Problem 1

Consider 3 mutually exclusive alternatives (X is variable):

Alternative / A / B / C
Year
0 / -X / -3X / $0
1 / $0 / $200 / $0
2 / $100 / $200 / $0
3 / $200 / $200 / $0

If X=$145, which alternative should be selected (Hint: compare present values of alternatives):

  1. if MARR=2%
  2. if MARR=10%
  3. if MARR=40%

If X is unknown and MARR=10%:

  1. over which range of X is A a preferred alternative?

Solution

Compute incremental cash flow B-A:

Alternative / A / B / B-A
Year
0 / -X / -3X / -2X
1 / 0 / 200 / 200
2 / 100 / 200 / 100
3 / 200 / 200 / 0

If interest rate is i%, the present values of the alternatives are:

,

,

;

MARR / A / B / B-A / Decision
a. / 2% / 139.58 / 141.78 / 2.20 / Accept B
b. / 10% / 87.91 / 62.37 / -25.54 / Accept A
c. / 40% / -21.09 / -117.22 / -96.12 / Accept C

d.The alterative A is preferred if the following conditions are satisfied simultaneously:

So,

,

So X should satisfy:

Problem 2

Consider three mutually exclusive alternatives :

A / B / C
First Cost / 560 / 340 / 200
Uniform Annual Benefits / 140 / 90 / 50

Each alternative has a 6 year useful life and assume that MARR is 10%

Which alternative should be selected?

a) Use Benefit-Cost Ratio Analysis (compare benefit-cost ratios of individual projects,

do not consider incremental analysis);

b) Use Pay-Back period analysis;

c) Is Pay-Back period analysis consistent with Present Worth Analysis? Explain;

d) Based on part a), if you used Future Worth Analysis which alternative would be selected (You do not have to apply future worth analysis, just state your answer and explain)

a.

Benefit-Cost ratio computations

A B/C=

B B/C=

C B/C=

b.Pay -Back period

A PAYBACK=

B PAYBACK==3.77 years

C PAYBACK==4 years

Conclusion: select B

  1. No, Pay-Back period analysis does not always select the alternative with the largest present worth. It is an approximate economic analysis.
  2. Select B: FWA and B/C are consistent.

Problem 3

Consider a $100,000 truck, with a three-year depreciable life and an estimated $10,000 salvage value. The utilization of the truck is shown below.

Year / Miles
1 / 30,000
2 / 40,000
3 / 20,000

(15%) i. Compute the truck depreciation schedule by each of the following methods

(3%) a. Straight Line;

(3%) b. Sum-of-years digits;

(3%) c. Double Declining Balance;

(3%) d. Unit of production;

(3%) e. Modified accelerated cost recovery system;

(5%) ii. If interest rate is 10%, arrange the schedules in order of decreasing preference

(Hint: compare present values of depreciation schedules).

i. Depreciation Schedules are:

Year / SL / SOYD / DDB
Depreciation / Depreciation / Depreciation / BV
1 / 30000 / 45000 / 66666.66667 / 33333.33333
2 / 30000 / 30000 / 22222.22222 / 11111.11111
3 / 30000 / 15000 / 1111.111111 / 10000
NPV / 74 605.56$ / 76 972.20$ / 79 806.33$
Year / Un. Prod. / MACRS
Depreciation / Depr(%) / Depreciation / Ordinary Losses / BV
1 / 30000 / 20% / 20000 / 80000
2 / 40000 / 32% / 32000 / 48000
3 / 20000 / 9.60% / 9600 / 28400 / 38400
NPV / 75 356.87$ / 77 658.90$

ii. The most desirable schedule is DDB, than MACRS, SOYD, Un.Prod, SL.

The least desirable schedule is by SL depreciation.

Problem 4

You are considering buying a device for $50,000 with useful life 2 years and a salvage value of $10,000. This device will produce an additional annual benefit at the end of each year during its useful life. Combined federal and state tax rate is 55%; your MARR is 20%. The device is depreciated by straight-line depreciation.

a). What is break even for annual benefit, that make the investment acceptable?

b). What is break even for annual benefit in presence of inflation 10%?( inflation affects annual benefit and salvage value, MARR is expressed in inflated dollars)

Solution.

a). Depreciation is for two years.

If X is the annual benefit generated by the device, present worth of the investment is:

.

To find the break even we solve the equation: .

b). In the case when the inflation is present, the depreciation does not change, but benefits and salvage value become greater:

To find the break even point we solve the equation: .

,

So, .