Quick question

Time Magazine

Hi all,

I'm doing an transport economics project for college. I want to add something into it that's not covered on the course, namely what rate of growth is needed to justify an investment. As it's not on the course it's not a case of me avoiding work, it's just me not remembering the method from LC Maths. I always hated those nonlinear/log/e^x problems.

Anyway, here's the problem.

Dublin Bus carried 146m passengers in 2006. For their troubles, they got a subsidy of €113m or about 75c per passenger. The government are considering spending €50m on a new scheme that will attract new customers and will spread the cost over twenty years (€2.5m per year). So they need 68m new passengers over the twenty years to justify the cost (68m*€0.75 = €51m). This equals 3.3m new customers a year if the scheme attracts people immediately, an increase of once-off but constant thereafter 2.3% in the 146m passengers. This is all good. All simple calculations.

Now what if it doesn't attract people immediatley? What if there's continuous growth? If that's the case, what growth rate is needed to justify it?

Is the equation I need the one below?

146(x)^20 = 146+68
x^20 = 1.47
log x = (1/20)(log 1.47)
x = 1.019
Growth rate needed = 1.9%?

If this is right, can I express this as "A 2% year-on-year increase in passenger numbers will justify this investment"?

Podge_irl

Yeah, that looks grand to me

Time Magazine

Okay I just ran a simulation of that in Excel and see that I was doing that wrong entirely . What that does is provide 68m new passengers in the final year, 65m in the 19th year etc...

What I'm looking for, I think, is more like

20
Σ x^i = 68.
i=1

What's x?

Edit: I think I'm looking for 1.0022 = 0.22% passenger growth per year. This fits far better in my head.

Podge_irl

Ibid wrote: »

Okay I just ran a simulation of that in Excel and see that I was doing that wrong entirely . What that does is provide 68m new passengers in the final year, 65m in the 19th year etc...

I think your notation is a bit dodgy, but the method is sound.

If you have 146m passengers this year, you will have 146(1+x) next year if your growth is x. Similarly, in the third year you will have (146(1+x))(1+x). And in the final year you will have 146((1+x)^20), which should be equal to 214.

Solving for x through logs you get 0.0193, or 1.93%.

Time Magazine

You're getting 146+68=214 extra passengers in year 20. But at the subsidy rates you need an extra 68m passenger trips in total over twenty years to justify the investment. So you need SFA increase year on year.

Podge_irl

Ah, ****e it, I misunderstood what you were saying. Never mind so!

Fremen

Ibid wrote: »

Okay I just ran a simulation of that in Excel and see that I was doing that wrong entirely . What that does is provide 68m new passengers in the final year, 65m in the 19th year etc...

What I'm looking for, I think, is more like

20
Σ x^i = 68.
i=1

What's x?

Edit: I think I'm looking for 1.0022 = 0.22% passenger growth per year. This fits far better in my head.

Your maths skills are pretty sharp, but I'm afraid your excel skills leave something to be desired, I think that's where you made the mistake.

Between the second-last year and the last year, a 2% year-on-year increase only adds four million people (it adds 68 million over the whole 20 years).
You get roughly 2.8 million people joining the first year, which grows to 4 million in the final year.

If you start with P people, and grow at two per cent per annum, then in year N you have
P*((1.02)^N) people. Notice that in year 20, you have just over 216 million, so you're better to stick with 1.9%

I think the problem you're looking at is analogous to simple Vs. compound interest in savings, though it depends how you want to model the uptake caused by the new advertising.