Archive for the ‘Geometric series’ Category

The concept of Present Value

June 20, 2011 Leave a comment

Economists are always evaluating the value of the costs and benefits of a particular project, often over a long time scale. When you do this you don’t just add up the total sum of an expected future income stream, you have to discount it as you look into the future. This is the idea of working out the ‘present value’ of some form of return in the future.

The discount rate is basically the opportunity cost of directing resources elsewhere, so for private investment projects a good benchmark to use is the market interest rate. This is logical because it represents an alternative use of the investment funds – the firm could have just lent it to someone else (or invested it in a bank to lend it on their behalf) and earned the market rate. So if the investment is being used to finance an investment project which is forecast to earn income streams in the future, it is only worthwhile if those income streams will end up being better than what the firm could have earned by simply investing the money at the market rate of interest and taking the returns from that.

It will be easier to see with an example.

Lets say a firm is considering making an investment in a new piece of machinery which will cost £10000 now. The projections are that this piece of machinery will last for six years, and will increase the firm’s profits by £3000 in each of the first two years, by £2000 in each of the next two years, and by £1000 in the final two years before it needs to be replaced. That means in total it will bring in £12000 over 6 years.

Total value of income stream:3000+3000+2000+2000+1000+1000=12000

So it costs £10000 and brings in £12000 over 6 years…sounds like it is worth it?

Not necessarily. Suppose the market interest rate is 4%.

If you use a discount rate of 4%, then the present value of the income stream is \frac {3000}{1.04}+\frac {3000}{1.04^2}+\frac {2000}{1.04^3}+\frac {2000}{1.04^4}+\frac {1000}{1.04^5}+\frac {1000}{1.04^6} = 10758.13

Now given that the cost of the investment is 10000 now, the net present value of the project is -10000 + 10758.13 = 758.13

So yes, it is worth it. The present value of your forecast income streams is worth more than the cost of the investment, but not by the £2000 that you would get simply by adding up the values.

If the net present value of a project > 0, then it is a worthwhile investment.

When it comes to calculating the present value of a government project, the opportunity cost is the net present value of what the funds would otherwise have been used to produce. In the UK most public sector investments take 3.5% as a benchmark discount rate, this is the social time preference rate. The danger with this is that discounting can effectively disenfranchise future generations, if you are looking at a long term project that will yield gains over many years. We are still today benefiting from the investment that our ancestors made in the railways and canals etc, but if you discount every project at 3.5% then the present value of something in 100 years time is effectively the value divided by 31, and something 200 years in the future gets divided by 972. This is pretty controversial when it comes to evaluating the gains of environmental projects. So the UK government uses a declining discount rate, the 3.5% counts only for the first 30 years, and then gradually declines over different time horizons.

Categories: Geometric series, Maths

Geometric series

June 20, 2011 Leave a comment

Here’s a technique that is well worth mastering, because geometric series come up over and over again in anything to do with multipliers, investments and also in intermediate macro. Typically you are faced with a problem which has some sort of series in it like this: S=1+x+ x^2 +x^3 +x^4 +.....x^n You have to find the sum of the series.

There’s an easy technique which sorts this out. You just multiply both sides by 1-x. Here’s why it works:
S(1-x)=[1+x+ x^2 +x^3 +x^4 +.....x^n]-[x+ x^2 +x^3 +x^4 +.....x^{n+1}]

Look at what happens now to the right hand side of the equation, pretty much everything cancels out, and you are just left with S(1-x)=1-x^{n+1} so S=\frac{1-x^{n+1}}{(1-x)}. You can multiply top and bottom of the right hand side by -1 to express it in a form which might be a bit more convenient to use for calculations: S=\frac{x^{n+1}-1}{(x-1)}

If x>1 then the series will ‘explode’ (just keep getting bigger), if -1<x<1 then the series will converge around a value.

Here's an example. If you have a fixed rate mortgage for £100000, with a constant interest rate of 5% compounded annually, over a 25 year term, what would be the monthly repayment (if it is the same every month?)

You can do this by working out what you will owe at the end of each year and seeing if you can spot the pattern.

At the end of the first year, you owe 100000(1.05)-12x. Your original debt of 100000 has increased in value by 5% but you have made 12 monthly payments of value x.

At the end of the second year, you owe (100000(1.05)-12x)(1.05)-12x. This is basically
your debt at the end of the first year multiplied by 1.05 minus the 12 monthly payments you made in the second year. This can be rewritten as 100000(1.05^2)-12x(1.05)-12x

At the end of the third year, you owe 100000(1.05^3)-12x(1.05^2)-12x(1.05)-12x

At the end of the fourth year, you owe 100000(1.05^4)-12x(1.05^3)-12x(1.05^2)-12x(1.05)-12x

Now you can see the pattern developing.
At the end of the 25th year, you owe 100000(1.05^{25})-12x(1.05^{24})-12x(1.05^{23})-......-12x. Of course as this loan is a 25 year term, at the end of the 25th year you owe nothing, so we have an expression in terms of x: 0 = 100000(1.05^{25})-12x(1.05^{24})-12x(1.05^{23})-......-12x

We can rearrange this to say 100000(1.05^{25})=12x(1.05^{24})+12x(1.05^{23})+......+12x and take a factor of 12x out of the right hand side to get 100000(1.05^{25})=12x(1.05^{24}+1.05^{23}+......+1).

When I see a series like this I like to write the terms in the bracket in increasing order, starting with 1, so I would re-write that equation as 100000(1.05^{25})=12x(1+1.05+1.05^2 + 1.05^3 + .....1.05^{24}). Call this equation (1).

Now it is starting to look a bit more manageable. We can use that trick I explained above, to sort out the series in the bracket. Define the series as S, so that S=(1+1.05+1.05^2 + 1.05^3 + .....1.05^{24}).

Now you just multiply both sides by (1-1.05) to get S(1-1.05)=(1+1.05+1.05^2 + 1.05^3 + .....1.05^{24})-(1.05+1.05^2 + 1.05^3 + .....1.05^{25}).

This cancels down beautifully to getS(1-1.05)=(1-1.05^{25}) and hence S=\frac{(1-1.05^{25})}{(1-1.05)}

Now we can sub that into (1) above, 100000(1.05^{25})=12x \frac{(1-1.05^{25})}{(1-1.05)}

A bit of rearranging gets it in terms of x, \frac {100000(1.05^{25})(1-1.05)}{12(1-1.05^{25})}=x so x works out as £591.27

Categories: Geometric series, Maths