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Example: LOM Calculation

The following example illustrates how an LOM calculation is made for four classes of traffic: ct0, ct1, ct2, and ct3.

The class types have been assigned the following values:

ct0 = 40ct1 = 30ct2 = 20ct3 = 10

These class type values yield the following bandwidth constraints:

BC0 = (ct3 + ct2 + ct1 + ct0) = 100BC1 = (ct3 + ct2 + ct1) = 60BC2 = (ct3 + ct2) = 30BC3 = (ct3) = 10

LSPs from class type ct0 can take up to 100 percent of bandwidth on the link. LSPs from class type ct1 can take up to 60 percent of the bandwidth on the link, and so on.

If you assume for this example that the class types have the following LOM values:

LOM(ct0) = 8LOM(ct1) = 4LOM(ct2) = 2LOM(ct3) = 1

In the absence of any other reservation, LSPs from class type ct0 can take up to 800 percent of the available bandwidth (8 x 100 = 800). In the absence of any other reservation, LSPs from class type ct1 can take up to 240 percent of the available bandwidth (4 x 60 = 240). and so on.

The maximum amount of bandwidth that can be reserved is:

ct0 = LOM(ct0) x BC0 = 800ct1 = LOM(ct1) x BC1 = 240ct2 = LOM(ct2) x BC2 = 60ct3 = LOM(ct3) x BC3 = 10

For the undersubscribed class type ct3, the maximum reservable bandwidth is the same as the bandwidth constraint. For the overbooked class types, these values are not the values of the bandwidth constraint-taking into account the oversubscription for each class type separately. The oversubscription per class type in the sum is not taken into account because ultimately the entire bandwidth constraint can be filled with the bandwidth reservation of just one class type, so you have to account for that class type’s bandwidth oversubscription only.

When calculating the available bandwidth for CTc, you need to express reservations from other classes as if they were from CTc. The reservation from class ctx is normalized with the LOM of ctx, but it is then multiplied by the LOM of CTc.

For the previous example, assume that LSP1 has class type ct3 configured with bandwidth of 10 and a priority of 0.

The values for the reservable bandwidth will be:

ct0 = 8 x (100 - 10) = 720ct1 = 4 x min((100-10), (60-10)) = 200ct2 = 2 x min((100-10), (60-10), (30-10)) = 40ct3 = 1 x min((100-10), (60-10), (30-10), (10-10)) = 0

These numbers can be rationalized as follows: the normalized reservation is 10 percent. If this bandwidth came from class type ct0, it would be equivalent to an overbooked reservation of 80 percent. You can see that 720 percent (800 – 80 = 720) of the bandwidth remains available for other LSPs.


Published: 2010-04-28

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