# Terminology ¶

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Learn about the terminology for scheduling problems (optimal cost, early/last start time, and total/free/certain margin).

optimal cost/duration

Also called durée optimale/coût optimal, this is the least number of days the project will last. This is the last task (usually called END) early start time.

Note: we are using the notation $A(10)$, for the task $A$ having a duration/cost of $10$.

 early start time In French, it's called date au plus tôt. It's the number of days/... you will have to wait before doing this task. for the first one, it's 0 for the next ones, it's the maximum result of the sum of the previous early start time and the task duration/cost Ex: If a task $C$ need the task $A(10)$ and $B(20)$, and they are both starting after 30 days, then the task C while start after at least $\max(30+20, 30+10)=50$. The early start time for $C$ is $50$. last start time In French, it's called date au plus tard. This is the last date for a task and if we pass this date then the optimal duration of the project will increase. You will start from the end. for the last one, it's "early start time" value for the previous ones, it's the minimum result of the subtraction of the predecessor last start time their duration Ex: If a task $C(last=???)$ got two predecessors $D(cost=4, last=45)$ and $E(cost=7, last=45)$ then the $\min(45-4,45-7)=38$, so we have $C(last=38)$.
margin
total

Also called marge totale. This is the maximum delay that we can take for a task without affecting the optimal cost.

Operation: early - last

Ex: If $D(early=42, last=45)$ then the total margin is $45-42=3$.

free

Also called marge libre. Same as the total margin, but without changing the next task starting date.

Operation: solve $x$ @ x + \text{early_start} + \text{cost} \le \forall_{successor}\hspace{0.3cm} \text{early_start_successor} @

Ex: the free margin for $I(cost=7, early=30)$ having one successor $J(early=69)$ is $x + 30 + 7 \le 69 \Leftrightarrow x = 69-37=32$

certain

Also called marge certaine. Same as the free margin, but considering that every task started with the maximum delay.

Operation: $a$ is the certain margin value of the task $A$ if for all predecessors $p$ of $A$, $\text{last_start}_p - (\text{last_start}_A + cost(A \to p)) \ge a \ge 0$.

Ex: If a task $C(last=38)$ got two predecessors $D(cost=4, last=45)$ and $E(cost=7, last=45)$ then we have

• $45-38+4=3$
• $45-38+7=0$
• certain margin=0

Instead of margin, you can use

• "optimistic time estimate" instead of total margin
• "normal time estimate" instead of free margin
• "pessimistic time estimate" instead of certain margin

but I will use a French-friendly kind of name.

## Alternative example ¶

The early start time for C is

$\text{early start for C} = \max \sum_\text{i predecessor of C} \text{early_start}_i + cost(i \to C)$

If you got a task C that can only be done after the task "A" and the task "B", then simply check what's the task that you will have to wait for (=maximum), and the "wait" value is the "cost=duration + the start".

The last start time for C, it's the same as the early start time because C seems to be the last task.

The last start time for A, it's

$\text{last start for A} = \min \sum_\text{i successor of A} \text{last_start}_i - cost(A \to i) = min( 30 - 30 ) = 0$

The last start time for B, it's

$\text{last start for B} = \min \sum_\text{i successor of B} \text{last_start}_i - cost(B \to i) = min( 30 - 20 ) = 10$

The optimal cost/duration is simply the last task early start + cost. Since we don't have a cost for C, we can assume that the optimal duration is $30$. C task's name should have been END.