# 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.

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## 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:

$$\bar{C}_w(\sigma) = \frac{\sum_{k=1}^{n} p_{\sigma(k)} \cdot C_{\sigma(k)}}{\sum_{k=1}^{n} p_{\sigma(k)}}$$

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## 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 $\bar{C}(\sigma)$. $\blacksquare$

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## 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**. $\blacksquare$

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## 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.

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## 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.

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## 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 |

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## 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.**

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*This proof was developed conversationally and formalized on 2026-03-28.*