+r)n

Given Find Fn = 6 years F = (1,500)

(1+0.06)6

P = $ 1,500 F = $ 2,128r = 6.0 %

Future Value


Method #2: Tables

The value of (1+i)n = (F/P,i,n) has been tabulated for various i and n.

From the handout, use the table with interest rate of 6% to find the appropriate factor. The first step is to layout the problem as follows:

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

F = 1500 (F/P, 6%, 6)

F = 1500 ( ) =

Future Value


Method #2: Tables

The value of (1+i)n = (F/P,i,n) has been tabulated for various i and n.

Obtain the F/P Factor, then calculate F:

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

F = 1500 (F/P, 6%, 6)

F = 1500 (1.4185) = $2,128

Present ValueIf you want to find the amount needed at present in order to accrue a certain amount in the future, we just solve Equation 1 for P and get:

P = F / (1+r)n (2)

Example: If you will need $25,000 to buy a new truck in 3 years, how much should you invest now at an interest rate of 10%

compounded annually?

(P/F,i,n)

Given Find PF = n = i =

Present ValueIf you want to find the amount needed at present in order to accrue a certain amount in the future, we just solve Equation 1 for P and get:

P = F / (1+r)n (2)

Example: If you will need $25,000 to buy a new truck in 3 years, how much should you invest now at an interest rate of 10%

compounded annually?

Given Find PF = $25,000 P = F / (1+r)n

n = 3 years P = (25,000) /(1 + 0.10)3

i = 10.0% = $18,783

Present Value

If you want to find the amount needed at present in order to accrue a certain amount in the future, we just solve Equation 1 for P and get:

P = F / (1+r)n (2)

Example: If you will need $25,000 to buy a new truck in 3 years, how much should you invest now at an interest rate of 10% compounded annually?

What is the factor to be used? ____________

Solve:

Present Value

If you want to find the amount needed at present in order to accrue a certain amount in the future, we just solve Equation 1 for P and get:

P = F / (1+r)n (2)

Example: If you will need $25,000 to buy a new truck in 3 years, how much should you invest now at an interest rate of 10% compounded annually?

What is the factor to be used? 0.7513

P = F(P/F,i,n) = (25,000)(0.7513) = $ 18,782

Present ValueExample: If you will need $25,000 to buy a new truck in 3 years, how much

should you invest now at an interest rate of 9.5% compounded annually?

Method #1: Direct Calculation: Straight forward - Plug and Crank

Method #2: Tables: Interpolation

Which table in the appendix will be used? Tables i = 9% & 10%

What is the factor to be used? A13 (9%): 0.7722

A14 (10%): 0.7513

Assume 9.5%: 0.7618

P = F(P/F,i,n) = (25,000)(0.7618) = $ 19,044

Engineering EconomicsEIT ReviewUniform Series & Effective Interest

Annuities

Uniform series are known as the equal annual payments made to an interest bearing account for a specified number of periods to obtain a future amount.

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P

A1 A2 A3 A4 A5 A6

F

Time

Annuities Formula The future value (F) of a series of payments (A) made during (n)

periods to an account that yields (i) interest:

F = A [ (1+i)n – 1 ] (5)

i

Where F = Future sum of moneyn = number of interest periodsA = Series of n equal payments made at the end of each

periodi = Effective interest rate per period

Derivation of this formula can be found in most engineering economics texts & study guides

Notation: (F/A,i,n) or if using tables F = A (F/A,i,n)

Example:

What is the future value of a series payments of $10,000 each, for 5 years, if deposited into a savings account yielding 6% nominal interest compounded yearly?

Draw the cash flow diagram.

F = A [ (1+i)n – 1 ] = i

Check with Factor Values:

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

Example:

What is the future value of a series payments of $10,000 each, for 5 years, if deposited in a savings account yielding 6% nominal interest compounded yearly?

Draw the cash flow diagram.

F = A [ (1+i)n – 1 ] = 10,000 [ (1+0.06)5 – 1 ] = 10,000 [ 1.3382 – 1 ] i 0.06 0.06

= $ 56,370

Checking with Factor Values:

F = A (F/A,i,n) = 10,000 (F/A, 6%, 5) = 10,000 ( 5.6371 ) = $ 56,370

Sinking Fund We can also get the corresponding value of an annuity (A) during (n)

periods to an account that yields (i) interest to be able to get the future value (F) :

Solving for A: A = i F / [ (1+i)n – 1 ] (6)

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

Example:How much money would you have to save annually in order to buy a car in 4 years which has a projected value of $18,000? The savings account offers 4.0% yearly interest.

Sinking Fund

Example:

How much money do we have to save annually to buy a car 4 years from now that has an estimated cost of $18,000? The savings account offers 4.0 % yearly interest.

A = i F / [ (1+i)n – 1 ]

A = (0.04 x 18,000) / [ (1.04)4 -1 ] = 720 / 0.170 = $4,239

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

A = (18,000)(A/F,4.0,4) = (18,000)(0.2355) = $4,239

Present Worth of an Uniform Series Sometimes it is required to estimate the present value (P) of a series of

equal payments (A) during (n) periods considering an interest rate (i)

From Eq. 1 and 5

P = A [ (1+i)n – 1 ] (7)

i (1+i)n

Notation: P = A (P/A,i,n)

Example:What is the present value of a series of royalty payments of $50,000 each for 8 years if nominal interest is 8%?

P =

Present Worth of an Uniform Series

Example:

What is the present value of a series of royalty payments of $50,000 each for 8 years if nominal interest is 8%?

P = A [ (1+i)n – 1 ] = 50,000 [ (1+0.08)8 – 1 ] = 50,000 [ 1.8509 – 1 ]

i (1+i)n 0.08 (1.08)8 0.1481

= $ 287,300

P = A (P/A,i,n) = (50,000)(P/A,8,8) = (50,000)(5.7466) = $ 287,300

Uniform Series Capital Recovery This is the corresponding scenario where it is required to estimate the value of a

series of equal payments (A) that will be received in the future during (n) periods considering an interest rate (i) and are equivalent to the present value of an investment (P)

Solving Eq. 7 for A

A = i P (1+i)n (8)

(1+i)n -1

Notation: A = P (A/P,i,n)

Example:If an investment opportunity is offered today for $5 Million, how much must it yield at the end of every year for 10 years to justify the investment if we want to get a 12% interest?

A =

Uniform Series Capital Recovery

Example:

If an investment opportunity is offered today for $5 Million, how much must it yield at the end of every year for 10 years to justify the investment if we want to get a 12% interest?

A = i P (1+i)n = 0.12 x 5 (1+0.12)10 = 0.6 [ 3.1058 ]

(1+i)n -1 (1.12)10 - 1 2.1058

= 0.8849 Million $ 884,900 per year

Engineering EconomicsEIT ReviewVarying Compounding PeriodsEffective Interest


Example:

An investment opportunity is available which will yield $1000 per year for the next three years and $600 per year for the following two years. If the interest is 12% and the investment has no terminal salvage value, what is the present value of the investment?

What is Step #1?


Example:

An investment opportunity is available which will yield $1000 per year for the next three years and $600 per year for the following two years. If the interest is 12% and the investment has no terminal salvage value, what is the present value of the investment?

Step #1

What are we trying to solve? P

What are the known variables? A, i, n


Step #2: Draw a Cash Flow Diagram

Receipts

Time

PV (?)

A1 A2 A3

A4 A5

A1 + A2 + A3 = $1000

A4 + A5 = $600


Step #2: Draw a Cash Flow Diagram – Method #1

Time

A1

A2 A3A1

A1 = A2 = A3 = A4 = A5 = $600 A1 = A2 = A3 = $400

A2 A3 A4 A5

Time+

P = A1 (P/A1, i, n1) + A2 (P/A2, i, n2) = ($600)(P/A1, 12, 5) + ($400)(P/A2, 12, 3) = $3,124


Step #2: Draw a Cash Flow Diagram – Method #2

Time

A1

A4 A5

A1 = A2 = A3 = $1000 A4 = A5 = $600

A2 A3

Time+

P = A1 (P/A1, i, n1) + A2 (P/A2, i, n2)(P/F, i, n3) = ($1000)(P/A1, 12, 3) + ($600)(P/A2, 12, 2)(P/F, 12, 3) = $3,124

P

Varying Payment and Compounding Intervals

Thus far, problems involving time value of money have assumed annual payments and interest compounding periods

In most financial transactions and investments, interest compounding and/or revenue/costs occur at frequencies other than once a year (annually)

An infinite spectrum of possibilities Sometimes called discrete, periodic compounding

In reality, the economics of project feasibility are simply complex annuity problems with multiple receipts & disbursements

Compounding Frequency

Compounding can be performed at any interval (common: quarterly, monthly, daily)

When this occurs, there is a difference between nominal and effective annual interest rates

This is determined by:

i = (1 + r/x)x – 1

where: i = effective annual interest rate r = nominal annual interest rate x = number of compounding periods per year


Example: If a student borrows $1,000 from a finance company which charges interest at a compound rate of 2% per month:

What is the nominal interest rate:

r = (2%/month) x (12 months) = 24% annually

What is the effective annual interest rate:

i = (1 + r/x)x – 1

i = (1 + .24/12)12 – 1 = 0.268 (26.8%)

Nominal and EffectiveAnnual Rates of Interest

The effective interest rate is the rate compounded once a year which is equivalent to the nominal interest rate compounded x times a year

The effective interest rate is always greater than or equal to the nominal interest rate

The greater the frequency of compounding the greater the difference between effective and nominal rates. But it has a limit Continuous Compounding.

Frequency Periods/year Nominal Rate Effective Rate

Annual 1 12% 12.00%Semiannual 2 12% 12.36%Quarterly 4 12% 12.55%Monthly 12 12% 12.68%Weekly 52 12% 12.73%Daily 365 12% 12.75%Continuously ∞ 12% 12.75%


It is also important to be able to calculate the effective interest rate (i) for the actual interest periods to be used.

The effective interest rate can be obtained by dividing the nominal interest rate by the number of interest payments per year (m)

i = (r/m)

where: i = effective interest rate for the period r = nominal annual interest rate

When Interest Periods Coincide with Payment Periods

When this occurs, it is possible to directly use the equations and tables from previous discussions (annual compounding)

Provided that:

(1) the interest rate (i) is the effective rate for the period

(2) the number of years (n) must be replaced by the total number of interest periods (mn), where m

equals the number of interest periods per year


Example: An engineer plans to borrow $3,000 from his company credit union, to be repaid in 24 equal monthly installments. The credit union charges interest at the rate of 1% per month on the unpaid balance. How much money must the engineer repay each month?

A = P (A/P, i, mn) = i P (1+i)n

(1+i)n -1

A = ($3000) (A/P, 1%, 24) = $141.20


Example: An engineer wishes to purchase an $80,000 lakeside lot (real estate) by making a down payment of $20,000 and borrowing the remaining $60,000, which he will repay on

a monthly basis over the next 30 years. If the bank charges interest at the rate of 9½% per year, compounded monthly, how much money must the engineer repay each month?

i = (r/m) = (0.095/12) = 0.00792 (0.79%)

A = P (A/P, i, mn) = i P (1+i)n (1+i)n -1

A = ($60000) (A/P, 0.79%, 360) = $504.50

Total amount repaid to the bank?

When Interest Periods are Smaller than Payment Periods

When this occurs, the interest may be compounded several times between payments.

One widely used approach to this type of problem is to determine the effective interest rate for the given interest period, and then treat each payment separately.


Example: Approach #1

An engineer deposits $1,000 in a savings account at the end of each year. If the bank pays interest at the rate of 6% per year, compounded quarterly, how much money will have accumulated in the account after 5 years?

Effective Interest Rate: i = (6%/4) = 1.5% per quarter

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

F = $1000(F/P,1.5%,16) + $1000(F/P,1.5%,12) + $1000(F/P,1.5%,8) + $1000(F/P,1.5%,4) + $1000(F/P,1.5%,0)

Using formulas or tables: F = $5,652


Another approach, often more convenient, is to calculate an effective interest rate for the given payment period, and then proceed as though the interest periods and the payment periods coincide.

i = (1 + r/x)x – 1


Example: Approach #2

An engineer deposits $1,000 in a savings account at the end of each year. If the bank pays interest at the rate of 6% per year, compounded quarterly, how much money will have accumulated in the account after 5 years?

i = (1 + r/x)x – 1 = (1 + 0.06/4)4 – 1 = 0.06136 (6.136%)

F = $1,000 (F/A,6.136%,5)

Using formula: F = $5,652

When Interest Periods are Larger than Payment Periods

When this occurs, some payments may not have been deposited for an entire interest period. Such payments do not earn any interest during that period.

Interest is only earned by those payments that have been deposited or invested for the entire interest period.

Situations of this type can be treated in the following manner:

Consider all deposits that were made during the interest period to have been made at the end of the interest period (i.e., no interest earned during the period)

Consider all withdrawals that were made during the interest period to have been made at the beginning of the interest period (i.e., earning no interest)

Then proceed as though the interest periods and the payment periods coincide.


Example: A person has $4,000 in a savings account at the beginning of a calendar year; the bank pays interest at 6% per year, compounded quarterly. Given the transactions presented in the following table (next slide), find the account balance at the end of the calendar year.

Effective Interest Rate (i) = 6%/4 = 1.5% per quarter


Example: Date Deposit Withdrawal Effective DateJan. 10 $175 Jan. 1st (beginning 1st Q)Feb. 20 $1,200 Mar. 31th (end of 1st Q)Apr. 12 $1,800 April 1st (beginning 2nd Q)May 5 $65 June 30 (end of 2nd Q)May 13 $115 June 30 (end of 2nd Q)May 24 $50 April 1st (beginning 2nd Q)June 21 $250 April 1st (beginning 2nd Q)Aug. 10 $1,600 Sept. 30 (end of 3rd Q)Sept. 12 $800 July 1st (beginning 3rd Q)Nov. 27 $350 Oct. 1 (beginning 4th Q)Dec. 17 $2,300 Dec. 31 (end of 4th Q)Dec. 29 $750 Oct. 1 (beginning 4th Q)


Example: A person has $4,000 in a savings account at the beginning of a calendar year; the bank pays interest at 6% per year, compounded quarterly. Given the transactions presented in the following table (next slide), find the account balance at the end of the calendar year.

F = ($4000-$175)(F/P,1.5%,4) + ($1200-$2100)(F/P,1.5%,3) +

($180-$800)(F/P,1.5%,2) + ($1600-$1100)(F/P,1.5%,1) + $2300

Using formula: F = $5,287

Continuous Compounding

Continuous Compounding can be thought of as a limiting case example, where the nominal annual interest rate is held constant at r, the number of interest periods becomes infinite, and the length of each interest period becomes infinitesimally small.

The effective annual interest rate in continuous compounding is expressed by the following equation:

i = limm→∞[(1 + r/m)m – 1] = er - 1


Example: A savings bank is selling long-term savings certificates that pay interest at the rate of 7 ½% per year, compounded continuously. What is the actual annual yield of these certificates?

i = er – 1 = e0.075 – 1 = 0.0779 (7.79%)


Discrete payments:

If interest is compounded continuously but payments are made annually, the following equations can be used:

F/P = ern A/P = (er – 1) / (1 – e-rn)

P/F = e-rn P/A = (1 – e-rn) / (er – 1)

F/A =(ern – 1) / (er – 1) A/F =(er – 1) / (ern – 1)

Where: n = the number of years

r = nominal annual interest rate

Continuous Compounding (Discrete payments)

Example: A savings bank offers long-term savings certificates at 7 ½% per year, compounded continuously. If a 10-year certificate costs $1,000, what will be its value upon maturity?

F = P x (F/P,r,n) = P x ern

F = ($1,000) x e(0.075)(10) = $2,117

Continuous Compounding (Discrete payments) If interest is compounded continuously but payments are

made (x) times per year, the previous formulas remain valid as long as r is replaced by r/x and with n being replaced by nx.

Example: A person borrows $5,000 for 3 years, to be repaid in 36 equal monthly installments. The interest rate is 10% per year, compounded continuously. How much must be repaid at the end of each month?

(A/P,r/x,nx) (A/P,10/12,36)

A = (P) [(er – 1) / (1 – e-rn)]

= ($5,000) [(e0.10/12 – 1) / (1 – e-(0.10/12)(12x3))]

= $161.40

Gradient Series Thus far, most of the course discussion has focused on uniform-

series problems

A great many investment problems in the real world involve the analysis of unequal cash flow series and can not be solved with the annuity formulas previously introduced

As such, independent and variable cash flows can only be analyzed through the repetitive application of single payment equations

Mathematical solutions have been developed, however, for two special types of unequal cash flows:

Uniform Gradient Series Geometric Gradient Series

Uniform Gradient Series

A Uniform Gradient Series (G) exists when cash flows either increase or decrease by a fixed amount in successive periods.

In such cases, the annual cash flow consists of two components: (1) a constant amount (A1) equal to the cash flow in the first period(2) a variable amount (A2) equal to (n-1)G

As such: AT = (A1) + (A2)

A2 = G [(1/i) – (n/i)(A/F,i,n)]

where: [(1/i) – (n/i)(A/F,i,n)] is called the uniform gradient factorand is written as (A/G,i,n)

Therefore: AT = (A1) + G(A/G,i,n))


Example: An engineer is planning for a 15 year retirement. In order to supplement his pension and offset the anticipated effects of inflation and increased taxes, he intends to withdraw $5,000 at the end of the first year, and to increase the withdrawal by $1,000 at the end of each successive year. How much money must the engineer have in this account at the start of his retirement, if the money earns 6% per year, compounded annually?

Want to Find: PGiven: A1, G, i, and n

T = 0

P

1

$5000

$6000

2 14 15

$18000

$19000

$7000

$8000

3 4


Example:

AT = (A1) + G(A/G,i,n)

A2 = G(A/G,i,n) = $1000 (A/G,6%,15) = $1000 (5.926) = $5926

AT = $5000 + $5926 = $10,926

P = AT (P/A,i,n) = $10,926 (P/A,6%,15) = $10,926 (9.7123) = $106,120

Geometric Gradient Series

Since receipts and expenditures rarely increase or decrease every period by a fixed amount, Uniform Gradient Series (G) problems have limited applicability

With Geometric Gradients, the increase or decrease in cash flows between periods is not a constant amount but a constant percentage of the cash flow in the preceding period.

Like Uniform Gradients, Geometric Gradients limited applicability but are sometimes used to account for inflationary cost increases

AK = A (1 + j)K-1

Where: j equals the percent change in the cash flow between periods

A is the cash flow in the initial periodAK is the cash flow in any subsequent period

Geometric Gradient Series

Present Value of the series equals:

P = A (1 + j)-1 ∑ [(1 + j)/(1 + i)]-K

For i = j: P = (n x A ) / (1 + i)

For i ≠ j: P = A [1–(1-j)n (1+i)-n] / (i - j)

Nomenclature: P = A (P/A,i,j,n)

n

K = 1

Engineering Economics Fundamentals: EIT Review Hugh Miller Colorado School of Mines Mining...

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