NivaarExam PrepOfficial exam papers ↗

18-Env-B5 Industrial & Hazardous Waste Management · Undated paper

Question 9 of 10: Present-Worth Comparison of Two Waste-Minimization Processes

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2019 — 18-Env-B5: Industrial & Hazardous Waste Management (3 hours, open book). Marks are indicated beside each question for a total of 100 marks; all ten questions are answered in full below as a complete study resource.

Reference texts: LaGrega, Buckingham & Evans, Hazardous Waste Management (2nd ed.); Nemerow & Dasgupta, Industrial and Hazardous Waste Treatment (2nd ed.); Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.); Davis & Cornwell, Introduction to Environmental Engineering (6th ed.).

Question 9: Present-Worth Comparison of Two Waste-Minimization Processes (15 point)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Given data
QuantityProcess AProcess B
Initial machinery costUSD 1,200,000USD 1,500,000
Maintenance, years 1–10USD 150,000/yrUSD 100,000/yr (all 20 yr)
Maintenance, years 11–20USD 180,000/yr
Salvage value00
Interest rate, $i$15 % per year
Analysis horizon, $n$20 years

Find. Which process has the lower present worth of costs over the 20-year horizon.

02468101214161820$1.2M$150k/yr (yrs 1-10)$180k/yr (yrs 11-20)Process A cash flow (i = 15%)period (year)
Process A cost cash-flow: initial outlay at year 0, a lower maintenance annuity for years 1–10, then a higher one for years 11–20.

Approach. Since both alternatives are cost-only, zero-salvage, equal-length (20-year) options, compare them on a present-worth-of-cost basis: whichever has the smaller PW is more economical. Process A's maintenance is a two-tier annuity, handled as a 10-year annuity for years 1–10 plus a second 10-year annuity for years 11–20 discounted back through year 10.

  1. Annuity/single-payment factors at $i=15\%$. $$(P/A,15\%,10) = \frac{1-(1.15)^{-10}}{0.15} = 5.019 \qquad (P/F,15\%,10)=(1.15)^{-10}=0.2472 \qquad (P/A,15\%,20)=\frac{1-(1.15)^{-20}}{0.15}=6.259$$
  2. Present worth, Process A. $$PW_A = \$1{,}200{,}000 + \$150{,}000\,(P/A,15\%,10) + \$180{,}000\,(P/A,15\%,10)(P/F,15\%,10)$$ $$PW_A = 1{,}200{,}000 + 150{,}000(5.019) + 180{,}000(5.019)(0.2472) = 1{,}200{,}000 + 752{,}800 + 223{,}300 = \boxed{\$2{,}176{,}100}$$
  3. Present worth, Process B. $$PW_B = \$1{,}500{,}000 + \$100{,}000\,(P/A,15\%,20) = 1{,}500{,}000 + 100{,}000(6.259) = \boxed{\$2{,}125{,}900}$$

$PW_B < PW_A$ by about USD 50,200 (roughly 2 % of either total), so Process B is more economical at 15 % interest over the 20-year horizon — its higher initial cost is more than offset by its flat, lower maintenance charge compared with Process A's step up to USD 180,000/yr in the second decade. The margin is modest relative to the total present worth, so the ranking would be worth re-checking if the interest rate or the second-decade maintenance estimate for Process A carries meaningful uncertainty; at a lower interest rate (which weights the distant, cheaper Process A maintenance years less) the gap would narrow further, while a higher rate would favour Process B by even more.

Question 9 — final results
QuantityValue
Present worth of costs, Process AUSD 2,176,100
Present worth of costs, Process BUSD 2,125,900
More economical optionProcess B (lower PW by ≈ USD 50,200)