Planning Under a Recruitment Budget

Why a recruitment budget needs more than one number

A MAMS design may recruit to later stages only for arms that continue after interim decisions. Its total recruitment is therefore uncertain before the trial begins. A planning budget can refer to expected recruitment, averaged over possible stopping scenarios, or to maximum capacity, the number of participants that would be needed if every arm continued. These quantities answer different operational questions. An expected total can help forecast typical resource use; a maximum total helps determine whether the trial can be completed even in the largest recruitment scenario.

The distinction matters especially with shared control. If any experimental arm continues, the design may recruit the next-stage control group, even when other arms have stopped. Counting a full control group for each continuing arm would overstate recruitment when control participants are shared; omitting control recruitment whenever one arm stops can understate it.

Definition: Let $N_s$ be the total number of participants recruited under scenario $s$, and let $w_s$ be that scenario's probability, with $w_s\geq 0$ and $\sum_s w_s=1$. The expected enrollment is $E(N)=\sum_s w_sN_s$. The maximum enrollment across the specified scenarios is $N_{\max}=\max_s N_s$. An expected-enrollment budget constrains $E(N)$; a hard capacity limit constrains $N_{\max}$.

The scenario probabilities must refer to a clearly defined set of assumptions about treatment effects, outcome variability, interim decision rules, and other design features. They need not imply that scenario $s$ is a literal state of nature known in advance: in planning, they can be weights used to summarize resource use across specified scenarios. The calculation does not itself establish that the design has acceptable power, type I error, or other operating characteristics. Those design constraints must still be checked separately.

Count recruitment by stage and stopping pattern

For each stage, specify the number recruited to each experimental arm and to control. Let $n_{jk}$ be the stage-$k$ increment for experimental arm $j$, and let $n_{0k}$ be the control increment. Define $I_{jk}=1$ if arm $j$ is active for recruitment at stage $k$, and $I_{jk}=0$ otherwise. With a shared control recruited whenever at least one experimental arm is active, define $I_{0k}=1$ if $\sum_j I_{jk}>0$, and $I_{0k}=0$ otherwise. Then recruitment at stage $k$ is

N_k=n_{0k}I_{0k}+\sum_j n_{jk}I_{jk}.

The total enrollment is the sum across stages. In particular, stage 1 is commonly recruited for every arm, while later increments depend on the interim stopping pattern. Taking expectations of the stage indicators gives a convenient calculation:

E(N)=\sum_k\left[n_{0k}P(I_{0k}=1)+\sum_j n_{jk}P(I_{jk}=1)\right].

This result follows from linearity of expectation: the expected value of each fixed increment multiplied by its recruitment indicator is that increment times the probability that the indicator equals one. No independence assumption among the arms' continuation indicators is needed for this formula. However, dependence does matter when calculating the probability that shared control is recruited from separate arm-specific continuation probabilities. That union probability must be derived from a joint distribution or from an explicit dependence assumption.

Worked Example: Expected Enrollment With Shared Control

Consider a hypothetical two-stage design with experimental arms A and B and a shared control. Stage 1 recruits 30 participants to each of the three groups, for 90 participants. At stage 2, each active experimental arm recruits 24 participants, and a shared control group of 24 is recruited if at least one arm continues. Suppose each arm has probability 0.60 of continuing, and the two continuation decisions are independent under the planning scenario.

The probability that neither arm continues is $0.40\times0.40=0.16$. Thus the probability of recruiting the stage-2 shared control is

P(I_{02}=1)=1-P(\text{neither continues})=1-(1-0.60)^2=1-0.16=0.84.

The expected stage-2 enrollment is 24 times 0.84 for control, plus 24 times 0.60 for each of the two experimental arms:

$$ E(N_2)=24(0.84)+24(0.60)+24(0.60)=20.16+14.40+14.40=48.96. $$

Therefore $E(N)=90+48.96=138.96$ participants. This fractional result is an expected value, not a claim that a realized trial enrolls a fraction of a participant. If both arms continue, the maximum is $90+24+24+24=162$ participants. The design's expected enrollment is below its maximum because some scenarios stop one or both arms before stage 2.

Compare each budget with the quantity it constrains

A budget stated as “no more than 145 participants on average” is an expected-use constraint: the relevant comparison is $E(N)\leq145$. It does not guarantee that every realization will be at most 145. A site or funder that cannot recruit more than 145 participants under any permitted stopping pattern has a hard capacity constraint: the relevant comparison is $N_{\max}\leq145$. The same design can satisfy the first and fail the second.

If costs differ by stage or treatment group, replace participant counts with costs. Let $c_{jk}$ be the cost per participant in group $j$, stage $k$, including control as $j=0$, and let $F$ be any fixed cost. Scenario cost is

C_s=F+\sum_k\left[c_{0k}n_{0k}I_{0k,s}+\sum_j c_{jk}n_{jk}I_{jk,s}\right].

Expected cost is $\sum_s w_sC_s$, while maximum cost is $\max_s C_s$. Fixed costs are included in every scenario, so they are added once to each scenario cost, not multiplied by the number of stages. Where a monetary budget is meant to guarantee completion, compare it with maximum cost rather than expected cost.

Worked Example: An Expected Cost Budget and a Capacity Limit

Use the hypothetical design above. Suppose stage-1 participants cost $700 each in control and $900 each in either experimental arm. At stage 2, the same per-participant costs apply. Assume a fixed trial cost of $10,000. The stage-1 variable cost is

30(700)+2[30(900)]=21{,}000+54{,}000=\$75{,}000.

Under the stated independent continuation probabilities, expected stage-2 control recruitment is $24(0.84)=20.16$, while expected stage-2 recruitment in the two experimental arms is $24(0.60)=14.40$ in each arm. Thus expected stage-2 variable cost is

20.16(700)+2[14.40(900)]=14{,}112+25{,}920=\$40{,}032.

Expected total cost, including fixed cost, is $75{,}000+40{,}032+10{,}000=\$125{,}032$. A $130,000 expected-cost budget is met. But if both arms continue, stage-2 variable cost is $24(700)+2[24(900)]=16{,}800+43{,}200=\$60{,}000$, so maximum total cost is $75{,}000+60{,}000+10{,}000=\$145{,}000$. A hard $130,000 cost cap is not met. The conclusion depends on the budget's meaning: the design meets the expected-cost limit but cannot guarantee staying under the same amount.

A systematic check for candidate stage sizes

When several candidate stage-size sets have already been developed, evaluate each against the same scenarios, probabilities, and budget definition. The scenario weights should be applied consistently across candidates. The following calculation sequence reduces the risk of mixing maximum and expected enrollment or counting shared control incorrectly.

Specify increments and recruitment rules.
Record each stage's group-specific sample sizes and state exactly when shared control is recruited.

Enumerate stopping patterns.
For each pattern, calculate the actual total recruitment or cost. Do not treat a stopped arm as recruiting later increments.

Calculate expected use and maximum capacity separately.
Weight scenario totals by their probabilities for an expectation; take the largest scenario total for the maximum.

Apply the stated budget rule.
Compare expected use with an expected-use budget, and maximum use with a hard cap. Retain all statistical design constraints as separate requirements.

Worked Example: Comparing Candidate Stage Sizes

Suppose the same two-arm stopping assumptions apply: each arm independently continues with probability 0.60, and stage-2 shared control is recruited if either continues. Consider candidate designs with equal stage-1 group sizes $a$, and equal stage-2 increments $b$ per active group, including the shared control increment. Their expected total enrollment is

$$ E(N)=3a+b(0.84+0.60+0.60)=3a+2.04b, $$

and their maximum is $N_{\max}=3a+3b$, when both arms continue. For candidate 1, $a=25$ and $b=20$: $E(N)=3(25)+2.04(20)=75+40.8=115.8$, and $N_{\max}=75+60=135$. For candidate 2, $a=28$ and $b=22$: $E(N)=3(28)+2.04(22)=84+44.88=128.88$, and $N_{\max}=84+66=150$. For candidate 3, $a=30$ and $b=24$: $E(N)=90+48.96=138.96$, and $N_{\max}=90+72=162$.

Under an expected-enrollment budget of 130 participants, candidates 1 and 2 meet the budget, while candidate 3 does not. Under a hard cap of 150 participants, candidates 1 and 2 meet the cap, while candidate 3 does not. These are resource-feasibility conclusions only. They do not show that candidate 1, 2, or 3 achieves a desired power or preserves other design properties; those must be checked before selecting a design.

Diagnostics, alternatives, and limitations

A useful diagnostic is to compare the expected total with the scenario range. With nonnegative scenario probabilities summing to one, $E(N)$ must lie between the smallest and largest scenario totals. If a calculated expectation is outside that range, check the scenario weights, stage increments, and whether a shared-control increment has been included correctly. Also verify that stage-1 recruitment appears in every scenario when the design requires it, and that later enrollment is zero for arms that have stopped.

Check the union probability for shared control explicitly. Under independence, if two arms each continue with probability 0.60, the probability that at least one continues is $1-(1-0.60)^2=0.84$, not 0.60 and not 0.64. If the decisions are dependent, this independence calculation is not justified; use the joint stopping-pattern probabilities or the probability of neither continuing. A scenario table is often the clearest audit trail when there are few arms or when stopping decisions are dependent.

Expected-use planning is often more informative than a maximum alone when later recruitment is unlikely in many scenarios, while a maximum is essential for guaranteed capacity. Neither replaces the other. A risk-averse operational plan might use an expected-use target for budgeting and a separate contingency plan for high-recruitment scenarios. A further alternative is to report selected scenario quantiles, but these depend on a specified probability model and do not provide the guarantee of a hard cap.

The main limitation is that expected resource use is only as credible as the scenario probabilities and recruitment assumptions. If continuation probabilities are misspecified, the expected total can be misleading. Moreover, a budget feasibility calculation is not a design-selection rule by itself: a smaller candidate might cost less but fail the trial's statistical requirements. Expected enrollment also does not describe the variability of enrollment; two designs can have the same expectation but very different maximums or scenario distributions. Assessment of that variability belongs in a fuller scenario analysis.

Key takeaway: Calculate scenario-specific enrollment or cost first. Use probability-weighted totals for expected resource use and the largest relevant scenario for maximum capacity. State which quantity the budget constrains, and evaluate statistical design requirements separately.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

Two arms independently continue with probabilities 0.70 and 0.50. What is the probability that a shared-control stage is recruited?
A design has scenario totals of 100, 140, and 180 participants with probabilities 0.20, 0.50, and 0.30. Find expected and maximum enrollment. Does it meet an expected budget of 150?
In the preceding scenario set, would meeting an expected budget of 150 guarantee a hard cap of 150? Explain.
A two-arm design recruits 20 participants per group at stage 1. At stage 2 it recruits 10 per active experimental arm and 10 shared controls if any arm continues. If each arm independently continues with probability 0.40, calculate expected total enrollment.

Solutions. (1) The probability neither continues is $(1-0.70)(1-0.50)=0.30(0.50)=0.15$, so shared control is recruited with probability $1-0.15=0.85$. (2) Expected enrollment is $0.20(100)+0.50(140)+0.30(180)=20+70+54=144$ participants; the maximum is 180. Since $144\leq150$, the expected budget is met. (3) No. The 180-participant scenario exceeds 150, so the expected-budget result does not guarantee the hard cap. (4) Stage 1 enrolls $3(20)=60$. The probability of any continuation is $1-(0.60)^2=0.64$, and expected stage-2 recruitment is $10(0.64)+2[10(0.40)]=6.4+8=14.4$. Expected total enrollment is $60+14.4=74.4$ participants.