Title: Engineering Economics
1Engineering Economics
Hugh Miller Colorado School of Mines Mining
Engineering Department Fall 2007
2Basics
 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)
 YearEnd 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)
3Concept of Interest
 If you won the lotto, would you rather get 1
Million now or 50,000 for 20 years?  What about automobile and home financing? What
type of financing makes more sense?  Interest Money paid for the use of borrowed
money.  If we put money in the perspective of an asset,
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 
4Why Interest exist?
 Taking the lenders 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 !
5Simple Interest
 Simple 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 get after 3 years?
 1,000 1,000(0.06) 1,000(0.06)
1,000(0.06) 1,180  Note that each year the interest are calculated
only over 1,000.  Does it means that you could draw the 60 at the
end of each year?
6Terms
 If the percentage is not paid at the end of the
period, then, this amount is added  to the original amount (principal) to calculate
the interest for the second term.  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?
7Solving 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
8Solving Interest Problems
 Step 2 Draw a Cash Flow Diagram
Receipts
Disbursements
9Interest Formulas
 The compound interest relationship may generally
be expressed as  F P (1r)n (1)
 Where F Future sum of money
 P Present sum of money
 r Nominal rate of interest
 n number of interest periods
 Other variables to be introduced later
 A Series of n equal payments made at the end
of each period  i 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
10Interest Formulas
 r Nominal rate of interest
 i Effective interest rate per period
 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 gt r
11Future 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 (1r)n
 Given Find F
 n
 P
 i
12Future 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 (1r)n
 Given Find F
 n 6 years F (1,500)(10.06)6
 P 1,500 F 2,128
 r 6.0
13Future 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 2 Tables
 The value of (1i)n (F/P,i,n) has been
tabulated for various i and n.  From the handout, you will find tables with
interest rates from ½ to 18. The first step is
to layout the problem as follows  F P (F/P,i,n)
 F 1500 (F/P, 6, 6)
 F 1500 ( )
14Future 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 2 Tables
 The value of (1i)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
15Present 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 / (1r)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 P
 F
 n
 i
16Present 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 / (1r)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 P
 F 25,000 P F / (1r)n
 n 3 years P (25,000) /(1 0.10)3
 i 10.0 18,783
17Present 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 / (1r)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
18Present 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 / (1r)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
19Present Value
 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 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
20Engineering EconomicsEIT ReviewUniform Series
Effective Interest
21Annuities
 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.
22Annuities Formula
 The future value (F) of a series of payments (A)
made during (n) periods to an account that yields
(i) interest  F A (1i)n 1 (5)
 i
 Where F Future sum of money
 n number of interest periods
 A Series of n equal payments made at the end
of each period  i 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)
23Example
 What is the future value of a series of savings
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 (1i)n 1
 i
 Check with Factor Values
 F A (F/A,i,n)

24Example
 What is the future value of a series of savings
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 (1i)n 1 10,000 (10.06)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
25Sinking 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 / (1i)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. 

26Sinking 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 / (1i)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
27Present 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 (1i)n 1 (7)
 i (1i)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
28Present 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 (1i)n 1 50,000 (10.08)8 1
50,000 1.8509 1  i (1i)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 
29Uniform 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 (1i)n (8) 
(1i)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
30Uniform 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 (1i)n 0.12 x 5 (10.12)10
0.6 3.1058  (1i)n 1 (1.12)10  1
2.1058  0.8849 Million ? 884,900 per year
31Engineering EconomicsEIT ReviewVarying
Compounding PeriodsEffective Interest
32Solving Interest Problems
 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?
33Solving Interest Problems
 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? PV
 What are the known variables? A, i, n
34Solving Interest Problems
 Step 2 Draw a Cash Flow Diagram
PV (?)
Receipts
A1
A2
A3
A4
A5
Time
A1 A2 A3 1000
A4 A5 600
35Solving Interest Problems
 Step 2 Draw a Cash Flow Diagram Method 1
P A1 (P/A1, i, n1) A2 (P/A2, i, n2)
(600)(P/A1, 12, 5) (400)(P/A2, 12, 3)
3,124
A1
A2
A3
A4
A5
A2
A3
A1
Time
Time
A1 A2 A3 A4 A5 600
A1 A2 A3 400
36Solving Interest Problems
 Step 2 Draw a Cash Flow Diagram Method 2
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
A1
A2
A3
A4
A5
Time
Time
A1 A2 A3 1000
A4 A5 600
37Varying 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 mine feasibility are
simply complex annuity problems with multiple
receipts disbursements
38Compounding 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
39Compounding Frequency
 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)
40Nominal 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 8 12
12.75
41Compounding Frequency
 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

42When 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
43When Interest Periods Coincide with Payment
Periods
 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 (1i)n

(1i)n 1  A (3000) (A/P, 1, 24) 141.20
44When Interest Periods Coincide with Payment
Periods
 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 (1i)n

(1i)n 1  A (60000) (A/P, 0.79, 360) 504.50
 Total amount repaid to the bank?
45When 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.
46When Interest Periods are Smaller than Payment
Periods
 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
47When Interest Periods are Smaller than Payment
Periods
 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
48When Interest Periods are Smaller than Payment
Periods
 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
49When 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.
50When Interest Periods are Larger than Payment
Periods
 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
51When Interest Periods are Larger than Payment
Periods
 Example
 Date Deposit Withdrawal
Effective Date  Jan. 10 175 Jan. 1st
(beginning 1st Q)  Feb. 20 1,200 Mar. 30th (end of 1st
Q)  Apr. 12 1,500
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)
52When Interest Periods are Larger than Payment
Periods
 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 (4000175)(F/P,1.5,4) (12001800)(F/P,1
.5,3)  (180800)(F/P,1.5,2) (16001100)(F/P,1.5,1
) 2300  Using formula F 5,601
53Continuous 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?8(1 r/m)m 1 er  1
54Continuous Compounding
 Example A savings bank is selling longterm
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)
55Continuous Compounding
 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 ern)
 P/F ern P/A (1 ern) / (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
56Continuous Compounding (Discrete payments)
 Example A savings bank offers longterm savings
certificates at 7 ½ per year, compounded
continuously. If a 10year 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
57Continuous 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 ern)
 (5,000) (e0.10/12 1) / (1
e(0.10/12)(12x3))  161.40
58Gradient Series
 Thus far, most of the course discussion has
focused on uniformseries 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
59Uniform 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 (n1)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 factor  and is written as (A/G,i,n)
 Therefore AT (A1) G(A/G,i,n))
60Uniform Gradient Series
 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 P Given A1, G, i, and
n
19000
18000
8000
7000
6000
5000
T 0
1
2
14
15
3
4
P
61Uniform Gradient Series
 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
62Geometric 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)K1 
 Where j equals the percent change in the
cash flow between periods
 A is the cash flow in the initial period
 AK is the cash flow in any subsequent period
63Geometric 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(1j)n (1i)n / (i  j)
 Nomenclature P A (P/A,i,j,n)
n
K 1