Planning the Whole Design
A MAMS enrollment plan has to work as a whole. Stage sizes determine the maximum number of participants, while interim decisions determine how often the trial reaches later stages and how many experimental arms remain. Scenario weights then summarize expected enrollment across plausible treatment-effect patterns. A plan with attractive expected enrollment is not suitable if it misses its power target or exceeds the recruitment limit.
This workshop combines those quantities for a hypothetical trial with two experimental arms and a shared control. The aim is not to find a universally best allocation: it is to make the decision criteria explicit and select a plan that satisfies them.
The hypothetical primary outcome is continuous, normally distributed, and measured in outcome units. For each participant, its variance is \(\sigma^2=100\), so the standard deviation is 10 outcome units. The planning target is a mean difference of 5 outcome units for an effective treatment. The one-sided familywise type-I error rate is \(\alpha=0.025\). Interim stopping is for futility only; final treatment claims use a prespecified stagewise combination test, with its final critical value calibrated for the shared-control comparisons to control familywise error under the global null.
These assumptions define the setting in which the planning calculations below are interpreted. In particular, the numerical operating characteristics used in the examples are hypothetical outputs from evaluating the specified stopping rule and combination test under this normal model. They are not observed results from a real trial.
From Stage Sizes to Enrollment
At stage 1, all three groups are recruited, so enrollment is \(3n_1\). At stage 2, the shared control is recruited once, regardless of whether one or two experimental arms continue. Thus, given \(K\), stage-2 enrollment is \(n_2(1+K)\). The maximum enrollment occurs when both arms continue, while expected enrollment depends on the probability of each continuation path.
Here \(s\) indexes a treatment-effect scenario. If \(q_{js}\) is the probability that arm \(j\) is retained at the interim analysis in scenario \(s\), then linearity of expectation gives \(E(K\mid s)=q_{1s}+q_{2s}\). The arms’ retention indicators need not be independent for this equality to hold. If the scenario weights are \(w_s\), with \(w_s\geq0\) and \(\sum_s w_s=1\), the scenario-weighted expected enrollment is
The maximum and expected enrollment answer different questions. The maximum is a bound on the number of participants the plan could require if both arms proceed. Expected enrollment is an average under the specified scenario weights and continuation probabilities; it is not a guarantee that recruitment will stay below that average.
Worked Example: Calculating Scenario-Weighted Expected Enrollment
Consider plan B with \(n_1=36\) and \(n_2=48\). Suppose the planning analysis gives the following expected numbers of retained arms: 0.65 in scenario 1, 1.00 in scenario 2, and 1.30 in scenario 3. Assign scenario weights 0.50, 0.30, and 0.20, respectively.
First calculate expected enrollment in each scenario. In scenario 1, it is \(3(36)+48(1+0.65)=108+79.2=187.2\) participants. In scenario 2, it is \(108+48(2)=204\) participants. In scenario 3, it is \(108+48(2.30)=218.4\) participants.
The weighted expected number of retained arms is \(0.50(0.65)+0.30(1.00)+0.20(1.30)=0.325+0.300+0.260=0.885\). Therefore, \(E_w(N)=108+48(1+0.885)=108+90.48=198.48\), or about 198.5 participants.
The maximum is \(N_{\max}=3(36)+3(48)=252\) participants. The expected enrollment is below this maximum, as it must be, but plan B can still require more than its expected enrollment in a particular trial.
Power and Feasibility Must Be Checked Together
For a retained arm, the stage-specific treatment-control contrast uses the shared control data. Under the stated normal model and equal per-arm sample sizes, the variance of a stage-specific estimated difference is \(2\sigma^2/n_k\). If the true mean difference is \(\delta\), the corresponding standardized contrast has mean \(\delta\sqrt{n_k/(2\sigma^2)}\) and variance 1. Stagewise contrasts for the same arm are independent under independent recruitment across stages; contrasts for different arms are correlated because they share control observations.
In a two-stage combination test, the stage statistics are combined using prespecified information weights, and the final test is compared with a critical value calibrated for the joint null distribution of the two arm comparisons. The interim rule also affects operating characteristics: an arm that fails the futility criterion does not proceed to the final test. For this workshop, assume the rule and calibration have been fixed in advance. The reported power is the probability of declaring at least one treatment with a true effect of 5 outcome units effective in that scenario. It is evaluated separately for each scenario, not averaged using the scenario weights.
A plan is eligible only if it meets the target power of 0.80 in every prespecified scenario with at least one effective treatment, controls familywise type-I error at 0.025 under the global null, and respects the recruitment cap. Scenario weights are used to compare expected enrollment among eligible plans; they do not replace these scenario-specific power constraints.
Worked Example: Comparing Candidate Plans
The following hypothetical calculations use scenario weights \(w_1=0.50\), \(w_2=0.30\), and \(w_3=0.20\). Scenario 1 has arm 1 effective at a 5-unit difference and arm 2 null; scenario 2 has both arms effective at 5 units; scenario 3 has arm 1 effective at 5 units and arm 2 at 3 units. In each scenario, the table gives the evaluated probability that at least one effective treatment is declared, and the expected number of retained arms.
| Plan | \(n_1,n_2\) | Power: scenarios 1, 2, 3 | \(E(K\mid s)\): scenarios 1, 2, 3 | Maximum enrollment |
|---|---|---|---|---|
| A | 30, 45 | 0.78, 0.86, 0.75 | 0.75, 1.10, 1.40 | 225 |
| B | 36, 48 | 0.84, 0.90, 0.82 | 0.65, 1.00, 1.30 | 252 |
| C | 42, 50 | 0.88, 0.93, 0.86 | 0.60, 0.95, 1.20 | 276 |
Assume the global-null calibration gives familywise type-I error 0.025 for each candidate and the recruitment cap is 280 participants. Plan A has maximum enrollment \(3(30)+3(45)=225\), but it fails the power target in scenarios 1 and 3 because 0.78 and 0.75 are below 0.80.
Plan B has maximum enrollment \(3(36)+3(48)=252\), below the cap. Its power values, 0.84, 0.90, and 0.82, all meet the target. From the preceding example, its weighted expected enrollment is 198.48 participants.
Plan C has maximum enrollment \(3(42)+3(50)=276\), also below the cap, and its power values all exceed 0.80. Its weighted expected number of retained arms is \(0.50(0.60)+0.30(0.95)+0.20(1.20)=0.300+0.285+0.240=0.825\). Thus, its weighted expected enrollment is \(3(42)+50(1+0.825)=126+91.25=217.25\) participants.
Both B and C satisfy the stated power, type-I error, and recruitment requirements. If the objective is to minimize scenario-weighted expected enrollment subject to those constraints, plan B is preferred: 198.48 is less than 217.25 participants. Plan A is not eligible, even though its expected enrollment could be lower, because it does not meet the power requirement.
A Practical Planning Sequence
An integrated comparison is easiest to audit when the constraints and objective are handled separately. In particular, do not use a favorable weighted average to conceal weak power in a scenario that the design is required to handle.
State the outcome distribution, variance, clinically relevant effect, allocation, stopping rule, and treatment-effect patterns to be evaluated. Assign scenario weights only when a weighted expected-enrollment objective is wanted.
For each candidate \((n_1,n_2)\), evaluate the prespecified test and interim rule under each scenario. Record scenario-specific power and continuation probabilities, as well as the global-null type-I error.
Exclude a candidate if it misses a required power target, fails type-I error control, or has maximum enrollment above the recruitment cap. Do not round or alter stage sizes without reevaluating these conditions.
Calculate scenario-weighted expected enrollment for each remaining candidate. Select according to the stated objective, and report both expected and maximum enrollment with the assumptions that produced them.
Scenario Weights, Diagnostics, and Alternatives
Scenario weights describe how much each scenario contributes to the planning average. They do not change the scenario-specific probabilities themselves. If the weights shift toward scenarios with more arms likely to continue, expected enrollment can rise even though the design and its power estimates are unchanged.
Worked Example: Checking the Effect of Scenario Weights
For plan B, change the weights to 0.20, 0.20, and 0.60, giving greater weight to scenario 3. The weighted expected number of retained arms becomes \(0.20(0.65)+0.20(1.00)+0.60(1.30)=0.13+0.20+0.78=1.11\). Expected enrollment is then \(108+48(1+1.11)=108+101.28=209.28\) participants.
This is higher than the 198.48 participants under the original weights. The power values for plan B remain 0.84, 0.90, and 0.82: changing weights changes the planning average, not the operating characteristics within a scenario. The plan still satisfies the power target, but its expected enrollment comparison with other plans should use the same weights.
Several checks can reveal an error in an integrated calculation. First, verify that scenario weights are nonnegative and sum to 1. Second, check that \(0\leq E(K\mid s)\leq2\), and that \(3n_1+n_2\{1+E(K\mid s)\}\) lies between the minimum \(3n_1+n_2\) and the maximum \(3n_1+3n_2\). Third, ensure that the control group is counted once at each stage. Counting a separate control group for each active arm overstates stage-2 enrollment in a shared-control design. Finally, confirm that every required scenario passes its power constraint before comparing weighted expected enrollment.
The main alternative objectives answer different planning questions. Minimizing maximum enrollment prioritizes a hard upper bound; minimizing weighted expected enrollment prioritizes average recruitment under specified scenario weights. A recruitment cap is a constraint, not an objective: expected enrollment below the cap does not make a plan acceptable if its maximum exceeds that limit. Conversely, a plan under the cap may still be unsuitable if its power is inadequate.
These calculations depend on the assumed outcome variance, effects, stopping rule, and scenario probabilities. If those inputs are uncertain, a plan’s ranking may change. Report the assumptions and the sensitivity of the comparison rather than presenting one weighted expected enrollment as an unconditional forecast. Any simulation-based operating-characteristic estimates should also be checked for adequate numerical precision; small differences between candidate plans may not be meaningful if simulation error is of similar size.
Common Planning Mistakes
- Choosing the smallest expected enrollment before checking power. Expected enrollment is an objective only among plans that meet the required statistical constraints. A low-enrollment candidate that misses a target is not an acceptable solution.
- Confusing the average with the cap. A plan with expected enrollment below the recruitment limit can still exceed the limit when both arms continue. Check maximum enrollment directly.
- Counting shared controls twice. With one shared control group at stage 2, enrollment is \(n_2(1+K)\), not \(2n_2K\).
- Applying scenario weights to power requirements. A weighted average power can be high while one required scenario has inadequate power. Check each constrained scenario separately.
- Treating the retention events as independent without justification. Shared controls induce dependence between treatment-control contrasts. Use operating characteristics that account for their joint distribution and the prespecified multiplicity calibration.
Check Your Understanding
Answer each question using the definitions and planning setup above. Solutions follow the questions.
- A two-stage design has \(n_1=24\), \(n_2=40\), and one arm continues. What is its total enrollment?
- For a plan with \(n_1=30\), \(n_2=50\), and scenario-specific \(E(K)=0.8\), calculate expected enrollment in that scenario and maximum enrollment.
- A candidate has powers 0.83, 0.79, and 0.91 in three required scenarios, with a target of 0.80 in each. Is it eligible on power grounds, even if its weighted average power is above 0.80?
- Why can a plan with expected enrollment of 210 participants still fail a recruitment cap of 250?
Solutions. (1) Stage 1 enrolls \(3(24)=72\) participants. With one arm continuing, stage 2 enrolls \(40(1+1)=80\), for \(72+80=152\) participants. (2) Expected enrollment is \(3(30)+50(1+0.8)=90+90=180\) participants. Maximum enrollment is \(3(30)+3(50)=240\) participants. (3) No. The power of 0.79 is below the required 0.80 in one scenario, so the candidate fails the stated constraint regardless of its weighted average. (4) Expected enrollment is an average, not an upper bound. If both arms continue, the plan may recruit more than 250; its maximum enrollment must be checked against the cap.