Proof: unweighted avg completion time is a biased metric

Mathematical proof that unweighted average task completion time
is gameable by scheduling policy (SPT), while work-weighted
completion time is schedule-invariant. Demonstrates that SPT's
apparent advantage is an artifact of the metric, not genuine
throughput improvement.

Co-Authored-By: Claude Opus 4.6 (1M context) <noreply@anthropic.com>

README.md

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+# Unweighted Average Completion Time Is Not a Fair Metric for Task Scheduling
+
+A mathematical proof that unweighted average task completion time is a biased
+statistic that incentivizes cherry-picking easy work, and that any scheduling
+advantage it appears to reveal is an artifact of the metric — not a reflection
+of genuine throughput or service quality.
+
+---
+
+## 1. Definitions
+
+Let there be **n** tasks with processing times $p_1, p_2, \ldots, p_n$.
+
+A **schedule** $\sigma$ is a permutation of $\{1, 2, \ldots, n\}$ assigning
+tasks to execution order on a single executor.
+
+The **completion time** of task $\sigma(k)$ under schedule $\sigma$ is:
+
+$$C_{\sigma(k)} = \sum_{j=1}^{k} p_{\sigma(j)}$$
+
+The **unweighted mean completion time** is:
+
+$$\bar{C}(\sigma) = \frac{1}{n} \sum_{k=1}^{n} C_{\sigma(k)}$$
+
+The **work-weighted mean completion time** is:
+
+$$\bar{C}_w(\sigma) = \frac{\sum_{k=1}^{n} p_{\sigma(k)} \cdot C_{\sigma(k)}}{\sum_{k=1}^{n} p_{\sigma(k)}}$$
+
+---
+
+## 2. SPT Is Optimal for the Unweighted Statistic
+
+**Theorem 1.** The schedule that minimizes $\bar{C}(\sigma)$ is Shortest
+Processing Time first (SPT): sort tasks so that $p_{\sigma(1)} \le p_{\sigma(2)} \le \cdots \le p_{\sigma(n)}$.
+
+**Proof (exchange argument).**
+
+Consider any schedule $\sigma$ in which two adjacent tasks $i, j$ satisfy
+$p_i > p_j$ with task $i$ scheduled immediately before task $j$. Let $t$ be the
+start time of task $i$.
+
+| | Task $i$ finishes | Task $j$ finishes | Sum |
+|---|---|---|---|
+| **Before swap** ($i$ then $j$) | $t + p_i$ | $t + p_i + p_j$ | $2t + 2p_i + p_j$ |
+| **After swap** ($j$ then $i$) | $t + p_j$ | $t + p_j + p_i$ | $2t + p_i + 2p_j$ |
+
+The change in the sum of completion times is:
+
+$$(2p_i + p_j) - (p_i + 2p_j) = p_i - p_j > 0$$
+
+Every swap of a longer-before-shorter adjacent pair strictly reduces the total.
+Any non-SPT schedule contains such a pair. Repeated swaps converge to SPT.
+Therefore SPT uniquely minimizes $\bar{C}(\sigma)$. $\blacksquare$
+
+---
+
+## 3. The Work-Weighted Statistic Is Schedule-Invariant
+
+**Theorem 2.** The work-weighted mean completion time $\bar{C}_w(\sigma)$ is
+the same for every schedule $\sigma$.
+
+**Proof.**
+
+Expand the numerator:
+
+$$\sum_{k=1}^{n} p_{\sigma(k)} \cdot C_{\sigma(k)} = \sum_{k=1}^{n} p_{\sigma(k)} \sum_{j=1}^{k} p_{\sigma(j)}$$
+
+Reindex by letting $a = \sigma(k)$ and $b = \sigma(j)$. The double sum counts
+every ordered pair $(a, b)$ where $b$ is scheduled no later than $a$:
+
+$$= \sum_{\substack{a, b \\ b \preceq_\sigma a}} p_a \, p_b$$
+
+For any pair $(a, b)$ with $a \ne b$, exactly one of $\{b \preceq_\sigma a\}$
+or $\{a \prec_\sigma b\}$ holds. The diagonal terms ($a = b$) contribute $p_a^2$
+regardless of order. Therefore:
+
+$$\sum_{\substack{a, b \\ b \preceq_\sigma a}} p_a \, p_b = \sum_{a} p_a^2 + \sum_{\substack{a \ne b \\ b \prec_\sigma a}} p_a \, p_b$$
+
+Now consider the complementary sum:
+
+$$\sum_{\substack{a \ne b \\ a \prec_\sigma b}} p_a \, p_b$$
+
+Together the two off-diagonal sums cover all unordered pairs $\{a, b\}$:
+
+$$\sum_{\substack{a \ne b \\ b \prec_\sigma a}} p_a \, p_b + \sum_{\substack{a \ne b \\ a \prec_\sigma b}} p_a \, p_b = \sum_{a \ne b} p_a \, p_b$$
+
+The right-hand side is schedule-independent. By symmetry of $p_a p_b$, both
+off-diagonal sums are equal:
+
+$$\sum_{\substack{a \ne b \\ b \prec_\sigma a}} p_a \, p_b = \frac{1}{2} \sum_{a \ne b} p_a \, p_b$$
+
+Therefore:
+
+$$\sum_{k=1}^{n} p_{\sigma(k)} \cdot C_{\sigma(k)} = \sum_a p_a^2 + \frac{1}{2} \sum_{a \ne b} p_a \, p_b = \frac{1}{2}\left(\sum_a p_a\right)^2 + \frac{1}{2}\sum_a p_a^2$$
+
+This expression contains no reference to $\sigma$. Since the denominator
+$\sum p_a$ is also schedule-independent:
+
+$$\bar{C}_w(\sigma) = \frac{\frac{1}{2}\left(\sum p_a\right)^2 + \frac{1}{2}\sum p_a^2}{\sum p_a}$$
+
+is **constant across all schedules**. $\blacksquare$
+
+---
+
+## 4. Concrete Example
+
+Two tasks: $A$ with $p_A = 1$ hour, $B$ with $p_B = 10$ hours.
+
+### SPT order (A first)
+
+| Task | Completion time |
+|------|----------------|
+| A | 1 |
+| B | 11 |
+
+- Unweighted mean: $(1 + 11) / 2 = 6.0$
+- Work-weighted mean: $(1 \times 1 + 10 \times 11) / 11 = 111/11 \approx 10.09$
+
+### Reverse order (B first)
+
+| Task | Completion time |
+|------|----------------|
+| B | 10 |
+| A | 11 |
+
+- Unweighted mean: $(10 + 11) / 2 = 10.5$
+- Work-weighted mean: $(10 \times 10 + 1 \times 11) / 11 = 111/11 \approx 10.09$
+
+SPT appears **4.5 hours better** on the unweighted metric but provides
+**zero improvement** on the work-weighted metric. The apparent advantage exists
+only because the unweighted statistic lets a 1-hour task "vote" equally with
+a 10-hour task.
+
+---
+
+## 5. Connection to Little's Law
+
+Little's Law states $L = \lambda W$, where $L$ is the average number of tasks
+in the system, $\lambda$ is the arrival rate, and $W$ is the average time a
+task spends in the system.
+
+For a stable system, $L$ and $\lambda$ are determined by arrival and service
+rates — not by scheduling policy. Therefore $W = L / \lambda$ is
+**schedule-invariant** when measured correctly (i.e., weighted by the quantity
+being served).
+
+SPT appears to violate this only because the unweighted statistic counts
+*completions* rather than *work*, systematically underweighting large tasks.
+
+---
+
+## 6. Consequences
+
+**Theorem 3 (Metric Bias).** Any scheduling policy that minimizes unweighted
+mean completion time necessarily maximizes the completion time of the largest
+task relative to other schedules.
+
+**Proof.** SPT places the largest task last. Its completion time equals the
+total processing time $\sum p_i$, which is the maximum possible completion
+time for any individual task. Meanwhile, FIFO or any non-SPT order would
+allow the large task to finish earlier. $\blacksquare$
+
+This creates a **starvation incentive**: rational agents optimizing the
+unweighted statistic will indefinitely defer large tasks in favor of
+small ones.
+
+### Real-world manifestations
+
+| Domain | Gameable metric | Perverse outcome |
+|--------|----------------|------------------|
+| Support desks | Tickets closed / day | Complex issues ignored |
+| Sprint planning | Story count velocity | Work split into trivial pieces |
+| Emergency rooms | Average wait time | Critical patients deprioritized |
+| Academic publishing | Papers per year | Incremental work favored over deep research |
+
+---
+
+## 7. Conclusion
+
+The unweighted average completion time is a **biased statistic** that:
+
+1. **Can be gamed** by scheduling policy (Theorem 1), unlike work-weighted
+   completion time which is schedule-invariant (Theorem 2).
+2. **Incentivizes starvation** of large tasks (Theorem 3).
+3. **Contradicts Little's Law** unless tasks are uniformly sized.
+
+A metric that can be improved by reordering work — without doing any
+additional work — is measuring the scheduling policy, not the system's
+capacity or effectiveness.
+
+**Unweighted average completion time is not a fair or accurate measurement
+of task execution performance.**
+
+---
+
+*This proof was developed conversationally and formalized on 2026-03-28.*