E02 CashFlow

Centre for Management of

Technology and

Entrepreneurship

University of Toronto

Copyright: Joseph C. Paradi 1996-2004

Categories of Cash Flows

 _{First cost - expense to build or to buy and install}  _{Operations and maintenance (O&M) - annual}

expense, can include ; electricity, labour, repairs, etc.

 _{Salvage value(s) - receipt at project termination for}

disposal of the equipment (can be a salvage cost)

 _{Revenues - annual receipts due to sale of products or}

services

 _{Overhauls - major capital expenditure that occurs part}

way through the life of the asset

 _{Prepaid expenses - annual expenses, such as leases}

Economic Equivalence

 _{How do we measure and compare economic worth of} various cash flow profiles? We need to know:

 _{Magnitude of cash flows}

 _{Their direction (receipt or disbursement)}  _{Timing (when does transaction occur)}

 _{Applicable interest rate(s) during time period under consideration}  _{We should be economically indifferent to choosing}

between two alternatives that are economically

equivalent and could therefore be traded for one another in the financial marketplace.

 _{Any cash flow can be converted to an equivalent cash} flow at any point in time.

An Example of Equivalence

 _{Suppose you are offered the alternative of receiving}

either $3,000 at the end of 5 yrs (guaranteed) or P dollars today. You would deposit the P dollars into an account that pays 8% interest. What value of P would make you indifferent to your choice between P dollars today and the $3,000 at the end of 5 yrs?

 _{Determine the present amount that is economically}

equivalent to $3,000 in 5 yrs given the investment potential of 8% per year.

F = $3,000, N = 5 years, i = 8% per year Find P

Economic Equivalence:

General Principles

 _{Equivalence calculations to compare alternatives}

require a common time basis

 _{Equivalence depends on the interest rate}

 _{May require conversion of multiple payment cash flow}

to a single cash flow

 _{Equivalence is maintained regardless of the individual's}

Equivalence - A Factor

Approach

 _{Now we can look at organizing the approach to}

Engineering Economy by defining its language and

notations.

 _{Engineering Economy Factors apply compound interest}

to calculate equivalent cash flow values. The tabulated values at the end of the book (or you can use the

functions in spreadsheets) convert from one cash flow quantity (P, F, A, or G) to another.

 _Assumptions:

1. Interest is compounded once per period 2. Cash flow occurs at the end of the period 3. Time 0 is period 0 or the start of period 1 4. All periods are the same length

Definitions

 _{The following are the definitions for the variables used:}

i interest rate per period, expressed as a decimal

N Number of periods (also called the study horizon)

P Present cash flow, or value equivalent to a cash flow series

F Future cash flow at the end of period N, or future worth at the end of period N equivalent to a cash flow series

A Uniform periodic cash flow (annuity) at the end of every period from 1 to N. Also a uniform constant amount

equivalent to a cash flow series.

G Gradient or constant period-by-period change in cash flows from period 1 to N (arithmetic series)

Factor Notation

 _{These have their roots in the pre-computer age when}

prepared tables were used by engineers for many design needs. However, they still serve to state the problem and can be used to solve it too.

 _{The format of engineering economy factors is:}

 _{(X/Y, i%, N) where X and Y are chosen from the cash}

flow symbols P,F,A and G.

 _{So, if you have Y multiplied by a factor, you get the}

equivalent value of X: P = A(P/A, i ,N) e.g. convert from a cash flow (F) in year 10 to an equivalent present

value (P), the factor is:

Names of the EE Factors -

see pp 83 in text

 _{So now we know how all this works together, but these factors}

have names as follows:

(P/F,i,N) Present worth factor

(F/P,i,N) Compound amount factor (P/A,i,N) Series present worth factor

(A/P,i,N) Capital recovery factor, how much an investment has to return to recover its cost - no salvage value

(A/F,i,N) Sinking fund factor - where a savings account is used to accumulate funds for future investment

(F/A,i,N) Series compound amount factor

 _{Note that the first letter is the variable you are seeking and the}

second the one you have P/A want First Cost (investment), have annuity payment

Compound Interest Factors

for Discrete Compounding

 _{a single disbursement or receipt}

 _{a set of equal amounts in/out over a sequence of periods -}

annuity

 _{a set of equal amounts in/out that change by a constant}

amount from one period to the next in a sequence of periods -

arithmetic gradient series

 _{a set of equal amounts in/out that change by a constant}

proportion from one period to the next in a sequence of periods

- geometric gradient series

 _{The following assumptions apply:}

 _{compounding periods are equal}

 _{cash flows at the end of the period (consider payment at 0 to}

be at period -1

The Basic Single Payment

Factors

 _{P and F are the basic single payment quantities and}

they are related as follows:

F = P(1 + i)N can be developed as:

Fn= P(1+ in) - where: i is the annual interest rate -

simple interest - but this is not very useful, so we develop the compound interest model:

Fn = Fn-1 (1+ i)

 _{So we can see this in the EE notation as:}

(F/P,i,N) = (1 + i)N and for the reverse

(P/F,i,N) = (1 + i)-N

 _{The limits of P/F and F/P factors are:}

 _{1 when i and N approach 0}

The Relationship between P

and F

P

F

0 1 2 N-1 N

F occurs N periods after P

If you had $2,000 now and invested it at 10%, how much would it be worth in 8 years?

P = $2,000, i = 10% per year, N = 8 years F = P(1+i)N _{=$2,000(1+0.10)}8_{= $4,287.18}

Another P and F Example

P

F

0 1 2 N-1 N

F occurs N periods after P

If $1,000 is to be received in 5 years. At an annual interest rate of 12%, what is the P of this amount? F=$1000, i = 12%, N = 5 years

P = F/(1+i)N= $1000/(1+.12)5 = $567.40

Discrete Compounding,

Combining Factors

 _{We can combine these factors in various ways to create}

a model for the real problem at hand:

 _{delayed income stream because of startup time}

 _{another need may be prepaid expenses - also a way to deal}

with startup delays or construction delays.

 _{We can link formulas together to model the actual}

proposition - in fact deriving one formula’s factor from an other’s

 _{Also, many times the problem has to be defined in}

more than one part

 _{Then, there are many approaches to a specific problem}

A Simple Example

 _{We invest $120,000 [P=-120,000]}

 _{We pay for 5 years $30,000/year}

[P=-30,000(P/A,12%,5)]

 _{Then pay $35,000 at year 3 [P=-35,000(P/F,12%,3)]}  _{Receive $40,000 at year 4 [P=40,000(P/F,12%,4)]}



Compound Interest Factors

for Annuities: Uniform Series

1 2 N-2 N-1 N

A A A A A

P _{P = A(1 + i)-1 + A(1 + i)-2 + …. + A(1 + i)-(N-1) + A(1 + i)-N}

= A(1 + i)-K

P = A

Series Present Worth Factor P = A(P/A i, N) p.62 Text



Note: P occurs 1 period

Before 1st A, F would occur

Compound Interest Factors

for Annuities: Uniform Series

 _{Then to get (F/A, i, N) we can convert this to a future}

single payment by multiplying both sides by (1+i)N

 _(1+i)N * P = A[(1+i)N-1]/i resulting in

 _{F = A[(1+i)}N-1]/i p.60 text

Bonds

 _{Financial instrument used by large firms and}

government to raise funds to finance projects, debt obligations,

 _{A special form of loan, usually with a long term}

 _{The creditor (firm or government) promises to pay a}

stated amount of interest at specified intervals for a defined period and then to repay the principal at a specific date (maturity date)

 _{Canada savings bonds usually represent the lowest}

interest because they are the safest (set tone for interest rates), then are provincial bonds, next are “Blue-Chip” corporates (banks, large firms)

Bond Terminology

P = Purchase price of a bond

F = Sales price (or redemption value) of a bond

V = Par (face) value – the stated value on the bond r = annual interest shown (or coupon rate)

n = compounding period (quarterly, six months, etc.) i = the yield rate per annum (usually the market rate) N = number of remaining years to maturity

Redemption – when the issuer pays for it in cash

Maturity Date – date on which the par value is to be repaid (date bond expires)

Market Value – the price one has to pay to buy a bond A = V(r/n) = the value of a single interest payment (a coupon)

Bond Valuation

 _{Yield to maturity (return on investment) = actual}

interest earned from a bond over the holding period (equivalent to IRR calculations)

 _{When considering buying or selling bonds, the yield is}

the fundamental consideration and this depends on a number of issues:

 _{Current yields for that type of security (economy dependent)}  _{Inflation rate (usually included in the yield by market forces)}  _{Your tax position}

 _{Foreign exchange risk if you invest in other currencies}

 _{The actual price of a bond, even at issue, is very}

Example of Arithmetic

Gradients

 _{We have two propositions for investment:}

 _{New Multi Processor Computer}  _{Natural Gas Pipeline}

 _{Cash Flow Profiles}

End of Year Computer Natural Gas

0 -$50,000 -$50,000

1 +$20,000 +$5,000

2 +$15,000 +$10,000

3 +$10,000 +$15,000

4 +$5,000 +$20,000

 _{Which is preferable, the computer or natural gas? Both}

CF Diagrams for Computer &

Natural Gas Options

0 1 2 3 4 $50,000 $20,000 $15,000 $10,000 $5,000 (+) (-) 0 1 2 3 4 $50,000 $20,000 $15,000 $10,000 $5,000 (+) (-) Computer Option Natural Gas Option

Arithmetic Gradients

 _{An easy definition of an arithmetic gradient is: revenue} grows by $10,000 (G) each year.

 _{But the rate is not always constant}

 _{There are four possibilities besides G=0}

1. A>0 and G>0 - means positive and increasing 2. A>0 and G<0 - means positive but decreasing

3. A<0 and G>0 - means negative but becoming less so 4. A<0 and G<0 - means negative and becoming more so

0 1 2 3 N G 2G 3G 4 (N-1)G

Note: No cash flow at end of period 1 Each successive cash flow increases by a fixed amount equal to G

Arithmetic Gradients

cont'd...

 _{A series of receipts or disbursements that start at 0 at}

the end of the 1st period and then increase by a

constant amount from period to period.

 _{e.g. increasing operating costs for an aging machine}

 _{The sum of an annuity plus an arithmetic gradient}

series is a common pattern.

 _{The gradient formulas can be derived just as the}

annuity formulas were. See Text pages 68-70.

 _{To convert gradient to uniform series:}

 _{A = G (A/G, i, N)}

Arithmetic Gradient with

A>0

Geometric Gradient Series

 _{Very useful to model inflation, change in productivity,}

shrinkage of market size

 _{Each successive cash flow increases (or decreases) by}

a constant % each period.

 _{The base value of the series is A. The “growth” rate}

(rate of increase or decrease) is referred to as g.

 _{If g is positive, the terms are increasing in value.}  _{If g is negative, terms are decreasing in value.}

Arithmetic Gradient with

A>0 and G>0 = Geometric

Series

 _{See Factor development in the text on pp 71.}

0 1 2 3 N-1 A(1+g) A(1+g)2 4 A N-2 A(1+g)3 A(1+g)N-1

Note: There is a cash flow at end of period 1

Geometric Gradients

 _{We can define a growth adjusted interest rate i}0:

 _{There are 4 cases}

 _{i > g > 0 Growth is + but less than interest rate i}0 is +

 _{g > i > 0 Growth is + and greater than interest rate i}0 is –

 _{g = i > 0 Growth is + and = to interest rate}

 _{g < 0 Growth is negative (series is decreasing i}0 is +

0 ₀ 1 A i P N g      _{ } _ 0

1

1

1

1

g

i

i









Geometric Gradients

Formulas

g

1

1

)

i

1

(

i

1

)

i

1

(

)

N

,i

,

g

,

A

/

P

(

or

)

g

1

(

)

N

,

i

,

A

/

P

(

)

N

,i

,

g

,

A

/

P

(































Deferred Annuity

 _{If the cash flow does not begin until some later date,}

the annuity is known as deferred annuity.

 _{Annuity is deferred for J periods.}  P₀ = A (P/A, i%, N-J) ( P/F, i%, J)

0 1 2 J _J+1J+2 _J+3 N-1 N

P = ?

Non-standard Annuities and

Gradients

 _{Treat each cash flow individually}

 _{Convert the non-standard annuity or gradient to}

standard form by changing the compounding period

 _{Convert the non-standard annuity to standard by}

finding an equal standard annuity for the compounding period

 _{How much is accumulated over 20 years in a fund that}

pays 4% interest, compounded yearly, if $1,000 is deposited at the end of every fourth year?

0 4 8 12 16 20

$1000

Different Solutions

 _{Method 1: consider each cash flows separately}

F = 1000 (F/P,4%,16) + 1000 (F/P,4%,12) + 1000 (F/P,4%,8) + 1000 (F/P,4%,4) + 1000 = $7013

 _{Method 2: convert the compounding period from annual}

to every four years

ie = (1+0.04)4-1 = 16.99%

F = 1000 (F/A, 16.99%, 5) = $7013

 _{Method 3: convert the annuity to an equivalent yearly}

annuity

A = 1000(A/F,4%,4) = $235.49 F = 235.49 (F/A,4%,20) = $7012

PW Computations when N





 _{These cases are also called "perpetuities" - e.g.}

scholarships, endowments, etc.

 _{There are many engineering projects where the life of}

the project is so long as to be considered ‘forever’.

Usually 50 years as a rule of thumb, but assume cash flows continue indefinitely.

 _{There are other projects where a sum of money is}

provided at the beginning of the project and it is then invested to yield the amount used for the project

(annuities for 50 years)

 _{For both, the PW of the infinitely long uniform series of}

PW when N





 _{Until now, we assumed that the horizon N is a fixed}

number of years.

 _{The present worth of a very long and uniform series of}

cash flows is calculated as:

          _{ }     _   _   _{ }        lim ( / , , ) (1 ) 1 lim (1 ) 1 1 (1 ) lim N N N N N N P A P A i N i P A i i i P A i A P