simple spanning tree 802.1d, we have three ether switch on gns3

In this article of simple spanning tree 802.1d, we have three ether switch on gns3.

Switch AAA has vlan bia(built-in-address)  c400.0b14.0000 , switch BBB has vlan 1 bia c402.0b14.0000 and switch CCC has vlan bia c403.0b14.0000. Hence, assuming the AAA and BBB came up first, they sent their BPDU to each other with fields like Protocol ID,  Protocol version, BPDU type, flag, TCN, Root Bridge ID, Sender's bridge ID, Root Path cost, Port ID, Hello time, Max age, Message age, Forward delay.

Some overview about STP:

1. Switches send BPDU at every 2 sec interval by default.

2. Message age and max age is 20 sec by default.

3. Forward delay is 15 sec which is used in interval.

4. Any port takes maximum time of 50 seconds to reach the forwarding state from blocking. This can be broken into (Blocking to listening->20 sec, listening to learning MAC-> 15 sec, learning to Forwarding packets-> 15 sec)

5. BPDU types are: Configuration bpdu, TCN BPDU, TCN acknowledgement bpdu. A BPDU has 13 fields and it is important to know them each, that will help in understanding RSTP and other protocols further.

6. Switch A if is root and switch B connected to it over 100mb bandwidth than its designated cost is 0 to reach root A and if one more switch C is connected to B over 100 MB again then its DC cost will 19 as the cost is pre-specified for all link capacity.

** Designated cost is compared when the cost capacity of link is equal. If both links are 100mbps then DC can be taken into account. If bandwidth is different or altered like one is 10 and other is 100 Mbps then DC might not show the expected result even if it is 0 and the port will not become root port.

Continuing with our two switches AAA and BBB, both will claim each other as root bridge and will have their own sender id in bpdu.

AAA: I am the root bridge with my bpdu packet as below. Similarly,

BBB: I am the root bridge with my bpdu as just having my own MAC address(right).

Both switch now will compare each others and switch BBB found that it's MAC is of higher ID as sum of bridge priority+ mac ( mac being higher of BBB). So, now switch BBB will use all bpdu and will put mac add of AAA as it is lower in Root bridge ID field. On Ethernet switches lower is always the better. So here Root bridge is AAA.

Now CCC also has come up in similar fashion connected to AAA F1/1 and BBB F1/3.

CCC: Hello, I am CCC with mac c403.0b14.0000 and I am the root bridge in my BPDU(sent to AAA and BBB).

AAA sends it BPDU with self as root bridge & sender bridge ID and BBB also sends its BDPU with root ID that of AAA and self sender ID. Switch CCC compares its bridge ID with AAA and CCC, founds itself in poor to both. Similarly, BBB and AAA also compares and found CCC's bridge ID is higher and poor so no need for new root bridge election. Thus from now onward CCC and BBB will send all BPDUs with Root bridge ID that of AAA and self sender's ID. Ex:

BBB bpdu

**Root Bridge facilitates in mitigation of any bridging loop but remember root bridge AAA  itself doesn't thwarts the loop. 

Loop is broken when any non root bridge finds itself in a position where it received BPDUs with same root bridge ID from two of its ports and then it compares this with the initial results of comparison with its neighbors mac addresses, of course after comparing the cost to reach the root. It also means in technical jargon that a switch has two root ports which is a reason for loop. A switch can have as many ports to reach all friendly non root switches and we call those ports as designated forwarding ports but a switch must not have more than 1 port to reach the root. This is how a switch calculates Spanning-tree.

AAA spanning tree status;

BBB spanning tree status;

CCC spanning tree status below; please see the F1/3  here is in blocking state for #3 reasons, first the CCC lost the root bridge by all means because of higher mac + bridge priority(32768), secondly the cost to reach AAA root is less via f1/1 which is 0 (designated cost) than via f1/3 which is 19, thirdly F1/3 is greater than F1/1 and as I said lower is always better in three switches. 

The situation would look something like below:

CCC interface f1/3 is in BLK state

 

Now, we will extend couple of ether switches to CCC and BBB and will try to understand what their behavior will be one by one on the tongue tip.

Shown to the right, I have connected a DDD name switch by interface f1/4 to CCC. It is obvious DDD is now like a leaf and doesn't have anything next to it. So since AAA is the root, DDD will have f1/4 as the only way to reach the root. The problem occurs when switch DDD sends a BPDU with better and lower bridge ID. This will cause to CCC then to BBB and AAA to consider DDD root if it succeeds.

**Root bridge is a hallmark entity in any L2 architecture through which all other switches try to identify themselves in the positional arrangement under VLAN.

Similarly, if Switch EEE is connected to BBB as DDD is to CCC then it will show similar effect with EEE also acting as leaf bridge.

The output for EEE is,

EEE#sh spanning-tree brief

VLAN1

  Spanning tree enabled protocol ieee

  Root ID    Priority    32768

             Address     c400.0b14.0000

             Cost        38

             Port        45 (FastEthernet1/4)

             Hello Time   2 sec  Max Age 20 sec  Forward Delay 15 sec

  Bridge ID  Priority    32768

             Address     c40a.07e8.0000

             Hello Time   2 sec  Max Age 20 sec  Forward Delay 15 sec

             Aging Time 300

Interface                                                                   Designated

Name                           Port ID Prio    Cost  Sts        Cost      Bridge ID                         Port ID

-------------------- ------- ---- ----- --- ----- -------------------- -------

FastEthernet1/4           128.45   128    19      FWD    19        32768 c402.0b14.0000     128.45

Now, I will connect DDD and EEE on a common FE link f1/5 and a loop is bound to occur between AAA, BBB, EEE, DDD, CCC in a circle:

A tongue tip calculation involves calculation spanning tree the time you understood their bridge ID and the pattern of already blocked ports. Since f1/3 of CCC was already blocked so a topology was loop free until f1/5 on DDD and EEE was brought in. The moment I introduced f1/5  on DDD 7 EEE all switches could hear root bpdu on two of their ports. If to believe that AAA, BBB and CCC already had their understanding of what to be blocked, it was now turn of newly connected EEE and DDD over F1/5  to exchange BPDU and negotiate what to be blocked.

DDD and EEE didn't block f1/4 ports because it has least cost to reach the root AAA with 19(DC). If 1/4 is blocked, the cost could escalate to 19x2=38. That's DDD would reach AAA via EEE and then via BBB. Same is applicable for EEE should 1/4 was to be blocked. Thus we can conclude that given the least cost parameters one can reach the root AAA, the bridges can choose the best path depending on links bandwidth  and port priority solely than by administrative changes(least recommended).

Below is the EEE spanning tree output:

EEE#sh spanning-tree brief

VLAN1

  Spanning tree enabled protocol ieee

  Root ID    Priority    32768

             Address     c400.0b14.0000

             Cost        38

             Port        45 (FastEthernet1/4)

             Hello Time   2 sec  Max Age 20 sec  Forward Delay 15 sec

  Bridge ID  Priority    32768

             Address     c40a.07e8.0000                                                        ** EEE vlan 1 MAC address

             Hello Time   2 sec  Max Age 20 sec  Forward Delay 15 sec

             Aging Time 300

Interface                                                           Designated

Name                         Port ID Prio Cost  Sts    Cost       Bridge ID                         Port ID

-------------------- ------- ---- ----- --- ----- -------------------- -------

FastEthernet1/4        128.45   128    19  FWD  19         32768 c402.0b14.0000   128.45

FastEthernet1/5        128.46   128    19  BLK    38        32768 c409.0b14.0000   128.46

Now, let's make one more test just to check where we have reached with  our understanding.

We will connected a FE between BBB and DDD with bandwidth 100mbps.

The answer will be simple to at present prevalent topology. A new loop will be introduced with f1/6 between BBB & DDD.

So to thwart that again, who will block depends upon whose BID is what? DDD has poor bridge ID to CCC and BBB thought cost equates to 19(DC) for both. But since BBB has lower BID to CCC so DDD's f1/4 will go down to blocking. But if I reduce the cost of f1/6 link of DDD to 10 mbps, then certainly F1/4 will come and up and take the role of root port to reach the root bridge.