Skip to content
GATETECH
Official PYQ · 2004Q57Fragmentation and IP Addressing, IPv4, CIDRMCQ1 mark

GATE CS 2004 Q57 — Fragmentation and IP Addressing, IPv4, CIDR

Explain with AI

Consider three IP networks A,BA, B and CC. Host HAH_A in network AA sends messages each containing 180180 bytesbytes of application data to a host HCH_C in network CC. The TCP layer prefixes 2020 byte header to the message. This passes through an intermediate network BB. The maximum packet size, including 2020 byte IP header, in each network, is:

  • A:A: 10001000 bytes\text{bytes}
  • B:100B: 100   bytes\text{bytes}
  • C:1000C: 1000 bytes\text{bytes}

The network AA and BB are connected through a 11 MbpsMbps link, while BB and CC are connected by a 512512 KbpsKbps link (bps = bits per second).

What is the rate at which application data is transferred to host HCH_C? Ignore errors, acknowledgments, and other overheads.

Choose one option
View worked solution

Answer: B

To find the application data transfer rate, we need to determine the total number of bytes transmitted over the bottleneck link (between BB and CC) for every 180180 bytes of application data.

  1. At Host HAH_A:
    • Application data = 180180 bytes
    • TCP segment size = 180 bytes (data)+20 bytes (TCP header)=200180 \text{ bytes (data)} + 20 \text{ bytes (TCP header)} = 200 bytes
  2. In Network BB:
    • The MTU of network BB is 100100 bytes, which includes a 2020-byte IP header. Thus, the maximum IP payload size per fragment is 100−20=80100 - 20 = 80 bytes.
    • Since 8080 is a multiple of 88, it is a valid fragment payload size.
    • The 200200-byte TCP segment is fragmented into:
      • Fragment 1: 8080 bytes payload + 2020 bytes IP header = 100100 bytes
      • Fragment 2: 8080 bytes payload + 2020 bytes IP header = 100100 bytes
      • Fragment 3: 4040 bytes payload + 2020 bytes IP header = 6060 bytes
    • Total bytes transmitted for one message = 100+100+60=260100 + 100 + 60 = 260 bytes.

The link between BB and CC has a bandwidth of 512512 Kbps and acts as the bottleneck. The efficiency of the transmission is:

Efficiency=Application DataTotal Data Transmitted=180260\text{Efficiency} = \frac{\text{Application Data}}{\text{Total Data Transmitted}} = \frac{180}{260}

Therefore, the application data transfer rate is:

Rate=512 Kbps×180260≈354.46 Kbps\text{Rate} = 512 \text{ Kbps} \times \frac{180}{260} \approx 354.46 \text{ Kbps}

This matches option B.

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

Published
Report a correction
Keep exploringMore Fragmentation and IP Addressing, IPv4, CIDR questions