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23-Chem-A5 Chemical Plant Design and Economics · May 2016

Question 2 of 6: Net Present Worth of Competing Projects

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

Notes on this paper

National Exams — May 2016 — 04-Chem-A5 Chemical Plant Design and Economics. Three-hour, closed-book exam; one two-sided aid sheet and an approved calculator permitted. Six equally weighted (20-mark) questions are posed and the candidate answers any five; only the first five are marked. All six are answered below for completeness. Questions 1–3 are numerical (distillation heat integration, after-tax net present worth, and a coagulant-dosage cost optimisation); questions 4–6 are qualitative essays on materials of construction, plant startup/shutdown safety, and equipment-selection factors.

Reference texts: M.S. Peters, K.D. Timmerhaus & R.E. West, Plant Design and Economics for Chemical Engineers (5th ed., McGraw-Hill) — the exam's named primary text (interest and profitability Ch. 7–10, materials of construction Ch. 12, plant safety and loss prevention Ch. 3, equipment selection throughout Ch. 14–22); R. Smith, Chemical Process Design and Integration (2nd ed., Wiley) and B. Linnhoff et al., A User Guide on Process Integration (IChemE) — pinch analysis and column heat integration behind Question 1; R.K. Sinnott & G. Towler, Chemical Engineering Design (Coulson & Richardson vol. 6) — materials selection and equipment sizing; supporting Canadian practice from CCOHS, provincial OH&S process-safety-management regulations and the CSA Z767 (PSM) framework.

Question 2: Net Present Worth of Competing Projects (20 marks)

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. Two mutually exclusive projects, each with a 20-year life, straight-line depreciation to zero salvage on 100 % of the investment, a 52 % tax rate and a 12.5 % before-tax minimum acceptable rate of return.

SymbolInorganic plantTextile expansion
Revenue $R$$33.70 M$30.90 M
Capital $TCI$$52.50 M$57.30 M
Annual cost $C$$25.10 M$21.50 M
Life $n$20 yr20 yr

Find. The net present worth (NPW) of each project on a consistent after-tax basis, and hence which project the firm should prefer.

Approach

Because the return is quoted before tax while the cash flows are taxed, we discount after-tax cash flow at the equivalent after-tax rate $i = i_{bt}(1-t)$. Depreciation is not a cash outflow but it shields tax, so the annual after-tax cash flow is the taxed operating profit plus the depreciation add-back; NPW is that uniform series brought to present worth minus the capital.

02468101214161820-$52.5M (TCI)+$5.49M/yr+$5.49M/yrProject 1 after-tax cash flow (M$)period (year)
Figure 2.1 — After-tax cash-flow diagram for the inorganic-chemicals project: a $52.5 M capital outlay at year 0 followed by a uniform $5.49 M/yr after-tax cash flow for 20 years, discounted at the after-tax rate of 6 % to give NPW = +$10.5 M.
  1. Convert the hurdle rate to an after-tax basis. A before-tax return $i_{bt}$ is equivalent, on taxed money, to $$i = i_{bt}(1 - t) = 0.125\,(1 - 0.52) = 0.060 \;\;(6.0\%)$$ This is the rate at which after-tax cash flows must be discounted for consistency with the 12.5 % before-tax criterion.
  2. Straight-line depreciation. With zero salvage over 20 years, $D = TCI/n$: for the inorganic plant $D_1 = 52.5\text{M}/20 = \$2.625$ M/yr; for the textile expansion $D_2 = 57.3\text{M}/20 = \$2.865$ M/yr.
  3. After-tax cash flow (ATCF). Taxable income is revenue minus cash cost minus depreciation; tax is levied on that, and depreciation is added back because it is a non-cash charge: $$\text{ATCF} = (R - C - D)(1 - t) + D$$ Inorganic plant: taxable $= 33.70 - 25.10 - 2.625 = \$5.975$ M, so $\text{ATCF}_1 = 5.975(0.48) + 2.625 = \$5.493$ M/yr.
    Textile expansion: taxable $= 30.90 - 21.50 - 2.865 = \$6.535$ M, so $\text{ATCF}_2 = 6.535(0.48) + 2.865 = \$6.002$ M/yr.
  4. Uniform-series present-worth factor at 6 %, 20 yr. $$\left(\frac{P}{A}\right)_{6\%,\,20} = \frac{1 - (1.06)^{-20}}{0.06} = 11.470$$
  5. Net present worth. $\text{NPW} = -TCI + \text{ATCF}\,(P/A)$: $$\text{NPW}_1 = -52.50 + 5.493(11.470) = +\$10.5\text{ M}$$ $$\text{NPW}_2 = -57.30 + 6.002(11.470) = +\$11.5\text{ M}$$ ==**Both projects are profitable; the textile-fibers expansion has the higher NPW (+$11.5 M vs. +$10.5 M) and is preferred.**==
QuantityInorganic plantTextile expansion
Depreciation $D$$2.625 M/yr$2.865 M/yr
Taxable income$5.975 M/yr$6.535 M/yr
After-tax cash flow$5.493 M/yr$6.002 M/yr
Net present worth @6 %+$10.5 M+$11.5 M (preferred)
Check: the after-tax hurdle rate $i = i_{bt}(1-t) = 6\%$ is the consistent way to reconcile a before-tax MARR with taxed cash flows. Discounting the same after-tax cash flows at the nominal 12.5 % instead would drive both NPWs strongly negative (−$12.7 M and −$13.8 M), which is the classic error of mixing a before-tax rate with after-tax money; the two projects remain rank-ordered the same way in both treatments.