Add HOW_TO_MATHS.md: guide to reading the mathematical notation
Explains every symbol, operator, and proof technique used in the paper, assuming no math background beyond basic arithmetic. Covers summation notation, fractions, subscripts/superscripts, Greek letters, set notation, piecewise functions, limits, correlation, variance, mutual information, and the exchange argument proof technique. Includes a worked walkthrough of Theorem 1's proof step-by-step, translating each mathematical statement into plain English. Co-Authored-By: Claude Opus 4.6 (1M context) <noreply@anthropic.com>
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# How to Read the Mathematics in This Paper
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This guide explains every piece of mathematical notation used in the paper,
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in the order you will encounter it. No prior math background beyond basic
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arithmetic is assumed.
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---
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## The Basics
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### Variables and Subscripts
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A variable is a letter that stands for a number. In this paper:
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- $n$ = the total number of tasks
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- $p_i$ = the processing time (how many hours of work) for task $i$
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- $C_i$ = the completion time (when task $i$ is finished) under a given schedule
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- $q_i$ = the priority class of task $i$ (1 = Critical, 4 = Low)
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- $w(q)$ = the weight assigned to priority class $q$
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The subscript $i$ is just a label. $p_1$ means "the processing time of
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task 1," $p_2$ means "the processing time of task 2," and so on. When we
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write $p_i$, we mean "the processing time of task $i$, for any $i$."
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### Lists of Things
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$p_1, p_2, \ldots, p_n$ means "a list of processing times, from task 1
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up to task $n$." The $\ldots$ just means "and so on." If there are 5 tasks,
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this is $p_1, p_2, p_3, p_4, p_5$.
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---
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## Symbols You Will See
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| Symbol | Name | What it means | Example |
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|--------|------|---------------|---------|
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| $=$ | Equals | Left side is the same as right side | $2 + 3 = 5$ |
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| $\ne$ | Not equal | These two things are different | $3 \ne 4$ |
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| $>$ | Greater than | Left is bigger | $5 > 3$ |
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| $<$ | Less than | Left is smaller | $3 < 5$ |
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| $\le$ | Less than or equal to | Left is smaller or the same | $3 \le 5$, $3 \le 3$ |
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| $\ge$ | Greater than or equal to | Left is bigger or the same | $5 \ge 3$, $5 \ge 5$ |
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| $\approx$ | Approximately equal | Close but not exact | $10/3 \approx 3.33$ |
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| $\cdot$ | Multiplication | Same as $\times$ | $3 \cdot 4 = 12$ |
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| $\perp$ | Independent of | No relationship between them | $p_i \perp q_i$ means task size and priority are unrelated |
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| $\forall$ | For all | "This is true for every..." | $S_i = S_j \; \forall \, i, j$ means "slowdown is equal for all tasks" |
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| $\in$ | Is a member of | Belongs to a set | $q_i \in \{1, 2, 3, 4\}$ means priority is one of 1, 2, 3, or 4 |
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| $\blacksquare$ | End of proof | "The proof is complete" | Appears at the end of every proof |
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---
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## The Big Sigma: Summation ($\sum$)
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This is the most important symbol in the paper. It means "add up a list
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of things."
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$$\sum_{i=1}^{n} p_i$$
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Read this as: "Add up $p_i$ for every $i$ from 1 to $n$."
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If there are 3 tasks with processing times $p_1 = 2$, $p_2 = 5$, $p_3 = 3$:
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$$\sum_{i=1}^{3} p_i = p_1 + p_2 + p_3 = 2 + 5 + 3 = 10$$
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The letter under the $\sum$ ($i = 1$) tells you where to start counting.
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The number on top ($n$ or $3$) tells you where to stop. The expression
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after the $\sum$ ($p_i$) tells you what to add up.
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### Variations You Will See
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**Summing a product:**
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$$\sum_{i=1}^{n} w(q_i) \cdot C_i$$
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Add up $w(q_i) \times C_i$ for each task. If task 1 has weight 8 and
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completion time 3, and task 2 has weight 1 and completion time 5:
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$$= 8 \cdot 3 + 1 \cdot 5 = 24 + 5 = 29$$
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**Summing with a condition:**
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$$\sum_{\substack{a \ne b}} p_a \, p_b$$
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Add up $p_a \times p_b$ for every pair of tasks $a$ and $b$ where $a$
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and $b$ are different tasks. The condition ($a \ne b$) filters which
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terms to include.
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**Summing up to a variable position:**
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$$\sum_{j=1}^{k} p_{\sigma(j)}$$
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Add up processing times for the first $k$ tasks in the schedule. This is
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how completion time is defined — it is the total work done up to and
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including position $k$.
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---
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## Fractions
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$$\frac{a}{b}$$
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This means $a \div b$. The top is the numerator, the bottom is the
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denominator. When you see:
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$$\bar{C}(\sigma) = \frac{1}{n} \sum_{k=1}^{n} C_{\sigma(k)}$$
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This means: add up all the completion times, then divide by $n$ (the
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number of tasks). This is an **average** — the same operation as
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calculating a mean in everyday life.
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### Nested Fractions
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$$\bar{C}_w(\sigma) = \frac{\sum_{k=1}^{n} p_{\sigma(k)} \cdot C_{\sigma(k)}}{\sum_{k=1}^{n} p_{\sigma(k)}}$$
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The top (numerator) is: add up (processing time $\times$ completion time)
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for each task. The bottom (denominator) is: add up all the processing
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times. This is a **weighted average** — tasks with more work count more
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heavily.
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---
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## Schedules and Permutations ($\sigma$)
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$\sigma$ (the Greek letter "sigma") represents a **schedule** — the order
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in which tasks are done.
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If you have 3 tasks and decide to do them in order 2, 3, 1, then:
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- $\sigma(1) = 2$ — the first task you do is task 2
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- $\sigma(2) = 3$ — the second task you do is task 3
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- $\sigma(3) = 1$ — the third task you do is task 1
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When you see $p_{\sigma(k)}$, it means "the processing time of whichever
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task is in position $k$ of the schedule."
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### Ordering Symbols
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- $b \preceq_\sigma a$ means "task $b$ is scheduled at the same time as
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or before task $a$" — i.e., $b$ comes first (or they are the same task)
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- $a \prec_\sigma b$ means "task $a$ is scheduled strictly before task $b$"
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These are just ways of saying "which task comes first in the schedule."
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---
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## The Bar: Averages ($\bar{C}$)
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A bar over a letter means "average." So:
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- $\bar{C}$ = average completion time
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- $\bar{C}_w$ = weighted average completion time (the subscript $w$ distinguishes it)
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- $\bar{p}$ = average processing time
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---
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## The Delta: Change ($\Delta$)
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$\Delta$ (capital delta) means "the change in" or "the difference."
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- $\Delta_i = C_i - p_i$ — the delay for task $i$ (how much longer it
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waited beyond its own processing time)
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- $\Delta D = w(q_j) \cdot p_i - w(q_i) \cdot p_j$ — how much the
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priority-weighted delay cost changes when you swap two tasks
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---
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## Subscripts and Superscripts
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**Subscripts** (below) are labels or indices:
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- $p_i$ = processing time of task $i$
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- $p_{\max}$ = the largest processing time
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- $p_{\min}$ = the smallest processing time
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- $\bar{C}_{\text{SPT}}$ = average completion time under the SPT schedule
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- $\bar{C}_{\text{priority}}$ = average completion time under priority scheduling
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**Superscripts** (above) indicate context or power:
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- $p_a^2$ = $p_a \times p_a$ (squared)
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- $\bar{C}^{\text{SPT}}$ = completion time under SPT (same idea as subscript, just different formatting)
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---
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## Set Notation
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$$q_i \in \{1, 2, 3, 4\}$$
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The curly braces $\{ \}$ define a **set** — a collection of possible
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values. $\in$ means "is a member of." So this reads: "the priority
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$q_i$ is one of the values 1, 2, 3, or 4."
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---
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## The Piecewise Function (Cases)
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$$w(q) = \begin{cases} 8 & q = 1 \text{ (Critical)} \\ 4 & q = 2 \text{ (High)} \\ 2 & q = 3 \text{ (Medium)} \\ 1 & q = 4 \text{ (Low)} \end{cases}$$
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This defines a function that gives different outputs depending on the
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input. Read it as: "If $q$ is 1, then $w$ is 8. If $q$ is 2, then $w$
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is 4." And so on. It is just a lookup table written in math notation.
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---
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## Limits and Convergence
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$$\lim_{T \to \infty} \frac{W(T)}{T} = \mu$$
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Read as: "As $T$ gets larger and larger (approaching infinity), the value
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of $W(T) / T$ gets closer and closer to $\mu$." In this paper, it means:
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over a long enough time horizon, the throughput settles to a fixed rate
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$\mu$ regardless of scheduling order.
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- $\to$ means "approaches" or "goes toward"
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- $\infty$ means infinity (without bound)
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---
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## Mutual Information ($I$)
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$$I(\sigma_{\text{SPT}}) = 0 \quad \text{when } p_i \perp q_i$$
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$I$ here measures **how much knowing one thing tells you about another**.
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$I = 0$ means "knowing a task's position in the SPT schedule tells you
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absolutely nothing about its priority." The schedule and the priority
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system are statistically independent — they contain no information about
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each other.
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---
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## Functions
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$f(\bar{C})$ means "some function of $\bar{C}$." A function takes an
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input and produces an output. When the paper writes:
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$$U_{\text{client}} = f\!\left(\bar{C}(\sigma)\right), \quad f' < 0$$
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It means: client satisfaction ($U$) depends on the average completion
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time, and $f' < 0$ means the function is **decreasing** — when the
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average goes up, satisfaction goes down. The $'$ (prime) notation refers
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to the derivative, which indicates the direction of change.
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---
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## Correlation ($\text{Corr}$)
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$$\text{Corr}(p_i, q_i) \approx 0$$
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Correlation measures whether two quantities move together. A correlation
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of 0 means they are unrelated — knowing the size of a task tells you
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nothing about its priority. A positive correlation means they increase
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together; a negative correlation means one goes up when the other goes
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down.
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---
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## Variance ($\text{Var}$) and Standard Deviation ($\sigma_p$)
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Variance measures how spread out a set of numbers is. If all tasks take
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about the same time, variance is low. If some are 15 minutes and others
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are 40 hours, variance is high.
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The **coefficient of variation** $CV = \sigma_p / \bar{p}$ normalizes
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the spread by dividing by the average. Here $\sigma_p$ is the standard
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deviation (the square root of variance) of the processing times, and
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$\bar{p}$ is the mean. A $CV$ above 1 means the spread is wider than
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the average — the data is highly variable.
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(Note: $\sigma_p$ uses the same Greek letter as schedule $\sigma$, but
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they are different things. Context tells you which is which — $\sigma_p$
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with a subscript $p$ always means standard deviation of processing times;
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$\sigma$ alone or $\sigma(k)$ always means a schedule.)
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---
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## How to Read the Proofs
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### The Exchange Argument
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The most common proof technique in this paper works like this:
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1. Assume you have any schedule.
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2. Find two adjacent tasks that are in the "wrong" order.
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3. Swap them and show that the metric improves (or stays the same).
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4. Conclude: since every "wrong" pair can be improved by swapping,
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the best schedule has no "wrong" pairs — i.e., everything is in
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the "right" order.
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This is like proving that a sorted list is optimal by showing that
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any out-of-order pair can be fixed by swapping, and each swap makes
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things better.
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### Reading a Proof Step by Step
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Take the simplest proof in the paper (Theorem 1):
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> *"Consider any schedule in which two adjacent tasks $i, j$ satisfy*
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> *$p_i > p_j$ with task $i$ scheduled immediately before task $j$."*
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Translation: Pick any two neighboring tasks where the bigger one comes
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first.
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> *"Let $t$ be the start time of task $i$."*
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Translation: Call the clock time when we start working on task $i$ by
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the name $t$. We don't need to know what $t$ actually is.
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> *"The change in the sum of completion times is:*
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> *$(2p_i + p_j) - (p_i + 2p_j) = p_i - p_j > 0$"*
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Translation: When the bigger task comes first, the total completion time
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is $p_i - p_j$ hours **more** than if the smaller task came first. Since
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$p_i > p_j$, this difference is positive — swapping them makes it better.
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> *"Therefore SPT uniquely minimizes $\bar{C}(\sigma)$."* $\blacksquare$
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Translation: Since every swap of big-before-small improves the metric,
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the best possible schedule is all tasks sorted from smallest to largest.
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Proof complete.
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---
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## Quick Reference Card
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| When you see... | It means... |
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|----------------|-------------|
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| $\sum_{i=1}^{n}$ | Add up for each task from 1 to $n$ |
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| $\frac{a}{b}$ | $a$ divided by $b$ |
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| $\bar{X}$ | The average of $X$ |
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| $\Delta X$ | The change in $X$ |
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| $\sigma$ | A schedule (the order tasks are done) |
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| $\sigma(k)$ | Which task is in position $k$ |
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| $p_i$ | How long task $i$ takes |
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| $C_i$ | When task $i$ finishes |
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| $q_i$ | Priority class of task $i$ |
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| $w(q_i)$ | How much priority class $q_i$ is weighted |
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| $\le, \ge$ | Less/greater than or equal to |
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| $\ne$ | Not equal to |
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| $\in$ | Is a member of |
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| $\forall$ | For all |
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| $\perp$ | Independent of |
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| $\blacksquare$ | End of proof |
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---
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## A Note on Reading Math
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Mathematical notation is a compression format. The formula:
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$$\bar{C}(\sigma) = \frac{1}{n} \sum_{k=1}^{n} C_{\sigma(k)}$$
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says in 15 characters what takes a full sentence in English: "The average
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completion time under schedule $\sigma$ equals the sum of all individual
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completion times divided by the number of tasks."
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If a formula looks intimidating, break it apart:
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1. Find the main operation (usually $=$ splitting left from right)
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2. Identify the pieces on each side
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3. Replace the symbols with words using the tables above
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4. Read it as a sentence
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Every formula in this paper can be understood this way. The math is not
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doing anything more complicated than arithmetic — it is just doing it
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for a general case rather than specific numbers.
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---
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*This guide accompanies the paper "Unweighted Average Completion Time
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Is Not a Fair Metric for Task Scheduling" (2026-03-28).*
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