Doc. Add information about more advanced ordering coefficients
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Evgeniy L
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@ -361,7 +361,7 @@ Which can be represented as next inequality.
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\begin{cases}
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x_1 + x_2 \le 100 \\
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x_3 + x_4 \le 200 \\
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x_3 + x_4 \le 100 \\
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x_1 + x_3 = 100 \\
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x_2 + x_4 = 100
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\end{cases}
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@ -373,13 +373,13 @@ And objective function.
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Maximize: x_1 + x_2
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So we may have two solutions here:
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So we may have two obvious solutions here:
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#. var for 1st disk, root for 2nd
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#. root for 1st disk, var for 2nd
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Objective function is being used by the algorithm to decide, which solution
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"better". Currently all elements in coefficients vector are equal to 1
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is "better". Currently all elements in coefficients vector are equal to 1
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.. math::
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c = \begin{bmatrix}
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@ -400,7 +400,153 @@ We can change coefficients in a way that first volume has higher coefficient tha
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1
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\end{bmatrix}^{T}\\[2ex]
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Now Linear Pgroamming solver will try to maximize the solution with respect to specified order of spaces.
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Now Linear Programming solver will try to maximize the solution with respect to specified order of spaces.
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But it's not so simple, if we take a closer look at the results we may get.
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Lets consider two solutions and calculate the results of objective function.
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.. math::
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cx = \begin{bmatrix}
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100 &
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0 &
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0 &
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100
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\end{bmatrix}
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\begin{bmatrix}
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4 \\
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3 \\
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2 \\
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1
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\end{bmatrix} =
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\begin{bmatrix}
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400 \\
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0 \\
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0 \\
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100
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\end{bmatrix}\\[2ex]
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sum\{cx\} = 500
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The result that objective function provides is **500**, if **root** is allocated on the first disk and **var** on second one.
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.. math::
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cx = \begin{bmatrix}
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50 &
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50 &
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50 &
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50
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\end{bmatrix}
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\begin{bmatrix}
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4 \\
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3 \\
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2 \\
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1
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\end{bmatrix} =
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\begin{bmatrix}
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200 \\
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150 \\
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100 \\
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50
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\end{bmatrix}\\[2ex]
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sum\{cx\} = 500
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The result that objective function provides is **500**, if **root** and **var** are allocated equally on both disks.
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So we need a different monolitically increasing sequence of integers, which is increasing as slow as possible.
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Also sequence must not violate next requirements.
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.. math::
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n_{i+1} \gt n_i \\[2ex]
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n_{i} + n_{j+1} \gt n_{i+1} + n_{j}$ where $i+1 < j
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It means that sum of ranges taken on the "lower side" of a sequence has to be always smaller than sum of range on "higher side".
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In our example this requirement is violated
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.. math::
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i = 1\\
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j = 3\\
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1 + 4 = 2 + 3
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Such sequence has been `found <http://math.stackexchange.com/questions/1596496/finding-a-monotonically-increasing-sequence-of-integers/1596812>`_
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.. math::
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1,2,4,6,9,12,16,20,25,30,36,42\cdots
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there are `many ways <https://oeis.org/A002620>`_ to caculate such sequence, in our implementation next one is being used
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.. math::
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a_n=\lfloor\frac {n+1}2\rfloor\lfloor\frac {n+2}2\rfloor
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Lets use this sequence in our examples
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.. math::
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cx = \begin{bmatrix}
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100 &
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0 &
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0 &
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100
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\end{bmatrix}
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\begin{bmatrix}
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6 \\
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4 \\
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2 \\
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1
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\end{bmatrix} =
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\begin{bmatrix}
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600 \\
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0 \\
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0 \\
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100
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\end{bmatrix}\\[2ex]
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sum\{cx\} = 700
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And when **root** and **var** are allocated on both disks equally
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.. math::
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cx = \begin{bmatrix}
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50 &
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50 &
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50 &
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50
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\end{bmatrix}
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\begin{bmatrix}
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6 \\
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4 \\
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2 \\
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1
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\end{bmatrix} =
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\begin{bmatrix}
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300 \\
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200 \\
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100 \\
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50
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\end{bmatrix}\\[2ex]
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sum\{cx\} = 650
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Soo :math:`700 > 650` first function has greater maximization value, that is exactly what we need.
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Weight
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~~~~~~
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@ -444,7 +590,7 @@ Each space can have **weight** variable specified (**1** by default), which is u
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x_1 + x_2 \le 100 \\
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x_3 + x_4 \le 200 \\
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x_1 + x_3 = 100 \\
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x_2 + x_4 = 100
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x_2 + x_4 = 100 \\
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x_2 * (1 / weight) + x_4 * (-1 / weight) = 0
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\end{cases}
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@ -498,18 +644,13 @@ So in solver we get a list of **ssd** disks, if there are any.
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Lets adjust coefficients to make ceph-journal to be allocated on ssd, as a second priority ordering should be respected.
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In order to do that lets make order coefficient :math:`0 < order_coefficient < 1`.
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In order to do that lets make order coefficient :math:`0 < order coefficient < 1`.
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.. math::
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c = \begin{bmatrix}
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1 + (1/5) &
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1 + (1/9) &
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0 + (1/6) &
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0 + (1/4) &
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0 + (1/3) &
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1 + (1/2)
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\end{bmatrix}^{T}\\[2ex]
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Advacned ordering
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~~~~~~~~~~~~~~~~~
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