A good number needs a good question.

1. Compute ten matched differences B_i - A_i. Their mean is 1.2 and their sample standard deviation is 0.408248.
2. Divide the difference standard deviation by sqrt(10), giving 0.129099. Treating the same observations as unpaired gives 0.632456.
3. The paired calculation retains the shared disturbance as covariance; it cancels in each difference. The unpaired expression discards that relationship.

The variance ratio is 24 in this constructed batch. Matching is useful because the common disturbance is genuinely shared. These ten deliberately chosen observations do not validate a population distribution or establish a significance level.

1. Compute ten matched differences B_i - A_i. Their mean is 1.95 and their sample standard deviation is 0.506228.
2. Divide the difference standard deviation by sqrt(10), giving 0.160083. Treating the same observations as unpaired gives 2.367902.
3. The paired calculation retains the shared disturbance as covariance; it cancels in each difference. The unpaired expression discards that relationship.

The variance ratio is 218.793965 in this constructed batch. Matching is useful because the common disturbance is genuinely shared. These ten deliberately chosen observations do not validate a population distribution or establish a significance level.

1. Compute ten matched differences B_i - A_i. Their mean is 2.85 and their sample standard deviation is 0.702187.
2. Divide the difference standard deviation by sqrt(10), giving 0.222051. Treating the same observations as unpaired gives 4.467241.
3. The paired calculation retains the shared disturbance as covariance; it cancels in each difference. The unpaired expression discards that relationship.

The variance ratio is 404.737155 in this constructed batch. Matching is useful because the common disturbance is genuinely shared. These ten deliberately chosen observations do not validate a population distribution or establish a significance level.

1. A is weighted using 55% easy work; B uses 40%. Those are different test distributions.
2. Use the shared 20% easy reference for both: A = 0.562, B = 0.582.
3. The raw difference is -4.9 percentage points. The shared-workload difference is 2 percentage points.

The direction reverses after standardization. B was defined to improve both strata by 2 percentage points. Aggregation cannot identify that fact without the workload mix. These rates are stipulated probabilities, not estimates with sampling intervals.

1. A is weighted using 70% easy work; B uses 32.5%. Those are different test distributions.
2. Use the shared 45% easy reference for both: A = 0.677, B = 0.707.
3. The raw difference is -14.25 percentage points. The shared-workload difference is 3 percentage points.

The direction reverses after standardization. B was defined to improve both strata by 3 percentage points. Aggregation cannot identify that fact without the workload mix. These rates are stipulated probabilities, not estimates with sampling intervals.

1. A is weighted using 88% easy work; B uses 23.5%. Those are different test distributions.
2. Use the shared 75% easy reference for both: A = 0.815, B = 0.857.
3. The raw difference is -25.47 percentage points. The shared-workload difference is 4.2 percentage points.

The direction reverses after standardization. B was defined to improve both strata by 4.2 percentage points. Aggregation cannot identify that fact without the workload mix. These rates are stipulated probabilities, not estimates with sampling intervals.

1. The continuous solution allocates 333.340886, 164.99532, 107.333695 shots. It minimizes variance under the time constraint before integer rounding.
2. Round down, then spend affordable remaining units by the largest immediate variance reduction per unit time. The resulting counts are 335, 165, 107. This greedy rounding is not claimed to solve every integer allocation problem exactly.
3. Variance is sum((coefficient × standard deviation)^2 / shots) = 0.0108. Equal time across groups yields 0.01143.

The rounded allocation uses 1,200 of 1,200 time units. Its variance is 0.001118% above the continuous lower bound. The calculation shows why a costly measurement may deserve fewer shots even when its contribution is noisy.

1. The continuous solution allocates 529.53979, 275.156946, 192.141339 shots. It minimizes variance under the time constraint before integer rounding.
2. Round down, then spend affordable remaining units by the largest immediate variance reduction per unit time. The resulting counts are 531, 275, 192. This greedy rounding is not claimed to solve every integer allocation problem exactly.
3. Variance is sum((coefficient × standard deviation)^2 / shots) = 0.009629. Equal time across groups yields 0.011011.

The rounded allocation uses 2,700 of 2,700 time units. Its variance is 0.000186% above the continuous lower bound. The calculation shows why a costly measurement may deserve fewer shots even when its contribution is noisy.

1. The continuous solution allocates 656.836967, 334.870921, 309.829776 shots. It minimizes variance under the time constraint before integer rounding.
2. Round down, then spend affordable remaining units by the largest immediate variance reduction per unit time. The resulting counts are 662, 335, 309. This greedy rounding is not claimed to solve every integer allocation problem exactly.
3. Variance is sum((coefficient × standard deviation)^2 / shots) = 0.01043. Equal time across groups yields 0.012277.

The rounded allocation uses 4,500 of 4,500 time units. Its variance is 0.001247% above the continuous lower bound. The calculation shows why a costly measurement may deserve fewer shots even when its contribution is noisy.

1. Covariance = correlation × s_A × s_B = -0.54. Insert it in all three variance terms.
2. Differentiate the variance with respect to A’s weight, with B’s weight fixed to 1 - w. The stationary point gives w = 0.710526.
3. The standard uncertainty is sqrt(0.554211) = 0.744453. An independence-based weight is 0.764151, but its real variance must still include covariance.

Both weights lie between zero and one in this regime. The estimate is unbiased only because both inputs were assumed unbiased and the weights sum to one. A gain error or wrong covariance can reverse this result.

1. Covariance = correlation × s_A × s_B = 0.5. Insert it in all three variance terms.
2. Differentiate the variance with respect to A’s weight, with B’s weight fixed to 1 - w. The stationary point gives w = 0.767123.
3. The standard uncertainty is sqrt(1.315068) = 1.146764. An independence-based weight is 0.719101, but its real variance must still include covariance.

Both weights lie between zero and one in this regime. The estimate is unbiased only because both inputs were assumed unbiased and the weights sum to one. A gain error or wrong covariance can reverse this result.

1. Covariance = correlation × s_A × s_B = 2.48. Insert it in all three variance terms.
2. Differentiate the variance with respect to A’s weight, with B’s weight fixed to 1 - w. The stationary point gives w = 1.053726.
3. The standard uncertainty is sqrt(2.398336) = 1.548656. An independence-based weight is 0.624756, but its real variance must still include covariance.

The optimal solution uses a negative weight, which subtracts shared noise while increasing sensitivity to model error. The estimate is unbiased only because both inputs were assumed unbiased and the weights sum to one. A gain error or wrong covariance can reverse this result.

1. All 20 null hypotheses are assumed true. Independence gives the probability of no alert as (1 - 0.01)^20.
2. Subtract from one: the campaign alert probability is 0.182093, even though each test uses 0.01.
3. Dividing the desired family bound 0.05 by 20 gives 0.0025. The union bound does not require independent tests; the displayed exact corrected probability does.

At least one false alert occurs with probability 18.209306% under the independent all-null scenario. The correction is a declared campaign rule, not a substitute for replication or a complete analysis of statistical power. A laboratory must count the comparisons actually searched, including hidden tuning decisions.

1. All 120 null hypotheses are assumed true. Independence gives the probability of no alert as (1 - 0.01)^120.
2. Subtract from one: the campaign alert probability is 0.70062, even though each test uses 0.01.
3. Dividing the desired family bound 0.05 by 120 gives 0.000417. The union bound does not require independent tests; the displayed exact corrected probability does.

At least one false alert occurs with probability 70.061961% under the independent all-null scenario. The correction is a declared campaign rule, not a substitute for replication or a complete analysis of statistical power. A laboratory must count the comparisons actually searched, including hidden tuning decisions.

1. All 240 null hypotheses are assumed true. Independence gives the probability of no alert as (1 - 0.012)^240.
2. Subtract from one: the campaign alert probability is 0.944835, even though each test uses 0.012.
3. Dividing the desired family bound 0.05 by 240 gives 0.000208. The union bound does not require independent tests; the displayed exact corrected probability does.

At least one false alert occurs with probability 94.483461% under the independent all-null scenario. The correction is a declared campaign rule, not a substitute for replication or a complete analysis of statistical power. A laboratory must count the comparisons actually searched, including hidden tuning decisions.

1. The constructed eleven observations are 8, 9, 10, 11, 12, 9, 10, 11, 10, 8, 17. Only the final entry receives the declared extra disturbance.
2. The arithmetic mean uses every magnitude and gives 10.454545. The median is the sixth sorted value, 10.
3. Remove one observation from each sorted tail and average the remaining nine to obtain 10. This is a specified transformation, not permission to discard inconvenient measurements after seeing a result.

The disturbance shifts the mean by 0.454545 from the stipulated target. The comparison demonstrates influence, not a claim that a median estimates every physical observable correctly. For an expectation value defined as a mean, robust summaries can change the estimand and must be justified by a contamination model.

1. The constructed eleven observations are 9.5, 11, 12.5, 14, 15.5, 11, 12.5, 14, 12.5, 9.5, 35.5. Only the final entry receives the declared extra disturbance.
2. The arithmetic mean uses every magnitude and gives 14.318182. The median is the sixth sorted value, 12.5.
3. Remove one observation from each sorted tail and average the remaining nine to obtain 12.5. This is a specified transformation, not permission to discard inconvenient measurements after seeing a result.

The disturbance shifts the mean by 1.818182 from the stipulated target. The comparison demonstrates influence, not a claim that a median estimates every physical observable correctly. For an expectation value defined as a mean, robust summaries can change the estimand and must be justified by a contamination model.

1. The constructed eleven observations are 11.3, 13.4, 15.5, 17.6, 19.7, 13.4, 15.5, 17.6, 15.5, 11.3, 57.7. Only the final entry receives the declared extra disturbance.
2. The arithmetic mean uses every magnitude and gives 18.954545. The median is the sixth sorted value, 15.5.
3. Remove one observation from each sorted tail and average the remaining nine to obtain 15.5. This is a specified transformation, not permission to discard inconvenient measurements after seeing a result.

The disturbance shifts the mean by 3.454545 from the stipulated target. The comparison demonstrates influence, not a claim that a median estimates every physical observable correctly. For an expectation value defined as a mean, robust summaries can change the estimand and must be justified by a contamination model.

1. Raw MSE = bias squared + sampling variance = 0.0025 + 0.0025.
2. Mitigated MSE = residual bias squared + inflated sampling variance = 0.0001 + 0.005.
3. Equating the MSEs gives N = 104.166667. Integer shot counts above that value favor mitigation in this model.

At 100 shots, the raw estimate has lower modeled MSE. This is an accuracy comparison over repeated hypothetical experiments, not a judgment based on whichever single answer looks closer to an expected result. Biases must be independently characterized before using the model to plan an experiment.

1. Raw MSE = bias squared + sampling variance = 0.0025 + 7.8125e-05.
2. Mitigated MSE = residual bias squared + inflated sampling variance = 0.0004 + 0.000156.
3. Equating the MSEs gives N = 119.047619. Integer shot counts above that value favor mitigation in this model.

At 3,200 shots, mitigation has lower modeled MSE. This is an accuracy comparison over repeated hypothetical experiments, not a judgment based on whichever single answer looks closer to an expected result. Biases must be independently characterized before using the model to plan an experiment.

1. Raw MSE = bias squared + sampling variance = 0.0025 + 1.2207e-06.
2. Mitigated MSE = residual bias squared + inflated sampling variance = 0.001024 + 3.66211e-06.
3. Equating the MSEs gives N = 338.753388. Integer shot counts above that value favor mitigation in this model.

At 204,800 shots, mitigation has lower modeled MSE. This is an accuracy comparison over repeated hypothetical experiments, not a judgment based on whichever single answer looks closer to an expected result. Biases must be independently characterized before using the model to plan an experiment.

1. An interval L leaves fraction L/(L + calibration time) for acquisition. At L = 13, that is 0.619048.
2. Uniform age since a reset gives mean squared drift (r L)^2 / 3 = 5.63333e-07. Shot variance adds 2.80449e-06.
3. Scan integer intervals 1 through 180; the best is 13. The reference interval 120 gives MSE 4.98519e-05.

The best tested cadence is 13 time units. It is conditional on deterministic linear drift, a perfect reset and long-campaign duty-cycle accounting. If calibration leaves an offset or the drift is stochastic, both the age term and the optimum change. This model uses fractional expected shots; an actual schedule would need integer acquisitions and explicit start/end effects.

1. An interval L leaves fraction L/(L + calibration time) for acquisition. At L = 4, that is 0.181818.
2. Uniform age since a reset gives mean squared drift (r L)^2 / 3 = 1.92e-06. Shot variance adds 4.24383e-06.
3. Scan integer intervals 1 through 180; the best is 4. The reference interval 120 gives MSE 0.001729.

The best tested cadence is 4 time units. It is conditional on deterministic linear drift, a perfect reset and long-campaign duty-cycle accounting. If calibration leaves an offset or the drift is stochastic, both the age term and the optimum change. This model uses fractional expected shots; an actual schedule would need integer acquisitions and explicit start/end effects.

1. An interval L leaves fraction L/(L + calibration time) for acquisition. At L = 2, that is 0.0625.
2. Uniform age since a reset gives mean squared drift (r L)^2 / 3 = 1.92e-06. Shot variance adds 7.40741e-06.
3. Scan integer intervals 1 through 180; the best is 2. The reference interval 120 gives MSE 0.006913.

The best tested cadence is 2 time units. It is conditional on deterministic linear drift, a perfect reset and long-campaign duty-cycle accounting. If calibration leaves an offset or the drift is stochastic, both the age term and the optimum change. This model uses fractional expected shots; an actual schedule would need integer acquisitions and explicit start/end effects.

1. The error model gives p_L(d) = A × ratio^((d + 1)/2). Multiply by logical registers × base operations × distance to count the declared fault opportunities.
2. At d = 17, the resulting union bound is min(1, 1,360,000 × 2.06391e-10) = 0.000281.
3. The illustrative rotated-patch allocation is 8 × (2 × 17² - 1) = 4,616 physical qubits. The serial duration is 0.17 seconds. Routing space, factories and extra control patches are excluded.

Distance 17 is the smallest tested odd distance meeting the stated target. The gate count alone is not the exposure count: this model makes each logical operation last d rounds. Its scaling prefactor and threshold ratio are invented. No threshold theorem or device forecast is inferred from the scan.

1. The error model gives p_L(d) = A × ratio^((d + 1)/2). Multiply by logical registers × base operations × distance to count the declared fault opportunities.
2. At d = 29, the resulting union bound is min(1, 31,320,000 × 2.75138e-11) = 0.000862.
3. The illustrative rotated-patch allocation is 18 × (2 × 29² - 1) = 30,258 physical qubits. The serial duration is 2.61 seconds. Routing space, factories and extra control patches are excluded.

Distance 29 is the smallest tested odd distance meeting the stated target. The gate count alone is not the exposure count: this model makes each logical operation last d rounds. Its scaling prefactor and threshold ratio are invented. No threshold theorem or device forecast is inferred from the scan.

1. The error model gives p_L(d) = A × ratio^((d + 1)/2). Multiply by logical registers × base operations × distance to count the declared fault opportunities.
2. At d = 41, the resulting union bound is min(1, 147,600,000 × 1.35082e-10) = 0.019938.
3. The illustrative rotated-patch allocation is 30 × (2 × 41² - 1) = 100,830 physical qubits. The serial duration is 10.332 seconds. Routing space, factories and extra control patches are excluded.

The tested distance range does not meet the target. The gate count alone is not the exposure count: this model makes each logical operation last d rounds. Its scaling prefactor and threshold ratio are invented. No threshold theorem or device forecast is inferred from the scan.

1. Create 60 arrivals separated by 1 time units. The first two jobs in every 10-job block use burst service 1.6.
2. For each arrival, select the first available worker; start at max(arrival time, worker-free time). Completion minus arrival is the response time.
3. The one-worker queue misses 0 deadlines. The two-worker schedule misses 0 under exactly the same arrival and service trace.

The average utilization proxy is 0.68, while the maximum one-worker response is 2.2. A mean below one does not rule out burst-related deadline failures. The second worker is a scheduling experiment; real decoding may have dependencies, memory contention and communication costs omitted here.

1. Create 60 arrivals separated by 1 time units. The first two jobs in every 12-job block use burst service 2.5.
2. For each arrival, select the first available worker; start at max(arrival time, worker-free time). Completion minus arrival is the response time.
3. The one-worker queue misses 20 deadlines. The two-worker schedule misses 0 under exactly the same arrival and service trace.

The average utilization proxy is 0.895833, while the maximum one-worker response is 4. A mean below one does not rule out burst-related deadline failures. The second worker is a scheduling experiment; real decoding may have dependencies, memory contention and communication costs omitted here.

1. Create 60 arrivals separated by 1 time units. The first two jobs in every 12-job block use burst service 3.58.
2. For each arrival, select the first available worker; start at max(arrival time, worker-free time). Completion minus arrival is the response time.
3. The one-worker queue misses 60 deadlines. The two-worker schedule misses 10 under exactly the same arrival and service trace.

The average utilization proxy is 1.200833, while the maximum one-worker response is 15.8. A mean below one does not rule out burst-related deadline failures. The second worker is a scheduling experiment; real decoding may have dependencies, memory contention and communication costs omitted here.

1. Supply is zero in slots 10, 11 and 20; all other slots produce 10 tokens. Demand includes the declared periodic bursts.
2. Cumulative net production reaches a minimum of -5. An initial reserve of max(0, -minimum) = 5 prevents a negative end-of-slot balance.
3. Add that reserve to every cumulative value. The largest start/end boundary inventory is 25; final stock is 20. Supply and demand are netted simultaneously at each boundary.

The toy schedule needs 5 stored tokens before it begins, even though net production across the campaign is 15. A reserve addresses timing, not state quality. If tokens represent prepared quantum resources, finite storage lifetime and extra storage qubits require a separate physical model.

1. Supply is zero in slots 10, 11 and 20; all other slots produce 11 tokens. Demand includes the declared periodic bursts.
2. Cumulative net production reaches a minimum of -63. An initial reserve of max(0, -minimum) = 63 prevents a negative end-of-slot balance.
3. Add that reserve to every cumulative value. The largest start/end boundary inventory is 83; final stock is 0. Supply and demand are netted simultaneously at each boundary.

The toy schedule needs 63 stored tokens before it begins, even though net production across the campaign is -63. A reserve addresses timing, not state quality. If tokens represent prepared quantum resources, finite storage lifetime and extra storage qubits require a separate physical model.

1. Supply is zero in slots 10, 11 and 20; all other slots produce 13 tokens. Demand includes the declared periodic bursts.
2. Cumulative net production reaches a minimum of -99. An initial reserve of max(0, -minimum) = 99 prevents a negative end-of-slot balance.
3. Add that reserve to every cumulative value. The largest start/end boundary inventory is 129; final stock is 0. Supply and demand are netted simultaneously at each boundary.

The toy schedule needs 99 stored tokens before it begins, even though net production across the campaign is -99. A reserve addresses timing, not state quality. If tokens represent prepared quantum resources, finite storage lifetime and extra storage qubits require a separate physical model.

1. With no common outage, sum the binomial probabilities for 10 through 12 active modules.
2. The independent capacity probability is 0.629787. Multiply by 0.998, the chance the shared service is available.
3. Even infinitely many otherwise adequate independent modules cannot exceed the common-mode ceiling 0.998 in this model.

The installed system has modeled usable capacity probability 0.628527. Extra modules protect against independent outages while the shared failure mode persists. The common outage is an explicit invented mechanism; the independence calculation must not be used as a measured reliability claim for a real modular quantum machine.

1. With no common outage, sum the binomial probabilities for 11 through 17 active modules.
2. The independent capacity probability is 0.997678. Multiply by 0.983, the chance the shared service is available.
3. Even infinitely many otherwise adequate independent modules cannot exceed the common-mode ceiling 0.983 in this model.

The installed system has modeled usable capacity probability 0.980717. Extra modules protect against independent outages while the shared failure mode persists. The common outage is an explicit invented mechanism; the independence calculation must not be used as a measured reliability claim for a real modular quantum machine.

1. With no common outage, sum the binomial probabilities for 13 through 23 active modules.
2. The independent capacity probability is 1. Multiply by 0.965, the chance the shared service is available.
3. Even infinitely many otherwise adequate independent modules cannot exceed the common-mode ceiling 0.965 in this model.

The installed system has modeled usable capacity probability 0.965. Extra modules protect against independent outages while the shared failure mode persists. The common outage is an explicit invented mechanism; the independence calculation must not be used as a measured reliability claim for a real modular quantum machine.

1. Start empty. Each slot gives independent probabilities 0.08 and 0.05 for the two links. Simultaneous success ends the request.
2. A lone success is held for 1 additional partner-attempt slots. On an unsuccessful final slot it expires; the next slot begins empty. The original link is not re-attempted while its herald is held.
3. Propagate the exact state probabilities through 15 slots. Completed plus unfinished probability remains one after every update. The completion probability is 0.138833.

Finite memory raises deadline completion from 5.834877% to 13.883327%. The hard lifetime cutoff grants perfect usefulness before expiry and none after. It is not a physical fidelity-decay law. Storage write/read loss, swapping, classical latency and quantum-state verification are excluded.

1. Start empty. Each slot gives independent probabilities 0.13 and 0.08 for the two links. Simultaneous success ends the request.
2. A lone success is held for 6 additional partner-attempt slots. On an unsuccessful final slot it expires; the next slot begins empty. The original link is not re-attempted while its herald is held.
3. Propagate the exact state probabilities through 30 slots. Completed plus unfinished probability remains one after every update. The completion probability is 0.803002.

Finite memory raises deadline completion from 26.921336% to 80.300185%. The hard lifetime cutoff grants perfect usefulness before expiry and none after. It is not a physical fidelity-decay law. Storage write/read loss, swapping, classical latency and quantum-state verification are excluded.

1. Start empty. Each slot gives independent probabilities 0.19 and 0.116 for the two links. Simultaneous success ends the request.
2. A lone success is held for 12 additional partner-attempt slots. On an unsuccessful final slot it expires; the next slot begins empty. The original link is not re-attempted while its herald is held.
3. Propagate the exact state probabilities through 48 slots. Completed plus unfinished probability remains one after every update. The completion probability is 0.994875.

Finite memory raises deadline completion from 65.690656% to 99.487518%. The hard lifetime cutoff grants perfect usefulness before expiry and none after. It is not a physical fidelity-decay law. Storage write/read loss, swapping, classical latency and quantum-state verification are excluded.

1. The first successful link arrives with per-slot probability a + b - a b = 0.0938. Its mean wait is 10.660981 slots.
2. For positive-integer geometric waits X and Y, max(X,Y) = X + Y - min(X,Y). The expected synchronized wait is 68.505686 slots.
3. The earlier herald waits |X - Y| slots. Using |X-Y| = 2 max(X,Y) - X - Y gives 57.844705. Simultaneous-only operation instead waits 833.333333 slots on average.

Perfect storage shortens the modeled mean wait by a factor of 12.16444, but the older state waits an average of 14.461176 milliseconds. A mean age does not specify a tail probability or surviving fidelity; use the finite-memory lab when a deadline or cutoff defines usefulness.

1. The first successful link arrives with per-slot probability a + b - a b = 0.1873. Its mean wait is 5.339028 slots.
2. For positive-integer geometric waits X and Y, max(X,Y) = X + Y - min(X,Y). The expected synchronized wait is 19.985647 slots.
3. The earlier herald waits |X - Y| slots. Using |X-Y| = 2 max(X,Y) - X - Y gives 14.646619. Simultaneous-only operation instead waits 129.87013 slots on average.

Perfect storage shortens the modeled mean wait by a factor of 6.49817, but the older state waits an average of 7.323309 milliseconds. A mean age does not specify a tail probability or surviving fidelity; use the finite-memory lab when a deadline or cutoff defines usefulness.

1. The first successful link arrives with per-slot probability a + b - a b = 0.293164. Its mean wait is 3.41106 slots.
2. For positive-integer geometric waits X and Y, max(X,Y) = X + Y - min(X,Y). The expected synchronized wait is 11.014659 slots.
3. The earlier herald waits |X - Y| slots. Using |X-Y| = 2 max(X,Y) - X - Y gives 7.603599. Simultaneous-only operation instead waits 45.795933 slots on average.

Perfect storage shortens the modeled mean wait by a factor of 4.157726, but the older state waits an average of 6.082879 milliseconds. A mean age does not specify a tail probability or surviving fidelity; use the finite-memory lab when a deadline or cutoff defines usefulness.

1. For each candidate, multiply independent stage yields, then the declared age eligibility exp(-delay/5). This is a generic survival flag, not an entangled-state fidelity.
2. Divide the shared attempt-resource budget 6,000 by the number of stages consumed per attempt. Multiplying by eligibility gives a resource-normalized rate.
3. The highest rate is 1,001.690998 per second through Two stages; the highest individual-attempt eligibility belongs to Three stages.

A longer route can improve per-attempt delivery while consuming enough shared capacity to lose on rate. The route comparison is an original stochastic workflow abstraction. Real entanglement routing needs link scheduling, state-quality composition and correlated-resource constraints that this multiplication does not describe.

1. For each candidate, multiply independent stage yields, then the declared age eligibility exp(-delay/15). This is a generic survival flag, not an entangled-state fidelity.
2. Divide the shared attempt-resource budget 6,000 by the number of stages consumed per attempt. Multiplying by eligibility gives a resource-normalized rate.
3. The highest rate is 1,307.812948 per second through Two stages; the highest individual-attempt eligibility belongs to Three stages.

A longer route can improve per-attempt delivery while consuming enough shared capacity to lose on rate. The route comparison is an original stochastic workflow abstraction. Real entanglement routing needs link scheduling, state-quality composition and correlated-resource constraints that this multiplication does not describe.

1. For each candidate, multiply independent stage yields, then the declared age eligibility exp(-delay/27). This is a generic survival flag, not an entangled-state fidelity.
2. Divide the shared attempt-resource budget 6,000 by the number of stages consumed per attempt. Multiplying by eligibility gives a resource-normalized rate.
3. The highest rate is 1,387.655311 per second through Two stages; the highest individual-attempt eligibility belongs to Three stages.

A longer route can improve per-attempt delivery while consuming enough shared capacity to lose on rate. The route comparison is an original stochastic workflow abstraction. Real entanglement routing needs link scheduling, state-quality composition and correlated-resource constraints that this multiplication does not describe.

1. Exactly one signal among m independent channels has probability m s (1-s)^(m-1). Require zero independent backgrounds by multiplying by (1-b)^m.
2. The scan over 1 to 80 channels peaks at m = 64, with clean-event probability 0.3588.
3. Any click has probability 0.631856, which also includes unwanted multiplets and backgrounds. Background-only frames account for 1.895079% of click frames.

The useful model rate is 358.800348 frames per second at 64 channels. More clicks do not necessarily mean more single-signal events. This calculation assumes the desired event class is physically meaningful; it does not assert that an ordinary detector can identify and reject every background-only frame.

1. Exactly one signal among m independent channels has probability m s (1-s)^(m-1). Require zero independent backgrounds by multiplying by (1-b)^m.
2. The scan over 1 to 80 channels peaks at m = 21, with clean-event probability 0.357002.
3. Any click has probability 0.639221, which also includes unwanted multiplets and backgrounds. Background-only frames account for 3.046199% of click frames.

The useful model rate is 1,249.505799 frames per second at 21 channels. More clicks do not necessarily mean more single-signal events. This calculation assumes the desired event class is physically meaningful; it does not assert that an ordinary detector can identify and reject every background-only frame.

1. Exactly one signal among m independent channels has probability m s (1-s)^(m-1). Require zero independent backgrounds by multiplying by (1-b)^m.
2. The scan over 1 to 80 channels peaks at m = 11, with clean-event probability 0.362716.
3. Any click has probability 0.625885, which also includes unwanted multiplets and backgrounds. Background-only frames account for 3.318567% of click frames.

The useful model rate is 2,357.654505 frames per second at 11 channels. More clicks do not necessarily mean more single-signal events. This calculation assumes the desired event class is physically meaningful; it does not assert that an ordinary detector can identify and reject every background-only frame.

1. Sum the binomial loading distribution from 12 successes through 20. The chance of enough atoms is 0.595599.
2. Assume exactly 12 independent required moves, each surviving with probability 0.99. Their joint survival is 0.886385.
3. Multiply the independent loading and movement stages to obtain 0.52793. Search 12 through 92 sites for the declared target 0.75.

The tested minimum is 24 loading sites. Extra sites fix loading shortages but cannot remove the movement ceiling 0.886385. Real rearrangement may require a variable number of moves; then its movement survival must be conditioned on the actual loading pattern.

1. Sum the binomial loading distribution from 22 successes through 35. The chance of enough atoms is 0.92533.
2. Assume exactly 22 independent required moves, each surviving with probability 0.9885. Their joint survival is 0.77533.
3. Multiply the independent loading and movement stages to obtain 0.717437. Search 22 through 102 sites for the declared target 0.75.

The tested minimum is 37 loading sites. Extra sites fix loading shortages but cannot remove the movement ceiling 0.77533. Real rearrangement may require a variable number of moves; then its movement survival must be conditioned on the actual loading pattern.

1. Sum the binomial loading distribution from 34 successes through 53. The chance of enough atoms is 0.999997.
2. Assume exactly 34 independent required moves, each surviving with probability 0.9867. Their joint survival is 0.634299.
3. Multiply the independent loading and movement stages to obtain 0.634298. Search 34 through 114 sites for the declared target 0.75.

No tested loading-site count reaches the target. Extra sites fix loading shortages but cannot remove the movement ceiling 0.634299. Real rearrangement may require a variable number of moves; then its movement survival must be conditioned on the actual loading pattern.

1. The installed idle load is fixed heat + channels × idle heat = 34.4 microwatts. All installed channels contribute even while their group waits.
2. Each active channel adds 2.4 microwatts above that idle load. Floor the remaining margin divided by the increment, then cap at 24 installed channels.
3. Read 2 groups in sequence and add other operations: 20 microseconds.

This is a steady-load scheduling abstraction. It illustrates why additional wires can add idle heat while providing little extra simultaneous capacity. Pulsed thermal transients, unequal channels, heat sinking and maximum permissible temperatures require a real thermal analysis before applying the result to a device.

1. The installed idle load is fixed heat + channels × idle heat = 53 microwatts. All installed channels contribute even while their group waits.
2. Each active channel adds 2.85 microwatts above that idle load. Floor the remaining margin divided by the increment, then cap at 44 installed channels.
3. Read 3 groups in sequence and add other operations: 24 microseconds.

This is a steady-load scheduling abstraction. It illustrates why additional wires can add idle heat while providing little extra simultaneous capacity. Pulsed thermal transients, unequal channels, heat sinking and maximum permissible temperatures require a real thermal analysis before applying the result to a device.

1. The installed idle load is fixed heat + channels × idle heat = 83.24 microwatts. All installed channels contribute even while their group waits.
2. Each active channel adds 3.39 microwatts above that idle load. Floor the remaining margin divided by the increment, then cap at 68 installed channels.
3. Read 6 groups in sequence and add other operations: 36 microseconds.

This is a steady-load scheduling abstraction. It illustrates why additional wires can add idle heat while providing little extra simultaneous capacity. Pulsed thermal transients, unequal channels, heat sinking and maximum permissible temperatures require a real thermal analysis before applying the result to a device.

1. Multiply channels × samples × bytes per complex sample to obtain 16,384 raw bytes. MB uses 1,000,000 bytes in this case.
2. Divide by link bandwidth: transfer requires 0.08192 milliseconds per shot. The other stage times are 0.2 and 0.4 milliseconds.
3. A perfectly overlapped three-stage pipeline fills once, then completes one shot per slowest-stage interval. Total = sum(stage times) + (shots - 1) × max(stage times) = 0.400282 seconds.

The pipeline model shortens this batch from 0.68192 to 0.400282 seconds. It assumes distinct resources, sufficient buffers and no next-shot dependence on the previous result. Mid-circuit feedback can impose dependencies that prevent this overlap. A compressed result may omit waveform information needed for later calibration checks.

1. Multiply channels × samples × bytes per complex sample to obtain 71,680 raw bytes. MB uses 1,000,000 bytes in this case.
2. Divide by link bandwidth: transfer requires 0.28672 milliseconds per shot. The other stage times are 0.25 and 0.65 milliseconds.
3. A perfectly overlapped three-stage pipeline fills once, then completes one shot per slowest-stage interval. Total = sum(stage times) + (shots - 1) × max(stage times) = 0.650537 seconds.

The pipeline model shortens this batch from 1.18672 to 0.650537 seconds. It assumes distinct resources, sufficient buffers and no next-shot dependence on the previous result. Mid-circuit feedback can impose dependencies that prevent this overlap. A compressed result may omit waveform information needed for later calibration checks.

1. Multiply channels × samples × bytes per complex sample to obtain 186,368 raw bytes. MB uses 1,000,000 bytes in this case.
2. Divide by link bandwidth: transfer requires 0.745472 milliseconds per shot. The other stage times are 0.31 and 0.95 milliseconds.
3. A perfectly overlapped three-stage pipeline fills once, then completes one shot per slowest-stage interval. Total = sum(stage times) + (shots - 1) × max(stage times) = 0.951055 seconds.

The pipeline model shortens this batch from 2.005472 to 0.951055 seconds. It assumes distinct resources, sufficient buffers and no next-shot dependence on the previous result. Mid-circuit feedback can impose dependencies that prevent this overlap. A compressed result may omit waveform information needed for later calibration checks.

1. For one bosonic mode, n(T) = 1/(exp(h f / kT) - 1). The stipulated input temperature gives occupation 1,562.246488.
2. Each ideal matched attenuator mixes transmitted input noise with its own thermal noise: n_out = eta n_in + (1 - eta)n_stage. The two stages give 174.531205 and 0.174599.
3. Classical incident signal power dissipates as P_in(1 - eta) at each stage, after accounting for preceding attenuation. The signal-power balance sums to 10 microwatts.

Moving attenuation changes both the final noise occupation and where the signal power is dissipated. This calculation accounts only for the declared traveling mode and incident signal. It omits reflections, broadband thermal transport, cable conduction and nonlinear components, so it is an explanatory balance rather than a refrigerator wiring design.

1. For one bosonic mode, n(T) = 1/(exp(h f / kT) - 1). The stipulated input temperature gives occupation 1,249.697214.
2. Each ideal matched attenuator mixes transmitted input noise with its own thermal noise: n_out = eta n_in + (1 - eta)n_stage. The two stages give 28.509523 and 0.285162.
3. Classical incident signal power dissipates as P_in(1 - eta) at each stage, after accounting for preceding attenuation. The signal-power balance sums to 15 microwatts.

Moving attenuation changes both the final noise occupation and where the signal power is dissipated. This calculation accounts only for the declared traveling mode and incident signal. It omits reflections, broadband thermal transport, cable conduction and nonlinear components, so it is an explanatory balance rather than a refrigerator wiring design.

1. For one bosonic mode, n(T) = 1/(exp(h f / kT) - 1). The stipulated input temperature gives occupation 1,007.723589.
2. Each ideal matched attenuator mixes transmitted input noise with its own thermal noise: n_out = eta n_in + (1 - eta)n_stage. The two stages give 13.576839 and 2.151841.
3. Classical incident signal power dissipates as P_in(1 - eta) at each stage, after accounting for preceding attenuation. The signal-power balance sums to 21 microwatts.

Moving attenuation changes both the final noise occupation and where the signal power is dissipated. This calculation accounts only for the declared traveling mode and incident signal. It omits reflections, broadband thermal transport, cable conduction and nonlinear components, so it is an explanatory balance rather than a refrigerator wiring design.

1. Generate y(L) = reference + a/L + b/L² at L = 6, 10, 16, 24. These are exact synthetic function values, not noisy observations.
2. Fit a straight line in 1/L while omitting curvature. Using all sizes gives intercept 0.582319; the largest two give 0.594792.
3. At the held-out size 40, the all-size fit predicts 0.630441 and the generating curve gives 0.63875. Their difference is -0.008309.

The known intercept is 0.6, so the fitted intercept error can be measured directly here. A small residual over the observed window does not guarantee an accurate extrapolation outside it. On a real experiment the infinite-size answer is unknown; varying the fit window and testing held-out sizes becomes evidence about model sensitivity, not proof of the limit.

1. Generate y(L) = reference + a/L + b/L² at L = 11, 15, 21, 29. These are exact synthetic function values, not noisy observations.
2. Fit a straight line in 1/L while omitting curvature. Using all sizes gives intercept 0.621764; the largest two give 0.636864.
3. At the held-out size 50, the all-size fit predicts 0.66398 and the generating curve gives 0.6752. Their difference is -0.01122.

The known intercept is 0.65, so the fitted intercept error can be measured directly here. A small residual over the observed window does not guarantee an accurate extrapolation outside it. On a real experiment the infinite-size answer is unknown; varying the fit window and testing held-out sizes becomes evidence about model sensitivity, not proof of the limit.

1. Generate y(L) = reference + a/L + b/L² at L = 17, 21, 27, 35. These are exact synthetic function values, not noisy observations.
2. Fit a straight line in 1/L while omitting curvature. Using all sizes gives intercept 0.682791; the largest two give 0.693915.
3. At the held-out size 62, the all-size fit predicts 0.714273 and the generating curve gives 0.723954. Their difference is -0.009681.

The known intercept is 0.71, so the fitted intercept error can be measured directly here. A small residual over the observed window does not guarantee an accurate extrapolation outside it. On a real experiment the infinite-size answer is unknown; varying the fit window and testing held-out sizes becomes evidence about model sensitivity, not proof of the limit.

1. The deliberately assumed objective is a/r^q + b r. Differentiate: -q a/r^(q+1) + b = 0, giving r* = 44.72136.
2. Scan positive integers up to 120. The minimum occurs at r = 45, where the terms are 0.008889 and 0.009.
3. The summed quantity is an illustrative error objective. Its coefficients must refer to a compatible error measure before terms can be physically added. It is not a general Trotter bound for an arbitrary Hamiltonian.

The numerical optimum is 45 steps. Increasing steps beyond it reduces one term while increasing the other. The cited product-formula literature motivates the existence of approximation error, but the additive linear implementation term and all coefficients are assumptions created for this decision exercise.

1. The deliberately assumed objective is a/r^q + b r. Differentiate: -q a/r^(q+1) + b = 0, giving r* = 13.700134.
2. Scan positive integers up to 120. The minimum occurs at r = 14, where the terms are 0.004592 and 0.0098.
3. The summed quantity is an illustrative error objective. Its coefficients must refer to a compatible error measure before terms can be physically added. It is not a general Trotter bound for an arbitrary Hamiltonian.

The numerical optimum is 14 steps. Increasing steps beyond it reduces one term while increasing the other. The cited product-formula literature motivates the existence of approximation error, but the additive linear implementation term and all coefficients are assumptions created for this decision exercise.

1. The deliberately assumed objective is a/r^q + b r. Differentiate: -q a/r^(q+1) + b = 0, giving r* = 13.214761.
2. Scan positive integers up to 120. The minimum occurs at r = 13, where the terms are 0.008876 and 0.0169.
3. The summed quantity is an illustrative error objective. Its coefficients must refer to a compatible error measure before terms can be physically added. It is not a general Trotter bound for an arbitrary Hamiltonian.

The numerical optimum is 13 steps. Increasing steps beyond it reduces one term while increasing the other. The cited product-formula literature motivates the existence of approximation error, but the additive linear implementation term and all coefficients are assumptions created for this decision exercise.

1. Fold the actual frequency into the interval from zero to fs/2: |((f + fs/2) mod fs) - fs/2| = 11 Hz.
2. Evaluate zero-phase cosines at t_k = k/fs for k = 0,...,31. The largest difference between actual and aliased sample values is 2.29816e-14, arising from floating-point arithmetic.
3. The spacing of the discrete Fourier transform bins is fs/N = 2.5 Hz. This is a grid spacing, not a universal resolution limit or a proof that adjacent lines are separable.

The 91 Hz and 11 Hz zero-phase cosines give the same ideal sample sequence. A better-fitting sampled curve cannot identify which continuous signal was present. Filtering, an independently justified bandwidth limit or a changed sampling rate supplies information absent from the original samples.

1. Fold the actual frequency into the interval from zero to fs/2: |((f + fs/2) mod fs) - fs/2| = 21 Hz.
2. Evaluate zero-phase cosines at t_k = k/fs for k = 0,...,39. The largest difference between actual and aliased sample values is 1.00031e-13, arising from floating-point arithmetic.
3. The spacing of the discrete Fourier transform bins is fs/N = 3.25 Hz. This is a grid spacing, not a universal resolution limit or a proof that adjacent lines are separable.

The 281 Hz and 21 Hz zero-phase cosines give the same ideal sample sequence. A better-fitting sampled curve cannot identify which continuous signal was present. Filtering, an independently justified bandwidth limit or a changed sampling rate supplies information absent from the original samples.

1. Fold the actual frequency into the interval from zero to fs/2: |((f + fs/2) mod fs) - fs/2| = 33 Hz.
2. Evaluate zero-phase cosines at t_k = k/fs for k = 0,...,39. The largest difference between actual and aliased sample values is 9.54792e-14, arising from floating-point arithmetic.
3. The spacing of the discrete Fourier transform bins is fs/N = 4.75 Hz. This is a grid spacing, not a universal resolution limit or a proof that adjacent lines are separable.

The 413 Hz and 33 Hz zero-phase cosines give the same ideal sample sequence. A better-fitting sampled curve cannot identify which continuous signal was present. Filtering, an independently justified bandwidth limit or a changed sampling rate supplies information absent from the original samples.

1. Complex128 amplitudes use 16 × 2^24 bytes = 0.25 GiB. A GiB is 2^30 bytes.
2. Workspace plus resident checkpoint copies multiply that state footprint by 2.5, giving 0.625 GiB per job. Floor available RAM divided by that footprint.
3. The 24 jobs require 1 waves, each taking 22 seconds.

Resident checkpoints can reduce concurrency at a discrete memory boundary, making the batch slower even before save time is counted. They may still be valuable for recovering failed work, but this no-failure schedule does not credit that benefit. Disk checkpoints, shared buffers, distributed simulation and restart probabilities need different accounting.

1. Complex128 amplitudes use 16 × 2^26 bytes = 1 GiB. A GiB is 2^30 bytes.
2. Workspace plus resident checkpoint copies multiply that state footprint by 4.5, giving 4.5 GiB per job. Floor available RAM divided by that footprint.
3. The 24 jobs require 1 waves, each taking 57 seconds.

Resident checkpoints can reduce concurrency at a discrete memory boundary, making the batch slower even before save time is counted. They may still be valuable for recovering failed work, but this no-failure schedule does not credit that benefit. Disk checkpoints, shared buffers, distributed simulation and restart probabilities need different accounting.

1. Complex128 amplitudes use 16 × 2^29 bytes = 8 GiB. A GiB is 2^30 bytes.
2. Workspace plus resident checkpoint copies multiply that state footprint by 4.5, giving 36 GiB per job. Floor available RAM divided by that footprint.
3. The 24 jobs require 8 waves, each taking 94.2 seconds.

Resident checkpoints can reduce concurrency at a discrete memory boundary, making the batch slower even before save time is counted. They may still be valuable for recovering failed work, but this no-failure schedule does not credit that benefit. Disk checkpoints, shared buffers, distributed simulation and restart probabilities need different accounting.

1. Subtract setup before counting complete runs: A fits 17; B fits 4. An unfinished run contributes no result.
2. For independent repeats, the probability all answers lie below a quality level is the single-run cumulative probability raised to the number of runs. This gives the distribution of the best answer.
3. The top-quality probabilities are 0.500413 and 0.7599. Summing the quality-level increments times their exceedance probabilities gives expected best values 0.850118 and 0.927387.

B has the higher expected best quality under the declared deadline. The quality scale, distributions and runtimes are invented. This comparison is meaningful only if both methods solve the same fixed instance and use the same scoring rule; repeated attempts must not silently receive new tuning information.

1. Subtract setup before counting complete runs: A fits 26; B fits 8. An unfinished run contributes no result.
2. For independent repeats, the probability all answers lie below a quality level is the single-run cumulative probability raised to the number of runs. This gives the distribution of the best answer.
3. The top-quality probabilities are 0.84845 and 0.983204. Summing the quality-level increments times their exceedance probabilities gives expected best values 0.954535 and 0.994961.

B has the higher expected best quality under the declared deadline. The quality scale, distributions and runtimes are invented. This comparison is meaningful only if both methods solve the same fixed instance and use the same scoring rule; repeated attempts must not silently receive new tuning information.

1. Subtract setup before counting complete runs: A fits 44; B fits 10. An unfinished run contributes no result.
2. For independent repeats, the probability all answers lie below a quality level is the single-run cumulative probability raised to the number of runs. This gives the distribution of the best answer.
3. The top-quality probabilities are 0.992775 and 0.999351. Summing the quality-level increments times their exceedance probabilities gives expected best values 0.997832 and 0.999805.

B has the higher expected best quality under the declared deadline. The quality scale, distributions and runtimes are invented. This comparison is meaningful only if both methods solve the same fixed instance and use the same scoring rule; repeated attempts must not silently receive new tuning information.

1. The ideal success after k iterations is sin²((2k+1)theta), where theta = asin(sqrt(0.015625)).
2. The declared attempt time is preparation + (2k+1) × oracle-equivalent time. The factor 2k+1 is this exercise’s accounting convention, not a universal decomposition of a Grover circuit.
3. At k = 6, 5 complete independent attempts fit. Their combined success is 1-(1-p)^attempts = 1.

The best tested campaign choice is k = 6; the best single-attempt probability occurs at k = 6. They optimize different quantities. A real advantage claim must also establish the oracle construction, state preparation, verification and fair classical search cost; this ideal rotation model supplies none of those performance facts.

1. The ideal success after k iterations is sin²((2k+1)theta), where theta = asin(sqrt(0.004464)).
2. The declared attempt time is preparation + (2k+1) × oracle-equivalent time. The factor 2k+1 is this exercise’s accounting convention, not a universal decomposition of a Grover circuit.
3. At k = 11, 2 complete independent attempts fit. Their combined success is 1-(1-p)^attempts = 0.999999.

The best tested campaign choice is k = 11; the best single-attempt probability occurs at k = 11. They optimize different quantities. A real advantage claim must also establish the oracle construction, state preparation, verification and fair classical search cost; this ideal rotation model supplies none of those performance facts.

1. The ideal success after k iterations is sin²((2k+1)theta), where theta = asin(sqrt(0.002404)).
2. The declared attempt time is preparation + (2k+1) × oracle-equivalent time. The factor 2k+1 is this exercise’s accounting convention, not a universal decomposition of a Grover circuit.
3. At k = 16, 2 complete independent attempts fit. Their combined success is 1-(1-p)^attempts = 0.999995.

The best tested campaign choice is k = 16; the best single-attempt probability occurs at k = 16. They optimize different quantities. A real advantage claim must also establish the oracle construction, state preparation, verification and fair classical search cost; this ideal rotation model supplies none of those performance facts.

1. Enumerate every subset of 8 tasks: 2^8 = 256 candidates. Keep only those whose total time does not exceed 35.
2. Maximize the declared value sum, breaking ties by less time. The selected tasks are 2, 4, taking 34 time units.
3. Input-order packing gives value 41; greedy value-per-time packing gives 51. The exact search gives 51.

This small classical calculation makes an experimental planning objective explicit. The learning values are invented preferences, not measured information gain or a ranking of scientific importance. Dependencies between experiments, uncertain duration and complementary results violate the additive model. No quantum optimization advantage is claimed.

1. Enumerate every subset of 8 tasks: 2^8 = 256 candidates. Keep only those whose total time does not exceed 50.
2. Maximize the declared value sum, breaking ties by less time. The selected tasks are 1, 2, 3, 5, 7, taking 49 time units.
3. Input-order packing gives value 87; greedy value-per-time packing gives 90. The exact search gives 90.

This small classical calculation makes an experimental planning objective explicit. The learning values are invented preferences, not measured information gain or a ranking of scientific importance. Dependencies between experiments, uncertain duration and complementary results violate the additive model. No quantum optimization advantage is claimed.

1. Enumerate every subset of 8 tasks: 2^8 = 256 candidates. Keep only those whose total time does not exceed 68.
2. Maximize the declared value sum, breaking ties by less time. The selected tasks are 1, 3, 4, 5, 7, 8, taking 68 time units.
3. Input-order packing gives value 130; greedy value-per-time packing gives 149. The exact search gives 149.

This small classical calculation makes an experimental planning objective explicit. The learning values are invented preferences, not measured information gain or a ranking of scientific importance. Dependencies between experiments, uncertain duration and complementary results violate the additive model. No quantum optimization advantage is claimed.

1. For one parameter, the estimator is (mean_plus - mean_minus)/2. Independent equal-variance shifted samples give variance v/(2n).
2. The expected squared Euclidean sampling error sums over 12 coordinates. Set parameters × v/(2n) <= target, giving 240 shots at each of the two shifts.
3. The active-subset estimate leaves known synthetic omitted squared norm 0.001. Only 0.009 remains for sampling error; if this is nonpositive, no shot count repairs the omission.

Measuring fewer coordinates can reduce work or consume the error budget through omitted directions. The omitted-gradient norm is provided by construction here; an actual optimizer normally does not know it for free. Parameter-shift identities require an appropriate gate generator and shift rule. Correlations within or between the two shifted sample means can change each coordinate variance; cross-coordinate covariance alone does not change the trace-based squared-error sum.

1. For one parameter, the estimator is (mean_plus - mean_minus)/2. Independent equal-variance shifted samples give variance v/(2n).
2. The expected squared Euclidean sampling error sums over 22 coordinates. Set parameters × v/(2n) <= target, giving 440 shots at each of the two shifts.
3. The active-subset estimate leaves known synthetic omitted squared norm 0.0055. Only 0.0045 remains for sampling error; if this is nonpositive, no shot count repairs the omission.

Measuring fewer coordinates can reduce work or consume the error budget through omitted directions. The omitted-gradient norm is provided by construction here; an actual optimizer normally does not know it for free. Parameter-shift identities require an appropriate gate generator and shift rule. Correlations within or between the two shifted sample means can change each coordinate variance; cross-coordinate covariance alone does not change the trace-based squared-error sum.

1. For one parameter, the estimator is (mean_plus - mean_minus)/2. Independent equal-variance shifted samples give variance v/(2n).
2. The expected squared Euclidean sampling error sums over 34 coordinates. Set parameters × v/(2n) <= target, giving 680 shots at each of the two shifts.
3. The active-subset estimate leaves known synthetic omitted squared norm 0.0109. Only -0.0009 remains for sampling error; if this is nonpositive, no shot count repairs the omission.

Measuring fewer coordinates can reduce work or consume the error budget through omitted directions. The omitted-gradient norm is provided by construction here; an actual optimizer normally does not know it for free. Parameter-shift identities require an appropriate gate generator and shift rule. Correlations within or between the two shifted sample means can change each coordinate variance; cross-coordinate covariance alone does not change the trace-based squared-error sum.

1. Test 300 interrogation times from T2/50 to 4 T2. At each time, floor campaign/(interrogation + dead time) to count complete cycles.
2. Use contrast C(t) = initial contrast × exp(-t/T2). At the selected time 0.339452, C = 0.511139 and 3081 cycles fit.
3. Near the most sensitive Ramsey fringe point, the stipulated independent projection-noise model gives frequency uncertainty 1/(2 pi C t sqrt(cycles × atoms)) = 0.001653 Hz.

Long interrogation increases phase accumulation while reducing visibility and the number of cycles. Dead time changes the balance. The grid optimum is a result of the declared local linear-response model; oscillator noise, phase wrapping, correlations and systematic shifts are excluded. It is not a measured clock accuracy or a universal optimum.

1. Test 300 interrogation times from T2/50 to 4 T2. At each time, floor campaign/(interrogation + dead time) to count complete cycles.
2. Use contrast C(t) = initial contrast × exp(-t/T2). At the selected time 0.838635, C = 0.422578 and 864 cycles fit.
3. Near the most sensitive Ramsey fringe point, the stipulated independent projection-noise model gives frequency uncertainty 1/(2 pi C t sqrt(cycles × atoms)) = 0.000817 Hz.

Long interrogation increases phase accumulation while reducing visibility and the number of cycles. Dead time changes the balance. The grid optimum is a result of the declared local linear-response model; oscillator noise, phase wrapping, correlations and systematic shifts are excluded. It is not a measured clock accuracy or a universal optimum.

1. Test 300 interrogation times from T2/50 to 4 T2. At each time, floor campaign/(interrogation + dead time) to count complete cycles.
2. Use contrast C(t) = initial contrast × exp(-t/T2). At the selected time 1.418488, C = 0.377374 and 467 cycles fit.
3. Near the most sensitive Ramsey fringe point, the stipulated independent projection-noise model gives frequency uncertainty 1/(2 pi C t sqrt(cycles × atoms)) = 0.00054 Hz.

Long interrogation increases phase accumulation while reducing visibility and the number of cycles. Dead time changes the balance. The grid optimum is a result of the declared local linear-response model; oscillator noise, phase wrapping, correlations and systematic shifts are excluded. It is not a measured clock accuracy or a universal optimum.

1. The true sensor fields are B ± G L/2. Multiply A by (1 + gain error), subtract B’s reading, then divide by separation.
2. The resulting bias is gB/L + gG/2 = 0.053393. The common field cancels only when the gains match.
3. Random variance of the difference is s_A² + s_B² - 2 rho s_A s_B. Dividing its square root by L gives 0.765942, and combining variance with squared bias gives RMSE 0.7678.

The common-field leakage contributes 0.053333 gradient units. A quiet difference can still be biased if gain mismatch converts a large shared field into an apparent gradient. Increasing separation changes both noise conversion and spatial interpretation; the model assumes a constant gradient across every tested baseline.

1. The true sensor fields are B ± G L/2. Multiply A by (1 + gain error), subtract B’s reading, then divide by separation.
2. The resulting bias is gB/L + gG/2 = 0.223217. The common field cancels only when the gains match.
3. Random variance of the difference is s_A² + s_B² - 2 rho s_A s_B. Dividing its square root by L gives 0.274048, and combining variance with squared bias gives RMSE 0.353451.

The common-field leakage contributes 0.222857 gradient units. A quiet difference can still be biased if gain mismatch converts a large shared field into an apparent gradient. Increasing separation changes both noise conversion and spatial interpretation; the model assumes a constant gradient across every tested baseline.

1. The true sensor fields are B ± G L/2. Multiply A by (1 + gain error), subtract B’s reading, then divide by separation.
2. The resulting bias is gB/L + gG/2 = 0.387161. The common field cancels only when the gains match.
3. Random variance of the difference is s_A² + s_B² - 2 rho s_A s_B. Dividing its square root by L gives 0.112428, and combining variance with squared bias gives RMSE 0.403154.

The common-field leakage contributes 0.386441 gradient units. A quiet difference can still be biased if gain mismatch converts a large shared field into an apparent gradient. Increasing separation changes both noise conversion and spatial interpretation; the model assumes a constant gradient across every tested baseline.

1. If all comparisons reflected consistent offsets, (B-A) + (C-B) - (C-A) would equal zero. The observed closure is 0 nanoseconds.
2. With independent edge errors, closure uncertainty is sqrt(s_AB²+s_BC²+s_AC²) = 0.948683. The standardized closure is 0.
3. Fix clock A as the arbitrary zero and solve weighted least squares for B and C. The fitted offsets are 4 and 7; residuals are 0, -8.88178e-16, -8.88178e-16.

The loop detects inconsistent comparisons but cannot identify the responsible clock or link by itself. A shared calibration offset can evade the loop entirely, and correlated transfer errors change its uncertainty. Fixing A to zero is a coordinate choice, not evidence that clock A is accurate against an external time standard.