Practice Example #1C: 255.255.255.128 (/25) Network 192.168.10.0 Solve this Pls follow this step In picture /27 But in need in /25 192.168.1.0 255.255.255.224 (127) 111 00000 d. I Subnets 5 2. 2 = 32 hosts was S. 256-224 (32) N Ne NN ₂ 0, 32,64, 96, 128, 160, 191,224 N₂. 0 Subs First hast 1 best hast 30 Broadst 31 dast host+1= Subnet address No 32 65 33 62 94 63 95 64 اله حره ته د من الفنى ويزداد فقدار حدود hots . الرجال * First host=' Suburt address +1 * Last host Subnet - 2 address Broadcast - Lasthost +1 acolas NG NA 4 94 121 158 126 124 159 128 فوق العام الموارے) N 160 161 190 والا تو ان کے موجود 191 (214) 1921 193 2.22 215 254 255
Practice Example #1C: 255.255.255.128 (/25) Network 192.168.10.0 Solve this Pls follow this step In picture /27 But in need in /25 192.168.1.0 255.255.255.224 (127) 111 00000 d. I Subnets 5 2. 2 = 32 hosts was S. 256-224 (32) N Ne NN ₂ 0, 32,64, 96, 128, 160, 191,224 N₂. 0 Subs First hast 1 best hast 30 Broadst 31 dast host+1= Subnet address No 32 65 33 62 94 63 95 64 اله حره ته د من الفنى ويزداد فقدار حدود hots . الرجال * First host=' Suburt address +1 * Last host Subnet - 2 address Broadcast - Lasthost +1 acolas NG NA 4 94 121 158 126 124 159 128 فوق العام الموارے) N 160 161 190 والا تو ان کے موجود 191 (214) 1921 193 2.22 215 254 255
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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