Ace Gate practice booklet Volume 1

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Shivam_j asked in Computer Networks Oct 16, 2022 retagged Oct 17, 2022 by makhdoom ghaya

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Abhrajyoti00 · Answer 1 · 2022-10-17T14:18:17+0000

Answer: 1022

Class B Subnet Mask given = $255.255.119.0$ = $11111111.11111111.01110111.00000000$

Here it is a non-contiguous subnet mask, which is possible only theoretically, not practically.

In this type of SM, we calculate only no. of $1’s$ as part of NID + SID and $0’s$ as HID.

Thus here, no. of $1’s$ = $22 \ bits$

NID = Class B = $16 \ bits$

SID = $22-16 = 6 \ bits$

HID = $32-22 = 10 \ bits$

Thus no. of hosts/subnet possible = $2^{10}-2 = 1022$