Question

What is the size of an ARP packet when the protocol is IPv4 and hardware is Ethernet?

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Is the size of ARP packet fixed? Explain.

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Assume we have an isolated link (not connected to any other link) such as a private network in a company. Do we still need addresses in both network layer and data link layer? Explain.

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In a block code, a dataword is 20 bits, the corresponding codeword is 25 bits. What are the values of k, r, n according to the definition in the textbook? How many redundant bits are added to each dataword?

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In a codeword, we add two redundant bits to each 8 bit data word. How many (a) valid code word and (b) invalid code words are possible?

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Answer 1

1)
In the case of IPv4 networks running on Ethernet. In this scenario, the packet has 48-bit fields for the
sender hardware address and target hardware address, and 32-bit fields for the corresponding
sender and target protocol addresses . Thus, the ARP packet size in this case is 28 bytes.

   28 bytes = 8 bytes of header info plus12 bytes hardware +8 bytes network data

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2)

-->No
-->The ARP packet size must vary because it contains 2 Hardware/MAC addresses in it and 2 different
protocol addresses in it. Depending on the datalink and network protocol used the size addresses vary.

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4) dataword(K)=20 bits
codeword(N)=25 bits
Redundant bits(r)=n-k=25-20=5bits

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5)dataword - k=8
codeword-n
redundant r=n-k=2
       n=2+k=2+8
       n=10
a)valid codewords=2^k
       =2^8

b)non valid codewords=2^n - 2^k
           =2^10 - 2^8