Why constraints belong inside the design problem
A mathematically attractive sampling schedule is not useful if patients cannot attend at those times, two draws are too close together, or the assay cannot quantify the resulting samples. Constrained robust design treats such requirements as part of the optimization problem, rather than as practical details to check only after choosing a schedule. The objective is to find the best design among those that can actually be implemented and measured.
The objective can be an expected information score under specified parameter scenarios or a worst-case score across those scenarios. The choice between these criteria is made before optimization; the constraints then define which schedules each criterion is allowed to consider. In particular, a high score cannot compensate for violating a genuine safety, operational, or assay requirement.
Writing practical requirements as constraints
A constraint should express a requirement in measurable terms. If draw \(j\) must occur within a visit window \([L_j,R_j]\), write \(L_j\leq t_j\leq R_j\). If successive draws must be separated by at least \(\Delta\), require \(t_{j+1}-t_j\geq\Delta\). These conditions also enforce temporal order when \(\Delta>0\). A maximum total number of visits is a constraint on \(m\); a maximum blood volume can be written as \(\sum_j v_j\leq V_{\max}\), where \(v_j\) is the blood volume assigned to draw \(j\).
Assay restrictions need a similarly explicit definition. For instance, if a sample is usable only when its predicted concentration falls within a quantifiable range, a design may be required to satisfy that range under specified parameter scenarios. The exact mathematical condition depends on the assay and model. A design that produces a value below the quantification limit is not automatically feasible just because a numerical information calculation returned a score for it; the analysis must specify how such samples are treated.
It is important to decide which requirements are truly hard constraints. A clinic’s maximum permitted blood volume or a validated visit window may be non-negotiable. A preference for fewer visits can instead be expressed as a soft penalty or as a burden component of the objective. Mixing the two without saying so can yield an uninterpretable compromise—for example, an optimizer may trade a small increase in information for a violation that clinicians consider unacceptable.
Define the allowed time ranges, the parameter scenarios, and the utility calculation before comparing schedules.
Remove schedules that violate visit windows, spacing, assay-validity rules, or patient-burden limits.
Calculate expected utility or worst-case utility only for feasible schedules, using the specified scenario weights or scenario set.
Confirm its constraints and scenario-specific utilities, then assess whether small changes or plausible operational deviations would make it infeasible.
Worked examples
Worked Example: Choosing Among Schedules With Spacing and Assay Limits
Consider three sampling times, in hours, with visit windows \(t_1\in[1,2]\), \(t_2\in[2,4]\), and \(t_3\in[4,5]\). The minimum spacing is 1.5 hours. A separate assay-validity check has already identified which candidate schedules produce usable samples under the scenarios being considered. Utilities below are hypothetical information scores; larger values are better.
| Schedule | Times (hours) | Scenario 1 | Scenario 2 | Scenario 3 | Assay valid? |
|---|---|---|---|---|---|
| A | (1, 3, 5) | 8 | 6 | 9 | Yes |
| B | (1, 2, 5) | 12 | 10 | 11 | No |
| C | (2, 4, 5) | 10 | 8 | 9 | Yes |
| D | (1.5, 3.5, 5) | 7 | 8 | 7 | Yes |
First check the operational constraints. Schedule A has gaps \(3-1=2\) and \(5-3=2\) hours, so it satisfies the minimum spacing and all three visit windows. Schedule B has a first gap of \(2-1=1\) hour, less than 1.5 hours; it also fails the assay-validity check. Its higher scores do not make it eligible. Schedule C has a final gap of \(5-4=1\) hour and is therefore infeasible. Schedule D has gaps of 2 hours and 1.5 hours, so it is feasible.
For the worst-case criterion, take the minimum of the three scenario utilities for each feasible schedule: $$ U_{\min}(A)=\min(8,6,9)=6,\qquad U_{\min}(D)=\min(7,8,7)=7. $$ Thus D is preferred: its worst scenario score is 7, compared with 6 for A. This conclusion concerns only the stated scenarios and constraints; it does not establish that D is best for every possible parameter value or operational setting.
Worked Example: Expected Utility With a Blood-Volume Limit
Suppose three scenarios have specified weights \(0.50\), \(0.30\), and \(0.20\). The maximum allowed total blood volume is 12 mL. Three candidate schedules have the following utilities and total blood volumes:
| Schedule | Scenario utilities | Total blood volume |
|---|---|---|
| A | (9, 5, 8) | 10 mL |
| B | (7, 8, 6) | 11 mL |
| C | (8, 7, 9) | 14 mL |
Schedules A and B satisfy the 12 mL limit; C does not. Compute expected utility for the two feasible designs. For A, $$ \sum_s w_s U(A,\theta_s) =(0.50)(9)+(0.30)(5)+(0.20)(8) =4.5+1.5+1.6=7.6. $$ For B, $$ \sum_s w_s U(B,\theta_s) =(0.50)(7)+(0.30)(8)+(0.20)(6) =3.5+2.4+1.2=7.1. $$ Among the feasible candidates, A has the greater expected utility, 7.6 rather than 7.1, and uses 10 mL. Schedule C’s expected utility would be \((0.50)(8)+(0.30)(7)+(0.20)(9)=7.9\), but it exceeds the hard limit by \(14-12=2\) mL. It cannot be selected under the stated problem, even though its expected score is higher.
Worked Example: A Continuous Worst-Case Optimum Within an Assay Window
Suppose the first draw is fixed at 0.5 hours and a second draw is chosen at time \(t\). The visit window is 1 to 6 hours, the minimum spacing is 1.5 hours, and assay checks require \(2\leq t\leq5\). The spacing constraint gives \(t-0.5\geq1.5\), or \(t\geq2\). Combining all restrictions, the feasible range is \(2\leq t\leq5\).
For two hypothetical parameter scenarios, let the utility scores be \(U_1(t)=10-(t-2)^2\) and \(U_2(t)=8-(t-5)^2\). The robust objective is to maximize \(\min\{U_1(t),U_2(t)\}\) over the feasible range. The two curves are equal when $$ 10-(t-2)^2=8-(t-5)^2. $$ Expanding and collecting terms gives \(6+4t=-17+10t\), so \(23=6t\) and \(t=23/6\approx3.833\) hours. At this time, $$ U_1(23/6)=10-(11/6)^2=239/36\approx6.639, \qquad U_2(23/6)=8-(7/6)^2=239/36\approx6.639. $$
This equality point is the maximum of the minimum on the feasible interval. On \([2,5]\), \(U_1(t)\) decreases as \(t\) increases, while \(U_2(t)\) increases. Before their intersection, \(U_2\) is the smaller score, so the minimum increases with \(t\); after the intersection, \(U_1\) is the smaller score, so the minimum decreases. Therefore the intersection maximizes the lower of the two utilities. The selected time is inside the assay-valid interval and satisfies the spacing and visit-window constraints.
Justification, diagnostics, and alternatives
The order of operations matters. If a schedule violates a hard constraint, optimizing its score is irrelevant to the constrained problem. Formally, if the feasible set is \(\mathcal{F}\), the chosen design must satisfy \(d^\ast\in\mathcal{F}\) as well as having an objective value at least as large as that of any other design in \(\mathcal{F}\). This also explains why feasibility must be checked before ranking, as in the examples: ranking all candidates first could select a design that is not available for implementation.
A practical implementation should report more than the winning score. Check each constraint at the selected times, inspect utilities scenario by scenario, and identify which constraints are binding or nearly binding. For continuous optimization, compare the returned solution with a fine grid or several starting points when feasible; a solver’s reported optimum does not by itself demonstrate that the global optimum was found. For a discrete candidate set, an exhaustive comparison is transparent, while a larger search can be checked by repeating it with a denser candidate set.
Common warning signs include an empty feasible set, a selected time just outside a visit window due to rounding, and a design whose assay validity changes under small parameter perturbations. If \(\mathcal{F}\) is empty, there is no constrained optimum: the requirements conflict, and the design team must revisit which requirements can be relaxed or whether another schedule structure is needed. If a design is feasible only under exact timing, consider whether the visit-window constraint should represent realistic scheduling variation rather than an ideal appointment time.
Hard constraints are not the only way to represent practical trade-offs. One alternative is to include burden in the objective, for example by subtracting a prespecified penalty for extra visits. This permits explicit trade-offs but requires a defensible penalty scale and may permit burden levels that stakeholders would have preferred to rule out. A hard cap is clearer when exceeding it is unacceptable. Similarly, expected utility emphasizes average performance under the assigned weights, whereas a worst-case criterion protects against the least favorable included scenario. Neither criterion protects against scenarios omitted from the analysis.
Other limitations remain even when constraints and the objective are correctly specified. A candidate set can miss a better schedule between its grid points; an assay rule based on inaccurate predictions can misclassify usable samples; and patient burden may not be captured by a single volume or visit-count limit. The optimization is therefore conditional on the model, scenarios, utility, and operational assumptions supplied to it. Report these choices alongside the resulting schedule so that its practical meaning is clear.
Check Your Understanding
Use the stated constraints and criterion in each question. Solutions are included for checking.
- A two-draw design has \(t_1\in[1,2]\), \(t_2\in[3,5]\), and a minimum spacing of 1.5 hours. Is \((1.5,3)\) feasible with respect to windows and spacing? Solution: Yes. Both times are within their windows, and \(3-1.5=1.5\) hours exactly meets the minimum.
- Two feasible schedules have scenario utilities \((4,9)\) and \((6,6)\). Which has the larger worst-case utility? Solution: The first has \(\min(4,9)=4\); the second has \(\min(6,6)=6\). The second is preferred by the worst-case criterion.
- With scenario weights \(0.70\) and \(0.30\), a feasible schedule has utilities 5 and 10. Calculate its expected utility. Solution: \((0.70)(5)+(0.30)(10)=3.5+3=6.5\) utility units.
- A schedule has the highest expected utility but exceeds a hard blood-volume limit. Can it be the solution to the constrained optimization problem? Solution: No. It is outside the feasible set. The objective is optimized only over schedules that satisfy the limit.