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# Project 11K Capital Equipment

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Project 11K Capital Equipment document sample

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```									                                                       Engr 360 Engineering Econ. 5.1

Economic Analysis Based on Present Worth
The key application here is to interpret or resolve feasible
alternatives into equivalent present consequences (i.e.,
present worth).

For now, in our present-worth analysis we will assume that
prices/costs remain constant over time. In other words, we will
ignore the effects of inflation and deflation.

Our criterion to select the best alternative depends on the situation:

For fixed input, we want to maximize PW of outputs (benefits or revenues)
For fixed output, we want to minimize PW of inputs (costs or expenses)
For neither fixed input nor output, we want to maximize the net PW
Note: Net PW = PW(benefits) – PW(costs)
Engr 360 Engineering Econ. 5.2
Present Worth Analysis (contin.)

The most common application of PW analysis is to estimate the
present value of expected future cash flows that include revenues
and expenses (or receipts and disbursements). This allows us to
quantify the present value of incoming-producing property and,
thus, establish a fair price at which the property could be sold.

When considering various alternatives, the consequences of each
one (i.e., the cash flows) must be considered for a fixed time
period, known as the analysis period (say, n = 10 yr).

There are three common situations to analyze:

1. The useful life of each alternative equals the analysis period.
2. The alternatives have useful lives different from the analysis period.
3. The analysis period is considered infinite (here, PW = “capitalized cost”).
Engr 360 Engineering Econ. 5.3

Analysis Period Equivalent to Useful Life

This is the most straight-forward comparison and relies on using
an analysis period that equals the useful life of each alternative
(where all alternatives have the same useful life).

Example:
We are considering the purchase of a robot to speed-up assembly and cut costs
at our small electronics plant. Options Rob1 and Rob2 both have a useful life
of 10 yr. and no salvage cost. Rob1 costs \$40k and will result in uniform
annual savings of \$7500. Rob2 costs \$45k and will provide declining cost
savings of \$12k in Yr.1, \$11k in Yr.2, \$10k in Yr.3, and so on. Assuming an
annual interest rate of 6%, which is the more economical option?
7.360
Rob1: P = -\$40,000 + \$7500[P/A, 6%, 10] = \$15,200

7.360                  29.602
Rob2: P = -\$45,000+\$12,000[P/A, 6%, 10] – \$1000[P/G, 6%, 10] = \$13,718
We recommend Rob1.
Engr 360 Engineering Econ. 5.4

Analysis Period Differs from Useful Life

To work with a consistent analysis period when dealing with
alternatives with different useful lives, we commonly either:

1. Set the analysis period equal to the shorter useful life, and
then assign an appropriate salvage value (at this time) to the
longer useful life.

2. Set the analysis period equal to the least common multiple of
the useful lives, and then rely on replacement of alternatives
to carry us out to the end of that analysis period.

3. Set the analysis period to a reasonable time period less than
the useful life, at which terminal (salvage, market) values can
be estimated and used for consistent comparisons.
Engr 360 Engineering Econ. 5.5

Example:
Consider a new version of Rob2 (in the previous example) that has a 15-yr
useful life with a “used” value of \$8000 at 10 years. How does it now compare
economically with Rob1? (Recall: Rob1 costs \$40k and will result in uniform
annual savings of \$7500. Rob2 costs \$45k and will provide declining cost
savings of \$12k in Yr.1, \$11k in Yr.2, \$10k in Yr.3, and so on. Assume annual
interest rate of 6%.) We still use an analysis period of 10 yr.
7.360
Rob1: P = -\$40,000 + \$7500[P/A, 6%, 10] = \$15,200

7.360                 29.602
Rob2: P = -\$45,000+\$12,000[P/A, 6%, 10] – \$1000[P/G, 6%, 10]

0.5584
+ \$8000[P/F, 6%, 10]    = \$13,718 + \$4467 = \$18,185

We recommend new Rob2.
Engr 360 Engineering Econ. 5.6

Infinite Analysis Period

A facility, service, infrastructure, etc. is to be maintained for an
infinite period. This type of present worth analysis is called
“capitalized cost”.

Without diminishing the initial sum, the end-of-period payments A
must equal iP for each period. Thus,

for n = infinity, A = Pi

and the Capitalized Cost, P = A/i
Engr 360 Engineering Econ. 5.7

Example:

Our telecommunications company plans to install a fiber-optic network in an
urban area that will be in service indefinitely. The installation cost is \$1.5M and
the estimated annual cost for maintenance and upgrades to the system is
\$45,000. Assuming 5% annual interest, what is the capitalized cost of the
network?

Capitalized cost, P = Po + A/i = \$1,500,000 + \$45,000/0.05

= \$1,500,000 + \$900,000 = \$2.4M
Engr 360 Engineering Econ. 6.1

Economic Analysis Based on Annual Cash Flow
The key application here is to interpret or resolve feasible
alternatives into equivalent annual cash flow amounts.

For now, in our annual cash flow analysis we will assume that
prices/costs remain constant over time. In other words, we will
ignore the effects of inflation and deflation.

Our criterion to select the best alternative depends on computing one of the
following values:

Equivalent uniform annual benefit (EUAB), where we have fixed input
Equivalent uniform annual cost (EUAC), where we have fixed output
The difference between these two values (EUAB - EUAC), where neither
input nor output is fixed
Engr 360 Engineering Econ. 6.2

Capital Recovery Cost

In a typical engineering econ. analysis in which there is an initial
disbursement (usually a payment or purchase) known as P and a
future salvage value that can be realized (known as S), the
computed EUAC is known as the capital recovery cost of a
project.

It can be computed in several different ways:

EUAC = P[A/P, i, n] – S[A/F, i, n]

EUAC = (P - S)[A/P, i, n] + Si

EUAC = (P - S)[A/F, i, n] + Pi

Note: Any salvage value at the end of the analysis period will decrease the
EUAC.
Engr 360 Engineering Econ. 6.3

Important Relationships

1. An expenditure (pay-out or loss) of money during the analysis period will
increase the EUAC, whereas a receipt (income or gain) of money will
decrease the EUAC.

2. If there are irregular cash flows during the analysis period, it often is
helpful to first calculate the PW of cost, then apply the EUAC equation:

EUAC = (PW of cost)[A/P, i, n]

3. When there are cash flows according to an arithmetic gradient, then apply
the following gradient uniform series expression:

EUAC = G[A/G, i, n]

Note: Analysis periods typically are consistent in the same manner as in present-
worth analysis. However, we can compare EUAC values directly for different useful-
life periods when identical replacement (same costs, performance, etc.) is assumed
for the shorter-life alternative at the end of its useful life.
Engr 360 Engineering Econ. 6.4

Example:
Our consulting firm is evaluating two different lease options for a fleet of four
vehicles over the next three years. Option 1 requires a down-payment of
\$15,000 followed by annual payments of \$12,000. Option 2 requires \$11,000
up front followed by payments of \$14k, \$13k, \$12k over the three years. What
is our best choice based on EUAC using 6% interest?

Opt. 1: EUAC = Annual cost of initial cash payout + regular annual cost
0.3741
= \$15,000[A/P, 6%, 3] + \$12,000 = \$17,612

Opt. 2: EUAC = Annual cost of initial payout + annual cost of gradient pymts.

= \$11,000[0.3741] + PWgrad[0.3741]
2.673                2.569
where: PWgrad = \$14,000[P/A, 6%, 3] - \$1,000[P/G, 6%, 3] = \$34,853

Thus, EUAC = \$11,000[0.3741] + \$34,853[0.3741] = \$17,154

Choose Lease Option 2.
Engr 360 Engineering Econ. 6.5

Example:
Our firm is considering the purchase (\$80,000) of lab testing equipment with a
15-yr life. The annual maintenance costs are projected to be \$0 in the first
year followed by regular annual increases of \$600. Terminal (salvage) value at
the end of the useful life is estimated to be \$30,000 and is based on uniform
depreciation in value (i.e., \$5000 loss in value per year). What is the EUAC,
given an analysis period of 10 years and interest of 4%?
| 30,000
|
0…...1…...2…...3…...4…...5…...6…...7…...8…...9…...10
|     0    |    | |         | |       | |       |     |
|        600    |     |     | |       | |       |     |
|            1200     |     | |        | |       |    |
|                 1800       | |       | |       |    |
| 80,000                     | ................

0.1233              4.177                    0.0833
EUAC = \$80,000[A/P,4%,10] + \$600[A/G,4%,10] – \$30,000[A/F,4%,10]
= \$9871
Engr 360 Engineering Econ. 6.6

Infinite Analysis Period

For a decision alternative considered to have an infinite life (e.g.,
facility, service, infrastructure), we can conduct the EUAC
analysis over an infinite analysis period.
In this case, we see that: EUACinf = Pi + Any other annual costs

Example:
Recall from Chap. 5 notes: A fiber-optic network is installed in an urban
area and will be in service indefinitely. The installation cost is \$1.5M and the
estimated annual cost for maintenance and upgrades to the system is \$45,000.
Assuming 5% annual interest, what is the EUAC of the network?
EUACinf = \$1,500,000(0.05) + \$45,000 = \$120,000
0.0528
Now, assume useful life of 60 yr: EUAC60 = \$1,500,000[A/P,5%,60) + 45,000
= \$124,200

Note: Diff. in EUAC between long life & infinite life is small unless a very low interest is used

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