S.O.S. Mathematics CyberBoard

Hey guys i really need some help on this project I'm doing...i hope theres someone who could help me out and i'd really appreciate it...even if its some kinda hint its still fine...i just need some ideas...here's the question:

In the spreadsheet for problem1.xls you will nd the specications for two projects.
This includes how much will be received or invested and at which time. Some of the
values are unknown, specically X; t1; t2; t3. Investigate what values for these variables
will make project 2 preferable to project 1. You may impose restrictions on your
variables to attain closed form solutions if you like. Assume money is earned at the
compound interest rate given in the spreadsheet.


Project 1 Investment Time
-10000 0
-10000 1
-5000 2
3000 4
5000 5
10000 6
30000 7
30000 15

Project 2 Investment Time
-13000 0
-18000 1
10000 2
X 4
27000 t1
21000 t2
6000 t3

Compound interest rate 0.1

I dare do say, that is quite unclear, my good man!

Anyhow...in case this helps...

Project#1 seems to have year-end flows as following:
1: negative at first...start-up costs and the likes, I assume
2: then positives, thus revenue...
If so, then you will have $42,407 at end of 15th year (10% annual).
(the Future Value)

Problem seems to be to compare this to Project#2; so I assume that
we use 15 years again, to make it a sensible comparison...
We need to calculate x as the required amount at end of year#4,
that will result in an ending balance of $42,407 after taking in
consideration that 3 more flows will be received: 27000,21000,6000.

I see no reason for making that a $54,000 amount received at end
of 15th year: nothing prevents that a per problem's wording.

Right before x is received at end of year#4, the balance will be
$(30,891) ... in the red!
So we have:
1.10^11(x - 30891) + 54000 = 42407
x = 26,827

Well, that at least tells you that 2nd project will be equal to the
1st project after 15 years if $26,827 is 4th year flow, and if
27000+21000+6000 = $54,000 is the 15th year flow (none in between).

Hope that helps...

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