Doc. Complete best with disks section

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Evgeniy L 2016-01-11 16:07:34 +03:00
parent 812e57fa9a
commit 3aa3ef23dd
1 changed files with 27 additions and 6 deletions

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@ -588,7 +588,7 @@ Each space can have **weight** variable specified (**1** by default), which is u
x_3 + x_4 \le 200 \\
x_1 + x_3 = 100 \\
x_2 + x_4 = 100 \\
x_2 * (1 / weight) + x_4 * (-1 / weight) = 0
x_2 * (1 / weight) - x_4 * (1 / weight) = 0
\end{cases}
To satisfy last equality, spaces have to be equal in size.
@ -646,8 +646,29 @@ In order to do that lets make order coefficient :math:`0 < \textrm{order coeffic
.. math::
c = \begin{bmatrix}
1 + (1/9) &
0 + (1/6) &
0 + (1/4) &
1 + (1/2)
\end{bmatrix}^{T}\\[2ex]
1 + (1/2) \\
0 + (1/4) \\
0 + (1/6) \\
1 + (1/9)
\end{bmatrix}
#. build sets according to selected disks, in our case we have two sets, **hdd** and **ssd** disks
#. for spaces which belong to specific set of disks add **1** to a coefficient which represents this space on a disk from the set
#. if space does not belong to the set of disks, add **0**
To make sure that spaces are always (unless size constraints are not violated), going to be allocated on the disks which they best suited with,
we automatically add a special artificial volume **unallocated**, whose coefficient is always **1**, and in this case we should change
coefficient of space which belongs to set of disks to **2**.
.. math::
c = \begin{bmatrix}
2 + (1/2)\\
0 + (1/4)\\
1 \\
0 + (1/9)\\
2 + (1/12)\\
1
\end{bmatrix}
As the result if space has one or more **best_with_disks**, it will be allocated on the disks only.