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
Related questions
Question
Just answer please without explanation
![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= 2-2 = 27-2= 128-2=126 HOSTS
192.168.30.0/25
Subnet 1
192.168.30.0/25----Network Address
192.168.30.1----- First Useable Address 0+1=1
192.168.30.126----Last Useable Address 0+126=126
192.168.30.127.---Broadcast Address
126+1=127
Subnet 2
192.168.10.128/25-Network Address
0+128=128
192.168.10.129---
First Useable Address 128+1=129
192.168.10.254---Last Useable Address 128+126=254
192.168.10.255.-------Broadcast Address
254+1=255
Subnet Mask /25 > 255.255.255.128
Subnet Network First Host Last Host Broadcast
Address Address
Address
Address
1
What is the New Subnet Mask?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F72c5c1ee-c7ce-4765-92d2-3f0bdc1e0a72%2F5f588034-239a-4477-bfd0-425fd0468b6e%2F94cznvt_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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= 2-2 = 27-2= 128-2=126 HOSTS
192.168.30.0/25
Subnet 1
192.168.30.0/25----Network Address
192.168.30.1----- First Useable Address 0+1=1
192.168.30.126----Last Useable Address 0+126=126
192.168.30.127.---Broadcast Address
126+1=127
Subnet 2
192.168.10.128/25-Network Address
0+128=128
192.168.10.129---
First Useable Address 128+1=129
192.168.10.254---Last Useable Address 128+126=254
192.168.10.255.-------Broadcast Address
254+1=255
Subnet Mask /25 > 255.255.255.128
Subnet Network First Host Last Host Broadcast
Address Address
Address
Address
1
What is the New Subnet Mask?
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