Why Combination Doses Need a Partial Order
For a single drug with increasing dose levels, an escalation design can often assume that toxicity risk does not decrease as dose increases. With two drugs, that simple sequence may not exist. Increasing either drug while holding the other fixed may be assumed to increase or preserve toxicity risk, but two combinations that trade off the drugs need not have a known ordering. Treating every combination as if it belonged on one toxicity ladder can impose assumptions that the design does not support.
A partial order represents only the dose comparisons justified by the design’s monotonicity assumptions. It can constrain which moves count as upward escalation, while leaving incomparable combinations unranked. The order describes assumed relationships between toxicity probabilities; it does not itself estimate those probabilities or establish that a candidate is safe.
The order is strict because it relates distinct nodes, even though the toxicity inequalities on edges are non-strict. If \(u\prec v\), transitivity of \(\leq\) implies \(p_u\leq p_v\). The reverse inequality does not follow, so the order does not imply a strictly greater toxicity probability at the higher node.
Building the Graph From Monotonicity Assumptions
Suppose the first drug has dose levels \(A_0,A_1,\ldots\), and the second has levels \(B_0,B_1,\ldots\). A node is a particular pair \((A_i,B_j)\). If increasing drug A at a fixed B level is assumed not to reduce toxicity, encode a directed edge from the lower-A node to the higher-A node for the selected comparisons. Do the same for drug B where that assumption is justified. A path then carries the ordering through multiple steps.
The word “selected” matters. If monotonicity is supported only along some dose-specific directions, include only those comparisons. Do not add an edge between two combinations merely because one has a larger dose of one drug, or because their dose labels look numerically larger overall. A combination that increases one agent but decreases the other is often incomparable unless an additional, justified assumption orders it.
An edge records a structural assumption, not an observed outcome. It should be defined for the same endpoint, DLT assessment window, and target population across nodes. The assumptions also need to be scientifically defensible for the combinations under study. Observed DLT counts may inform risk estimates, but they do not automatically validate every monotonicity edge.
Worked Example: Ordering a Two-by-Two Dose Grid
Consider a hypothetical grid with four combinations: \(L=(A_0,B_0)\), \(X=(A_1,B_0)\), \(Y=(A_0,B_1)\), and \(H=(A_1,B_1)\). Assume toxicity is nondecreasing when either drug is increased while the other is held fixed. The specified edges are \(L\to X\), \(L\to Y\), \(X\to H\), and \(Y\to H\).
The paths give \(L\prec H\) by either route: \(L\to X\to H\) and \(L\to Y\to H\). Thus the encoded inequalities include \(p_L\leq p_X\leq p_H\) and \(p_L\leq p_Y\leq p_H\), so in particular \(p_L\leq p_H\).
There is no path from \(X\) to \(Y\) or from \(Y\) to \(X\). Therefore, \(X\) and \(Y\) are incomparable. For illustration, hypothetical probabilities \(p_L=0.10\), \(p_X=0.18\), \(p_Y=0.16\), and \(p_H=0.32\) satisfy all four edge inequalities. Their numerical values happen to put \(p_Y<p_X\), but that ordering is not entailed by the graph and would not be imposed by this partial-order assumption.
The edge graph is acyclic, so its path relation is a strict partial order. The comparison between \(X\) and \(Y\) remains open: neither combination can be called the safer one based on these monotonicity assumptions alone.
From the Order to Escalation Candidates
A partial order can constrain the direction of dose movement without selecting a unique next combination. From a current node \(x\), the upper set consists of all nodes \(y\) such that \(x\prec y\). These are combinations assumed to have toxicity probability at least as large as \(p_x\). The cover successors are the immediate upward neighbors: \(x\prec y\), with no node \(z\) satisfying both \(x\prec z\) and \(z\prec y\).
One transparent operational rule is to restrict upward moves to untreated cover successors of the current combination. This prevents a jump over an intermediate node in the specified order and makes the allowed moves auditable. It is a design choice, not a consequence that every partial-order design must use. A design may choose a different transition rule, but it should state it in advance and ensure it respects the encoded ordering.
Being an upward candidate is not the same as being safe to treat. The order only constrains relative toxicity probabilities. Any decision to assign patients still needs to use the design’s toxicity data, model, and safety rules. Likewise, if a candidate is incomparable with the current node, the monotonicity assumptions do not establish that it is an escalation or a de-escalation relative to the current combination.
Worked Example: Finding the Allowed Upward Moves
Consider five hypothetical dose combinations: \(L=(A_1,B_1)\), \(A=(A_2,B_1)\), \(B=(A_1,B_2)\), \(C=(A_2,B_2)\), and \(D=(A_3,B_1)\). Suppose the specified edges are \(L\to A\), \(L\to B\), \(A\to C\), \(B\to C\), and \(A\to D\). The graph is acyclic.
From the edges, \(L\prec C\) through both \(A\) and \(B\), and \(L\prec D\) through \(A\). The cover successors of \(A\) are \(C\) and \(D\): there is an edge to each, with no intermediate node on either path. \(B\) is not comparable with \(A\), \(C\), or \(D\) under the listed assumptions, except that \(B\prec C\) is specified directly.
Suppose hypothetical DLT probabilities are \(p_L=0.08\), \(p_A=0.14\), \(p_B=0.12\), \(p_C=0.23\), and \(p_D=0.19\). Each edge inequality holds: \(0.08\leq0.14\), \(0.08\leq0.12\), \(0.14\leq0.23\), \(0.12\leq0.23\), and \(0.14\leq0.19\). If the current combination is \(A\) and the design uses the cover-successor rule, its upward candidates are \(C\) and \(D\), subject to the design’s other eligibility and safety criteria. Although these illustrative probabilities have \(p_D<p_C\), the graph does not order \(C\) and \(D\); that numerical pattern cannot be used to claim the graph establishes their relative risk.
Why Acyclicity Matters
Transitivity of the path relation follows by joining paths: a path from \(u\) to \(v\) followed by a path from \(v\) to \(w\) gives a path from \(u\) to \(w\). Irreflexivity, however, requires that there be no path from a node back to itself. That is exactly why the specified-edge graph must be acyclic before its path relation is called a strict partial order.
A cycle can arise from inconsistent assumptions, an accidental edge, or a deliberate set of non-strict inequalities that implies equality. It should not be silently treated as a valid strict ordering. For example, edges \(P\to Q\), \(Q\to R\), and \(R\to P\) imply \(p_P\leq p_Q\leq p_R\leq p_P\). The inequalities force equality, not a strict ordering of these three nodes.
Worked Example: Detecting and Interpreting a Cycle
Suppose three hypothetical combinations have specified edges \(P\to Q\), \(Q\to R\), and \(R\to P\). The encoded assumptions are \(p_P\leq p_Q\), \(p_Q\leq p_R\), and \(p_R\leq p_P\). Combining them gives \(p_P\leq p_Q\leq p_R\leq p_P\).
Because the first and last quantities bound each other, \(p_P=p_Q=p_R\). The path relation also includes \(P\prec P\), since the cycle is a path back to \(P\); therefore it is not irreflexive and is not a strict partial order. The same applies to \(Q\) and \(R\).
If these non-strict inequalities are genuinely intended, one representation is to treat \(P,Q,R\) as a single strongly connected component, meaning that every node in the component can reach every other node. The cycle forces their toxicity probabilities to be equal under the encoded assumptions. Collapsing each strongly connected component produces a quotient graph; if all such components are collapsed, the quotient graph is acyclic and can be ordered by its path relation. If equality was not intended, the cycle is a diagnostic signal to revisit the specified directions rather than to claim a strict dose ranking.
Diagnostics, Alternatives, and Limits
Before using the graph, check that each edge has a clear scientific rationale and the same toxicity endpoint and assessment window at both ends. Then check for cycles and confirm that the transitive comparisons make sense. A cycle may reveal conflicting directions or an assumption of equality that needs to be made explicit. Also check whether the graph leaves combinations incomparable. Incomparability is not a software error: it records that the supplied assumptions do not order that pair.
A total order ranks every pair, so it can offer a simple sequence of combinations but may assert unsupported comparisons when agents trade off. A partial order preserves uncertainty about those comparisons but can leave multiple upward candidates. At the other extreme, making no monotonicity assumptions leaves the combinations unordered and provides no order-based restriction on movement. The partial-order approach lies between these alternatives: it uses selected monotonicity knowledge without filling gaps by assumption.
The graph does not specify the size of a toxicity difference, account for patient-specific risk, or guarantee that toxicity will be monotone in every population or assessment setting. It also does not determine how to choose among multiple eligible candidates, when to stop, or what probability of excessive toxicity is acceptable. Those decisions require additional design rules and evidence. In particular, an ordering assumption should not be mistaken for a fitted toxicity model or a safety criterion.
Check Your Understanding
Answer each question using the definitions and rules above. Short solutions follow.
- In a two-drug grid, dose combination \(U\) has a directed path to \(V\), but there is no path between \(V\) and \(W\) in either direction. What can be concluded about their encoded toxicity ordering?
- Why does a directed cycle prevent the path relation from being a strict partial order?
- Given edges \(J\to K\), \(K\to M\), and \(J\to M\), which of these edges is a cover relation from \(J\)?
- Does \(x\prec y\) establish that \(y\) is safe to assign to the next cohort? Explain.
Solutions. 1. The assumptions imply \(p_U\leq p_V\), but they do not order \(V\) and \(W\); those nodes are incomparable if neither has a path to the other. 2. A cycle gives a path from a node back to itself, violating irreflexivity. 3. \(J\to K\) is a cover relation because no intermediate node is specified between them. \(J\to M\) is not a cover relation because \(J\prec K\prec M\). 4. No. The relation only asserts an ordering of toxicity probabilities; assignment still depends on observed data, the model, and the design’s safety rules.