FE Civil Engineering Economics practice problems

You have a series of equal annual payments ($A$) and you want their value today ($P$), so the factor is the uniform series present worth $(P/A)$. Look up $(P/A,\, 8\%,\, 6) = 4.6229$ and multiply by 18,000 to get about 83,200. The 108,000 option just adds up the six payments ($18{,}000 \times 6$) with no discounting, ignoring the time value of money. The 88,500 option uses the wrong interest rate, applying $(P/A,\, 6\%,\, 6) = 4.9173$ instead of the 8% factor. The 11,300 option treats the whole stream as a single future lump sum and discounts it with $(P/F)$ instead of $(P/A)$.

1. Identify: $A = 18{,}000$, $i = 8\%$, $n = 6$ years. Find $P$.
2. Use the uniform series present worth factor:
   $(P/A,\, 8\%,\, 6) = \frac{(1.08)^6 - 1}{0.08(1.08)^6}$
3. Compute the factor:
   $(1.08)^6 = 1.58687 \;\Rightarrow\; (P/A) = \frac{0.58687}{0.12695} = 4.6229$
4. Calculate the present worth:
   $P = 18{,}000 \times 4.6229 = 83{,}212 \approx 83{,}200$

FE Reference Handbook 10.6, p. 235, Uniform Series Present Worth; p. 241, Factor Table i = 8%

The cash flow is a flat base of 5,000 dollars per year plus a uniform gradient $G = 1{,}000$ that starts in year 2. Convert the gradient to an equivalent annual amount with the gradient-to-uniform factor $(A/G,\, 10\%,\, 8) = 3.0045$, then add it to the base: $A = 5{,}000 + 1{,}000 \times 3.0045 = 8{,}004 \approx 8{,}000$. The 8,500 option uses the simple arithmetic average of the gradient ($1{,}000 \times 7/2 = 3{,}500$ added to 5,000), which ignores the time value of money. The 3,000 option reports only the gradient component and forgets to add the 5,000 base. The 12,000 option is the final-year cost ($5{,}000 + 1{,}000 \times 7$), not the annual equivalent.

1. Identify the structure: base annuity $A_{\text{base}} = 5{,}000$ plus uniform gradient $G = 1{,}000$, with $i = 10\%$ and $n = 8$.
2. Gradient-to-uniform factor:
   $(A/G,\, 10\%,\, 8) = \frac{1}{i} - \frac{n}{(1+i)^n - 1} = \frac{1}{0.10} - \frac{8}{(1.10)^8 - 1}$
3. Compute:
   $(1.10)^8 = 2.14359 \;\Rightarrow\; (A/G) = 10 - \frac{8}{1.14359} = 10 - 6.9955 = 3.0045$
4. Equivalent uniform annual cost:
   $A = 5{,}000 + 1{,}000 \times 3.0045 = 8{,}004 \approx 8{,}000$

FE Reference Handbook 10.6, p. 235, Uniform Gradient (A/G); p. 241, Factor Table i = 10%

Capitalized cost is the present worth of a project that lasts forever. For a perpetual uniform cost, the present worth of the annual maintenance is simply $A/i$, so $40{,}000 / 0.05 = 800{,}000$. Add the one-time construction cost: $2{,}000{,}000 + 800{,}000 = 2{,}800{,}000$. The 2,040,000 option adds only one year of maintenance instead of capitalizing it over an infinite horizon. The 2,002,000 option multiplies the maintenance by the rate ($40{,}000 \times 0.05$) instead of dividing. The 800,000 option gives only the capitalized maintenance and forgets the initial construction cost.

1. Capitalized cost = initial cost + present worth of perpetual annual cost.
2. Present worth of a perpetual uniform series:
   $P_{\text{perp}} = \frac{A}{i} = \frac{40{,}000}{0.05} = 800{,}000$
3. Add the initial construction cost:
   $CC = 2{,}000{,}000 + 800{,}000 = 2{,}800{,}000$

FE Reference Handbook 10.6, p. 236, Capitalized Cost

The lives are 3 and 6 years, so present worth must be compared over the least common multiple, 6 years. Compactor M must be purchased again at the start of year 4 (end of year 3). Its PW is $40{,}000 + 40{,}000(P/F,\, 10\%,\, 3) + 10{,}000(P/A,\, 10\%,\, 6) = 40{,}000 + 30{,}053 + 43{,}553 = 113{,}606$. Compactor N runs the full 6 years with no repeat: $60{,}000 + 7{,}000(P/A,\, 10\%,\, 6) = 60{,}000 + 30{,}487 = 90{,}487 \approx 90{,}500$. N is cheaper. The "Compactor M, PW = 64,900" option compares only one 3-year cycle of M against nothing, an invalid unequal-life comparison. The "Compactor M, PW = 113,600" option is M's correct PW but M is the more expensive option. The "Compactor N, PW = 102,000" option forgets to discount M's replacement or mis-adds N.

1. LCM of 3 and 6 is 6 years. Compactor M repeats once (replaced at end of year 3); Compactor N runs the full 6 years.
2. Factors at 10%:
   $(P/F,\, 10\%,\, 3) = 0.7513 \quad (P/A,\, 10\%,\, 6) = 4.3553$
3. Compactor M present worth of costs:
   $PW_M = 40{,}000 + 40{,}000(0.7513) + 10{,}000(4.3553) = 40{,}000 + 30{,}053 + 43{,}553 = 113{,}606$
4. Compactor N present worth of costs:
   $PW_N = 60{,}000 + 7{,}000(4.3553) = 60{,}000 + 30{,}487 = 90{,}487 \approx 90{,}500$
5. Compactor N has the lower present worth of costs. Select Compactor N.

FE Reference Handbook 10.6, p. 235, Single Payment Present Worth + Uniform Series Present Worth; p. 241, Factor Table i = 10%

At break-even, total revenue equals total cost, so the contribution margin per panel (selling price minus variable cost) must cover the fixed cost. The margin is $450 - 270 = 180$ dollars per panel, and $90{,}000 / 180 = 500$ panels. The 200-panel option divides the fixed cost by the selling price alone and ignores the variable cost. The 333-panel option divides by the variable cost instead of the contribution margin. The 125-panel option divides by the sum of price and variable cost ($450 + 270 = 720$) instead of their difference.

1. Contribution margin per panel:
   $p - v = 450 - 270 = 180$
2. Set revenue equal to total cost: $450Q = 90{,}000 + 270Q$, which gives $180Q = 90{,}000$.
3. Solve for the break-even quantity:
   $Q^* = \frac{90{,}000}{180} = 500 \text{ panels}$

FE Reference Handbook 10.6, p. 236, Cost Estimation and Break-Even Analysis

By choosing to stage equipment on the lot, the firm gives up the 30,000 dollars per year it could have earned by leasing it out. The value of the best forgone alternative is the definition of an opportunity cost, and it must be counted as a cost of the staging decision even though no cash actually changes hands. The sunk-cost choice confuses ownership with sunk cost: the lease income is a future, recoverable alternative, not money already spent and lost. The fixed-cost choice is wrong because the forgone income does not arise from operating the yard; it exists regardless of activity level. The marginal-cost choice misapplies marginal cost, which is the cost of one more unit of output, not a forgone alternative use of an asset.

1. Using the lot for staging means the firm cannot lease it out, so it forgoes 30,000 dollars per year of income.
2. Opportunity cost is the value of the best alternative given up when a choice is made.
3. The forgone lease income is therefore an opportunity cost and should be charged against the staging-yard option.
4. It is not a sunk cost (the income is a future alternative, not a past irrecoverable expense) and not a fixed or marginal cost of operating the yard.

FE Reference Handbook 10.6, p. 236, Cost Estimation; p. 235, Engineering Economics concepts

Disbenefits are negative effects on the public, so they are subtracted from the benefits in the numerator: $B/C = (B - D)/C = (8{,}000{,}000 - 1{,}200{,}000)/5{,}500{,}000 = 6{,}800{,}000/5{,}500{,}000 = 1.24$. The 1.45 option ignores the disbenefits entirely ($8{,}000{,}000/5{,}500{,}000$). The 1.19 option incorrectly adds the disbenefits to the cost denominator ($8{,}000{,}000/6{,}700{,}000$). The 1.86 option incorrectly subtracts the disbenefits from the cost denominator ($8{,}000{,}000/4{,}300{,}000$) instead of from the benefits.

1. Subtract disbenefits from benefits in the numerator:
   $B - D = 8{,}000{,}000 - 1{,}200{,}000 = 6{,}800{,}000$
2. Divide by the present worth of agency costs:
   $B/C = \frac{6{,}800{,}000}{5{,}500{,}000} = 1.24$
3. Since B/C = 1.24 > 1, the project is economically justified.

FE Reference Handbook 10.6, p. 237, Benefit-Cost Analysis

At a chance node you weight each outcome by its probability and sum: $EV = 0.50(500{,}000) + 0.30(200{,}000) + 0.20(-100{,}000) = 250{,}000 + 60{,}000 - 20{,}000 = 290{,}000$. The 200,000 dollars choice takes the plain arithmetic average of the three outcomes ($(500{,}000 + 200{,}000 - 100{,}000)/3$) and ignores the probabilities. The 330,000 dollars choice treats the low-demand outcome as a gain instead of a loss (adds 20,000 instead of subtracting). The 310,000 dollars choice drops the unprofitable low-demand branch entirely and sums only the first two terms.

1. Confirm probabilities sum to 1.0: $0.50 + 0.30 + 0.20 = 1.00$.
2. Apply the expected value formula at the chance node:
   $EV = 0.50(500{,}000) + 0.30(200{,}000) + 0.20(-100{,}000)$
3. Compute each term:
   $EV = 250{,}000 + 60{,}000 - 20{,}000$
4. Sum:
   $EV = 290{,}000$

FE Reference Handbook 10.6, p. 237, Economic Decision Trees; Expected Value

Straight-line depreciation is constant each year: $D = (C - S_n)/n = (150{,}000 - 30{,}000)/8 = 15{,}000$ per year. After 5 years the accumulated depreciation is $5 \times 15{,}000 = 75{,}000$, so the book value is $150{,}000 - 75{,}000 = 75{,}000$. The 56,250 option forgets to subtract salvage when computing the annual depreciation ($150{,}000/8 = 18{,}750$ per year, then $150{,}000 - 93{,}750$). The 45,000 option wrongly subtracts the accumulated depreciation from the depreciable base ($120{,}000 - 75{,}000$) instead of from the full cost. The 30,000 option assumes the asset is already at salvage value, which only happens at the end of its full 8-year life.

1. Annual straight-line depreciation:
   $D = \frac{C - S_n}{n} = \frac{150{,}000 - 30{,}000}{8} = 15{,}000$
2. Accumulated depreciation through year 5:
   $\sum_{k=1}^{5} D_k = 5 \times 15{,}000 = 75{,}000$
3. Book value at end of year 5:
   $BV_5 = 150{,}000 - 75{,}000 = 75{,}000$

FE Reference Handbook 10.6, p. 236, Depreciation; Book Value

Converting an actual (inflated) future amount into today's purchasing power means dividing by the inflation index $(1 + f)^n$, not discounting at any interest rate: $50{,}000 / (1.03)^{10} = 50{,}000 / 1.3439 = 37{,}205 \approx 37{,}200$. The 67,200 distractor multiplies by the inflation factor ($50{,}000 \times 1.3439$), moving in the wrong direction. The 35,000 distractor applies a simple, non-compounded 30% reduction ($50{,}000 \times (1 - 0.30)$). The 38,500 distractor divides by a simple factor of 1.30 instead of the compounded $(1.03)^{10}$.

1. Identify: actual-dollar future cost $= 50{,}000$, inflation $f = 3\%$, $n = 10$ years.
2. Deflate by the inflation index to reach constant (real) dollars:
   $\text{Real} = \frac{50{,}000}{(1 + f)^n} = \frac{50{,}000}{(1.03)^{10}}$
3. Compute the index:
   $(1.03)^{10} = 1.34392$
4. Result:
   $\text{Real} = \frac{50{,}000}{1.34392} = 37{,}205 \approx 37{,}200$

FE Reference Handbook 10.6, p. 236, Inflation

Accept if IRR ≥ MARR. Here 18% ≥ 12%, so the project clears the hurdle and should be accepted. The "Rejected, because IRR exceeds MARR" choice reverses the rule. The "too high to be realistic" and "deferred until MARR rises above 18%" choices invent reasoning not part of the criterion.

1. Rule: accept if IRR ≥ MARR.
2. 18% ≥ 12%.
3. So the project is accepted.

FE Reference Handbook 10.6, p. 236