Path cost from ALSW1 to root bridge(DSW2) via DSW1 is 2 + 2 = cost of 4. ( 10gbps+10Gbps)
Path cost from ALSW1 to root bridge(DSW2) is 4 = cost of 4. ( 1Gbps)
Therefore the patch cost equal and DSW2 is root bridge Therefore it will via Gi0/2.
Answer is C.
IF let say the question change the ALSW1 to DSW2 Gi0/2 interface to 100Mbps (cost=19) , the answer will be different.
ALSW1 to DSW2 will be blocked.
Answer will be D.
STP Rule 1—All ports of the root switch must be in forwarding mode.
DSW2 is root because it has lowest priority 24576.
Therefore, ALSW1 will block port Gi0/1
The path will be PC1 to ALSW1 to DSW2 to DSW1
True, but forwarding ports on the root switch would be misleading if considered alone. Path cost is considered by the downstream switch based on the received BPDU, not at the root.
The answer is indeed C, here my explanation. First thing to check is the cost to the root from ALSW1, and the 2 path have equal cost (direct link = 4, non-direct link = 2+2 = 4). What do we check next? The bridge ID (which is the HEX representation of the priority + the bridge MAC), the lowest wins. A root bridge always has the lowest bridge ID. Conclusion: the root port for ASLW1 is the port directly connected to DSW2, hence traffic will go ALSW1 --> DSW2 --> DSW1
Cisco SWs use Rapid PVST by default which uses different cost values. That being the case , G0/1 becomes the root port because its cost to reach the root bridge (10g+10g= 2000+2000) is smaller than directly to the root thru g0/2 (1g = 20000), which enters the alternate state and doesn't forward traffic.
Guys, the trick is how to calculate path costs.
From ALSW1 to root bridge(DSW2) via DSW1 is 2 + 1 = cost of 3. ( 10gbps+20Gbps). This is because the link between DSW1 and DSW2 is Etherchannel 20Gbps. According to the table 2-2 from OSG, 20 Gbps gives us the cost 1, not 2!
after i read some comments i see many people confused about this.
don't make it hard on your self, just think simple.
we have 3 switches forming a circle, we all know that DSW2 is the root bridge and all RB ports are forwarding. in this case, we know for sure that the only connection left is the one between ALSW1 and DSW1, this link should be blocking on 1 of it's ports.
Yeah... I got this answer wrong because I assumed otherwise. But we are supposed to know that short method is the default.
SW-1#show spanning-tree pathcost method
Spanning tree default pathcost method used is short
I'm no sure but.
NOTE
The original IEEE specification did not account for links faster than 1 Gbps. Specifically, 1 Gbps links were assigned a port cost of 1, 100 Mbps link a cost of 10, and 10 Mbps links a cost of 100. Any link faster than 1 Gbps (i.e., 10 GE) was automatically assigned the same port cost of 1 Gbps links (i.e., port cost of 1).
refer link: https://www.ciscopress.com/articles/article.asp?p=2832407&seqNum=4#:~:text=The%20default%20port%20cost%20is,a%20port%20cost%20of%20100.
In the topology above, we see DSW2 has lowest priority 24576 so it is the root bridge for VLAN 10 so surely all
traffic for this VLAN must go through it. All of DSW2 ports must be in forwarding state. And:
+ The direct link between DSW1 and ALSW1 is blocked by STP.
+ The direct link between DSW1 and ALSW2 is also blocked by STP.
Therefore PC1 must go via this path: PC1 -> ALSW1 -> DSW2 -> DSW1
C. is the answer.
But NOT directly because DSW2 is root bridge. It does not matter, if ALSW1 is directly connected to the root bridge or not. The best path to the root is choosen and this has nothing to do with direct connection.
When a switch has several ways to the root bridge, the port is chosen based on
- The cost of the port (in this case both times 4, so it is equal)
- port priority (also equal, as no info given)
- switch ID (MAC)
Because DSW2 has a lower MAC, the path via Gi0/2 is chosen.
that's absolutely incorrect. it comes down to root port selection, which is the main issue in this question and depends on what calculation method and protocol are being used
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