ENSC 201: The Business of Engineering

Lecture 2: Some Minor Details

Simple Interest and Compound Interest

In the case of simple interest, we just charge intereston the principal amount.

In the case of compound interest, we add the interest to the principal at regular intervals – the compounding interval – and charge interest on the sum.

For example, suppose we borrow $100 at 10% interest.

After N years, the amount we owe is:

Years 1 2 10 100

Simple $110 $120 $200 $1 100

Compound $110 $121 $259 $1 378 061

No-one ever uses simple interest, and we will never speak of it again.

Any compound interest rate can be described as i %per time_period1, compounded every time_period2

For example, a bank may charge 12% interest per year,compounded every month.

If time_period1 is the same as time_period2, then the interest rate is an effective interest rate.

Otherwise it is a nominal interest rate.

The appendices and formulas at the back of the bookall assume that we are talking about effective interestrates. So if we are quoted a nominal interest rate, wehave to convert it to an effective interest rate beforedoing any calculations.

For example, if a bank charges 12% interest per year, compounded every month, we can convert this to 1% interest per month, compounded every month. This is now an effective interest rate, and we can do calculations with it.

I put $100 in the bank for one year.

Which interest rate gives me more money at the end of the year:

12% interest per year, compounded every year

1% interest per month, compounded every month?

12% interest per year, compounded every year

gives me $100(F/P,12%,1) = $112.00

1% interest per month, compounded every month

gives me $100(F/P,1%,12) = $112.68

(1+j) = (1+i)12

Suppose I have an effective interest rate – 10% per month, compounded monthly, say – and I want to transform it to an equivalent effective annual rate. How do I do this?

The two rates are equivalent if they give me the sameamount of money after the same period of time.

So if the effective monthly rate is i and the equivalenteffective yearly rate is j, we must have

(Where `biennial’ means `every two years’)

Exercise: what effective biennial interest rate isequivalent to an effective annual rate of 10%?

(1+j) = (1+0.1)2

=1.21

So j = 21%.

Answer:

Let the effective biennial interest rate be j.

Then:

Reassuring note:

Almost every interest rate you come across in reallife will be an effective annual rate.

Continuous compounding:

Suppose we keep the nominal yearly interest rate constant-- r, say -- and decrease the compounding interval towards zero, what happens to the effective interest rate, j?

Decrease to months: j = (1 + r/12)12 – 1

Decrease to weeks: j = (1 + r/52)52 – 1

Decrease to days: j = (1 + r/365)365 – 1

r j

10% 10.471%

10% 10.506%

10% 10.516%

10% 10.517%Decrease forever: j = er – 1

Further reassuring note: continuous compoundingrarely shows up in real life. When it does, turn toAppendix B.

Final minor detail:

If we compound monthly or annually, we cansum up weekly and daily cash flows and treatthem as occurring at the end of the month or year. But if we’re compounding continuously,we are better off treating these cash flows ascontinuous.

This is what Appendix C is for.

Example:

The SFU library charges you $1/day for an overduebook and continuously compounds the amount youowe at a nominal rate of 10% per year.

How much do you owe after two years?

Solution:

You will owe $F, where

F = A(F/A,r,2)

r is 10% per year. A is $365/year (note that we haveto express A in terms of the same time unit as r.). We look the conversion factor up in Appendix C.

So you will owe

F = 365(2.2140) = $808.11

The mid-period convention:

An alternative way of representing continuous cashflows is to suppose that they arrive in one lump in themiddle of the period you’re considering.

On this approximate model, you consider that youowe the library two years’s fines, which is $730,from halfway through the period, that is, after oneyear. So the approximate solution would be:

F = 730(F/P,0.1,1), or slightly better, F = 730(F/P,e0.1,1)

which give $803 and $806.77 respectively.

How much does it matter?

Appendix A

Appendix B

Appendix C

N=2, i=0.01

2.0100 2.0101 2.0201

N=2, i = 0.25

2.25 2.28 2.59

N=20, i=0.01

22.02 22.03 22.14

N=20,i=0.25

342.94 519.01 589.65

Compare the values for (F/A,i,N):

End of Digression into Minor Details

Recap of the Important Parts

Cash flows can only be usefully compared if theyare converted to equivalent cash flows at thesame time period.

This is accomplished through the use ofconversion factors.

Among the conversion factors we have met sofar are:

Some Important Conversion Factors

Present worth of a future cashflow: (P/F,i,N)

Future worth of a present cashflow: (F/P,i,N)

Present worth of an annuity: (P/A,i,N)

Future worth of an annuity: (F/A,i,N)

More Important Conversion Factors

`Capital Recovery Factor’: (A/P,i,N)

`Sinking Fund Factor’: (A/F,i,N)

(This is how you calculate mortgage payments,for example.)

(`I want to have a million dollars by the time I retire’.)

`Capitalised Cost’: P = A/i

(Present worth of an infinite annuity.)

The purpose of all the conversion factors is to help usmake choices.

For example:

A drafting company employs 10 drafters at $800/week each. The CEO considers three alternatives:

1.Buy 8 low-end workstations at $2 000 each. Give two drafters 1 year’s notice. At the end of the year they each get $5 000 severance pay. Train the remaining eight in AutoCAD. The first training course is available in a year, and costs $2 000 per participant. After completing the course, each drafter gets a $100/week raise.

2.Buy 5 high-end workstations at $5 000 each. All the drafters get a year’s notice and $5 000 severance pay at the end of the year. Five new graduates are hired at $1 200 per week. They are trained in Pro-Engineer; to keep current, they will need a $5 000 retraining session every six months.

3.Do nothing.

Any of these options would allow the company to service its current customers. Any money saved can be invested at 10%, which is also the cost of borrowing money. What should they do?

A drafting company employs 10 drafters at $800/week each. The CEO considers three alternatives:

1.Buy 8 low-end workstations at $2 000 each. Give two drafters 1 year’s notice. At the end of the year they each get $5 000 severance pay. Train the remaining eight in AutoCAD. The first training course is available in a year, and costs $2 000 per participant. After completing the course, each drafter gets a $100/week raise.

2. Buy 5 high-end workstations at $5 000 each. All the drafters get a year’s notice and $5 000 severance pay at the end of the year. Five new graduates are hired at $1 200 per week. They are trained in Pro-Engineer; to keep current, they will need a $5 000 retraining session every six months.

3. Do nothing.

Any of these options would allow the company to service its current customers. Any money saved can be invested at 10%, which is also the cost of borrowing money. What should they do?

What time frame should we use?

How do we represent `Do nothing’?

Sketch the cash-flow diagrams.

What non-monetary factors would matter?

Option 1

$16000 $16 000

$10 000

Option 2

$25 000

$50 000

$104 000

PW ($000) = - 25 – 50 (P/F,0.1,1) – 25 (P/A,j,2N) + 104(P/A,0.1,N)(P/F,0.1,1)