The Gold Standard of Wireless Communication | Choosing Payload Length in a Fixed-Overhead Independent-Bit-Error Model

Our objective is long-run unique accepted payload per second. Useful data counts once when accepted; headers, unsuccessful copies and repeated transmissions do not become additional useful payload merely because they occupy the channel.

This quantity is often called goodput, but the measurement boundary must remain explicit. Here it refers to our fictional serial attempt process. It does not include idle application periods, other users, queued work or unrelated network stages.

The existing lesson on retry time and unique payload goodput develops the time-and-reward account. We now add a specific length-dependent success model and ask which length maximises that conditional rate. The new question needs both a success relationship and a complete attempt duration. Optimising only the fraction of transmitted bytes that belong to the payload would omit the error process.

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Let n be payload length in bytes and h be transmitted overhead in bytes. The nominal serial transmission rate is B bytes per second. Each attempt also occupies a fixed interval τ seconds outside that byte-transmission time.

The total attempt duration is t(n) = (n + h)/B + τ. Successful and unsuccessful attempts occupy this same duration in the initial model. We stipulate that any acceptance resolution is included and that acknowledgements introduce no additional loss or identity problem.

All transmitted payload and overhead bits face the error process described next. The fixed interval τ does not add bits to that process. These definitions are artificial but precise. A real protocol may need separate preambles, coding, control transmissions, backoff and outcome-dependent timing. Such items belong in an appropriate revised ledger, rather than being silently squeezed into whichever number makes the result convenient.

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Let p be the probability that a particular transmitted bit is erroneous. We assume errors are independent across bits and attempts, with the same p throughout the calculation. A packet is accepted if and only if every one of its transmitted bits is correct.

There are 8(n + h) such bits. Therefore the acceptance probability is s(n) = (1 − p) raised to the power 8(n + h). Multiplication follows from independence, not merely from knowing an average error fraction.

The all-or-nothing rule also matters. We are not allowing correction of some errors, partial delivery or acceptance of an undetected corrupted packet. These are model assumptions, not claims about every wireless link. Keep them attached to the formula so that a familiar-looking success probability does not get carried into a technology with different coding or acceptance behaviour.

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A per-bit error probability is not a per-byte or per-packet error probability. Setting each to the same numerical value describes different processes. Under our independent-bit account, the probability that an eight-bit byte is error-free is (1 − p)⁸.

The ns-3 error-model documentation explicitly distinguishes configurable bit, byte and packet units, alongside other error-model types. It also notes that most wireless models have their own technology-specific error models. That is useful source context for keeping a teaching model separate from an operational prediction.

For a paper check, suppose p = 0.01 and a packet contains 100 exposed bits. Its all-correct probability is 0.99¹⁰⁰, approximately 0.366. It is not 0.99 merely because the error-rate entry was one per cent. The event being counted determines the exponent.

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Define λ = −8 ln(1 − p), measured per byte. Then s(n) = exp[−λ(n + h)]. This is exactly the same independent-bit relationship for 0 < p < 1, rewritten with byte length.

The exponential form is convenient for differentiation and for checking units. The product λ(n + h) is dimensionless. Do not multiply a per-bit parameter by a byte count without the conversion.

In the main invented example we supply λ = 0.001 per byte. The corresponding per-bit probability is p = 1 − exp(−0.001/8), approximately 0.000124992. Those numbers are linked descriptions of the same assumed process. They are not measured properties of Wi-Fi, mobile service or a particular radio environment. Supplying the relationship lets the learner reproduce the acceptance probabilities without relying on a rounded p.

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Multiply the attempt duration by B. We obtain Bt(n) = n + h + Bτ. Define a = h + Bτ, so t(n) = (n + a)/B.

The quantity a is an equivalent byte cost in the time account. The Bτ part is not an additional sequence of error-exposed transmitted bytes. It expresses how much transmission time the fixed interval consumes at rate B.

This distinction explains why h appears in both the success model and the time model, while Bτ appears only in the latter. Replacing h by a in the error exponent would invent bit exposure during a nontransmission interval. A useful worksheet therefore keeps the physical overhead and the time-equivalent overhead in separate columns, even though they combine neatly in the denominator of the final rate formula.

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One attempt delivers n useful bytes with probability s(n) and zero otherwise. Its expected useful reward is therefore ns(n). Its duration is the fixed t(n) for the selected length.

Across many independent equal-length attempts under these conditions, unique accepted payload per elapsed attempt time approaches G(n) = ns(n)/t(n). Substituting gives G(n) = Bn exp[−λ(n + h)]/(n + a).

The expression has units of bytes per second. Its factors have clear roles: B supplies the nominal byte rate; n/(n + a) describes the fraction of cycle time represented by useful payload length; the exponential describes successful acceptance. This decomposition makes the trade-off visible. Larger n improves the time fraction while reducing acceptance probability. Neither factor alone determines the optimum of their product.

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Take B = 100,000 bytes per second, h = 20 bytes and τ = 0.0003 seconds. The fixed interval is 0.3 milliseconds. The equivalent time cost is a = 20 + 100,000 × 0.0003 = 50 bytes.

Keep λ = 0.001 per byte. At n = 200, the transmitted length is 220 bytes, so transmission takes 2.2 milliseconds. Adding the fixed interval gives a 2.5-millisecond attempt.

The acceptance probability is exp(−0.22), approximately 0.802519. Expected useful reward is about 160.504 bytes per attempt. Dividing by 0.0025 seconds gives about 64,201.5 bytes per second. This is approximately 513,612 bits per second after multiplying by eight. The nominal bit rate is 800,000 bits per second. Preserving the unit conversion prevents a goodput comparison from acquiring a factor-of-eight error.

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Use the same supplied parameters for every row. Only the payload length changes. This keeps the comparison about length instead of mixing several changed conditions.

The 800-byte payload has a much better payload-to-time fraction than the 50-byte payload, but its lower success probability more than offsets that advantage here. The 200-byte candidate beats both.

A table of a few candidates is evidence about those candidates. It is not yet proof that 200 is the best permitted length. A complete optimisation argument must show where the rate increases, where it decreases and how the permitted discrete lengths relate to that change.

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For n > 0, a > 0 and λ > 0, G(n) is positive. Maximising G is equivalent to maximising ln G because the logarithm preserves order.

Write ln G = ln B + ln n − ln(n + a) − λ(n + h). Differentiating with respect to continuous n gives 1/n − 1/(n + a) − λ. Combining the first two terms yields a/[n(n + a)] − λ.

The constants B and exp(−λh) influence the rate level but do not appear directly in this derivative. B can still affect a through Bτ, so it would be wrong to conclude that changing B never changes the best length. The simplification holds after the supplied parameters have been fixed. Track which quantities depend on others before interpreting a missing symbol as a missing physical effect.

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Set a/[n(n + a)] = λ. Rearranging gives n² + an − a/λ = 0. The positive root is n* = [√(a² + 4a/λ) − a]/2.

For a = 50 and λ = 0.001, the equation is n(n + 50) = 50,000. The positive solution is exactly n* = 200 because 200 × 250 = 50,000.

The negative quadratic root lies outside the payload domain. As with the earlier success equation, the mathematical domain helps select the meaningful result. We have treated n as continuous temporarily to locate the peak. Actual byte lengths remain discrete and can have additional protocol or application restrictions. The continuous answer becomes a guide to the permitted-candidate check, not permission to transmit a fractional byte or exceed an actual size limit.

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For positive n and a, the denominator n(n + a) increases strictly as n grows. Therefore a/[n(n + a)] decreases strictly, from arbitrarily large values near zero towards zero at large n.

Subtracting positive λ means the logarithmic derivative crosses zero exactly once. G increases before n* and decreases afterwards. This proves a unique global maximum over positive continuous lengths in the stated model.

The proof adds information that the candidate table could not supply. It rules out an untested second peak at another length. It also justifies later constrained checks: on an interval entirely below n*, the largest permitted length is best; on an interval entirely above n*, the smallest permitted length is best. Those conclusions follow from the model’s shape, not from a general rule that packet sizes should always be large or always be small.

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If the continuous optimum falls between whole-byte lengths and every whole-byte length is permitted, compare the floor and ceiling of n*. Strict increase followed by strict decrease ensures no more distant integer can beat both.

Ordinary nearest-integer rounding is not the proof. The rate can be asymmetric around its peak, so directly compare the two values. With the main parameters, n* is already the integer 200, which makes the discrete check straightforward.

If permitted lengths are restricted to another set, find the closest permitted value on each side of the peak, where such values exist, and compare them. For example, a format allowing only multiples of sixteen would require checking 192 and 208 rather than inventing a 200-byte format. The admissible set belongs in the problem statement alongside the objective and success model.

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Suppose the fictional format allows only whole-byte payloads from 40 through 120. The unconstrained peak of 200 is outside the interval. Since G is increasing throughout the allowed range, 120 is the constrained optimum.

If instead the allowed interval is 300 through 600, G is decreasing throughout it, so 300 is best. When an interval contains the continuous peak, use the permitted neighbours around that peak.

These examples illustrate constrained optimisation without changing the underlying error model. A numerically attractive unrestricted answer can be unusable because it is not an admissible packet size. For a real system, obtain its actual framing and application constraints rather than borrowing the invented intervals here. A correct optimisation result includes the allowed set; otherwise another reader cannot determine what has genuinely been optimised.

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Keep h = 20 and λ = 0.001 but remove the fixed interval, so τ = 0 and a = 20. The continuous optimum becomes approximately 131.774 bytes. Comparing 131 and 132 selects 132 among whole-byte lengths.

The best length has become shorter because there is less fixed time to spread across a payload. This is a conditional comparative result, not a claim that removing a protocol wait is an available engineering action.

The example helps separate physical overhead from fixed time. The error-exposed overhead remains twenty bytes, while only the time cost has changed. A calculation that changed the success exponent as well would answer a different scenario. For learning, ask the student to identify which expression changes before doing the arithmetic. That is a stronger transfer check than merely inserting a new a into the quadratic formula.

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Restore a = 50 and change λ to 0.004 per byte. The continuous optimum is approximately 89.564 bytes. Direct comparison selects 90 bytes over 89, with the other parameters unchanged.

The shorter optimum follows because successful acceptance now decreases more rapidly with length. It does not mean that a particular measured wireless link can be described by this new λ. We have changed a supplied mathematical parameter.

The direction can also be seen without evaluating the quadratic. At a fixed length, increasing λ lowers the logarithmic derivative. Its zero therefore moves towards a smaller n. This explanation connects the numerical result with the competing effects. A learner should be able to predict the direction first, then verify it, while acknowledging that a different real error process might not preserve the assumed exponential success curve.

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If p = 0, then λ = 0 and every attempt succeeds. The rate reduces to G(n) = Bn/(n + a). With a > 0, this increases with n and approaches B as n grows.

There is no finite unrestricted optimum in this zero-error boundary case. Over a finite permitted range, the largest length is best. The formula containing a/λ must not be evaluated by dividing by zero and then reporting a finite answer.

This boundary is instructive because it isolates the overhead effect. An internal peak arose from the competition between fixed cost and decreasing success, not from overhead alone. When the decreasing-success effect disappears, the argument changes. A useful derivation identifies such limits rather than treating one algebraic expression as valid for every parameter value printed in a spreadsheet.

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If h = 0 and τ = 0, then a = 0. For λ > 0, G(n) = B exp(−λn), which decreases with positive n. The smallest permitted payload is best in that changed model.

This does not establish a useful zero-length packet. At n = 0 there is no payload reward, and the original ratio of reward to occupied time needs care because its duration also vanishes. The actual problem should state a positive permitted minimum.

The model’s interior optimum therefore depends on a positive fixed cost as well as a positive error parameter. Removing either side of the trade-off changes the shape. Comparing both boundary cases gives the learner a way to detect an implementation that blindly returns the same sort of optimum even when the mathematical conditions no longer hold.

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Suppose an additional fictional requirement limits one complete attempt to D = 0.0015 seconds. The timing inequality is (n + h)/B + τ ≤ D. Rearranging gives n ≤ B(D − τ) − h.

With the main parameters, the cap is 100 bytes. Since the unrestricted optimum is 200 and the rate is increasing below it, 100 is the best permitted positive whole-byte length if no other restriction is added.

The condition concerns one attempt’s occupied time. It does not guarantee successful message delivery by a deadline: an attempt can fail, and retries or queued work add other timing questions. The separate guide to deadline delivery and burst loss develops those distinctions. A time cap should retain the event it actually bounds instead of becoming a broader service promise.

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For a packet with N independent exposed bits, the expected number of erroneous bits is Np. The probability of at least one error is instead 1 − (1 − p)ᴺ. They are different quantities.

In the main 200-byte example, N = 1,760 and Np is approximately 0.220. The rejection probability is 1 − exp(−0.22), approximately 0.1975. A rejected packet can contain more than one error, which is why the two values need not match.

Using 1 − Np as the exact acceptance probability would introduce a different approximation and can even produce a negative value for longer packets. If an approximation is used, state its purpose and relevant regime. Our exponential formulation preserves the exact independent-bit relationship and therefore avoids that particular approximation throughout the optimisation.

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Consider a separate mathematical comparison with 100 bits and a marginal error probability of 0.01 for each bit. Under independence, all-correct probability is approximately 0.366.

Now construct another process: with probability 0.01 every bit in the packet is erroneous, and otherwise every bit is correct. Each bit still has marginal error probability 0.01, but packet acceptance is 0.99. The within-packet dependence changed.

This extreme construction is a counterexample, not a physical channel description. It proves that a per-bit marginal alone does not determine packet success. Under length-independent acceptance in that second construction, increasing payload could retain the overhead advantage instead of producing the same internal peak. Evidence for the independence assumption is therefore part of the optimisation account, not a minor note added after choosing a packet size.

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If a receiver can correct some error patterns, all bits being initially correct is no longer the only route to acceptance. Coding also changes transmitted length, rate relationships and possibly processing or timing costs.

The independent-bit formula cannot be preserved merely by calling p a better error rate while leaving every other quantity unchanged without justification. The success event and exposure must be defined for the actual coding account.

For the underlying mathematical ideas, continue with error-correcting codes, parity and reliable data transfer. Our lesson deliberately grants no error correction or partial useful reward from a failed packet. Keeping this boundary explicit allows the derivation to remain clear without implying that an uncoded teaching model is a complete account of a modern wireless link.

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In an unlimited independent retry version with the same fixed attempt duration, a selected payload eventually succeeds. Its expected number of attempts is 1/s(n), so expected occupied time per accepted payload is t(n)/s(n). Dividing n by that time recovers G(n).

This agreement depends on the supplied conditions. It does not establish identical message completion times or permit counting every retransmission as useful payload. Outcome-dependent durations, finite observation windows, acknowledgement loss and additional per-message costs require their own account.

Use the existing retry-goodput lesson for those detailed cases. Here the purpose is to check that length optimisation uses a consistent reward boundary. A larger retransmission count can consume more time without increasing the unique information accepted by the recipient.

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Suppose λ is supplied only as lying between 0.0005 and 0.002 per byte, while B, h and τ retain their main values. For every fixed positive n, G decreases as λ increases. The lowest conditional rate therefore occurs at λ = 0.002.

If the stated objective is to maximise that worst-case model rate, optimise the upper endpoint. The continuous optimum is approximately 135.078 bytes, and a direct integer comparison selects 135.

This is a robust choice for the specified parameter interval and objective. It is not a confidence guarantee for an unmeasured link or a claim that the true process belongs to this family. The interval needs its own evidential justification. Distinguish uncertainty within a supplied independent-error model from uncertainty about whether that model adequately describes the task at all.

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Maximising expected payload per occupied attempt time can conflict with other goals. A time-sensitive application may care about completion by a deadline, a battery-limited device may need an energy account, and a shared service may need fairness between users.

The best length for G(n) does not automatically optimise any of those quantities. They require explicit objectives, constraints and evidence. Even in the fictional model, a time cap already changed the admissible optimum from 200 to 100 bytes.

Begin a practical discussion by naming the task, observation window and consequence of failure. The Wireless Communication root keeps coverage, capacity, delivery, delay and service responsibility separate. The mathematics becomes useful when its objective fits that task; an elegant maximum is not sufficient evidence that the complete service has improved.

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A relevant approved test would need packet identity, payload length, transmitted overhead, outcome and complete occupied time. It should record the conditions under which each size was examined, including changes that might affect success or duration.

A comparison made at different times or loads can confound payload length with channel conditions. A fitting exercise should also examine whether the assumed success curve and timing relationship describe the relevant records, rather than reporting only the size with the highest observed rate.

This guide supplies no procedure for capturing private traffic or altering an operational radio system. Use fictional data, a suitable simulator or authorised supervised measurements. The paper model helps identify the evidence needed: an observed nominal rate or one overall loss fraction cannot independently establish all the parameters and assumptions required by the optimum formula.

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A fictional report says that the largest payload must be best because it wastes the smallest fraction on fixed overhead. The time-fraction statement is correct under the fixed-cost assumptions, but the conclusion has omitted acceptance probability.

Repair it by calculating both factors. In our main comparison, the 800-byte candidate has time fraction 800/850, approximately 0.9412, while the 200-byte candidate has 200/250 = 0.80. Their success probabilities are approximately 0.4404 and 0.8025 respectively.

The products reverse the ranking: about 41,452 versus 64,202 bytes per second. This correction preserves the useful overhead observation and adds the missing relationship. It is more helpful than replacing one absolute slogan with another, such as always choose small packets. The learner can now explain why the answer depends on the supplied error process and objective.

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Use a new fictional account with B = 50,000 bytes per second, h = 10 bytes, τ = 0.0006 seconds and λ = 0.002 per byte. Only whole-byte payloads from 20 through 300 are permitted.

Calculate the equivalent time cost a before applying the optimum formula. Then find the continuous optimum and compare the two neighbouring permitted integer lengths. Explain why a correct time conversion matters even though the nominal rate is smaller than in the main example.

Finally add a one-attempt time budget of 0.002 seconds. Determine the resulting length cap and the best permitted payload under that cap. Keep the deadline interpretation narrow. This transfer changes the rate, header, fixed interval and error parameter, so it tests whether the learner can reconstruct the account rather than recall that two hundred was best in an earlier invented link.

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The new equivalent time cost is a = 10 + 50,000 × 0.0006 = 40 bytes. The stationary equation is n(n + 40) = 40/0.002 = 20,000. Its continuous root is approximately 122.829 bytes.

Compare 122 and 123 directly in G(n) = 50,000n exp[−0.002(n + 10)]/(n + 40). The 123-byte candidate is slightly better. Both belong to the original permitted interval, so 123 is the discrete optimum for that version.

The added attempt-time budget gives n ≤ 50,000(0.002 − 0.0006) − 10 = 60 bytes. Since G is increasing up to the continuous peak, 60 is the best allowed payload under the added cap. The shorter result follows from a changed feasible set, not a contradiction in the unconstrained calculation. Neither answer promises eventual delivery within two milliseconds.

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Our main objective assumes an ongoing supply of equal-length payloads. A finite message can leave a short final fragment, so its completion-time question needs an explicit fragmentation account.

For a separate 450-byte message, consider fragments of 200, 200 and 50 bytes under the main parameters. Suppose each fragment is retried independently until accepted, in sequence, and all the earlier timing assumptions hold. Expected total time is the sum of t(n)/s(n) over the three fragments, approximately 0.007303 seconds.

Two 225-byte fragments instead give approximately 0.007027 seconds. Three equal 150-byte fragments give approximately 0.007112 seconds. A steady-stream optimum of 200 therefore does not automatically specify the best partition of this finite message. The target and the final-fragment rule changed.

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For a fixed fragment count, the expected until-success time of one fragment is f(n) = (n + a)exp[λ(n + h)]/B. Its second derivative is exp[λ(n + h)] multiplied by [2λ + λ²(n + a)]/B, which is positive for the present positive parameters.

This convexity means that equal continuous fragment lengths minimise the summed expected time for a fixed total and fixed count. With whole-byte constraints, fragments should be as equal as the permitted rules allow in this simplified problem.

That observation helps compare explicit partitions; it does not remove format restrictions or identify the best fragment count without checking it. It also assumes each fragment is separately accepted and retried. A rule that discards the whole message after one failed fragment would define a different reward-and-time process.

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When reviewing a payload recommendation, write down what has been held fixed. Was the total message size fixed, or was there an endless stream? Did every candidate pay the same header and fixed interval? Were failed attempts charged their full time? Did success depend on the entire packet or only one protected part?

These questions are a practical check against mixing pieces of different models. A calculation may use a success probability from one frame length and a duration from another, then appear to improve the rate without describing any permitted attempt.

Rebuild one candidate from the primitive quantities before comparing summaries. If its transmitted bits, acceptance event, useful reward and elapsed time do not belong to the same cycle, the resulting number is not ready to guide a decision.

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The repair route begins with bits, bytes, expected useful reward and complete attempt duration. Ask the learner to explain why the fixed-time equivalent enters the denominator but not the exposed-bit exponent.

The stabilisation route compares several payload sizes, derives the continuous peak and checks permitted neighbours. Change one parameter and require a direction prediction before calculation. Retain the proof of increasing and decreasing behaviour so that the result is more than a spreadsheet search.

The extension route examines dependence, coding, uncertain parameters and competing objectives. Independence appears when the learner can state which result survives a changed condition and which needs a new model. The Education parent guide connects this habit with practical transfer: a method remains useful when its assumptions and evidence travel with it.

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Is the largest payload always the most efficient?

It improves the fixed-overhead fraction in this model, but acceptance can fall with length. The product determines goodput. Our worked optimum is conditional on the supplied independent-bit process.

Is 200 bytes a recommended wireless setting?

No. It is the exact optimum for one invented parameter set. No named technology or operational network has been evaluated.

Does the highest expected goodput guarantee timely delivery?

No. Expected rate, individual completion time and deadline success are different questions. Queueing, retries and other stages can matter.

Can I use an overall bit-error fraction without checking dependence?

Not to justify the independent-bit packet-success formula. Our counterexample gives the same bit marginals and different packet acceptance probabilities.

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Return to Wireless Communication for the full service question. Use retry-time accounting for detailed reward and elapsed-time ledgers, and deadline delivery and burst loss for timing and dependence claims.

For another mathematical mechanism, read parity and error-correcting codes. That route should retain its different acceptance and redundancy account.

The official ns-3 error-model documentation, checked on 24 October 2026, supports the distinctions between error-model units, configured burst patterns and technology-specific wireless models. It does not validate our fictional parameters or prescribe the optimum derived here. Every numerical link, candidate table, optimisation, counterexample and transfer task is an original hypothetical construction. Explore How X Works for wider first-principles learning routes.

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