1. M. Veeraraghavan 1 Bridge 55 Bridge 41 Bridge 48 1 1 2 2 1 2 LAN A LAN B LAN C EXAMPLE

2. M. Veeraraghavan 2

3. 3 Bridge 55 Bridge 41 Bridge 48 1 1 2 2 1 2 At Time t=0 each bridge sends a BPDU thinking it is the root [55,0,55] LAN A LAN B [41,0,41] [48,0,48] LAN C

4. M. Veeraraghavan 4 Bridge 55 Bridge 41 Bridge 48 1 1 2 2 1 2 At time t=0+  each bridge receives BPDUs that were sent at t=0 [48,0,48] [41,0,41] LAN A LAN B [55,0,55] [48,0,48] [55,0,55] [41,0,41] Assume that it takes a small  amount of time to receive the BPDUs LAN C

5. M. Veeraraghavan 5 Actions at bridge 55 At t=0+ , bridge 55 has received two BPDUs [48,0,48] and [41,0,41]. Its own BPDU (that it sent at t=0) is [55,0,55]. Order these according to ordering rules: –Order is: [41,0,41] [48,0,48] [55, 0, 55] Smallest root ID received is 41: so bridge 55 now thinks root is 41.

6. M. Veeraraghavan 6 Actions at bridge 55 contd. It received the BPDU [41,0,41] on its port 2. So its root port is 2. The root path cost is the cost in the BPDU [41,0,41] + 1 = 1. The reason it uses this BPDU is that the min-cost BPDU from the root is [41, 0, 41].

7. M. Veeraraghavan 7 Actions at bridge 55 contd. The new BPDU computed at bridge 55 is [41, 1, 55] which it will send at t=1 Test ports to check if bridge 55 is the designated bridge for the LAN connected to each port –Port 1: bridge 55 received [48, 0, 48] and now it is ready to send [41, 1, 55]. Since its BPDU is lower in the order, it decides port 1 is a designated port. –Port 2: bridge 55 received [41, 0, 41] on this port. Since this is lower than its own BPDU, port 2 is not a designated port and its BPDU will not be sent on this port.

8. M. Veeraraghavan 8 Actions at other bridges (48, 41) Similar actions occur at these bridges 41, 48. Each bridge computes: –root bridge –root port and root path cost –its new BPDU (which it will send at t=1) –designated ports

9. M. Veeraraghavan 9

10. 10 Bridge 55 Bridge 41 Bridge 48 1 1 2 2 1 2 At time t=1 each bridge sends its newly computed BPDU [41,1,55] LAN A LAN B [41,0,41] [41,1,48] LAN C

11. M. Veeraraghavan 11 Bridge 55 Bridge 41 Bridge 48 1 1 2 2 1 2 [41,1,48] [41,0,41] LAN A LAN B [41,1,55] [41,0,41] Assume that it takes a small  amount of time to receive the BPDUs At time t=1+  each bridge receives BPDUs that were sent at t=1 LAN C

12. M. Veeraraghavan 12 Actions at bridge 55 At t=1+ , bridge 55 has received two BPDUs [41,1,48] and [41,0,41]. Its own BPDU (that it sent at t=1) is [41,1,55]. Order these according to ordering rules: –Order is: [41,0,41] [41,1,48] [41,1, 55] Smallest root ID received is 41: so bridge 55 now thinks root is 41.

13. M. Veeraraghavan 13 Actions at bridge 55 contd. It received the BPDU [41,0,41] on its port 2. So its root port is 2. The root path cost is the cost in the BPDU [41,0,41] + 1 = 1. The reason it uses this BPDU is that the min-cost BPDU from the root is [41, 0, 41].

14. M. Veeraraghavan 14 Actions at bridge 55 contd. The new BPDU computed at bridge 55 is [41, 1, 55] which it will send at t=2 Test ports to check if bridge 55 is the designated bridge for the LAN connected to each port –Port 1: bridge 55 received [41, 1, 48] and now it is ready to send [41, 1, 55]. Since its BPDU is higher in the order, it will decide that port 1 is a not a designated port and hence it will not send its BPDU on this port. –Port 2 was already decided not to be a designated port.

15. M. Veeraraghavan 15 Actions at other bridges (48, 41) Similar actions occur at these bridges 41, 48. Each bridge computes: –root bridge –root port and root path cost –its new BPDU (which it will send at t=1) –designated ports

16. M. Veeraraghavan 16

17. M. Veeraraghavan 17 Designated and Root ports Bridge 55 Bridge 41 Bridge 48 1 1 2 2 1 2 LAN A LAN B D D D R R LAN C

18. M. Veeraraghavan 18 Spanning Tree Bridge 55 Bridge 41 Bridge 48 LAN A LAN B LAN C

19. M. Veeraraghavan 19 A host on LAN A sends a MAC frame to a host on LAN B Bridge 55 Bridge 41 Bridge 48 1 1 2 2 1 2 LAN A LAN B D D D R R LAN C Source Host Destination Host