Depreciation, Bonds, Breakeven Analysis

LessonAbout 1.5 h2 of 2 in Engineering Economics

Engineering Economics

Engineering Economics is the application of economic principles and mathematical techniques to evaluate engineering projects and financial decisions.

3. Depreciation and Valuation

Depreciation is the reduction in the value of a physical asset over time due to usage, wear and tear, or functional obsolescence.

3.1 Straight-Line Method

Straight-line depreciation assumes that the value of an asset decreases uniformly over its useful service life.

  • First cost: total initial purchase and installation cost
  • Salvage value: expected resale or scrap value at retirement
  • Book value: remaining recorded value after \(m\) years

Straight-Line Depreciation

\[ d = \dfrac{FC - SV}{n} \]
\[ BV_m = FC - d(m) \]

EE Board April 1997

Example. A machine has an initial cost of \(50,000\ \text{pesos}\) and a salvage value of \(10,000\ \text{pesos}\) after \(10\ \text{years}\). Find the book value after \(5\ \text{years}\) using straight-line depreciation.

Solution.

\[ d = \dfrac{FC - SV}{n} = \dfrac{50000 - 10000}{10} = 4,000.00\ \text{pesos/year} \]
\[ BV_5 = FC - d(m) = 50000 - 4000(5) = 30,000.00\ \text{pesos} \]

Answer: \(30,000.00\ \text{pesos}\)

ME Board April 1992

Example. A unit of welding machine cost \(45,000\ \text{pesos}\) with an estimated life of \(5\ \text{years}\). Its salvage value is \(2,500\ \text{pesos}\). Find its depreciation rate by straight-line method.

Solution.

\[ d = \dfrac{FC - SV}{n} = \dfrac{45000 - 2500}{5} = 8,500.00\ \text{pesos/year} \]
\[ \text{Depreciation rate} = \dfrac{d}{FC} \times 100\% = \dfrac{8500}{45000} \times 100\% = 18.89\% \]

Answer: \(18.89\%\)

3.2 Sinking Fund Method

Sinking fund method assumes that annual depreciation payments are placed into an imaginary sinking fund that earns compound interest.

Sinking Fund Depreciation

\[ d = \dfrac{(FC - SV)i}{(1 + i)^n - 1} \]
\[ D_m = \dfrac{d[(1 + i)^m - 1]}{i} \]
\[ BV_m = FC - D_m \]

Example. An equipment costs \(10,000\ \text{pesos}\) with a salvage value of \(500\ \text{pesos}\) at the end of \(10\ \text{years}\). Calculate the annual depreciation cost by sinking fund method at \(4\%\) interest.

Solution.

\[ d = \dfrac{(FC - SV)i}{(1 + i)^n - 1} = \dfrac{(10000 - 500)(0.04)}{(1 + 0.04)^{10} - 1} = 791.26\ \text{pesos} \]

Answer: \(791.26\ \text{pesos}\)

Example. An equipment costing \(250,000\ \text{pesos}\) has an estimated life of \(15\ \text{years}\) with a book value of \(30,000\ \text{pesos}\) at the end of the period. Compute the book value after \(10\ \text{years}\) using sinking fund method assuming \(i = 8\%\).

Solution.

\[ d = \dfrac{(FC - SV)i}{(1 + i)^n - 1} = \dfrac{(250000 - 30000)(0.08)}{(1 + 0.08)^{15} - 1} = 8,102.50\ \text{pesos} \]
\[ D_{10} = \dfrac{d[(1 + i)^{10} - 1]}{i} = \dfrac{8102.50[(1 + 0.08)^{10} - 1]}{0.08} = 117,377.40\ \text{pesos} \]
\[ BV_{10} = FC - D_{10} = 250000 - 117377.40 = 132,622.60\ \text{pesos} \]

Answer: \(132,622.60\ \text{pesos}\)

3.3 Declining Balance Method

Declining balance method, or Matheson method, calculates depreciation as a fixed annual percentage \(k\) of the remaining book value.

  • Constant ratio: fixed rate applied to decreasing book value
  • Limitation: cannot be used if salvage value is zero

Declining Balance Method

\[ k = 1 - \sqrt[n]{\dfrac{SV}{FC}} \]
\[ BV_m = FC(1 - k)^m \]
\[ d_m = FC(1 - k)^{m-1}k \]

CE Board November 1995

Example. A machine costing \(720,000\ \text{pesos}\) is estimated to have a book value of \(40,545.73\ \text{pesos}\) when retired at the end of \(10\ \text{years}\). Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation?

Solution.

\[ BV_n = FC(1 - k)^n \implies 40545.73 = 720000(1 - k)^{10} \]
\[ (1 - k)^{10} = 0.05631 \implies k = 0.25 = 25\% \]

Answer: \(25\%\)

CE Board May 1996

Example. A machine costing \(45,000\ \text{pesos}\) is estimated to have a book value of \(4,350\ \text{pesos}\) when retired at the end of \(6\ \text{years}\). Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation?

Solution.

\[ BV_n = FC(1 - k)^n \implies 4350 = 45000(1 - k)^6 \]
\[ (1 - k)^6 = 0.09667 \implies k = 0.3225 = 32.25\% \]

Answer: \(32.25\%\)

3.4 Sum-of-the-Years’-Digits Method

Sum-of-the-years’-digits method is an accelerated depreciation method where yearly charges decrease linearly over service life.

Sum-of-the-Years’-Digits (SYD)

\[ \sum \text{years} = \dfrac{n(n + 1)}{2} \]
\[ d_m = (FC - SV)\dfrac{n - m + 1}{\sum \text{years}} \]

ME Board April 1998

Example. A company purchases an asset for \(10,000.00\ \text{pesos}\) and plans to keep it for \(20\ \text{years}\). If the salvage value is zero at the end of \(20\ \text{years}\), what is the depreciation in the third year using SYD method?

Solution.

\[ \sum \text{years} = \dfrac{20(20 + 1)}{2} = 210 \]
\[ d_3 = (FC - SV)\dfrac{n - 3 + 1}{\sum \text{years}} = (10000 - 0)\left(\dfrac{18}{210}\right) = 857.14\ \text{pesos} \]

Answer: \(857.00\ \text{pesos}\)

ME Board April 1998

Example. An asset is purchased for \(9,000.00\ \text{pesos}\). Its estimated life is \(10\ \text{years}\) after which it will be sold for \(1,000.00\ \text{pesos}\). Find the book value during the first year if sum-of-years’ digit (SYD) depreciation is used.

Solution.

\[ \sum \text{years} = \dfrac{10(10 + 1)}{2} = 55 \]
\[ d_1 = (FC - SV)\dfrac{10}{55} = (9000 - 1000)\left(\dfrac{10}{55}\right) = 1,454.55\ \text{pesos} \]
\[ BV_1 = FC - d_1 = 9000 - 1454.55 = 7,545.45\ \text{pesos} \]

Answer: \(7,545.00\ \text{pesos}\)

3.5 Valuation of Bonds and Property

Bond valuation computes the present worth of future dividend payments plus the discounted redemption price at maturity.

  • Par value: face value \(F\) printed on the bond
  • Redemption value: amount \(C\) paid at maturity date
  • Bond dividend: periodic payment \(D = Fr\)

Bond Valuation

\[ D = F \times r \]
\[ V_n = \dfrac{C}{(1 + i)^n} + \dfrac{D[(1 + i)^n - 1]}{(1 + i)^n i} \]

Where: \(C\) = redemption value, \(r\) = bond rate per period, \(i\) = yield rate per period, \(n\) = remaining periods

ECE Board November 1996

Example. A \(100,000\ \text{pesos}\), \(6\%\) bond pays dividends semi-annually and will be redeemed at \(110\%\) on July 1, 1999. Find its price if bought on July 1, 1996, to yield an investor \(4\%\) compounded semi-annually.

Solution.

\[ C = 1.10(100000) = 110,000\ \text{pesos}, \qquad\qquad r = \dfrac{0.06}{2} = 0.03, \qquad\qquad i = \dfrac{0.04}{2} = 0.02, \qquad\qquad n = 3(2) = 6 \]
\[ D = Fr = 100000(0.03) = 3,000\ \text{pesos} \]
\[ P = \dfrac{110000}{(1 + 0.02)^6} + \dfrac{3000[(1 + 0.02)^6 - 1]}{(1 + 0.02)^6(0.02)} = 114,481.14\ \text{pesos} \]

Answer: \(114,481.14\ \text{pesos}\)

ME Board April 1998

Example. A company issued \(50\) bonds of \(1,000.00\ \text{pesos}\) face value each, redeemable at par at the end of \(15\ \text{years}\). The firm established a sinking fund consisting of annual deposits at \(4\%\) interest rate. What was the principal in the fund at the end of the \(12\text{th}\) year?

Solution.

\[ F_{\text{total}} = 50(1000) = 50,000\ \text{pesos} \implies 50000 = \dfrac{A[(1 + 0.04)^{15} - 1]}{0.04} \implies A = 2,497.05\ \text{pesos} \]
\[ F_{12} = \dfrac{2497.05[(1 + 0.04)^{12} - 1]}{0.04} = 37,519.00\ \text{pesos} \]

Answer: \(37,519.00\ \text{pesos}\)

4. Economy Studies and Capital Investment

Economy Studies evaluate mutually exclusive investment opportunities based on profitability, cost efficiency, and time value of money.

4.1 Present Worth and Future Worth Method

Present worth method evaluates alternatives by discounting all future cash receipts and disbursements to the present time.

Net Present Worth

\[ PW = \sum \dfrac{R_t}{(1 + i)^t} - FC \]

Example. Consider a project involving an investment of \(100,000\ \text{pesos}\) now and \(100,000\ \text{pesos}\) at the end of one year. Revenues of \(150,000\ \text{pesos}\) are generated at the end of years 1 and 2. What is the net present value if effective annual interest is \(10\%\)?

Solution.

\[ PW_{\text{revenues}} = \dfrac{150000}{1 + 0.10} + \dfrac{150000}{(1 + 0.10)^2} = 260,330.58\ \text{pesos} \]
\[ PW_{\text{investments}} = 100000 + \dfrac{100000}{1 + 0.10} = 190,909.09\ \text{pesos} \]
\[ NPV = 260330.58 - 190909.09 = 69,421.50\ \text{pesos} \]

Answer: \(69,421.50\ \text{pesos}\)

REE Pre-Board August 2015

Example. An investment property is bought for \(100,000\ \text{pesos}\) and yields \(10,000\ \text{pesos}\) profit per year for \(5\ \text{years}\). If the property is sold after \(5\ \text{years}\), what is the minimum price to break even, with interest at \(6\%\)?

Solution.

\[ F_1 = 100000(1 + 0.06)^5 = 133,822.56\ \text{pesos} \]
\[ F_2 = \dfrac{10000[(1 + 0.06)^5 - 1]}{0.06} = 56,370.93\ \text{pesos} \]
\[ F = F_1 - F_2 = 133822.56 - 56370.93 = 77,451.63\ \text{pesos} \]

Answer: \(77,430.00\ \text{pesos}\)

4.2 Annual Cost Method

Annual cost method converts all initial capital investments and recurring expenses into an equivalent uniform annual cost.

Annual Cost Method

\[ AC = FC(i) + \dfrac{(FC - SV)i}{(1 + i)^n - 1} + OM \]

CE Board May 1997

Example. A corporation uses a motor truck costing \(5,000\ \text{pesos}\) with life of \(2\ \text{years}\) and salvage value of \(800\ \text{pesos}\). How much could the corporation afford to pay for another truck with life of \(3\ \text{years}\) and salvage value of \(1,000\ \text{pesos}\) if money is worth \(4\%\)?

Solution.

\[ AC_1 = 5000(0.04) + \dfrac{(5000 - 800)(0.04)}{(1 + 0.04)^2 - 1} = 2,258.82\ \text{pesos} \]
\[ AC_2 = FC_2(0.04) + \dfrac{(FC_2 - 1000)(0.04)}{(1 + 0.04)^3 - 1} = 2,258.82 \implies FC_2 = 7,164.36\ \text{pesos} \]

Answer: \(7,164.37\ \text{pesos}\)

ME Board October 1995

Example. Equipment for a loading dock costs \(20,000\ \text{pesos}\) with salvage value of \(8,000\ \text{pesos}\) after \(3\ \text{years}\). The rate of return is \(10\%\). Determine the capital recovery rate per year.

Solution.

\[ CR = FC(i) + \dfrac{(FC - SV)i}{(1 + i)^n - 1} = 20000(0.10) + \dfrac{(20000 - 8000)(0.10)}{(1 + 0.10)^3 - 1} = 5,625.00\ \text{pesos} \]

Answer: \(5,625.00\ \text{pesos}\)

4.3 Rate of Return Method

Rate of Return (ROR) is the interest rate that equates the present worth of all cash inflows to the total initial investment.

Rate of Return (ROR)

\[ ROR = \dfrac{\text{Net Annual Profit}}{\text{Capital Investment}} \times 100\% \]
\[ PW_{\text{inflows}} = PW_{\text{outflows}} \]

EE Board April 1997

Example. A small machine has an initial cost of \(20,000\ \text{pesos}\), a salvage value of \(2,000\ \text{pesos}\) and a life of \(10\ \text{years}\). Operating cost per year is \(3,500\ \text{pesos}\) and revenues per year are \(9,000\ \text{pesos}\). What is the approximate rate of return (ROR) on the investment?

Solution.

\[ \text{Net Revenue} = 9000 - 3500 = 5,500\ \text{pesos/year} \]
\[ 20000 = \dfrac{5500[(1 + i)^{10} - 1]}{(1 + i)^{10} i} + \dfrac{2000}{(1 + i)^{10}} \implies i = 24.8\% \]

Answer: \(24.8\%\)

EE Board October 1997

Example. An investment of \(350,000\ \text{pesos}\) is made to be followed by payments of \(200,000\ \text{pesos}\) each year for \(3\ \text{years}\). What is the annual rate of return on investment for the project?

Solution.

\[ 350000 = \dfrac{200000[(1 + i)^3 - 1]}{(1 + i)^3 i} \implies i = 32.7\% \]

Answer: \(32.7\%\)

4.4 Payback Period and Benefit-Cost Ratio

Payback period measures time required to recover initial investment, while Benefit-Cost Ratio (B/C) evaluates project feasibility (\(B/C > 1\)).

Payback Period and Benefit-Cost Ratio

\[ \text{Payback Period} = \dfrac{FC}{\text{Annual Net Cash Inflow}} \]
\[ B/C = \dfrac{PW(\text{Benefits})}{PW(\text{Costs})} \]

Example. A machine costs \(25,000\ \text{pesos}\) and generates annual savings of \(15,000\ \text{pesos}\). Given an effective interest of \(18\%\), what is the discounted payback period in years?

Solution.

\[ 25000 = \dfrac{15000[(1 + 0.18)^n - 1]}{(1 + 0.18)^n(0.18)} \implies 1.18^n = 1.4286 \implies n = \dfrac{\ln 1.4286}{\ln 1.18} = 2.155\ \text{years} \]

Answer: \(2.155\ \text{years}\)

ME Board October 1997

Example. A fixture costing \(700.00\ \text{pesos}\) saves \(0.06\ \text{pesos}\) per item produced. Maintenance is \(40.00\ \text{pesos}\) annually. If \(3,500\ \text{units}\) are produced annually, what is the payback period at \(10\%\) interest?

Solution.

\[ \text{Net Savings} = 3500(0.06) - 40 = 170.00\ \text{pesos/year} \]
\[ 700(1 + 0.10)^n = \dfrac{170[(1 + 0.10)^n - 1]}{0.10} \implies 1.10^n = 1.70 \implies n = 5.57\ \text{years} \]

Answer: \(5.57\ \text{years}\)

5. Break-Even and Cost Analysis

Break-Even Analysis identifies the production level where total sales revenue exactly equals total operating expenses.

5.1 Break-Even Analysis

Break-even point (BEP) is the output volume where total revenue equals total cost, resulting in zero profit or loss.

  • Break-even point: income equals total expenses
  • Profit region: volume above break-even quantity
  • Loss region: volume below break-even quantity

Break-Even Quantity

\[ \text{Total Revenue} = \text{Total Cost} \]
\[ pN = FC + vN \implies N = \dfrac{FC}{p - v} \]

Where: \(p\) = unit selling price, \(v\) = unit variable cost, \(FC\) = fixed cost, \(N\) = break-even quantity

EE Board October 1998

Example. The annual maintenance cost of a machine shop is \(69,994\ \text{pesos}\). If the cost of making a forging is \(56\ \text{pesos}\) per unit and its selling price is \(135\ \text{pesos}\) per unit, find the number of units to be forged to break even.

Solution.

\[ pN = FC + vN \implies 135N = 69994 + 56N \]
\[ 79N = 69994 \implies N = 886\ \text{units} \]

Answer: \(886\ \text{units}\)

CE Board May 1998

Example. A manufacturer produces items at a labor cost of \(115\ \text{pesos}\) each, material cost of \(76\ \text{pesos}\) each, and variable cost of \(2.32\ \text{pesos}\) each. If the item has a unit price of \(600\ \text{pesos}\), how many units must be manufactured per month to break even if monthly overhead is \(428,000\ \text{pesos}\)?

Solution.

\[ v = 115 + 76 + 2.32 = 193.32\ \text{pesos/unit} \]
\[ 600N = 428000 + 193.32N \implies 406.68N = 428000 \implies N = 1,052.42 \approx 1,053\ \text{units} \]

Answer: \(1,053\ \text{units}\)

5.2 Fixed and Variable Costs

Fixed costs remain constant regardless of output volume, whereas variable costs change in direct proportion to production quantity.

Total Cost and Profit

\[ TC = FC + VC = FC + vN \]
\[ \text{Profit} = pN - (FC + vN) \]

ME Board October 1990

Example. Compute the number of blocks of ice an ice plant must sell per month to break even with data: electricity cost = \(20.00\ \text{pesos/block}\), tax = \(2.00\ \text{pesos/block}\), real estate tax = \(3,500.00\ \text{pesos/month}\), salaries = \(25,000.00\ \text{pesos/month}\), others = \(12,000.00\ \text{pesos/month}\), selling price = \(55.00\ \text{pesos/block}\).

Solution.

\[ FC = 3500 + 25000 + 12000 = 40,500.00\ \text{pesos/month} \]
\[ v = 20 + 2 = 22.00\ \text{pesos/block} \]
\[ 55N = 40500 + 22N \implies 33N = 40500 \implies N = 1,227.27 \approx 1,228\ \text{blocks} \]

Answer: \(1,228\ \text{blocks}\)

ME Board April 1998

Example. A company manufacturing electric motors has a capacity of \(200\ \text{motors/month}\). Variable costs are \(150.00\ \text{pesos/motor}\). Selling price is \(275.00\ \text{pesos}\). Fixed costs amount to \(20,000.00\ \text{pesos/month}\). Find the number of motors to be sold monthly to break even.

Solution.

\[ pN = FC + vN \implies 275N = 20000 + 150N \]
\[ 125N = 20000 \implies N = 160\ \text{motors} \]

Answer: \(160\ \text{motors}\)

Formula Summary

Formula NameFormula
Straight-Line Annual Depreciation\(d = \dfrac{FC - SV}{n}\)
Straight-Line Book Value\(BV_m = FC - d(m)\)
Sinking Fund Annual Depreciation\(d = \dfrac{(FC - SV)i}{(1 + i)^n - 1}\)
Sinking Fund Accumulated Depreciation\(D_m = \dfrac{d[(1 + i)^m - 1]}{i}\)
Sinking Fund Book Value\(BV_m = FC - D_m\)
Declining Balance Depreciation Rate\(k = 1 - \sqrt[n]{\dfrac{SV}{FC}}\)
Declining Balance Book Value\(BV_m = FC(1 - k)^m\)
Declining Balance Annual Depreciation\(d_m = FC(1 - k)^{m-1}k\)
Sum of the Years’ Digits\(\sum \text{years} = \dfrac{n(n + 1)}{2}\)
SYD Annual Depreciation\(d_m = (FC - SV)\dfrac{n - m + 1}{\sum \text{years}}\)
Bond Periodic Dividend\(D = F \times r\)
Bond Valuation Price\(V_n = \dfrac{C}{(1 + i)^n} + \dfrac{D[(1 + i)^n - 1]}{(1 + i)^n i}\)
Net Present Worth\(PW = \sum \dfrac{R_t}{(1 + i)^t} - FC\)
Annual Cost\(AC = FC(i) + \dfrac{(FC - SV)i}{(1 + i)^n - 1} + OM\)
Rate of Return\(ROR = \dfrac{\text{Net Annual Profit}}{\text{Capital Investment}} \times 100\%\)
Payback Period\(\text{Payback Period} = \dfrac{FC}{\text{Annual Net Cash Inflow}}\)
Benefit-Cost Ratio\(B/C = \dfrac{PW(\text{Benefits})}{PW(\text{Costs})}\)
Break-Even Quantity\(N = \dfrac{FC}{p - v}\)
Total Cost\(TC = FC + vN\)
Total Operating Profit\(\text{Profit} = pN - (FC + vN)\)