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 interest on 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 Simple$110$120$200$1 100 Compound$110$121$259$

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_period 1, compounded every time_period 2 For example, a bank may charge 12% interest per year, compounded every month.

If time_period 1 is the same as time_period 2, 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 book all assume that we are talking about effective interest rates. So if we are quoted a nominal interest rate, we have to convert it to an effective interest rate before doing 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) = $ % 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 same amount of money after the same period of time. So if the effective monthly rate is i and the equivalent effective yearly rate is j, we must have

(Where `biennial’ means `every two years’) Exercise: what effective biennial interest rate is equivalent 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 real life 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 rj 10%10.471% 10%10.506% 10%10.516% 10%10.517% Decrease forever: j = e r – 1

Further reassuring note: continuous compounding rarely shows up in real life. When it does, turn to Appendix B.

Final minor detail: If we compound monthly or annually, we can sum up weekly and daily cash flows and treat them as occurring at the end of the month or year. But if we’re compounding continuously, we are better off treating these cash flows as continuous. This is what Appendix C is for.

Example: The SFU library charges you $1/day for an overdue book and continuously compounds the amount you owe 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 have to 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 cash flows is to suppose that they arrive in one lump in the middle of the period you’re considering. On this approximate model, you consider that you owe the library two years’s fines, which is $730, from halfway through the period, that is, after one year. So the approximate solution would be: F = 730(F/P,0.1,1), or slightly better, F = 730(F/P,e 0.1,1) which give $803 and $ respectively.

How much does it matter? Appendix AAppendix BAppendix C N=2, i= N=2, i = N=20, i= N=20,i= 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 they are converted to equivalent cash flows at the same time period. This is accomplished through the use of conversion factors. Among the conversion factors we have met so far 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 us make 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 $10 000

Option 2 $ $ $ 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)