PW(MARR) PROJ. COMP.
If the PW of any project under consideration is evaluated at the MARR,
denoted PW(MARR), then
the sign of the PW evaluated at the MARR will allow you to determine if the
projects rate of return satisfies this minimum condition:
Does your A/P land at the correct "one
period before first payment"?
Does your A/F land at the correct "at last
payment"?
Did you correctly shift any resulting
values forward/back?
1000/year, 6% comp./year
1000/year, 6% comp./semi-ann.
REPEATED LIVES APPROACH
To find IRR, use trial and error method
via given formula and solving for i*.
Notice that it’s PW = 0 rearranged. If
cannot be solved analytically (no factor
matches), use linear interpolation.
Linear interpolation:
Use if don’t know N value that
corresponds to i %. Find two N values
that straddle the value needed at i. Use
equation.
If i* = MARR, the project is acceptable. For INDEPENDENT projects, do
this for all projects.
For example, take a project with MARR of 20%. Bring all cash flows to
present:
MUTUALLY EXCLUSIVE IRR PROJ. COMP
Alt. 2 is preferable, cheaper than Alt. 1.
STUDY PERIOD APPROACH
If a project has a longer service life than what will be required of it, estimate its worth at
the cut off point. Consider Alt. 1 & 2 from above, but the required service life is 10
years. Compare 10 years worth of both alternatives (Alt. 2 will be cut down):
NOTE: FOR THIS APPROACH, DIFFERING SERVICE LIVES ARE NOT
REQUIRED. The only requisite for this approach to be appropriate is that a project
service life exceeds the required service period.
PAYBACK PERIOD APPROACH
Payback period can be used to compare two projects to each other. The project with the
shorter payback period is more ideal. For projects with a constant cash flow, use the
equation given.
Let’s suppose the current interest rate is
12% compounded semi-annually. What
is the worth of a bond maturing in 15
years with a face value of $5,000, a
semi-annual coupon and a nominal
coupon rate of 7%?
Coupon payments are unchanged since
they are determined by the coupon rate
and the face value.
If asked for current value of bond, it
means value of the bond with the time
left on it. If the first payment is
immediate, do not apply interest to it.
Payments left after 10 years, with i = 4%:
Turn PWr into x * PWoutp, solve for x.
Remember to bring all cash flows to
PW. Remember, A payments will be the
same during all the cash flows. BRING
EVERYTHING TO SAME TIME,
PERIOD 0.
Projects for whom the IRR = MARR have a marginally acceptable rate of
return. A project for whom IRR > MARR is associated with situations
where PW(MARR) > 0.
Setup the equation like so, then solve for i*:
- If PW(MARR) = 0, then the project earns the same rate or return as the
MARR, and it is marginally acceptable.
- If PW(MARR) > 0, then the project earns a higher rate of return than the
MARR and it is acceptable.
- If PW(MARR) < 0, then the project earns a lower rate of return than the
MARR and it is not acceptable.
If cash flows match with comp. rate, use
r/m
INDEPENDENT IRR PROJ. COMP.
PW(MARR) > 0, project is acceptable
When annual savings or inflows are not constant, the payback period is computed by
deducting each period of savings from the first cost, keeping track of the cumulative
balance of the project, until the first cost is fully recovered.
This method can be performed with AW(MARR) or FW(MARR). All rules
remain the same, but take the cash flows to be annuities or in the future,
respectively.
For example, consider a project that has a first cost of $20,000 and annual savings that
follow an arithmetic gradient series increasing by $1,000 increments with the first cash
flow in period 2. What is the payback period for this project?
When comparing different projects with IDENTICAL service lives, use the
above method for each project. With DIFFERING service lives, use the repeated
lives approach, study period approach, or simply AW(MARR), as the length
doesn’t matter when looking at AW (all payments are the same).
Repeated Lives Approach: repeat projects as needed until their lengths are the
same, then compare. For ex., if project one lasts 5 years and project two lasts 10,
repeat project one twice (common multiple) in PW(MARR) calculations. Can
be done with future worth too. Only compound first costs of coming projects.
Study Period Approach - cut down longer project length to that of the shorter
project by estimating value of the longer project after the cutoff date.
If the cash flows were constant, then: 5k/6k = 5/6 , so PP is 6 and 5/6 years.
For MUTUALLY EXCLUSIVE projects, do NOT do the above method and
simply pick the project with the best IRR. Use incremental IRR method:
- Rank each alternative project by their first cost
- Find the IRR for the lowest first-cost alternative
- If that project has an IRR that exceeds the MARR, then it is seen as the
“current best candidate.” Otherwise discard and move to next candidate.
Find the incremental investment by subtracting the cash flows, period by
period, of the current best candidate from the challenger. This is the
incremental investment.
- Find the IRR of the incremental investment. If the incremental
IRR>MARR, then the current defender is discarded and the challenger
becomes the new defender. If the incremental IRR<MARR, challenger is
discarded, and the new challenger is the next most costly project. The
comparisons between alternatives happen in pairs. The incremental
investment and then incremental IRR are calculated for only two
alternatives at a time.
ERR
Marginal Analysis
With atypical cash flows, sometimes a factor can appear more than once
when solving for IRR. This will result in a quadratic, which cannot apply.
Instead, find the ERR
Once an asset has been installed and has been operating, the
costs of installation, and all other costs incurred up to that time
which cannot be recovered, are no longer relevant to any
decision to replace the current asset. These costs are referred to
as sunk costs. Therefore, EAC must be modified to fit this new
situation:
PRECISE ERR METHOD - look at the cash flows linearly, and create two
blocks. A receipt at time 0 helps pay off a disbursement at time 1, so combine
these two into a single net disbursement/receipt. Basically, turn non-simple
cash flow into simple cash flow:
Our goal is to find the minimum value in the concave-up graph of EAC
values for both defender and challenger. It is important to TURN ALL
COSTS into positives, and TURN ALL BENEFITS into negatives in the
cash flow diagram. Remember to always assume you sell at salvage value
for that time period, and if not, add it back in as a cost in the next time
period for the cash flow diagram.
Then use same method as if solving for IRR.
APPROXIMATE ERR METHOD - net all receipts forwards to the final
cash flow at rate = MARR. Net all disbursements forwards to the final cash
flow at rate = i*ea. Set the two future worths equal to each other, then solve
for approximate ERR.
USE THIS METHOD PRINCIPALLY
When i∗ea>MARR, both ERR and approximate ERR indicate the project is
acceptable.
When i∗ea<MARR, both ERR and approximate ERR indicate the project is
not acceptable.
IF EAC at n+1 < EAC at n, n is not the economic life of the project. For
example if n = 1, then 1 is not the economic life, and n = 2 must now be
compared to n = 3.
See example question:
The Jiffy Printer Company produces printers for home use. Jiffy is
considering installing an automated plastic moulding system to produce
parts for the printers. The moulder itself costs $200k and the installation
costs are estimated to be $50k. Operating and maintenance costs are
expected to be $300k in the first year and to rise at 5% per year up to a
maximum of seven years. Jiffy estimates depreciation will follow the
declining balance model using a rate of 40%. The moulder’s costs are
fixed (do not vary with production levels) and do not include material
costs. Jiffy’s MARR is 15%. Jiffy has an indefinite need for the service.
How long should Jiffy keep a moulder before replacing it with a new
model?
EAC1=250,000(F/P,15%,1)−200,000(0.6)+300k=467,500
Balance Sheet
REMEMBER: if O&M costs do not increase monotonically, 1
year rule cannot be implemented.
Gives a snapshot of an enterprise’s financial position at a
particular point in time, normally the last day of an accounting
period. Data included referred to as “across variables”, which
are measures of the value of assets such as cash, inventory, and
equipment, the level of a company’s indebtedness, and the
amount of equity accruing to a company’s shareholders.
The left column lists the assets (current and long term), the
middle column shows pertinent numbers for a given asset (e.g.,
equipment less the accumulated depreciation of that
equipment), and the right column shows subtotals for each
asset category, with a total amount in the last row. Any
accumulated depreciation needs to be deducted from buildings
and equipment. Land is not subject to depreciation. The total
assets must be equal to the total liabilities and owners’
equity. Liabilities, claims on the business’s assets by
non-owners, are divided into current and long-term liabilities.
Current liabilities are debts that are normally paid within a
year. Examples include accounts payable, taxes, wages payable,
bank loan payable, etc. Long-term liabilities are debts that are
normally paid beyond the current year. The owners’ equity
measures the difference between assets and liabilities, and it is a
measure of the company’s worth to its owners, or shareholders.
Income Statement
Summaries an enterprise’s revenues and expenses over a
specified accounting period. The income statement is
comprised of “through variables” such as revenues, expenses,
income before taxes, taxes, and net income. In general, the
major components of the income statement are revenues,
expenses, and profits/incomes. When calculating EBIT, only
the operating expenses of the business are deducted from
revenues. Income tax and interest on loans are excluded from
this measure because they are not directly associated with the
sales and operating expenses that are within the control of
managers. Any sources of revenues and expenses are each
separately totaled, and then total expenses are deducted from
total revenues to derive income before taxes. Taxes are then
deducted from income before taxes to derive the net income.