Continuing The Phone Lines Example

So let’s convert these into some cash flow diagrams. This one is relatively simple, this is just $300,000 now, and then there’s going to be this constant series of $15,000 per year. So to just replace this with a diagram, I’m going to write these as if these are all constants. There’s our $300,000 constant in time period 0 all the way to time period 20. There’s a convention when that you’re making initial investments, that’s things that all happen in time period 0. Then you have when you have these additional costs in the other time periods. So what you’re going to have here is a long series of constant $15,000 per year costs out to time period 20.

Now this problem, based on the way it’s stated, guide you on how to break it up into its parts. You could turn both of these two sets of phone lines into a single cash flow diagram like that, but it’s not the best way to solve the problem. The best way to think about it is this is a line, literally a cash flow diagram line. This is also a line, literally a cash flow diagram line. So to get these guys on the board, we have the first year’s installation costing $150,000 in period 0 and we have extents of $10,000 per year all the way out to the 20th year. We have a case in year 5 where we have a 2nd installation of $150,000 with an additional maintenance expense associated with it. So this gives us our cash flow diagrams of our two options. This is the all now option, this is the part now option that’s right down here. So now after we have come to some agreement on what these cash flows are supposed to look like, the thing we should do next is start writing out what things are supposed to look like in factor notation. These can actually be something relatively simple

For this one, the present worth of all now is just equal to this initial expense, this negative $300,000, and then you have these additional expenses here of $15,000 every single year for 20 years. So it’s looking for P given you know A, whatever their interest rate is—we’ll figure that out later—with N = 20. That gives you the present worth of this option.

The second option is slightly more complex because it has a few more parts to it. This is the present worth of this part now option. What we’re going to have here is the present worth of this time installation, which is $150,000 with a negative sign in front of it, and here’s that $10,000 per year for 20 years. So that’s that first installation. This here is a little trickier because most of this stuff happens after year 5, so we’re going to have some complicated time shifting there. So I’ll have a plus sign right there, here’s a big fraction. What we have is our $150,000 installation. This is going to be in time period 5, so don’t let that disturb you too much. Here’s our maintenance expense, and this is only because this is time period 6 all the way to 20. So this is only 15 years. What we’re going to have in the denominator is 1 plus the interest rate raised to the power of 5. Essentially what’s going on here is that this time calculation already sits in time period 5—that installation cost—then this term right here, minus 10 times P given we know A interest rate and then 15 gathers all the stuff together that’s in periods 6 through 20 and dumps is all in time period 5. So we have a lot of wealth concentrated right now. What this term in the denominator does is it brings it all the way back to time period 0. So now what we have here is the present worth of that asset.

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