Choose the method.
Then do the math.

Set revenue equal to total cost: pq = F + vq.

1. Set revenue equal to total cost: pq = F + vq.
2. Contribution margin = $8.00 per unit.
3. Divide fixed cost by margin; round UP only when output must be whole units.

Use comparable technologies and price dates before scaling.

1. Use comparable technologies and price dates before scaling.
2. Capacity ratio = 2.5; exponent = 0.65.
3. Multiply the known cost by ratioᵇ. A cost index adjusts date; the exponent adjusts size.

Choose the borrower viewpoint: loan receipt positive at time 0, repayments negative.

1. Choose the borrower viewpoint: loan receipt positive at time 0, repayments negative.
2. Use one effective rate per payment period and payments at period end.
3. Each payment = interest on opening balance + principal reduction; the last balance should be zero.

Payments occur at years 5 through 10: 6 payments.

1. Payments occur at years 5 through 10: 6 payments.
2. At year 4, P = A(P/A,i,N) = $32,360.16.
3. Discount that single local P through 4 periods to year 0.

Separate the base uniform series from the arithmetic increments.

1. Separate the base uniform series from the arithmetic increments.
2. Sum A/(1+i)^t plus (t−1)G/(1+i)^t for t = 1,…,N.
3. At i = 0, P = NA + GN(N−1)/2. Equivalent annual gradient is G(A/G).

Convert nominal annual r into effective rate over one PAYMENT period.

1. Convert nominal annual r into effective rate over one PAYMENT period.
2. Number of payments N = years × payments/year = 16.
3. Use i_p and N together. Never insert an annual rate beside a monthly N.

The cash flows occur at deposit-period ends; this is not continuous cash inflow.

1. The cash flows occur at deposit-period ends; this is not continuous cash inflow.
2. A payment period is 1/2 year; effective rate is e^(r/p)−1.
3. Use a finite uniform series with 12 deposits. A single principal instead grows as Pe^(rN).

Draw the investment as negative at time 0 and net benefit as positive at years 1–N.

1. Draw the investment as negative at time 0 and net benefit as positive at years 1–N.
2. PW = −I + A(P/A,i,N) = $1,002.67.
3. FW = PW(F/P,i,N); AW = PW(A/P,i,N). Positive conversion factors preserve the sign and decision.

Capital recovery accounts for both the return OF capital and the required return ON capital.

1. Capital recovery accounts for both the return OF capital and the required return ON capital.
2. Annualize the first cost and subtract annualized terminal salvage.
3. Equivalent identity: CR=(I−S)(A/P)+Si. Add operating cost; a net-revenue AW would subtract this cost.

PW(12%) = $4,071.64 > 0; PW(18%) = $-3,092.43 < 0.

1. PW(12%) = $4,071.64 > 0; PW(18%) = $-3,092.43 < 0.
2. Interpolate i₁ = i_L + PW_L(i_H−i_L)/(PW_L−PW_H).
3. Evaluate PW(i₁); retain the endpoints of opposite sign and interpolate again: 15.248863%.

Separate positive and negative NET cash flows at each date.

1. Separate positive and negative NET cash flows at each date.
2. Discount all negatives to 0 at ε; accumulate positives to year 3 at ε.
3. Bridge these two equivalent totals over 3 periods. ERR can be negative and changes with the external rate.

Order alternatives by increasing initial investment and establish an acceptable defender.

1. Order alternatives by increasing initial investment and establish an acceptable defender.
2. Subtract A from B consistently at every date, including salvage.
3. Solve incremental PW = 0; for this conventional increment ΔIRR ≥ MARR agrees with ΔPW ≥ 0. If the increment is nonconventional, use incremental PW.

Assume identical replacement cycles and continuing service. A costs $10,000 with $1,000 salvage; B costs $18,000 with $2,000 salvage.

1. Assume identical replacement cycles and continuing service. A costs $10,000 with $1,000 salvage; B costs $18,000 with $2,000 salvage.
2. Compute each cycle EAC at the same MARR, or repeat full cycles to 15 years and compare cost PW.
3. For a fixed nonrepeatable study horizon, include terminal market values, replacements or service gaps explicitly. Never compare raw 3-year and 5-year PW as if horizons were equal.

The sign pattern −,+,− permits multiple positive roots but does not guarantee them.

1. The sign pattern −,+,− permits multiple positive roots but does not guarantee them.
2. Multiply by (1+i)² and solve the resulting quadratic; the financial domain is i>−1.
3. At cleanup cost $1,320, both 10% and 20% make PW zero. A single IRR ≥ MARR rule is unsafe.

Independent A and B may both be chosen if the budget permits.

1. Independent A and B may both be chosen if the budget permits.
2. Interdependency C ⇒ A imposes x_C ≤ x_A. Mutually exclusive alternatives would impose x_A+x_B≤1 instead.
3. Enumerate feasible combinations and maximize total PW under the budget; highest standalone return is not enough.

Discount each of the eight receipts at 10%, then accumulate from the initial investment.

1. Discount each of the eight receipts at 10%, then accumulate from the initial investment.
2. At year 4, $490.40 remains unrecovered. The next discounted receipt is $1,862.76.
3. Interpolate 0.263267 of the next year; compare simple payback I/A = 3.333333 years. Neither method counts all value after recovery.

The target F is at year 10, not time 0.

1. The target F is at year 10, not time 0.
2. Use A/F, the sinking-fund factor. F/A would convert a given annual deposit to a future value.
3. Each deposit earns interest until the target date; the last earns none before that date. The timeline is net saver cash flow, combining the last deposit with the target withdrawal.

Four withdrawals of $7,000.00 start at year 4, so their local present date is year 3.

1. Four withdrawals of $7,000.00 start at year 4, so their local present date is year 3.
2. Discount the withdrawal present worth to year 0: $18,404.91.
3. Three saving deposits must have the same PW, so A=P₀/(P/A,i,3). The diagram uses the saver viewpoint: deposits negative and withdrawals positive.

The first three benefits form a series anchored at year 0.

1. The first three benefits form a series anchored at year 0.
2. The four later benefits form a different series anchored at year 3.
3. Discount that second series by three years, then combine both with the negative initial investment.

A costs $10,000 and $2,000/year; B costs $18,000 and $1,000/year. Both provide required service through year 4.

1. A costs $10,000 and $2,000/year; B costs $18,000 and $1,000/year. Both provide required service through year 4.
2. Use their given year-4 market values, not salvage from their natural useful lives.
3. Compute positive cost PW = purchase + operating PW − terminal-value PW; compare B minus A and choose the smaller cost.

There are eight benefits, with A₁ at year 1 and A₁(1+g)⁷ at year 8.

1. There are eight benefits, with A₁ at year 1 and A₁(1+g)⁷ at year 8.
2. Use the geometric-series PW (or direct discounted summation); handle g=i by its limit.
3. Subtract the $10,000 initial investment and interpret the sign at MARR.