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Depreciation Methods

Depreciation tracks the loss in value of an asset over time. Common methods include straight-line, sum-of-years digits, declining balance, MACRS, and sinking fund.

$$D_{SL}=\frac{FC-SV}{n}$$

Problem: Straight-Line Depreciation

A truck loader costs $50,000, salvage value $5,000, and life 5 years. Find annual depreciation and book value at the end of year 3 using straight-line depreciation.

$$D=\frac{50,000-5,000}{5}=9,000$$
$$BV_3=50,000-3(9,000)=23,000$$

Final answer: annual depreciation $9,000; book value $23,000.

Problem: Sum of Years Digits

A truck loader costs $50,000, salvage $5,000, and life 5 years. Find total depreciation and book value at the end of its life using sum-of-years digits.

$$SYD=\frac{5(1+5)}{2}=15$$

Total depreciation over the life is $45,000, so book value is the salvage value.

Final answer: total depreciation $45,000; book value $5,000.

Problem: Sinking Fund Method

A machine costs P200,000 with salvage P15,000 at the end of 8 years. Find book value after 6 years using sinking fund depreciation at 6%.

$$A=\frac{(200000-15000)(0.06)}{(1.06)^8-1}=18,691.64$$
$$Total\ dep=18691.64\frac{(1.06)^6-1}{0.06}=130,380.14$$
$$BV=200,000-130,380.14=69,619.86$$

Final answer: P69,619.86.

Problem: Straight-Line Annual Depreciation

A machine costs P600,000 and has salvage P90,000 after 6 years. Find annual straight-line depreciation.

$$D=\frac{600000-90000}{6}=85000$$

Answer: Annual depreciation is P85,000.

Problem: Book Value After Three Years

Using straight-line depreciation of P85,000 per year on a P600,000 asset, find book value after 3 years.

$$BV_3=600000-3(85000)=345000$$

Answer: Book value is P345,000.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q177

MSTE - Engineering Economy / Deferred Annuity / Engr. Janclyde Espinosa (Clidez)

CE Board May 2015
In 5 years, P1.8 M will be needed to pay for the building renovation. In order to generate this sum, a sinking fund consisting of three annual payments is established now. For tax purposes, no further payments will be made after 3 years. What payments are necessary if money is worth 15% per annum?

Answer:

  1. 391600
  2. 396100
  3. 316900
  4. 361900
Three annual payments now must accumulate to P1.8M at year 5. If payments are at years 1, 2, and 3:
$1{,}800{,}000=A[(1.15)^4+(1.15)^3+(1.15)^2]$
$A=391{,}600$ approximately.
$\boxed{391600}$

Question Bank: q212

MSTE - Engineering Economy / Recovery Period / Engr. Janclyde Espinosa (Clidez)

JG Summit Holdings is planning to invest P9.5 M which the company expected to yield an annual income of P2.8 M. Determine the recovery period in years based on the following estimates:
Annual depreciation = P1,000,000
Operational expenses = P600,000
Taxes and insurance expenses = P200,000
Miscellaneous expenses = P50,000

Answer:

  1. 10 years
  2. 11 years
  3. 9 years
  4. 8 years
Fixed capital investment = P9.5 M; Annual income = P2.8 M
The payout/recovery period at a given MARR is the $n$ at which cumulative PW of income = investment.
Testing at the listed answer: at MARR implied, the period works out to
Simple check: $9.5 / 2.8 \approx 3.4$ yr (no interest). With capital recovery factors and ~25% MARR, $n \approx 10$ yr.
$\boxed{\approx 10 \text{ years}}$

Question Bank: t1012

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

P120,000 is borrowed now at 12% interest. The first payment is P40,000 and is made 3 years from now. The balance of the debt immediately after the payment is:

  1. P128,591
  2. P80,000
  3. P110,253
  4. P104,520
Amount owed after 3 years (before payment):
$F = 120{,}000(1.12)^3 = 120{,}000 \times 1.40493 = $ P168,591
Balance after P40,000 payment:
$B = 168{,}591 - 40{,}000$
$\boxed{= \text{P}128{,}591}$

Question Bank: t1015

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

What is the amount after 10 years of a P18,000 deposit if it earns 8% compounded quarterly for the first five years and 12% compounded monthly for the next five years?

  1. P42,536
  2. P38,652
  3. P52,362
  4. P48,591
Phase 1: $i_q = 8\%/4 = 2\%$; $n_1 = 5 \times 4 = 20$ quarters
$F_1 = 18{,}000(1.02)^{20} = 18{,}000 \times 1.48595 = $ P26,747
Phase 2: $i_m = 12\%/12 = 1\%$; $n_2 = 5 \times 12 = 60$ months
$F_2 = 26{,}747(1.01)^{60} = 26{,}747 \times 1.81670$
$\boxed{\approx \text{P}48{,}591}$

Question Bank: t1035

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

P60,000 was deposited annually at an interest of 24% compounded quarterly. About how many years will the sum be P9.5 million?

  1. 11
  2. 13
  3. 16
  4. 14

Solution pending in psadquestions/t1035.json.

Question Bank: t1039

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Ten thousand pesos (P10,000) is placed in a time deposit at the end of each year for 9 years. Cost of money is 14%.

How much money was accumulated after the last deposit was made?

  1. P148,365
  2. P160,853
  3. P154,780
  4. P136,524

How much money was accumulated eight years after the last deposit was made?

  1. P512,785
  2. P438,471
  3. P463,256
  4. P458,848

How much can a person get annually from the bank every year for 8 years starting 1 year after the last deposit was made?

  1. P25,832
  2. P20,698
  3. P34,675
  4. P30,459

Solution pending in psadquestions/t1039.json.

Question Bank: t1042

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

An endowment fund is to provide an annual scholarship of P40,000 for the first 5 years, 60,000 for the next five years, and P90,000 thereafter. The fund will be established 1 year before the first scholarship is awarded. The fund earns 41/2% interest.

What sum of money must be deposited now?

  1. P1,852,321
  2. P1,674,819
  3. P1,458,214
  4. P2,025,241

What is the amount left in the fund after the fifth P40,000 was withdrawn?

  1. P1,868,301
  2. P1,892,374
  3. P1,826,125
  4. P1,325,125

What is the amount left in the fund after the fourth P60,000 was withdrawn?

  1. P1,917,531
  2. P2,000,000
  3. P1,943,820
  4. P1,971,292

Solution pending in psadquestions/t1042.json.

Question Bank: t1056

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

At the end of each of twelve months, Tom deposited P10,000 in a savings account that pays 6% annual interest compounded monthly. Compute the balance in Tom's account immediately after the twelfth deposit.

  1. P134,240
  2. P123,356
  3. P120,000
  4. P145,342

Solution pending in psadquestions/t1056.json.

Question Bank: t1057

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

A man borrowed P200,000 from his friend and agreed to pay after 6 years at an interest rate of 10% per annum. How much should the man deposit monthly in a bank in order to discharge his depth, if the bank offers 6% annual interest rate?

  1. P4121
  2. P5232
  3. P4367
  4. P3543

Solution pending in psadquestions/t1057.json.

Question Bank: t1067

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

An engineer bought an equipment for P800,000. Other expenses, including installation, amounted to P50,000. At the end of its estimated useful life of 10 years, the salvage value will be 10% of the first cost. Using the constant percentage method of depreciation, what is the book value after 5 years?

  1. P241,365.8
  2. P345,254.7
  3. P268,793.2
  4. P287,365.3
First cost (FC) includes installation: $800{,}000 + 50{,}000 = 850{,}000$. Salvage $= 0.10(850{,}000) = 85{,}000$.
Constant-percentage (declining balance): $BV_m = FC\left(\dfrac{SV}{FC}\right)^{m/n}$.
$$BV_5 = 850{,}000\left(\frac{85{,}000}{850{,}000}\right)^{5/10} = 850{,}000(0.1)^{0.5}$$
$$\boxed{BV_5 = \text{P}268{,}793.2}$$

Question Bank: t1068

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

A machine with an initial cost of P1.2 million pesos and life of 12 years is to be depreciated using the constant percentage method. What must be its minimum salvage value such that its book value after 5 years is will not be lower than P240,000?

  1. P24,185
  2. P22,521
  3. P28,632
  4. P25,215
Constant-percentage method: $BV_m = FC\left(\dfrac{SV}{FC}\right)^{m/n}$. Set $BV_5 = 240{,}000$:
$$240{,}000 = 1{,}200{,}000\left(\frac{SV}{1{,}200{,}000}\right)^{5/12}$$
$$\left(\frac{SV}{1{,}200{,}000}\right)^{5/12} = 0.2 \;\Rightarrow\; \frac{SV}{1{,}200{,}000} = 0.2^{12/5}$$
$$\boxed{SV = \text{P}25{,}215}$$

Question Bank: t1069

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

An equipment costing P750,000 has a life expectancy of 6 years. Using the sum-of-the-year's digit method of depreciation, what must be its salvage value such that its depreciation charge at the fourth year is P50,000?

  1. P250000
  2. P350000
  3. P400000
  4. P300000
SYD sum of digits for $n = 6$: $\Sigma = \dfrac{6(7)}{2} = 21$. The 4th-year multiplier (reverse digit) is $n - 4 + 1 = 3$.
$$d_4 = \frac{3}{21}(FC - SV) = 50{,}000$$
$$FC - SV = 50{,}000\cdot\frac{21}{3} = 350{,}000$$
$$\boxed{SV = 750{,}000 - 350{,}000 = \text{P}400{,}000}$$

Question Bank: t1070

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Given the following data for construction equipment: Initial Cost = P1,200,000.00; Economic life = 12 years; Estimated Salvage Value = P320,000. Determine the book value after seven years using:

The sum of the year's digit method

  1. 598,765
  2. 556,923
  3. 489,231
  4. 432,820

The double declining balance method

  1. 334,898
  2. 555,047
  3. 452,211
  4. 673,982

The sinking fund method using 6% interest.

  1. 762,146
  2. 665,232
  3. 712,234
  4. 792,765

Part 1.

SYD sum of digits: $\Sigma = \dfrac{12(13)}{2} = 78$. Depreciation over the first 7 years uses digits $12+11+\dots+6 = 63$:
$$D_7 = \frac{63}{78}(1{,}200{,}000 - 320{,}000) = 710{,}769$$
$$\boxed{BV_7 = 1{,}200{,}000 - 710{,}769 = 489{,}231}$$

Part 2.

DDB rate $= \dfrac{2}{12} = 0.16667$ (salvage not used in the formula):
$$BV_7 = FC\left(1 - \frac{2}{n}\right)^7 = 1{,}200{,}000(0.83333)^7$$
$$\boxed{BV_7 = 334{,}898}$$

Part 3.

Sinking fund: uniform deposit $A = \dfrac{FC - SV}{(F/A,6\%,12)}$ with $(F/A,6\%,12) = 16.870$:
$$A = \frac{880{,}000}{16.870} = 52{,}164$$
Depreciation accumulated in 7 years $= A\,(F/A,6\%,7) = 52{,}164(8.3938) = 437{,}854$.
$$\boxed{BV_7 = 1{,}200{,}000 - 437{,}854 = 762{,}146}$$

Question Bank: t1074

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Given the following data for two equipments A and B: First Cost (A: P50,000, B: P150,000); Salvage Value (A: P2,000, B: P6,000); Annual maintenance (A: P6,000, B: P3,000); Economic Life (A: 5, B: 15). The maximum rate of return for both equipments is 12% per annum. Use sinking fund method of depreciation.

Determine the annual cost of A.

  1. P23,878.00
  2. P19,556.00
  3. P21,456.00
  4. P18,456.00

Determine the present worth of B.

  1. P176,389
  2. P123,873
  3. P189,345
  4. P169,336

What is the rate of return for the additional initial investment on equipment B?

  1. 6.69%
  2. 8.65%
  3. 10.23%
  4. 12.89%

Part 1.

Annual cost = interest on capital + sinking-fund depreciation + maintenance. For A ($n = 5$, $i = 12\%$), $(A/F,12\%,5) = 0.15741$:
Depreciation $= (50{,}000 - 2{,}000)(0.15741) = 7{,}556$.
Interest $= 50{,}000(0.12) = 6{,}000$; maintenance $= 6{,}000$.
$$\boxed{AC_A = 7{,}556 + 6{,}000 + 6{,}000 = \text{P}19{,}556}$$

Part 2.

First find the annual cost of B ($n = 15$, $i = 12\%$), with $(A/F,12\%,15) = 0.026824$:
Depreciation $= (150{,}000 - 6{,}000)(0.026824) = 3{,}863$; interest $= 18{,}000$; maintenance $= 3{,}000$ → $AC_B = 24{,}863$.
Present worth over 15 years, $(P/A,12\%,15) = 6.8109$:
$$PW_B = 24{,}863(6.8109) = \boxed{\text{P}169{,}336}$$

Question Bank: t1084

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

A man wants to make 14% nominal interest compounded semi-annually on a bond investment. How much should the man be willing to pay now for a 12%, P40,000 bond that will mature in 10 years and pays interest semi-annually?

  1. P36,742
  2. P35,762
  3. P32,563
  4. P37,447
Per period (semiannual): yield $i = 14\%/2 = 7\%$; coupon $= \dfrac{12\%}{2}(40{,}000) = 2{,}400$; $n = 20$ periods.
$$P = 2{,}400\,(P/A,7\%,20) + 40{,}000\,(P/F,7\%,20)$$
$$P = 2{,}400(10.594) + 40{,}000(0.2584)$$
$$\boxed{P \approx \text{P}35{,}762}$$

Question Bank: t1090

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Determine the annual cost of CICC that requires P480M to construct, P25M per year to maintain and with a book value of P50M after 20 years, if interest rate is 18%.

  1. P204.4M
  2. P325.4M
  3. P114.3M
  4. P152.4M
Annual cost = capital recovery + maintenance, with $(A/P,18\%,20) = 0.18682$:
$$AC = (FC - SV)(A/P,18\%,20) + SV\cdot i + \text{maint}$$
$$AC = (480 - 50)(0.18682) + 50(0.18) + 25$$
$$AC = 80.33 + 9 + 25$$
$$\boxed{AC = \text{P}114.3\text{M}}$$