The in-class section is closed-book and is worth 40% of the points for the mid-term, and the mid-term itself is worth 15% of the points for the course.
Questions 8 and 12 are worth 6 and 8 points respectively, all the other questions are worth 3 points.
Equity = Assets - Liabilities
The tricky question here is what rate of return you want on the $500,000 you're putting in. Some people wanted 15%, some wanted 10%, some were satisfied not to earn anything at all on it. I counted all these as correct, though I can't see why you'd be going into business if you don't want to make any money.
The Income Statement is an account of company cash flows over a fixed period, usually a year. It would normally include income from sales, and expenditures on labour, materials, etc.
The Balance Sheet is a snapshot of the company's state at an instant in time. It should list all assets and debts, and show the company's equity as the difference between them.
In the most general terms, the aspiration level for any one of the inputs to an economic analysis is the value of the input parameter at which the verdict of the analysis changes from `No Go' to `Go'. Identifying the aspiration level for the important parameters can help to focus market research -- instead of asking, ``Do we have any idea what the sales price might be?'', we can ask, ``Do we expect the sales price to be above or below $12.35?'' (which may be an easier question to answer.)
A particular example of an aspiration level is the break-even point for sales. This is the point where the verdict changes from ``We'll lose money'' to ``We'll make money''.
It's the ratio (cash + accounts receivable)/(total liabilities). It gives a measure of how capable the company is of meeting its debts from cash on hand, excluding resources tied up in inventory which may be difficult to liquidate at short notice. Its importance, and the value it needs to have for the company to be healthy, vary from industry to industry. In some industries, for example, inventory is easy to move and doesn't decay over time (for example, stockpiles of oil), whereas in other industries, inventory may be hard to move and perishable (stockpiles of milk; 1998-model cars in January 1999).
This question is explored in the last question of Assignment 7, due on March 5.
The problem I had in mind was that a ranking in order of IRR maximizes rate of return, whereas what we want to do is to maximize dollars of profit. This can be solved using an incremental analysis.
Some people mentioned that use of IRR can lead to multiple solutions. I also counted this as right, and this problem can be solved by using the ERR instead of the IRR. Suggesting incremental analysis as the answer to this problem is wrong, though -- we can still get multiple IRR's from an incremental analysis.
This is just 80,000 * 10 * (CCTF(d=20%) - CCTF(d=30%)).
(You can work out the solution one year at a time instead, but this is very tedious.)
A nominal annual rate of 12% is a nominal monthly rate of 1%. And we're compounding monthly, so this is also the effective monthly rate. So the effective annual rate is (1.01)12-1.
Let the amount be F. Then, equating present values, we have
F(P/F,0.07,5) = 1000(P/A,0.07,12)
from which we can find F.
A nominal rate of 8% per year is a nominal rate of 4% per six months, and the compounding interval is six months, so 4% is also the effective six-monthly interest rate. So the amount is
F = 4000(F/P,0.04,40)
Model A | Model B | |
---|---|---|
Cost | 350.00 | 120.00 |
Maintenance / year | 90.00 | 100.00 |
Life | 10 years | 4 years |
Alice's MARR is 5%.
The difficult part of this analysis is to come up with a fair basis for comparison. For the first part of the question, we need to consider a study period of 20 years. So whichever option we go with, we must make sure that the grass gets cut for 20 years. This means buying a new model A after 10 years, or new model B's after 4, 8, 12, and 16 years. Calculate the present value of these purchases:
PW(A) = 350 + 350(P/F,i,10)
PW(B) = 120 + 120(P/F,i,4) + 120(P/F,i,8) + 120(P/F,i,12) + 120 (P/F i,16)
(Note that I've left out the maintenance costs.) Now, we spread this cost evenly over the twenty years and add maintenance:
EUAC(A) = PW(A)(A/P,i,20) + 90
EUAC(B) = PW(B)(A/P,i,20) + 100
These come out to be almost exactly equal: $135 and $133 respectively. B is a bit cheaper, so we can go with B. (Though it would also be plausible to say that the two options are indistinguishable, given that the uncertainties in the analysis almost certainly amount to more than $2.)
For part (ii), we can go to a 4-year study period. To find the break-even point for the salvage value of A, solve the equation
350 + 90(P/A,i,4) - x(P/F,i,4) = 120 + 100(P/A,i,4)
for x.