GATE CS 2014 Set 2 — Question 35
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Computer Networks → Network Layer: Addressing → IPv4 Packet & Fragmentation
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Question
In the diagram shown below, L1 is an Ethernet LAN and L2 is a Token-Ring LAN. An IP packet originates from sender S and traverses to R, as shown. The links within each ISP and across the two ISPs, are all point-to-point optical links. The initial value of the TTL field is 32. The maximum possible value of the TTL field when R receives the datagram is __________.
Correct answer
26 to 26
Solution
The Time-To-Live (TTL) field in an IP packet is an 8-bit field that is decremented by one at each router the packet passes through. When the TTL value reaches zero, the packet is discarded. This mechanism prevents packets from looping indefinitely in the network.The initial TTL value is given as 32.To find the maximum possible value of the TTL field when the packet reaches the receiver R, we need to find the path with the minimum number of router hops from S to R. The TTL value at the receiver will be (Initial TTL) - (Number of hops).Let's trace the path and count the number of routers (hops) from S to R based on the provided diagram:
Final TTL value at R = 32 - 6 = 26.The maximum possible value of the TTL field when R receives the datagram is 26.
1.The packet originates at S in LAN L1 and is sent to the gateway router of LAN L1. Let's call this R1. (1st hop)
2.Router R1 forwards the packet to the first router in ISP1. Let's call this R2. (2nd hop)
3.Router R2 forwards the packet to the second router in ISP1. Let's call this R3. (3rd hop)
4.Router R3 forwards the packet to the first router in ISP2. Let's call this R4. (4th hop)
5.Router R4 forwards the packet to the second router in ISP2. Let's call this R5. (5th hop)
6.Router R5 forwards the packet to the gateway router of LAN L2. Let's call this R6. (6th hop)
7.Router R6 forwards the packet to the destination R within LAN L2.
The packet is forwarded by a total of 6 routers. Each router decrements the TTL value by 1.Total decrement in TTL = Number of hops = 6.Final TTL value at R = Initial TTL - Total decrementFinal TTL value at R = 32 - 6 = 26.The maximum possible value of the TTL field when R receives the datagram is 26.
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