Compare unknown duration cost with a first-buy threshold

An unknown duration cost question needs a decision rule that works before the final use count is revealed. In the stipulated exercise here, renting costs one fictional unit for each use, while buying costs ten units and covers that use and every later use. Renting for nine uses and buying before the tenth gives the smallest possible worst-case cost ratio among deterministic rules with the stated information. That ratio is nineteen tenths, or 1.9.

The numbers are invented mathematical inputs, without currency, a seller quote or an actual rental service behind them. No product lifetime or weather-vane performance is assumed. The exercise asks how a rule chosen without knowing the final duration compares with a benchmark that knows that duration. It establishes a result within a defined model, rather than recommending that someone rent or buy a particular real item.

Specify what is known before each decision

There will be a positive whole number of uses, called n. Before each use that actually occurs, you know its position in the sequence and that it is about to happen. You do not know whether another use will follow. A use cannot be skipped. Before buying, you may either rent for that one use at cost one or buy at cost ten. Once bought, the item covers all remaining uses without further charges.

Keep the timing precise. Buying before use ten means the first nine uses were rented. It does not mean paying for ten rentals and then buying afterward. The purchase itself covers use ten. If the sequence ends after use nine, that purchase never happens. Each decision is made only when its associated use is about to occur, so no charge is incurred for a future use that does not arrive.

There are no deposits, refunds, resale proceeds, financing charges, maintenance, storage costs or changes in the rental rate. Buying is irreversible. There is no discount, capacity limit or quality difference between the two ways of covering a use. The rule receives no signal about future duration beyond the continuing count. These restrictions are part of the mathematical question. Adding a real contractual term could change both costs and the best rule.

Calculate the benchmark with advance knowledge

Suppose a benchmark is told the complete value of n before paying anything. It can rent every use for a total n, or buy before the first use for ten. Buying after some rentals would cost more than ten, so it cannot improve on buying immediately when ownership is preferred. The smallest possible benchmark cost is therefore the smaller of n and ten, written min(n, 10).

At n equal to three, the benchmark rents and spends three. At n equal to twenty, it buys immediately and spends ten. At n equal to ten, the two options tie at ten. This benchmark has information unavailable to the rule being evaluated. Its role is to provide a consistent comparison at each possible final duration. It is not an action that the uninformed rule can imitate after learning that a sequence has already ended.

For each n, divide the rule’s total cost by this benchmark cost. The denominator is positive because n is positive. A ratio of one means equal totals for that duration; a ratio of two means the rule spent twice the benchmark. The worst-case ratio is the largest ratio over all allowed positive whole-number durations. No probability distribution or average duration is being assigned.

Describe a first-buy threshold before testing it

Let t be a positive whole number. A threshold-t rule rents before uses one through t minus one, then buys before use t if that use occurs. It takes no further purchasing action afterward. If n is less than t, every actual use was rented and total cost is n. If n is at least t, the rule paid t minus one in rentals and ten for ownership, giving total t plus nine.

This formula separates the two situations without guessing where the sequence ends. For t equal to ten, a six-use sequence costs six because the threshold is never reached. A ten-use sequence costs nineteen: nine rentals followed by one purchase. A hundred-use sequence also costs nineteen, since ownership covers every use after the purchase with no additional charge under the stipulated rules.

Write both the threshold and the before-use timing alongside a cost record. A phrase such as buy after ten uses describes a different threshold if it means ten paid rentals first. The arithmetic below applies to buying before the tenth use, after exactly nine rentals. That one-use difference changes the worst-case ratio and cannot be hidden inside a loose description of the rule.

Prove the nineteen-tenths bound for threshold ten

For durations one through nine, threshold ten rents throughout. Its cost is n, and the benchmark cost is also n. Every corresponding ratio equals one. There is no purchase to add at the end of such a record, because the tenth use never occurs. This handles every positive duration below the threshold, including the shortest allowed sequence.

For any duration at least ten, the threshold-ten cost is nineteen and the benchmark cost is ten. Every ratio in that range is nineteen divided by ten. The full set of possible ratios is therefore one for short sequences and nineteen tenths for sequences reaching the threshold. Nineteen tenths is an upper bound for every allowed duration, and the ten-use sequence attains it.

Attainment matters. A claimed worst-case bound could otherwise be a loose allowance rather than the exact maximum. Here a specific permitted duration gives cost nineteen against benchmark ten. The same ratio occurs for every later duration because both totals remain fixed after ten under their respective rules. Nothing about this conclusion requires an actual item to survive a measured number of uses.

Show why an earlier purchase has a worse maximum

Now consider any threshold t at most ten. Stop the sequence exactly at use t. The rule rents t minus one times and buys, so its total is t plus nine. The informed benchmark can rent all t uses for t, which is no greater than ten. The resulting ratio is (t + 9) / t, or 1 + 9/t. This supplies an allowed duration that any proposed earlier threshold must face.

It is also the maximum for that threshold. Before t the ratio is one. From t through ten, the rule’s numerator stays t plus nine while the benchmark denominator increases with n, so the ratio cannot rise after n equals t. Beyond ten the denominator remains ten, giving no larger ratio than at t. Thus the exact worst ratio for t at most ten is 1 + 9/t.

As a positive t increases up to ten, nine divided by t decreases. Threshold nine has worst ratio eighteen divided by nine, or two. Threshold five has worst ratio fourteen divided by five, or 2.8. Threshold one buys immediately and can face a single-use sequence, giving ten divided by one. Every integer threshold below ten consequently has a maximum strictly greater than nineteen tenths.

Show why waiting longer also increases the maximum

Consider instead a threshold t at least ten. If the duration reaches t, the rule costs t plus nine and the benchmark costs ten. The ratio is (t + 9) / 10. This increases with t because the denominator is fixed while each extra rental before purchase increases the numerator. Threshold eleven gives twenty divided by ten, or two; threshold twelve gives twenty-one divided by ten, or 2.1.

Check durations before the late purchase too. For n below ten, renting throughout ties the benchmark, giving ratio one. For n from ten up to t minus one, the rule has rented n times while the benchmark costs ten, so the ratio is n/10. Its largest such value is (t - 1)/10 when that interval exists. This is smaller than (t + 9)/10 by exactly one.

Durations at least t keep the rule’s cost at t plus nine and the benchmark at ten. The maximum for a threshold at least ten is therefore exactly (t + 9)/10. At t equal to ten, this agrees with the earlier branch and gives nineteen tenths. Beyond ten the expression strictly increases. Together the two branches identify ten as the unique minimizing positive integer threshold.

Explain which deterministic rules the proof covers

A deterministic rule makes the same decision whenever it receives the same information. Before any purchase in this model, its only changing information is the use count. There are no different observed prices, signals or outcomes creating alternative histories at a given count. Consequently, such a rule either has a first count at which it buys or it never buys. A finite first-buy count is precisely a threshold t as defined above.

A never-buy rule costs n at duration n. For n at least ten, its benchmark ratio is n/10. Allowed durations have no upper bound, so this ratio has no finite worst-case bound. That leaves the finite thresholds, already covered by the two formulas. Threshold ten is therefore best among all deterministic rules receiving only the stipulated information, not just among a few thresholds selected for a table.

A randomized rule would introduce probabilities and a different analysis. A rule with a trustworthy maximum duration, an advance signal or an option to resell would also receive different premises. This proof does not rank those alternatives. The narrow conclusion is enough: with the exact costs, timing and information specified here, every earlier or later deterministic first-buy count has a larger worst-case ratio.

Use exact controls without substituting an average

An exact finite control evaluates forty thresholds, from one through forty, against one hundred durations, from one through one hundred. All four thousand ratios use integer costs and exact fractions. Their maxima agree with the two derived branch formulas. Threshold ten is the unique best threshold within that checked range. The algebra above extends the result beyond the finite range, including arbitrarily late purchases and the never-buy case.

A favorable short sequence does not establish a better worst-case rule. Renting for nine uses costs nine and ties the benchmark, but the rule must also face the permitted tenth use. Similarly, describing twenty uses as typical would require a probability model that this exercise does not supply. Worst-case analysis keeps every allowed duration in the comparison rather than assigning a convenient forecast to one of them.

The reusable-liner cost worksheet uses receipts, usable yield and observed lifetime to compare per-batch costs. Those empirical inputs answer a different question from selecting a sequential rule before a stipulated duration is known.

The average demand pack budget example computes each case bill before averaging with declared weights. This exercise supplies no weights and seeks a worst-case ratio against an informed benchmark, rather than a mean bill.

Retain the actual native source limits

The SAVUK fisherman weather-vane listing has three images of a fisherman-and-boat silhouette above cardinal markers and a pole. Garden scenery and a fence are not established supplied contents. Setup, measurement and performance language remains unverified. The listing supplies no rental option or lifetime for this exercise.

The Zeckos sailboat weather-vane listing has three images of a distinct sailboat, cardinal cross, pole and dedicated roof bracket. The mount is part of the host arrangement. Images do not verify material, dimensions, finish, weather resistance or directional accuracy, and they establish no actual rent-or-buy transaction.

For the unknown duration cost model, retain the before-use rule, the benchmark min(n, 10) and the witness at ten uses. Nine rentals followed by purchase cost nineteen against an informed cost of ten. The two threshold formulas prove why moving the purchase earlier or later worsens the maximum ratio. A real decision needs its own applicable costs and terms before this fictional calculation could inform it.

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