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A class B network address 130.50.0.0 is submitted as follows. The last 10 bits of the host I'd are allotte for host number and the remaining 6 bits are reserved for subnet number.

Q- what are the first hosts address of 1st and 4th subnets?

  1. 130.50.4.1 and 130.50.16.1
  2. 130.50.1.1 and 130.50.4.1
  3. 130.50.0.0 and 130.50.3.0
  4. None of these
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Ans: Option D) None of the Above

A class B network address given:  130.50.0.0

Class B = 16 Bits NID (Network ID) and 16 bits HID (Host ID).

Given, 6 bits are reserved for subnet number. Therfore, total subnets = 2^6. 

Remaining 10 bits for HID. So, _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _  (are last 16 bits) , in which the least significant 10 bits will be 0 0. 0 0 0 0 0 0 0 1 (as we need the 1st valid host address of both 1st and 4th subnets), and the most significant 6 bits are for SIDs (Subnet IDs) [not marked in BOLD Places] 

Now the weights of last 4 bits in 3rd Octet = 8 , 4, 2 & 1. Out of which, 2 & 1 are in Host Address. So remaining Subnet ID weights to be filled are for 8 and 4 with combinations : (00, 01, 10, 11) 

Now filling the SIDs (Subnet IDs) [ → 

For subnet 1 : 0 0 0 0 0 0 (and rest bold ones) → 130.50.0.1 (Ans)

For subnet 2 : 0 0 0 0 0 1 (and rest bold ones) → 130.50.4.1

For subnet 3 : 0 0 0 0 1 0 (and rest bold ones) → 130.50.8.1

For subnet 4 : 0 0 0 0 1 1 (and rest bold ones) → 130.50.12.1 (Ans) 

 

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