You have an MPLS network and you have configured most-fill as a CSPF tiebreaker.Using the information in the exhibit, which path will be used to signal a new LSP requiring 12 Mbps?
A.
Path 1
B.
Path 2
C.
Path 3
D.
Path 4
Get 50% Discount on All Your Purchases
at PrepAway.com - Latest Exam Questions
This is ONE TIME OFFER
50%
why not option A ??
0
0
First it will tie on basis of fewest hop which is path 1 and 2. Now it will examine the most fill and thus path 4 will be selected.
0
0
Path 1 cannot be selected : 90% = 90 Mbps is already reserved of the 100 Mbps, so this path cannot be used for 12 Mbps required on the new LSP
Path 2, Path3 and Path 4 can be used but because of the most-fill as a CSPF tiebreaker.
Path 2 is the answer: because is it the less used at 10%
B. Path 2 is the right answer
0
0
Most Fill
———
The most-fill option prefers the path with the smallest minimum available bandwidth ratio. This rule
tries to fill a link before moving traffic to alternative link and might be preferred in certain usage-based billing environments where bulk discounts are gained by consolidating as much traffic onto as few links as possible. The most-fill option tends to fully pack your lower bandwidth links first.
such that your highest bandwidth links remain available for LSPs with large bandwidth requirements, which is another possible motivation for using this type of load balancing.
but also take in your consideration that we choose the smallest hops before compare the BW.
so we will choose D because A didn’t grantee the required BW 12 M.
0
0
The no specified on the links are available bandwidth ratio….
so the bottom link has the lowest bandwidth available and would be used for new lsp
correct answer is D.
0
0
Why 90%=90M? What is meant by ‘percentage reflect total bw reserved’
0
0
BW Reserved = Available BW
0
0
Path 1 can’t be used because it do not have avilable bandwidth
most-fill will not work, because CSPF will prefer path with fewest number of hops.
http://www.juniper.net/techpubs/en_US/junos9.6/information-products/topic-collections/nog-mpls-logs/cspf-understanding.html
0
0
To attempt to answer samis question; 90%=90M here because the diagram says all links are Fast E or 100M. Therefore 90% = 90M here because of the diagram.
0
0