Deriving Maximum Sample Size Under a MAMS Design

Why a MAMS Design Has a Maximum Enrollment

A multi-arm multi-stage (MAMS) design can stop an arm before the final stage, so the number enrolled may be below the design’s full planned total. A different quantity is the maximum planned enrollment: the total number of participants scheduled if every treatment arm remains in the design through every planned stage. It is a useful upper bound for recruitment planning, site capacity, and budget estimates. It is not a forecast of how many participants will usually be enrolled.

The key accounting choices are simple but consequential: count each participant once, include participants from every planned stage, and count a shared control group once rather than once per treatment comparison. The calculation uses the stage sizes specified in the design; it does not require a treatment effect, a stopping probability, or an error-rate calculation.

Definition: Suppose a design has \(K\) treatment arms and \(J\) stages. Let \(n_{kj}\) be the number of new participants assigned to treatment arm \(k\) in stage \(j\), and let \(n_{0j}\) be the number of new participants assigned to a shared control group in stage \(j\). If all arms continue through all stages, the maximum planned enrollment is the sum of these stage-specific participant counts. Each count refers to participants enrolled at that stage, not a cumulative total that includes earlier stages.

Deriving the Maximum-Enrollment Formula

Under the assumptions in the definition, stage \(j\) enrolls \(n_{0j}\) control participants and \(n_{kj}\) participants in each treatment arm \(k\). Thus, the total scheduled enrollment for that stage is \(n_{0j}+\sum_{k=1}^{K}n_{kj}\). The stages recruit different participants, so adding the stage totals gives the total for the full design. This yields

N_{\max}=\sum_{j=1}^{J}\left(n_{0j}+\sum_{k=1}^{K}n_{kj}\right) =\sum_{j=1}^{J}n_{0j}+\sum_{k=1}^{K}\sum_{j=1}^{J}n_{kj}.

The equality only rearranges a finite sum: control participants are added across stages, and treatment participants are added across arms and stages. If every treatment arm has the same planned stage size \(n_{Tj}\), the expression simplifies to

N_{\max}=\sum_{j=1}^{J}\left(Kn_{Tj}+n_{0j}\right).

The control term appears once per stage because there is one shared control group. If the protocol instead specifies distinct control groups for different comparisons, the control counts must reflect that design; the shared-control formula should not be imposed on a design that does not share controls.

Set the counting unit.
Confirm that each stage size is a count of new participants assigned at that stage, and determine whether the counts are randomized participants or evaluable participants.

Find each stage total.
Add the planned counts for all treatment arms and add the shared-control count once.

Sum across stages.
Add the stage totals for all \(J\) stages. This is the full-enrollment scenario in which no treatment arm stops before the final stage.

Reconcile the total.
Check that the result also equals the sum of each arm’s total and the shared control’s total, and verify that no cumulative count has been added more than once.

Worked Calculations

Worked Example: Equal Stage Sizes With Three Treatment Arms

Consider a hypothetical three-stage design with three treatment arms. At each stage, 40 new participants are assigned to each treatment arm and 60 new participants are assigned to the shared control group. Find the maximum planned enrollment.

There are \(K=3\) treatment arms and \(J=3\) stages. The stage total is

3(40)+60=120+60=180\text{ participants}.

The same stage sizes apply at all three stages, so

N_{\max}=3(180)=540\text{ participants}.

Checking by group, each treatment arm contributes \(3(40)=120\) participants, for \(3(120)=360\) treatment participants in total. The shared control contributes \(3(60)=180\). Therefore, \(360+180=540\). Counting 60 control participants separately for each treatment comparison would incorrectly give \(3(40+60)\times 3=900\), because it would count the shared control multiple times.

Worked Example: Unequal Stage Sizes Across Arms

In a hypothetical design, the planned new participants at each stage are shown below. The treatment arms have unequal sizes, and the shared-control count also changes by stage.

Stage	Arm 1	Arm 2	Arm 3	Shared control	Stage total
1	30	30	30	45	135
2	25	25	20	40	110
3	20	15	15	35	85

Add the counts within each stage. For stage 1, \(30+30+30+45=135\). For stage 2, \(25+25+20+40=110\). For stage 3, \(20+15+15+35=85\). Thus,

N_{\max}=135+110+85=330\text{ participants}.

As a separate check, Arm 1 contributes \(30+25+20=75\), Arm 2 contributes \(30+25+15=70\), and Arm 3 contributes \(30+20+15=65\). The shared control contributes \(45+40+35=120\). Their sum is \(75+70+65+120=330\), matching the stage-based calculation. Unequal sizes do not change the rule: include every arm-stage count and each shared-control stage count once.

Worked Example: Starting From Cumulative Targets

A hypothetical two-arm, two-stage design is described using cumulative targets rather than new participants per stage. At the end of stage 1, the cumulative counts are 50 in Arm 1, 40 in Arm 2, and 60 in the shared control. At the end of stage 2, the corresponding cumulative targets are 90, 75, and 110. Find the maximum enrollment without double-counting stage 1 participants.

The stage 1 increments are the stage 1 cumulative counts: 50, 40, and 60. The stage 2 increments are the stage 2 cumulative counts minus the stage 1 cumulative counts:

n_{1,2}=90-50=40,\qquad n_{2,2}=75-40=35,\qquad n_{0,2}=110-60=50.

The stage 1 total is \(50+40+60=150\), and the stage 2 total is \(40+35+50=125\). Therefore,

N_{\max}=150+125=275\text{ participants}.

Equivalently, add the final cumulative totals by group: \(90+75+110=275\). Adding the cumulative totals at both stages would give \(150+(90+75+110)=425\), which counts the stage 1 participants a second time. When cumulative targets are supplied, first take differences to identify the new stage counts, or use the final cumulative totals directly as the full-design total.

Assumptions, Checks, and Common Pitfalls

The calculation is an enrollment count, not a statistical approximation. Its assumptions are about what the counts represent and how the design is implemented:

All arms reach every planned stage. The maximum scenario includes all treatment arms at all stages; it does not apply a probability that an arm stops.
Stage counts refer to new participants. If the inputs are cumulative targets, convert them to increments before summing stage totals. Alternatively, use each group’s final cumulative target once.
Shared control is counted once. A participant in the common control group may contribute to comparisons with several arms, but remains one enrolled participant.
The unit is consistent. Do not add randomized participants in one group to evaluable-participant targets in another without an explicit conversion for the intended recruitment total.
Counts are feasible integers. If a sample-size calculation gives a fractional target, the design must specify how it is rounded. Use the actual integer counts in the enrollment sum, not unrounded targets that cannot be enrolled.

A useful diagnostic is to calculate the total in two ways: first, add the stage totals; second, add the full-design totals for each treatment arm and the shared control. The answers should agree. If they do not, check for a missing arm or stage, duplicated control participants, or cumulative counts added as though they were new participants.

Common mistake: “There are three treatment-control comparisons, so I should multiply the planned control count by three.” That is wrong when the comparisons use one shared control group. The correct calculation adds the control group’s planned count once at each stage; the same control data can inform multiple comparisons without representing multiple enrolled participants.

What the Maximum Does—and Does Not—Measure

The maximum planned enrollment answers a specific question: how many participants are scheduled if no treatment arm stops early and the design proceeds through its full set of stages? An expected sample size answers a different question because it averages enrollment over possible study paths. That calculation requires continuation probabilities, rather than simply summing every planned count, and is not the same as the full-enrollment total.

The maximum is also not automatically a guarantee of the number who will provide usable outcomes. If the stage sizes are targets for evaluable participants, the actual number recruited may need to be larger to allow for loss to follow-up or other unavailable outcomes. Any such adjustment depends on the protocol’s assumptions and rounding rule; it should be made explicitly and should not be silently folded into the stage-count formula.

Finally, this total presumes that the planned allocation and shared-control counts apply when all arms remain. If the protocol changes control allocation according to the number of active arms, use the counts specified for the all-arms-retained scenario. The formula is an accounting identity; it cannot replace the design’s allocation rules.

Check Your Understanding

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

A two-stage design has two treatment arms. At each stage, 24 new participants are assigned to each arm and 36 to a shared control. What is the maximum planned enrollment?
At one stage, a design enrolls 18 participants in Arm 1, 22 in Arm 2, and 30 in the shared control. Is the stage total 100 or 70? Explain.
A control group has cumulative targets of 45 after stage 1 and 80 after stage 2. How many new control participants are assigned at stage 2?
Why is the sum of all stage sizes not generally the expected sample size?

Solutions. (1) Each stage enrolls \(24+24+36=84\) participants, so \(N_{\max}=2(84)=168\). (2) The stage total is \(18+22+30=70\). The shared control is counted once, not once for each treatment comparison. (3) The stage 2 increment is \(80-45=35\) new control participants. (4) The sum of all stage sizes assumes every arm reaches every stage. Expected enrollment averages over possible paths, including paths with early stopping, and therefore requires continuation probabilities.