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Bandit Formulations · Tutorial 300 of 501

Constrained Bandits With Minimum Allocation Shares

Translate minimum allocation shares into integer quotas, then use a feasibility rule to identify which bandit actions preserve the possibility of meeting every quota.

Advanced 12 min read

What You'll Learn

  • Convert minimum arm shares into integer quotas for a fixed enrollment horizon.
  • Check whether rounded quotas are jointly feasible.
  • Derive a rule for deciding whether an allocation preserves future feasibility.
  • Apply that rule to index-based choices and multi-arm allocation paths.
  • Distinguish pathwise minimum shares from constraints on expected allocation.
  • Diagnose rounding, enrollment, and batch-allocation failures.

Why Put Minimum Shares Into the Policy?

A bandit policy uses accumulating outcomes to guide later treatment assignments. In a clinical trial, however, a reward-seeking rule may allocate too few participants to one or more treatments to satisfy a design requirement. A minimum-share requirement makes that concern explicit: each treatment must receive at least a specified fraction of the trial’s participants.

The important step is to translate a share into a count and then account for the remaining horizon. Checking only whether the current assignment seems reasonable is not enough. A policy must also avoid choices that leave too few enrollment slots to meet outstanding minimums. This is a feasibility constraint on the bandit policy, separate from the rule used to rank treatments by their observed or estimated rewards.

Definition: Consider \(K\) treatment arms and a fixed horizon of \(T\) participants, with one participant assigned at each time \(t=1,\ldots,T\). Let \(A_t\) denote the arm assigned at time \(t\), and let \(N_i(t)\) be the number assigned to arm \(i\) through time \(t\). A minimum share \(q_i\), where \(0\le q_i\le1\), requires \(N_i(T)/T\ge q_i\). The equivalent integer quota is \(m_i=\lceil q_iT\rceil\), so the requirement is \(N_i(T)\ge m_i\).

The ceiling is needed because participant counts are integers. For example, a requirement of at least 21% of a 10-person trial means at least \(\lceil 0.21(10)\rceil=3\) participants, not 2.1 participants. The quotas must be checked jointly: since each participant can be assigned to only one arm, a necessary condition for feasibility is \(\sum_{i=1}^{K}m_i\le T\). It is also sufficient when any arm can be assigned at any time and there are no additional restrictions on the assignment sequence.

Feasibility Across the Remaining Horizon

At time \(t\), define the remaining slots as \(r=T-t\). For each arm, its outstanding quota is the number of additional assignments still needed to reach its minimum. The total outstanding quota cannot exceed the number of remaining slots if completion is to remain possible.

$$ d_i(t)=\max\{0,m_i-N_i(t)\}, \qquad D(t)=\sum_{i=1}^{K}d_i(t). $$

A state is feasible when \(D(t)\le T-t\). The quantity \(D(t)\) counts the assignments that must still go to arms with unmet quotas; any remaining slots beyond that total are slack. This yields a direct rule for screening the next action.

Feasibility rule: At a feasible state with \(r=T-t\) slots remaining, an assignment to arm \(j\) is allowed only if the quotas can still be completed afterward. If arm \(j\) has an outstanding quota, the assignment reduces \(D(t)\) by one; otherwise it leaves \(D(t)\) unchanged. Thus the action is allowed exactly when the new outstanding total is at most \(r-1\), the number of slots left after the assignment.

In particular, when \(D(t)=r\), there is no slack: the next assignment must go to an arm with an outstanding quota. When \(D(t)<r\), there is at least one spare slot, so an assignment to an arm whose quota is already met can still be feasible. Minimum shares therefore do not necessarily prohibit assignments to other arms; they prohibit assignments only when those choices would make completion impossible.

Why the Feasibility Rule Works

The rule follows by checking what happens to the outstanding total after one assignment. Before the assignment, feasibility gives \(D(t)\le r\). If arm \(j\) has an outstanding quota, then \(d_j(t)\ge1\), so assigning the next participant to \(j\) reduces the total by one. If it does not, the total is unchanged. These are the only two possibilities.

$$ D(t+1)= \begin{cases} D(t)-1, & d_j(t)\ge1,\\ D(t), & d_j(t)=0. \end{cases} $$

The next state is feasible exactly when \(D(t+1)\le r-1\). If \(D(t)=r\), assigning to an arm with no outstanding quota leaves \(D(t+1)=r\), which exceeds the \(r-1\) slots left. Assigning to an arm with an outstanding quota instead gives \(D(t+1)=r-1\), which is feasible. If \(D(t)<r\), then the integer-valued total satisfies \(D(t)\le r-1\); even an assignment that does not reduce it leaves the next state feasible. This proves the rule for every feasible state.

Applying this check after each assignment maintains feasibility by induction: the initial state is feasible if \(\sum_i m_i\le T\), and every permitted action leads to another feasible state. If the policy continues assigning participants until the horizon, the quotas can therefore be met. This argument concerns quota feasibility, not whether the policy learns the arm rewards well or selects a clinically preferable treatment.

Worked Example: Converting Shares Into Feasible Counts

Suppose a hypothetical 40-participant trial has three arms with minimum shares \(q_1=0.20\), \(q_2=0.30\), and \(q_3=0.10\). Find the integer quotas, check feasibility, and assess a proposed final allocation of 10, 14, and 16 participants.

The quotas are

$$ m_1=\lceil 0.20(40)\rceil=8,\qquad m_2=\lceil 0.30(40)\rceil=12,\qquad m_3=\lceil 0.10(40)\rceil=4. $$

Their total is \(8+12+4=24\), which is no greater than the 40 available slots. The quota set is feasible, with 16 slots of slack beyond the minimums. The proposed allocation sums to \(10+14+16=40\), and each arm meets its quota: \(10\ge8\), \(14\ge12\), and \(16\ge4\). Its realized shares are \(10/40=0.25\), \(14/40=0.35\), and \(16/40=0.40\), respectively. Thus all three minimum-share requirements are satisfied.

Restricting a Bandit Rule, Not Replacing It

Once infeasible actions are excluded, the policy can use its chosen bandit criterion to select among the remaining actions. For example, it can rank arms using an index rule such as UCB, as in the earlier material on UCB allocation. The constraint does not prescribe how to estimate rewards or rank eligible arms. It only screens out choices that would make the minimum quotas impossible to finish.

Worked Example: An Index Choice With No Quota Slack

Consider a hypothetical five-participant trial with four arms and quotas \(m_1=2\), \(m_2=1\), \(m_3=2\), and \(m_4=0\). The quotas sum to 5, exactly the horizon. At the start, suppose the policy’s index scores are 0.72, 0.81, 0.66, and 0.94 for arms 1 through 4. Which arm should it assign?

Initially \(r=5\), and the outstanding quotas are \((2,1,2,0)\), so \(D(0)=5\). Since \(D(0)=r\), only an assignment to an arm with an outstanding quota is allowed. Arms 1, 2, and 3 are eligible; arm 4 is not. Among the eligible arms, the largest score is 0.81, so the constrained index policy assigns the participant to arm 2. The unconstrained largest score, 0.94 for arm 4, cannot be followed because it would leave all five required quota assignments for only four remaining slots.

After assigning to arm 2, the outstanding quotas become \((2,0,2,0)\), with total 4 and four slots left. Again there is no slack; the next assignment must go to arm 1 or arm 3. Continuing to choose only arms with outstanding quotas can meet the quotas, for example with the sequence 2, 1, 3, 1, 3. The final counts are \((2,1,2,0)\), exactly the required totals. The sequence is an illustration of feasibility, not a claim that these assignments maximize clinical benefit.

This example also shows why the constraint should be applied before selecting the highest-scoring arm. A score alone does not tell whether an action preserves the possibility of meeting the trial’s design requirements. Conversely, if there is slack, an arm with no minimum quota can remain eligible; the policy should not treat “minimum share” as “exclusive allocation.”

Worked Example: Identifying Slack at an Intermediate Time

A hypothetical 12-participant trial has three quotas \(m_1=3\), \(m_2=3\), and \(m_3=3\). After 7 assignments, the observed counts are \((2,2,3)\). Determine whether an assignment to arm 3 is feasible and what happens to feasibility if it is chosen.

The remaining slots are \(r=12-7=5\). The outstanding quotas are

$$ (d_1,d_2,d_3)=(1,1,0),\qquad D=1+1+0=2. $$

Because \(D=2<5=r\), there are three slots of slack. Arm 3 has no outstanding quota, but assigning the next participant to it leaves the outstanding total at 2 and leaves \(5-1=4\) slots. Since \(2\le4\), that choice is feasible. The policy can still meet the quotas by assigning one later participant to arm 1 and one to arm 2; the other two remaining slots are not needed for minimums. The conclusion is about feasibility only: it does not say whether assigning to arm 3 is the best reward-based decision.

Rounding, Diagnostics, and Failure Modes

Share constraints can appear feasible before they are converted to integer quotas but fail after rounding. For example, with \(T=10\), four arms each assigned a minimum share of \(0.21\) have shares summing to \(0.84\), yet each quota is \(\lceil2.1\rceil=3\). The integer quotas sum to 12, exceeding the 10 available assignments. No allocation can satisfy all four requirements. A design must therefore check \(\sum_i\lceil q_iT\rceil\le T\), not just \(\sum_iq_i\le1\).

At each decision point, a useful diagnostic is to calculate the remaining slots and outstanding quotas from the current assignment counts. If \(D(t)>T-t\), the state is already infeasible: some prior action, quota specification, or enrollment record is inconsistent with completion. If \(D(t)=T-t\), every remaining assignment must reduce an outstanding quota. If \(D(t)<T-t\), some allocation flexibility remains. These checks can be automated and logged alongside the arm counts.

The calculation assumes a fixed horizon, one assignment per participant, fixed quotas, and that each remaining participant can be assigned to any arm. A trial that stops early, changes its target enrollment, or introduces restrictions on which arms can be assigned at particular times needs a feasibility check adapted to those conditions. For batch enrollment of \(b\) participants, a proposed batch must leave total outstanding quotas no greater than the slots after the batch, \(T-t-b\); checking only the first assignment in the batch is not enough when the rest are committed simultaneously.

What the Constraint Guarantees—and What It Does Not

A pathwise minimum-share constraint requires the quotas to be met on every allocation path allowed by the policy. This is stronger than requiring only that the expected number assigned to an arm meet a target. For example, an expected count above a quota does not rule out some realized trials falling below that quota. The feasibility rule in this tutorial protects the possibility of meeting integer quotas on the realized assignment path.

Minimum shares also differ from forced exploration and minimum sampling-rate rules discussed in the earlier material on forced exploration. A fixed-horizon minimum quota is a terminal requirement stated in counts; it need not specify how often an arm is sampled at each intermediate time. Many different policies can satisfy the same quotas while producing different reward histories and allocations beyond the minimums.

Finally, a quota is a design constraint, not a guarantee of good statistical or clinical performance. Satisfying the minimum counts does not ensure precise treatment-effect estimates, balanced information at every interim time, or favorable outcomes. Nor does it ensure that the reward-based part of the policy will behave well. The constraint’s narrower guarantee is valuable: if the initial quotas are feasible and every action passes the remaining-horizon check, the policy will not make quota completion impossible.

Common Mistake

A frequent error is to say that, when outstanding quotas exactly equal the remaining slots, an assignment to an arm with no outstanding quota is still acceptable because that arm’s own minimum has already been met. That reasoning ignores the total remaining requirement. If \(D(t)=r\), all \(r\) slots are already needed to fill outstanding quotas. Assigning outside those quotas leaves \(r\) required assignments but only \(r-1\) slots. The correction is to compare the sum of outstanding quotas with the remaining horizon after every proposed action, not merely to check the chosen arm’s status.

Key takeaway: Convert each minimum share to an integer quota, verify the rounded quotas fit the horizon, and allow a proposed assignment only if the remaining outstanding quotas can still fit into the remaining slots. Apply the bandit’s reward-based choice among those feasible actions.

Check Your Understanding

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

  1. A 20-participant trial has minimum shares 0.25, 0.20, and 0.10. What are the integer quotas, and are they jointly feasible?
  2. At a state with 4 slots remaining, the outstanding quotas are \((2,1,1)\). Is an assignment to an arm with no outstanding quota feasible?
  3. At a state with 5 slots remaining, the outstanding quotas are \((1,1,0)\). Is assigning to the arm with zero outstanding quota feasible?
  4. Explain why an expected allocation above an arm’s quota does not establish that a pathwise minimum-share requirement has been met.

Solutions. (1) The quotas are \(\lceil0.25(20)\rceil=5\), \(\lceil0.20(20)\rceil=4\), and \(\lceil0.10(20)\rceil=2\). Their sum is 11, which is no greater than 20, so they are feasible. (2) No. The outstanding total is \(2+1+1=4\), equal to the 4 remaining slots. Assigning to an arm with no outstanding quota would leave 4 quota units but only 3 slots, making completion impossible. (3) Yes. The outstanding total is \(1+1=2\), less than the 5 remaining slots. Assigning to the arm with no outstanding quota leaves 2 quota units and 4 slots, so completion remains possible. (4) An expected count averages over possible allocation paths. Some paths can fall below the quota even when the average count exceeds it; a pathwise requirement must be satisfied on each permitted realized path.

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