[15] IP ADDRESS DECODING: IP: 139.182.148.50 NM: 255.255.254.0 128 64 32 1 1 1 NA: NETWORK ADDRESS 16 8 4 1 1 1 AND 255.255.254.0 = 11111111.11111111.11111110.00000000 10001010.10110110.10010100.00000000 139. 182. 148. 0 BA: BROADCAST ADDRESS 2 1 1 1 139.182.148.50=10001010.10110110.10010100.00110010 NA: BA: OR 255.255.254.0 = 00000000.00000000.00000001.11111111 10001010.10110110.10010101.11111111 139.182.148.50=10001010.10110110.10010100.00110010 139 182 PRACTICE: PLEASE CALCULATE [NA: Network Address] AND [BA: Broadcast Address] IP: 172.16.0.100 NM: 255.255.252.0 149 255 (First Address of the Network) (Last Address of the Network)
[15] IP ADDRESS DECODING: IP: 139.182.148.50 NM: 255.255.254.0 128 64 32 1 1 1 NA: NETWORK ADDRESS 16 8 4 1 1 1 AND 255.255.254.0 = 11111111.11111111.11111110.00000000 10001010.10110110.10010100.00000000 139. 182. 148. 0 BA: BROADCAST ADDRESS 2 1 1 1 139.182.148.50=10001010.10110110.10010100.00110010 NA: BA: OR 255.255.254.0 = 00000000.00000000.00000001.11111111 10001010.10110110.10010101.11111111 139.182.148.50=10001010.10110110.10010100.00110010 139 182 PRACTICE: PLEASE CALCULATE [NA: Network Address] AND [BA: Broadcast Address] IP: 172.16.0.100 NM: 255.255.252.0 149 255 (First Address of the Network) (Last Address of the Network)
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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