Finance-Theory · Unit 20 · Video 2 · Interactive Practice

Cash, After Tax, Incremental: Building a Project's Cash Flows

IKey Formulas

FormulaNameWhat it says
CFt=CFtwith−CFtwithoutCF_t = CF_t^{\text{with}} - CF_t^{\text{without}}Incremental cash flowOnly cash that differs between the firm with the project and without it
NPV=CF0+CF11+r1+⋯+CFT(1+rT)TNPV = CF_0 + \frac{CF_1}{1+r_1} + \cdots + \frac{CF_T}{(1+r_T)^T}Net present valueThe incremental after-tax cash flows are the numerators
tax savedt=τ×Dt\text{tax saved}_t = \tau \times D_tDepreciation tax shieldDepreciation DtD_t reaches cash only by lowering taxable income

Key Insight: Sunk costs stay out; opportunity costs, working capital and effects on the firm's other businesses go in. The schedule of depreciation changes when tax is saved, not how much.

IIVisualization 1 — With Minus Without: The Café Expansion

A bakery opens a café in its own idle warehouse: which candidate items belong in the café's cash flows?

💡 Every amount is an after-tax cash flow, and the café runs for five years at a discount rate of 8%8\% a year. The working capital is assumed to be recovered in full at t=5t = 5, when the café's inventory and receivables run down. The market study is spent whether or not the café opens; the warehouse rent is received only if it does not.

IIIVisualization 2 — Same Total Tax Saved, Different Timing

Three depreciation schedules for the café's $80,000 ovens at τ=20%\tau = 20\% save the same tax, at different dates.

💡 The totals agree only when the full cost is eventually deducted (no salvage or early sale), the tax rate stays the same, and the firm has enough taxable income to use each year's deduction. Which schedules a firm may use is set by the tax code; the three rows are side by side only to compare timing.

IVQuiz Questions

Problem 1 · One Year of the Café

Given: in year 2 of the bakery's café expansion, the café itself brings in $85,000 of after-tax cash. Without the café, the bakery would rent out the idle warehouse the café occupies for $25,000 a year after tax, and its existing shop loses $30,000 of after-tax cash a year to the café. The $40,000 market study was paid two years before the café opened. Find the project's year-2 cash flow.

✅ Correct! With the café the bakery has the café's $85,000 but loses the rent and the shop's $30,000; without it, it keeps both. The difference is $85,000 − $25,000 − $30,000 = $30,000.
❌ The warehouse is not free. Without the café the bakery would collect $25,000 of rent; the café gives that up, so the forgone rent is a cost of the project.
❌ The market study is sunk. It was paid before the decision and is gone whether or not the café opens, so it appears in neither year 2 nor any other year of the comparison.
❌ The existing shop is part of the firm. The $30,000 the shop loses to the café is cash the firm has without the café and not with it, so it comes off the café's cash flow.
Show solution

Compare the bakery's year-2 cash with and without the café, in thousands of dollars:

  • Café's own after-tax cash: 8585 with, 00 without, a difference of +85+85.
  • Warehouse rent: 00 with, 2525 without, a difference of −25-25 (opportunity cost).
  • Existing shop: 3030 less with the café, a difference of −30-30.
  • Market study: already paid in both cases, a difference of 00 (sunk).
CF2=85−25−30=30CF_2 = 85 - 25 - 30 = 30

The project's year-2 cash flow is $30,000.

Problem 2 · The Opening-Day Cash Flow

Given: the café's fit-out costs $110,000, paid at t=0t = 0. From opening day it needs $60,000 of extra inventory and receivables, also at t=0t = 0. The $40,000 market study was paid last month, before the decision. Find the project's cash flow at t=0t = 0.

✅ Correct! The fit-out and the working capital both leave the firm only if the café goes ahead; the study is already paid. $110,000 + $60,000 = $170,000 out.
❌ That adds the sunk market study. The $40,000 is spent whether or not the café opens, so it cancels out of with minus without: bygones are bygones.
❌ Working capital is a capital expenditure. Inventory and receivables are not expenses on the income statement, but they tie up $60,000 of cash from opening day, so they count at t=0t = 0.
❌ Both principles are reversed here. That figure counts the sunk $40,000 study and leaves out the $60,000 of working capital; it should be the other way round.
Show solution

At t=0t = 0, in thousands of dollars, the items that differ between the bakery with the café and without it are:

  • Fit-out: −110-110 (spent only with the café).
  • Working capital (inventory and receivables): −60-60, treated as a capital expenditure because the café ties up that cash.
  • Market study: 00, sunk.
CF0=−110−60=−170CF_0 = -110 - 60 = -170

The cash flow at t=0t = 0 is −$170,000. If, as is usual, the inventory and receivables are run down when the project ends, the $60,000 comes back then as a cash inflow.

Problem 3 · A Three-Year Pop-Up Café

Given: a three-year pop-up café needs a $95,000 fit-out and $60,000 of working capital at t=0t = 0; the working capital is recovered in full at the end of year 3. The café brings in $75,000 of after-tax cash in each of years 1 to 3, and the existing shop loses $30,000 of after-tax cash a year to it. The discount rate is 7%7\% a year. Find the project's NPV and the decision.

✅ Correct! The cash flows are −$155,000, $45,000, $45,000 and $105,000; their present values sum to about $12,072 above zero, so the café adds value.
❌ The working capital comes back. The $60,000 of inventory and receivables is released at the end of year 3, an inflow worth about $48,978 today. Leaving it out flips the decision to a wrong rejection.
❌ That leaves working capital out entirely. Paying $60,000 now and getting it back in three years is not free: it costs about $60,000 − $48,978 = $11,022 in present value.
❌ The lost shop cash is missing. The $30,000 a year the existing shop loses is cash the bakery has without the café and not with it; subtract it from each year's café cash.
Show solution

Step 1: incremental cash flows (thousands of dollars):

CF0=−95−60=−155,CF1=CF2=75−30=45,CF3=45+60=105CF_0 = -95 - 60 = -155, \qquad CF_1 = CF_2 = 75 - 30 = 45, \qquad CF_3 = 45 + 60 = 105

Step 2: discount at 7%7\%:

451.07≈42.0561,451.072≈39.3047,1051.073≈85.7113\frac{45}{1.07} \approx 42.0561, \qquad \frac{45}{1.07^2} \approx 39.3047, \qquad \frac{105}{1.07^3} \approx 85.7113

Step 3: add up:

NPV≈−155+42.0561+39.3047+85.7113=12.0721NPV \approx -155 + 42.0561 + 39.3047 + 85.7113 = 12.0721

The NPV is about +$12,072, so the bakery should accept. Forgetting the working-capital recovery would give about −$36,906 and a wrong rejection.

Problem 4 · Taking the Tax Shield Sooner

Given: the bakery buys a $50,000 delivery van. It can depreciate the van straight-line over 2 years ($25,000 a year) or take a 100% first-year deduction. The tax rate stays at 20%20\%, the bakery has ample taxable income, the full cost is deducted either way, and the discount rate is 6%6\% a year. Find the total tax saved over the van's life, and how much the 100% first-year deduction adds to the present value of the tax saved.

Total tax saved over the van's life

Gain in present value from the 100% first-year deduction

✅ Correct! Both schedules save 20%20\% of $50,000 = $10,000 in total, but the first-year deduction saves it all in year 1, worth about $9,434 today against about $9,167 for straight-line.
❌ That is the depreciation, not the tax saved. A deduction of $50,000 lowers taxable income by $50,000, which lowers the tax bill by τ×50,000\tau \times 50{,}000.
❌ Straight-line saves tax in both years. $5,000 is only one year's saving (20%20\% of $25,000); over two years it also totals $10,000.
❌ That counts the saving twice. Each schedule deducts the $50,000 only once in total, so each saves 0.20×50,000=10,0000.20 \times 50{,}000 = 10{,}000; straight-line's $5,000 a year runs for two years, not $10,000 a year.
❌ Check the total. The tax saved is τ\tau times the depreciation, summed over the van's life, and the full $50,000 is deducted under either schedule.
❌ Same total, different timing. The totals agree, but at any positive discount rate a dollar of tax saved in year 1 is worth more than one saved in year 2.
❌ That discounts the deductions, not the tax saved. Depreciation reaches cash only through τ×Dt\tau \times D_t; multiply that $1,335 gap by τ=20%\tau = 20\%.
❌ That compares against the undiscounted total. $10,000 − $9,434 is what discounting costs the first-year deduction; compare it with the present value of the straight-line savings instead.
❌ Check the present values. Discount each year's tax saved, τ×Dt\tau \times D_t, at 6%6\% and compare the two schedules.
Show solution

Tax saved each year is τ×Dt\tau \times D_t:

  • Straight-line: 0.20×25,000=5,0000.20 \times 25{,}000 = 5{,}000 in year 1 and again in year 2, a total of 10,00010{,}000.
  • 100% first-year: 0.20×50,000=10,0000.20 \times 50{,}000 = 10{,}000 in year 1, a total of 10,00010{,}000.

Present values at 6%6\%:

straight-line: 5,0001.06+5,0001.062≈4,716.98+4,449.98=9,166.96\text{straight-line: } \frac{5{,}000}{1.06} + \frac{5{,}000}{1.06^2} \approx 4{,}716.98 + 4{,}449.98 = 9{,}166.96 first-year: 10,0001.06≈9,433.96\text{first-year: } \frac{10{,}000}{1.06} \approx 9{,}433.96 9,433.96−9,166.96=267.009{,}433.96 - 9{,}166.96 = 267.00

Both schedules save $10,000 in total; taking the deduction in year 1 adds about $267 of present value, which is why firms accelerate when the tax code allows it.

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