Three Data Bundles: The Average-Cost Trap
A prepaid data provider offers three bundles. Option X: no monthly fee, R0,29 per megabyte used. Option Y: R30 per month for the first 200 MB, then R0,35 per megabyte for every megabyte beyond 200. Option Z: R120 per month for unlimited data. A subscriber's monthly usage swings between roughly 50 MB and 1 200 MB depending on how much she works from home; last month she used about 900 MB. She is convinced Option Y is the cheapest because, as she puts it, "the first 200 MB work out to only 15 cents a megabyte". She wants one bundle to commit to for the whole year.
1. Write the monthly cost of Options X and Y as functions of usage m (in MB) and find every usage level at which they cost the same.
Read the worked answer
Cost functions
- Option X: C_X(m) = 0,29m, for all m ≥ 0
- Option Y: C_Y(m) = 30 for 0 ≤ m ≤ 200; C_Y(m) = 30 + 0,35(m − 200) for m > 200 (which simplifies to C_Y(m) = 0,35m − 40 for m > 200)
Crossing point on the flat segment (m ≤ 200) 0,29m = 30 → m = 103,45 MB
Crossing point on the metered segment (m > 200) 0,29m = 0,35m − 40 → 40 = 0,06m → m = 666,67 MB
Reading the two solutions
- Below ≈103 MB, X is cheaper (e.g. at 50 MB: X = R14,50, Y = R30).
- Between ≈103 MB and ≈667 MB, Y is cheaper (e.g. at 400 MB: X = R116,00, Y = R100,00).
- Above ≈667 MB, X is cheaper again (e.g. at 1 000 MB: X = R290,00, Y = R310,00).
The relationship is not "Y is better or worse" - it flips twice, because Y's marginal rate (R0,35) is higher than X's flat rate even though Y's average rate starts lower.
2. At her current usage of 900 MB, determine which of the three options is cheapest and by how much, comparing against both alternatives.
Read the worked answer
Option X 0,29 × 900 = R261,00
Option Y 30 + 0,35 × (900 − 200) = 30 + 245 = R275,00
Option Z R120,00 flat
Comparison
- Z vs X: 261,00 − 120,00 = R141,00 saved
- Z vs Y: 275,00 − 120,00 = R155,00 saved
Answer: Option Z is cheapest at 900 MB, saving R141 a month against her current choice of X and R155 against her preferred Y. Over 12 months that is R1 692 versus Option X.
When Z wins
- Z ties X when 0,29m = 120 → m = 413,8 MB
- Z ties Y when 0,35m − 40 = 120 → m = 457,1 MB
So Z is cheapest for any month above about 457 MB - which covers all of her working-from-home-free months - and should be read as the default, with a cheaper bundle only in low-usage months.
3. Evaluate the subscriber's reasoning that Option Y is cheapest because the first 200 MB cost only 15 cents each. Then advise her on committing to one bundle for a year, given that her usage ranges from 50 MB to 1 200 MB.
Read the worked answer
The reasoning is an average-cost fallacy. R30 ÷ 200 MB = R0,15/MB is the average rate, and it is only achievable if she uses exactly 200 MB every month and stops there. Two things break it:
- The marginal rate above 200 MB (R0,35) is worse than Option X's flat R0,29. Every megabyte past 200 MB costs her more than the alternative, which is why Y reverses against X at 667 MB.
- The average rate drifts upward with usage. At 900 MB the effective price is 275 ÷ 900 = R0,31/MB - worse than X's R0,29 and far worse than Z's R0,13.
Mixing an average price (R0,15) with a marginal decision (how much to use) is the error.
Advice on a one-year commitment Her usage spans 50–1 200 MB, and no single metered bundle is cheapest across that range:
| Usage | Cheapest |
|---|---|
| below ≈103 MB | X |
| ≈103 – 457 MB | Y |
| above ≈457 MB | Z |
Since her usage is variable, a fixed choice will be wrong in some months. Two defensible strategies:
- Best single hedge: take Z (R120 unlimited) if most months are office months - it is never catastrophic and wins outright above 457 MB; or
- Better if the provider allows switching: Z in heavy months, X in light months, since Y's narrow winning band (≈103–457 MB) is easy to miss and its penalty above 667 MB is real.
What she should not do is commit to Y on the strength of a 15-cent figure that only holds at exactly 200 MB, a usage she almost never has.