Engineering Economics
Fundamentals: EIT/FE Review
Hugh Miller
Colorado School of Mines
Mining Engineering Department
Fall 2013
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 for your
particular situation?
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 borrower’s perspective ….
Interest represents a cost !
Simple 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 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?
Terms
In most situations, the percentage is not paid at the end of the period, where the interest
earned is instead added to the original amount (principal). In this case, interest earned from
previous 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
Solving Interest Problems
Step #2: Draw a Cash Flow Diagram
F
Receipts
A2
A3
A4
A5
A6
Cash Flow
+
A1
-
Time
P
Disbursements
Interest Formulas
The compound interest relationship may generally be expressed as:
F = P (1+r)n
Where
(1)
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
Interest Formulas
r = Nominal rate of interest
i = Effective interest rate per period

Nominal Interest is the annualized percentage of amount
borrowed.
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
n=
P=
i=
Find F
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
n =
P =
r =
Find F
6 years
$ 1,500
6.0 %
F = (1,500)(1+0.06)6
F = $ 2,128
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 #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
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 (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 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?
(P/F,i,n)
Given
F =
n =
i
=
Find P
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?
Given
F =
n =
i
=
Find P
$25,000
3 years
10.0%
P = F / (1+r)n
P = (25,000) /(1 + 0.10)3
= $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 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
Engineering Economics
EIT Review
Uniform Series & Effective Interest
Annuities
Uniform series are known as equal annual
payments made to an interest bearing account
for a specified number of periods to obtain a
future amount.

F
A2
A3
A4
A5
A6
Cash Flow
+
A1
-
Time
P
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 ]
i
(5)
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

Applications?
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?
Visualize the Cash Flow Diagram – then Select the Equation
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)
Applications?
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.
A=
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 ]
i (1+i)n
(7)
Notation: P = A (P/A,i,n)
Applications?
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
(1+i)n -1
(8)
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
(1+i)n -1
= 0.12 x 5 (1+0.12)10 =
(1.12)10 - 1
= 0.8849 Million 
0.6 [ 3.1058 ]
2.1058
$ 884,900 per year
Engineering Economics
EIT Review
Varying Compounding Periods
Effective Interest
Solving 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?
Solving 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?
P
What are the known variables?
A, i, n
Solving Interest Problems
Step #2: Draw a Cash Flow Diagram
PV (?)
Receipts
A1
A2
A3
A4
A1 + A2 + A3 = $1000
A4 + A5 = $600
A5
Time
Solving 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 A 5
A1 A2 A3
Time
A1 = A2 = A3 = A4 = A5 = $600
+
Time
A1 = A2 = A3 = $400
Solving 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
P A4 A5
Time
A1 = A2 = A3 = $1000
+
Time
A4 = A5 = $600
Solving Interest Problems
Step #2: Method #3 – Individual Cash Flow Contributions
PV (?)
Receipts
A1
A2
A3
A4
A5
Time
Treat A1 , A2 , A3 , A4 , A5 separately (P/F) and then add.
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
Compounding 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%)
Nominal and Effective
Annual 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
Annual
Semiannual
Quarterly
Monthly
Weekly
Daily
Continuously
Periods/year
1
2
4
12
52
365
∞
Nominal Rate
Effective Rate
12%
12%
12%
12%
12%
12%
12%
12.00%
12.36%
12.55%
12.68%
12.73%
12.75%
12.75%
Compounding 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
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
When 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 (1+i)n
(1+i)n -1
A = ($3000) (A/P, 1%, 24) = $141.20
When 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 (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.
When 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
When 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
When 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
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.
When 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
Quarter #1:
Quarter #2:
Quarter #3:
Quarter #4:
Jan. 1 – March 31
April 1 – June 30
July 1 – Sept. 30
Oct. 1 – Dec. 31
When Interest Periods are Larger than Payment Periods
Example:
Date
Jan. 10
Feb. 20
Apr. 12
May 5
May 13
May 24
June 21
Aug. 10
Sept. 12
Nov. 27
Dec. 17
Dec. 29
Deposit
Withdrawal
$175
$1,200
$1,800
$65
$115
$50
$250
$1,600
$800
$350
$2,300
$750
Effective Date
Jan. 1st (beginning 1st Q)
Mar. 31th (end of 1st Q)
April 1st (beginning 2nd Q)
June 30 (end of 2nd Q)
June 30 (end of 2nd Q)
April 1st (beginning 2nd Q)
April 1st (beginning 2nd Q)
Sept. 30 (end of 3rd Q)
July 1st (beginning 3rd Q)
Oct. 1 (beginning 4th Q)
Dec. 31 (end of 4th Q)
Oct. 1 (beginning 4th Q)
When 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 = ($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
Continuous Compounding
Example: An investment institution 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%)
Continuous 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 – 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 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
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 factor
and is written as (A/G,i,n)
Therefore:
AT = (A1) + G(A/G,i,n))
Uniform Gradient Series
Example: An engineer is planning on buying a boat with anticipated life of15
years. In order to offset the anticipated effects of inflation on maintenance and
increases in moorage fees and 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
before he buys the boat, if the money earns 6% per year, compounded
annually?
$19000
Want to Find: P
Given: A1, G, i, and n
$18000
$8000
$7000
$6000
$5000
T=0
1
P
2
3
4
14
15
Uniform 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
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 period
AK is the cash flow in any subsequent period
Geometric Gradient Series

Present Value of the series equals:
n
P = A (1 + j)-1 ∑ [(1 + j)/(1 + i)]-K
K=1
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)