Skip to content

GATE CS 2014 Set 2 Q25 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

Explain with AI

In the diagram shown below, L1L1 is an Ethernet LAN and L2L2 is a Token-Ring LAN. An IPIP packet originates from sender SS and traverses to RR, as shown. The links within each ISP\text{ISP} and across the two ISP\text{ISP}s, are all point-to-point optical links. The initial value of the TTL\text{TTL} field is 3232. The maximum possible value of the TTL\text{TTL} field when RR receives the datagram is _______.

View worked solution

Answer: 26

To find the maximum possible value of the Time-to-Live (TTL) field when the receiver RR receives the datagram, we need to understand how TTL is processed across different network layers:

  1. Layer 3 (Network Layer) Processing: The TTL field is part of the IP header. Every router (Layer 3 device) that forwards the IP packet decrements the TTL value by at least 11.
  2. Layer 2 (Data Link Layer) Processing: LAN devices such as switches, bridges, or hubs operating at the Data Link Layer (within Ethernet L1L1 and Token-Ring L2L2) do not modify or decrement the TTL field.

By tracing the path from sender SS to receiver RR in the given network diagram, the packet traverses exactly 66 routers:

  • Router 1: Connects Ethernet LAN L1L1 to ISP 1.
  • Router 2 & Router 3: Internal/boundary routers within ISP 1.
  • Router 4 & Router 5: Internal/boundary routers within ISP 2.
  • Router 6: Connects ISP 2 to Token-Ring LAN L2L2.

Since each of these 66 routers decrements the TTL by 11:

Final TTL=Initial TTL−Number of Routers\text{Final TTL} = \text{Initial TTL} - \text{Number of Routers} Final TTL=32−6=26\text{Final TTL} = 32 - 6 = 26

Thus, the maximum possible value of the TTL field when RR receives the datagram is 26.

Free practice · Your answers on this page are not saved.

Published
Report a correction
Keep exploringMore Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging questions