Computer Science - Networking - Discussion
Discussion Forum : Networking - Section 3 (Q.No. 40)
40.
How many class A, B, and C network IDs can exist?
Discussion:
3 comments Page 1 of 1.
Hardik said:
1 decade ago
class A 126
class B 16382
class c 2097150
so ans is 126+16382+2097150 = 2113658
class B 16382
class c 2097150
so ans is 126+16382+2097150 = 2113658
GHC - Gaurav said:
9 years ago
First octet
Class A : 0 to 127 = (128) - 2 = 126 (1st octet).
Class B : 128 to 191 = (64 *256) - 2 = 16382 (1st octet * 2nd octet).
Class C : 192 to 223 = (32 * 256 * 256) - 2 = 2097150 (1st octet * 2nd octet * 3rd octet).
Class A : 0 to 127 = (128) - 2 = 126 (1st octet).
Class B : 128 to 191 = (64 *256) - 2 = 16382 (1st octet * 2nd octet).
Class C : 192 to 223 = (32 * 256 * 256) - 2 = 2097150 (1st octet * 2nd octet * 3rd octet).
Shewangizaw Bogale said:
8 years ago
Not clear to me.
Why 126 is not multiplied by 256 as it is the 1st octet?
And also for Class B and C 1st octets.
Why 126 is not multiplied by 256 as it is the 1st octet?
And also for Class B and C 1st octets.
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