CIVL 2812—Project Appraisal (Semester 2, 2019)
CIVL 2812—Project Appraisal (Semester 2, 2021)
Page 1
Tutorial 5- Answers
ESSENTIAL QUESTIONS
Solution 6-41
Solution 6-47
Solution 7-1
The actual magnitude of depreciation cannot be determined until the asset is retired from
service (it is always paid or committed in advance). Also, throughout the life of the asset we
can only estimate what the annual or periodic depreciation cost is. Another difference is that
relatively little can be done to control depreciation cost once an asset has been acquired,
except through maintenance expenditures. Usually much can be done to control the ordinary
out-of−pocket expenses such as labor and material.
Solution 7-2
To be considered depreciable, a property must be:
• used in a business to produce income;
• have a determinable life of greater than one year; and
• lose value through wearing out, becoming obsolete, etc.
• not be inventory, stock in trade, or investment property
CIVL 2812—Project Appraisal (Semester 2, 2021)
Page 2
ESSENTIAL QUESTIONS
Solution 7-3
Personal property is generally any property that can be moved from one location to another, such as
equipment or furniture. Real property is land and anything erected or growing on it.
Solution 7-4
The cost basis is usually the purchase price of the property, plus any
sales taxes, transportation costs, and the cost of installation or
improving the property to make it fit for intended use. Salvage value
is not considered, nor is the cost of the land the property is on.
EXTENTION QUESTIONS
CIVL 2812—Project Appraisal (Semester 2, 2021)
Page 3
Solution 6-45
a) Assume repeatability and compute AW over useful life.
AWE1(15%)= -$14,000(A/P,15%,5) – $14,000 +$8,000(A/F,15%,5) = -$16,990
AWE2(15%) = -$65,000(A/P,15%,20) – $9,000 + $13,000(A/F,15%,20) = -$19,260
Select E1 to minimize costs.
b) Co-terminated assumption (5-year study period)
AWE1(15%) = −$16,990 ; unchanged from Part (a)
Compute the imputed market value of E2 at the end of year 5.
CRE2(15%) = $65,000 (A/P, 15%, 20) − $13,000 (A/F, 15%, 20) = $10,259.6
Imputed MV5 = $10,259.6 (P/A, 15%, 15) + $13,000 (P/F, 15%, 15)= $61,590
AWE2(15%) = −$65,000 (A/P, 15%, 5) − $9,000 + $61,590 (A/F, 15%, 5) = −$19,256 The small
difference in the value of AWE2(15%) in part (a) and part (b) is due to rounding.
Select E1 to minimize the costs.
The reason AWE2 in Part b is the same as in Part a) is the annual expenses are the same over the 20-year
period in Part a as they are over the 5-year period in Part b.
Solution 7-6
Basis = $100,000 + $12,000 = $112,000
a) d2 = (112,000-$8,000) / 10 = $10,400
b) B.V1 = $112,000 – $10,400 = $101,600
c) B.V8 = $112,000 – $10,400(8) = $28,800
Solution 7-7
Solution 7-17
Total units of production over the five year life = 133,000 m3
D4 = (24,000/133,000)($80,000 − $8,000) = $12,992.4
BV2 = $80,000 – (d1 + d2)
= $80,000 – [(20,000/133,000)($72,000) + (28,000/133,000)($72,000)]
= $80,000 – $25,985
= $54,015.