Applying Sample-Size Floors, Caps, And Rounding

From a Recalculated Requirement to an Enrollable Total

A sample-size re-estimation can produce a numerical requirement that is not itself an assignable enrollment. A calculation might call for 301.2 participants, but participants are enrolled as whole people, and a trial with a fixed allocation ratio may only permit certain totals. Cluster trials can impose an additional restriction: participants arrive in groups whose sizes and allocation must also be feasible. Minimum enrollment rules and maximum enrollment caps further restrict the choice.

The task here is not to redo the nuisance-parameter calculation. It is to translate its output into an enrollment that can be implemented, without silently rounding down, changing the allocation ratio, or pretending that a binding cap still meets the recalculated requirement. The examples use hypothetical trial-planning values.

Definition: Let $n_{\mathrm{req}}$ denote the recalculated required total number of randomized participants before integer and design constraints are applied. A feasible enrollment is an integer total that satisfies the prespecified allocation and unit constraints, as well as any applicable minimum and maximum. A feasible enrollment meets the recalculated requirement if it is at least $n_{\mathrm{req}}$, with the same design assumptions used to obtain that requirement.

This distinction matters: $n_{\mathrm{req}}$ describes what the calculation calls for, whereas the final planned enrollment describes what the design permits. If these differ because of a cap or structural constraint, report both.

Allocation Ratios Create a Set of Feasible Totals

Suppose two arms must be allocated in the ratio $a:b$, where $a$ and $b$ are positive integers. Write $g$ for their greatest common divisor and define the reduced allocation unit as

u=\left(\frac{a}{g},\frac{b}{g}\right), \qquad L=\frac{a}{g}+\frac{b}{g}.

An exact allocation ratio is achieved by assigning $k(a/g)$ participants to the first arm and $k(b/g)$ to the second, for some nonnegative integer $k$. Thus, the total must be a multiple of $L$. For a 2:1 ratio, for example, feasible totals are multiples of 3, with two-thirds assigned to the first arm and one-third to the second.

Key result: If only the total and a fixed two-arm ratio $a:b$ constrain allocation, the smallest feasible total at least $n_{\mathrm{req}}$ is $$ n_{\mathrm{up}}=L\left\lceil \frac{n_{\mathrm{req}}}{L}\right\rceil. $$ It is the smallest feasible total that does not fall below the recalculated requirement. This result assumes the requirement can be met by an enrollment at or above $n_{\mathrm{req}}$, under the same design model.

Here $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$. The formula rounds to the next allocation unit, not merely to the next whole participant. If the ratio is 1:1, the unit is 2 and the total must be even. If the ratio is 2:1, the unit is 3. A trial may have additional arm-specific restrictions, so the set of feasible totals must reflect all applicable constraints, not just this basic ratio.

A minimum and a cap should be applied to the feasible totals, rather than by changing the rounding rule. Let $n_{\min}$ be the minimum total and $n_{\max}$ the maximum total. A feasible total $n$ must satisfy

n_{\min}\leq n\leq n_{\max}.

If there is a feasible total at least $n_{\mathrm{req}}$ within these bounds, choose the smallest such total when the aim is to meet the requirement without unnecessary expansion. If none exists, the requirement and the constraints are incompatible. In particular, the largest feasible total under the cap may be smaller than $n_{\mathrm{req}}$. The cap does not make that smaller total adequate; it makes the recalculated requirement unattainable under the stated constraints.

Worked Example: Rounding a 2:1 Allocation Within Bounds

Suppose a hypothetical blinded re-estimation gives a required randomized total of $n_{\mathrm{req}}=301.2$. The trial uses a fixed 2:1 allocation, with a minimum total of 270 and a cap of 315. Find the smallest feasible enrollment that meets the recalculated requirement.

The allocation unit has $L=2+1=3$ participants. The smallest multiple of 3 that is at least 301.2 is

n_{\mathrm{up}}=3\left\lceil\frac{301.2}{3}\right\rceil =3\lceil100.4\rceil =3(101)=303.

The minimum and cap are both compatible with this choice: $270\leq303\leq315$. The 303 participants are allocated as $2(101)=202$ to the first arm and $1(101)=101$ to the second. The ratio is exactly 2:1, and the total exceeds the recalculated requirement by 1.8 participants, which is the smallest feasible increase under these constraints.

Simply rounding 301.2 to 302 would not preserve the 2:1 ratio, since 302 is not divisible by 3. Rounding to 303 is not arbitrary inflation; it is the least upward adjustment that permits the specified allocation.

Applying Floors and Caps Without Hiding a Conflict

A floor and a cap define an allowed interval, but allocation constraints can leave gaps inside that interval. Therefore, checking only whether $n_{\mathrm{req}}$ lies between the minimum and maximum is not enough. The correct check is whether the allowed interval contains a feasible total that is at least the requirement.

For a fixed allocation unit $L$, the largest feasible total not exceeding the cap is

n_{\mathrm{cap}}=L\left\lfloor\frac{n_{\max}}{L}\right\rfloor,

where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$. If $n_{\mathrm{cap}}<n_{\min}$, no total satisfies both the bounds and the allocation ratio. If $n_{\mathrm{cap}}\geq n_{\min}$ but $n_{\mathrm{cap}}<n_{\mathrm{req}}$, the cap prevents the requirement from being reached. If $n_{\mathrm{cap}}\geq n_{\mathrm{req}}$, the upward-rounded requirement can be checked against the floor and cap to select the smallest feasible total.

For designs with more than two arms, or with arm-specific minima or maxima, calculate the feasible allocations directly. A total might be a multiple of the sum of allocation units but still fail an arm-specific constraint. The general principle remains the same: identify the allowable integer allocations first, then choose the smallest one meeting the requirement if such an allocation exists.

Worked Example: When the Cap Makes the Requirement Unattainable

Suppose a hypothetical recalculation gives $n_{\mathrm{req}}=348.6$ randomized participants for a 1:1 trial. The prespecified minimum is 300 and the maximum is 340. Determine the largest feasible enrollment and whether it meets the recalculated requirement.

For a 1:1 ratio, $L=2$, so totals must be even. The largest even total not exceeding the cap is

n_{\mathrm{cap}}=2\left\lfloor\frac{340}{2}\right\rfloor =2(170)=340.

This total satisfies the minimum and cap, and it assigns 170 participants to each arm. But $340<348.6$, so no feasible enrollment under the stated cap can meet the recalculated requirement. Rounding the requirement upward would give an even total of 350, which exceeds the cap. The correct report is that the largest permitted allocation is 340, with 170 per arm, and that it is below the recalculated requirement; it is not correct to describe 340 as satisfying that requirement.

Cluster-Size Constraints

For a cluster-randomized trial, the enrollment unit may be a cluster rather than an individual participant. If every cluster contains exactly $q$ participants and clusters are allocated in a fixed ratio $a:b$, then each allocation unit consists of $a/g$ clusters in the first arm and $b/g$ in the second. Each such unit contains $qL$ participants. Consequently, feasible total participant counts are multiples of $qL$, provided all clusters have the same size $q$.

This calculation assumes the stated cluster size is fixed and that the ratio applies to the number of clusters. If cluster sizes vary, or if the trial permits partial clusters or different cluster sizes by arm, the simple multiple-of-$qL$ rule may not apply. In those settings, determine feasible cluster counts and participant totals from the actual enrollment rules rather than rounding as if every participant were independently assignable.

If the recalculated requirement is expressed as a number of analyzable participants but the protocol enrolls additional participants to account for expected attrition, convert to a randomized-participant requirement before applying the integer and cluster constraints. For a common anticipated attrition proportion $a_{\mathrm{loss}}$, a basic planning conversion is

n_{\mathrm{rand,req}}=\frac{n_{\mathrm{analyzable,req}}}{1-a_{\mathrm{loss}}}.

This conversion relies on the specified attrition assumption and is not itself a rounding rule. Apply the relevant allocation and cluster constraints to the resulting randomized requirement. If attrition differs by arm, a single overall inflation may not preserve the intended analyzable allocation; the arm-specific planning requirements need to be checked.

Worked Example: Rounding to Complete Cluster Allocation Units

A hypothetical cluster trial requires 430 randomized participants. Clusters contain exactly 12 participants, and the trial assigns clusters in a 1:1 ratio. The permitted total is between 360 and 420 participants. Find the largest feasible enrollment under the cap and determine whether it meets the requirement.

One allocation unit contains one cluster per arm, or 24 participants in total. Feasible totals are therefore multiples of 24. Rounding the requirement upward gives

24\left\lceil\frac{430}{24}\right\rceil =24\lceil17.9167\rceil =24(18)=432.

The 432-participant enrollment would require 36 clusters, 18 per arm, but it exceeds the cap of 420. The largest feasible total below the cap is

24\left\lfloor\frac{420}{24}\right\rfloor =24\lfloor17.5\rfloor =24(17)=408.

Thus the largest permitted design has 34 clusters, with 17 clusters per arm and 204 participants per arm. It satisfies the minimum and cap, but $408<430$, so it does not meet the recalculated requirement. Although 420 is below the cap, it is not an attainable total under these cluster and allocation constraints.

Why the Rounding Rule Works, and Where It Can Fail

The upward-rounding result follows directly from the structure of the feasible totals. With allocation unit $L$, every feasible total has the form $kL$ for an integer $k$. To meet the requirement it must satisfy $kL\geq n_{\mathrm{req}}$, or $k\geq n_{\mathrm{req}}/L$. The smallest integer $k$ with that property is $\lceil n_{\mathrm{req}}/L\rceil$. Multiplying by $L$ gives the smallest feasible total at or above the requirement. For fixed cluster size $q$, replace $L$ by $qL$ when total participants are counted.

The derivation also shows the assumptions behind the rule: the feasible totals really must form the stated multiples, the recalculated number must use the same unit as the enrollment being rounded, and a total at or above the requirement must remain sufficient under the planning criterion. If any of these conditions fails, the formula alone cannot certify a valid enrollment.

Check the calculation for practical warning signs. A proposed allocation may not sum to the proposed total, may not satisfy the stated ratio, may fall below the floor, or may exceed the cap after upward rounding. In a cluster trial, check both the number of clusters and the number of participants; a participant total that looks acceptable may not correspond to an assignable number of complete clusters. Also verify whether the recalculated requirement refers to randomized or analyzable participants before applying any rounding.

The main alternative is to round the total to the nearest integer and divide it between arms. That can be acceptable only when the design permits the resulting integer allocation and the rounded total still meets the requirement. Under an exact ratio or fixed cluster structure, it can fail either condition. Another alternative is to use the largest total allowed by a cap when the upward-rounded requirement exceeds it. That may be necessary operationally, but it is a constrained design choice, not a mathematical resolution of the shortfall. Report the gap and do not claim that the cap-limited total achieves the recalculated requirement.

The rounding step has limits. It does not validate the nuisance estimates, the original sample-size formula, or the assumptions about attrition and cluster size. Nor does rounding upward guarantee a desired operating characteristic if the underlying calculation or design model is inappropriate. Its role is narrower: given a requirement and clearly stated enrollment rules, identify an implementable integer design and make any incompatibility visible.

Common Mistake

Common mistake: “The calculation gives 301.2, so 302 is close enough; if the cap is 300, use 300 and report the planned sample size.” This reasoning can break the allocation ratio, round below the calculated requirement, and conceal a binding cap. Instead, find the smallest total that is feasible under the allocation and cluster rules and is at least the requirement. Then check the floor and cap. If no such total exists, report the largest feasible permitted enrollment and state plainly that it falls short.

Check Your Understanding

For each question, identify the feasible enrollment and explain whether it meets the stated requirement. Solutions follow each question.

A hypothetical 1:1 trial has $n_{\mathrm{req}}=257.1$, a floor of 220, and a cap of 280. What is the smallest feasible total that meets the requirement, and how many participants are assigned to each arm?
A hypothetical 3:2 trial requires 401 participants. What is the smallest total that preserves the exact allocation ratio and meets the requirement? Give the arm counts.
A hypothetical cluster trial has 10 participants per cluster, a 1:1 cluster allocation, a required total of 305 participants, and a cap of 320. What is the largest feasible total under the cap, and does it meet the requirement?
Explain why a cap-limited enrollment below $n_{\mathrm{req}}$ should not be described as meeting the recalculated requirement, even if it is the largest enrollment the trial can conduct.

Solutions. (1) Feasible totals are even. The smallest even total at least 257.1 is 258, which is within the bounds; assign 129 participants per arm. (2) The allocation unit has $L=3+2=5$, so $5\lceil401/5\rceil=5(81)=405$. The arm counts are $3(81)=243$ and $2(81)=162$. (3) One cluster per arm gives 20 participants per allocation unit. The largest multiple of 20 at or below 320 is 320, which is above 305; it uses 16 clusters total, 8 per arm, and meets the stated total requirement. (4) The requirement is the calculated total needed under the stated planning assumptions. A smaller enrollment may be operationally necessary, but it does not reach that calculated total; the shortfall must be reported rather than hidden by the cap.