IPV4 ADDRESS 8+8+8+8=32 BITS NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH 192.168.30.0/24-Subnetmask N- Network Portion H- Host Portion No of valid Hosts= 2-2 = 28-2= 256-2=254 HOSTS 192.168.30.0 ----Network Address 192.168.30.1- First Useable Address 0+1=1 192.168.30. 254 ---------Last Useable Address 0+254=254 192.168.30.255 --------Broadcast Address 254+1=255 Subnetting IP Networks - Given the address block 192.168.30.0/24, create an IP addressing scheme that would allow for 2 subnets with 126 host IP addresses per subnet. Compute for the first two subnets only IPV4 ADDRESS 8+8+8+8=32 BITS NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH 192.168.30.0/24-Subnetmask N- Network Portion H- Host Portion N= Bits Borrowed-----1 BIT BORROW NNNNNNNN.NNNNNNNN.NNNNNNNN.NHHHHHHH No of Subnets= 2" =2=2 SUBNETS Remaining Host Bits(h) 8-1-7 No of valid Hosts= 2h-2 = 27-2= 128-2=126 HOSTS 192.168.30.0/25 Subnet 1
IPV4 ADDRESS 8+8+8+8=32 BITS NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH 192.168.30.0/24-Subnetmask N- Network Portion H- Host Portion No of valid Hosts= 2-2 = 28-2= 256-2=254 HOSTS 192.168.30.0 ----Network Address 192.168.30.1- First Useable Address 0+1=1 192.168.30. 254 ---------Last Useable Address 0+254=254 192.168.30.255 --------Broadcast Address 254+1=255 Subnetting IP Networks - Given the address block 192.168.30.0/24, create an IP addressing scheme that would allow for 2 subnets with 126 host IP addresses per subnet. Compute for the first two subnets only IPV4 ADDRESS 8+8+8+8=32 BITS NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH 192.168.30.0/24-Subnetmask N- Network Portion H- Host Portion N= Bits Borrowed-----1 BIT BORROW NNNNNNNN.NNNNNNNN.NNNNNNNN.NHHHHHHH No of Subnets= 2" =2=2 SUBNETS Remaining Host Bits(h) 8-1-7 No of valid Hosts= 2h-2 = 27-2= 128-2=126 HOSTS 192.168.30.0/25 Subnet 1
Chapter12: Network Configuration
Section: Chapter Questions
Problem 13RQ
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