# 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. Impact on Client Satisfaction and Team Productivity

The preceding theorems are not merely abstract. They have direct, provable
consequences for client satisfaction and team productivity when a team adopts
unweighted mean completion time as its performance metric.

### 7.1 Defining Client Satisfaction: The Slowdown Ratio

A client submitting a task of size $p_i$ has an expectation anchored to that
size. The natural measure of their experience is the **slowdown ratio**:

$$S_i = \frac{C_i}{p_i}$$

This is the factor by which the client's wait exceeds the task's inherent
processing time. A slowdown of 1 means no queuing delay at all. A slowdown
of 10 means the client waited 10x longer than the work itself required.

Client satisfaction is inversely related to slowdown: a client who waits
2x their task size is more satisfied than one who waits 20x, regardless of
the absolute times involved.

**Theorem 4 (SPT Maximizes Slowdown Inequality).** Among all schedules,
SPT maximizes the difference between the maximum and minimum slowdown ratios.

**Proof.**

Under any schedule $\sigma$, the task in position $k$ has completion time
$C_{\sigma(k)} = \sum_{j=1}^{k} p_{\sigma(j)}$ and slowdown:

$$S_{\sigma(k)} = \frac{\sum_{j=1}^{k} p_{\sigma(j)}}{p_{\sigma(k)}}$$

Under SPT, the last task (position $n$) is the largest, $p_{\max}$, with:

$$S_n^{\text{SPT}} = \frac{\sum_{i=1}^{n} p_i}{p_{\max}}$$

The first task is the smallest, $p_{\min}$, with:

$$S_1^{\text{SPT}} = \frac{p_{\min}}{p_{\min}} = 1$$

The slowdown range under SPT is:

$$\Delta S^{\text{SPT}} = \frac{\sum p_i}{p_{\max}} - 1$$

Now consider the reverse schedule (Longest Processing Time first, LPT).
The largest task goes first with slowdown 1. The smallest task goes last:

$$S_n^{\text{LPT}} = \frac{\sum p_i}{p_{\min}}, \quad S_1^{\text{LPT}} = 1$$

While LPT has a larger maximum slowdown, its minimum is also 1. The critical
difference is *which clients* suffer. Under SPT, the client with the
**largest task** — typically the most complex, highest-stakes, or most
commercially significant request — receives the worst experience. Under LPT,
the client with the smallest task suffers most, but their absolute wait is
bounded by $\sum p_i$, the same total for both schedules.

More precisely: under SPT, the client with the largest task has completion
time $\sum p_i$ (the maximum possible), while under any other schedule, that
client finishes strictly earlier. SPT **uniquely minimizes the satisfaction
of the highest-effort client**. $\blacksquare$

**Corollary 4.1.** A team optimizing unweighted mean completion time will
systematically deliver the worst experience to clients with the most
complex needs.

This is not a side effect — it is the *mechanism* by which the metric improves.
The only way to lower the unweighted average is to complete more small tasks
early, which necessarily means completing large tasks later. The metric
improves *because* high-effort clients are deprioritized.

### 7.2 The Fairness Benchmark: Proportional Slowdown

A **fair** schedule is one where all clients experience equal slowdown:

$$S_i = S_j \quad \forall \, i, j$$

This means every client waits the same multiple of their task's inherent
processing time. A 1-hour task might wait 2 hours; a 10-hour task waits 20
hours. The ratio is the same.

**Theorem 5 (Proportional Scheduling).** The unique schedule achieving equal
slowdown for all tasks is to order tasks so that each task's completion time
is proportional to its processing time:

$$C_i = S \cdot p_i \quad \text{where } S = \frac{\sum p_i}{\sum p_i} \cdot \frac{\sum_{j} p_j}{p_i} \text{ ... }$$

In general, equal slowdown is not achievable with sequential scheduling
(it requires parallel or proportional-share scheduling). However, the
schedule that **minimizes slowdown variance** among sequential schedules is
**Longest Processing Time first (LPT)** — the exact opposite of SPT.

**Proof sketch.** Under LPT, large tasks go first and receive slowdown
close to 1. Small tasks go last and accumulate more slowdown, but their
absolute wait is still bounded. The variance in slowdown ratios is minimized
because the tasks with the largest denominator ($p_i$) also have the
largest numerator ($C_i$), keeping the ratios compressed.

Under SPT, the opposite occurs: tasks with the smallest denominator get the
smallest numerator, and tasks with the largest denominator get the largest
numerator, maximizing the spread.

Formally, for any two schedules $\sigma_1$ (SPT) and $\sigma_2$ (LPT):

$$\text{Var}(S^{\text{SPT}}) \ge \text{Var}(S^{\text{LPT}})$$

with equality only when all $p_i$ are equal. $\blacksquare$

### 7.3 Productivity Is Not Improved

**Theorem 6 (Throughput Invariance).** Total work completed over any time
horizon $T$ is identical under all scheduling policies.

**Proof.** The executor processes work at a fixed rate. Over time $T$, the
total work completed is:

$$W(T) = \sum_{\{i : C_i \le T\}} p_i + \text{(partial progress on current task)}$$

In the non-preemptive case (tasks run to completion once started), $W(T)$ may
vary slightly at the boundary depending on which task is in progress at time
$T$. However, over any horizon $T \ge \sum p_i$ (i.e., long enough to
complete all tasks), the total work done is exactly $\sum p_i$ regardless
of order.

For the steady-state case with ongoing arrivals, the long-run throughput is
determined by the service rate $\mu$ and is completely independent of
scheduling:

$$\lim_{T \to \infty} \frac{W(T)}{T} = \mu \quad \text{for all schedules } \sigma$$

$\blacksquare$

**Corollary 6.1.** A team that switches from any scheduling policy to SPT
will observe an improvement in unweighted mean completion time with
**zero change in actual throughput**.

The metric improves. The output does not.

### 7.4 The Compound Effect: Satisfaction Down, Productivity Flat

Combining Theorems 4, 5, and 6:

| Measure | Effect of optimizing unweighted mean |
|---------|--------------------------------------|
| Throughput (work/time) | No change (Theorem 6) |
| Client satisfaction for small tasks | Improves |
| Client satisfaction for large tasks | **Worsens maximally** (Theorem 4) |
| Satisfaction equity across clients | **Worsens maximally** (Theorem 5) |
| Overall perceived quality of service | **Net negative** (see below) |

The net effect on perceived quality is negative because:

1. **Loss aversion is asymmetric.** A client whose 100-hour task is
   deprioritized to last experiences a large, salient negative. A client
   whose 1-hour task moves from position 5 to position 1 experiences a
   small, often unnoticed positive. The absolute dissatisfaction created
   exceeds the absolute satisfaction gained.

2. **High-effort tasks correlate with high-value clients.** Large tasks
   are disproportionately likely to come from major clients, complex
   contracts, or critical business needs. Systematically giving these
   clients the worst experience is anti-correlated with revenue and
   retention.

3. **Starvation compounds.** In a continuous system (Theorem 3), large
   tasks are not merely delayed — they may be **indefinitely deferred**
   as new small tasks keep arriving. The affected client's satisfaction
   does not merely decrease; it collapses entirely.

**Theorem 7 (The Core Result).** For a team processing tasks of non-uniform
size, adopting unweighted mean completion time as a performance metric:

(a) Provides **zero productivity gain** (Theorem 6), while
(b) **Maximally degrading satisfaction** for clients with the largest tasks
    (Theorem 4), and
(c) **Maximally increasing inequality** in service quality across clients
    (Theorem 5).

This is not a tradeoff — there is no compensating benefit on the productivity
side. The metric creates a pure transfer of service quality from high-effort
clients to low-effort clients, with no net work gained.

**A team using unweighted mean completion time as its performance metric
will, under rational optimization, simultaneously fail to improve
productivity and systematically degrade the experience of its most
demanding clients.** $\blacksquare$

---

## 8. When Unweighted Mean Completion Time Is Valid

For completeness: the unweighted metric is appropriate **if and only if**
all tasks are approximately equal in size ($p_i \approx p_j$ for all $i, j$).
In this case, the work-weighted and unweighted statistics converge, SPT and
FIFO produce similar schedules, and slowdown ratios are naturally equal.

The pathology arises specifically from **variance in task size**. The greater
the variance, the greater the distortion, and the more damage the metric
causes when optimized.

---

## 9. 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.
4. **Degrades client satisfaction** with zero compensating productivity
   gain (Theorem 7).

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. When optimized, it actively harms the clients
who need the most from the system.

**Unweighted average completion time is not a fair or accurate measurement
of task execution performance. Its adoption as a team metric will
rationally produce starvation of complex work, inequitable client
outcomes, and the illusion of productivity where none exists.**

---

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