Distinguish growth from interest
A geometric cash-flow sequence grows by a constant percentage g. The discount rate i values those changing payments; it is a different rate.
Period 0: +$0.00. This is the time-zero cash flow. Periods use the payment interval stated in the model.
A geometric gradient multiplies each amount by 1+g rather than adding a fixed G.
Year 1 is A₁; year N is A₁(1+g)^(N−1).
Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.
What the model means
The first amount A₁ occurs at year 1; the Nth is A₁(1+g)ᴺ⁻¹. Summing discounted payments produces a geometric series. If g=i, every discounted contribution equals A₁/(1+i), so use the limit instead of dividing by zero.
Build the calculation
This topic is listed as an extension if time permits. It is useful for expenses that escalate and benefits that grow, but it is not a substitute for an arithmetic gradient.
Set up. Calculate. Explain.
Rebuild the starting scenario
First payment: 1000 $; Annual growth: 5 %; Interest rate: 10 % per year; Number of years: 5 years. Calculate the present worth of the growing series.
Setup hint
Year 1 is A₁; year N is A₁(1+g)^(N−1).
Open the visual solution
- Year 1 is A₁; year N is A₁(1+g)^(N−1).
- Discount each payment at i, which is distinct from growth g.
- For g ≠ i, P = A₁[1−((1+g)/(1+i))ᴺ]/(i−g).
Period 0: +$0.00. This is the time-zero cash flow. Periods use the payment interval stated in the model.
A geometric gradient multiplies each amount by 1+g rather than adding a fixed G.
Year 1 is A₁; year N is A₁(1+g)^(N−1).
Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.
Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.
Transfer the idea to a new scenario
Benefits begin at $1,800 in year 1 and grow 4% per year for ten years. Discount at 8% and find PW.
Setup hint
Year 1 is A₁; year N is A₁(1+g)^(N−1).
Open the visual solution
- Year 1 is A₁; year N is A₁(1+g)^(N−1).
- Discount each payment at i, which is distinct from growth g.
- For g ≠ i, P = A₁[1−((1+g)/(1+i))ᴺ]/(i−g).
Period 0: +$0.00. This is the time-zero cash flow. Periods use the payment interval stated in the model.
A geometric gradient multiplies each amount by 1+g rather than adding a fixed G.
Year 1 is A₁; year N is A₁(1+g)^(N−1).
Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.
Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.
If g=i in a finite geometric series, what is true?
Can you defend your setup?
Can you place each amount on the timeline, match the interest period, calculate the equivalent quantity, and state the decision or limitation?
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