BUSINESS INTELLIGENCE ITM - ITM 305 please help me answer this questions. 1) You
ID: 3886484 • Letter: B
Question
BUSINESS INTELLIGENCE ITM - ITM 305
please help me answer this questions.
1) You were given the IP address of 201.89.5.0 by your network admin and she requires 7 new subnets be created out of the new network IP address.
Subnet the new network according to her requirement and list the following info for all the subnets. /CIDR , The network new Subnet Mask, All Subnets/Networks IP , all hosts IP and all Broadcast IP Addresses.
2) You are given an IP address of 192.180.176.0 from your ISP. You wanted to create a subnet that can accommodate a minimum of 120 nodes. Subnet the new network according to your requirement and list the following for all the new subnets. /CIDR , the network's Subnet Mask, All Subnets/Networks IP , all hosts IP and all Broadcast IP Addresses.
Explanation / Answer
Dear Student,
here is the answer...
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The given IP address is: 201.89.5.0
The IP address 201.89.5.0 is a class C address
default subnet mask of class C is 255.255.255.0
We need to borrow 3 bits from the host bit to create 7 subnets.
if we borrow the three bit from the host bits in the given address then the CIDR notation of the IP address will be: 201.89.5.0/27 (so here 27 bits are the network bits)
bits remaining for the host = 32-27 = 5
number of subnets = 23 = 8
now host per subnets = we have only 5 bits left in the last octet hence
hosts = 25 = 32
we know that two host are reserved
1) network address
2) broadcast address
hence valid hosts per subnets = 32-2 = 30
block size = 0, 31, 63, ....255
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below i have created a table which shows the All subnets, Network IP and host IP and the broadcast address.
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Answer 2:
The given IP address is: 197.180.176.0
The IP address 197.180.176.0 is a class C address
default subnet mask of class C is 255.255.255.0
We need to borrow 2 bits from the host bit to create 2 subnets and these subnets will have 128 hosts per subnet. And our requirement is minimum 120 hosts per subnet, hence our requirement fulfills here.
if we borrow the 1 bit from the host bits in the given address then the CIDR notation of the IP address will be: 197.180.176.0/25 (so here 25 bits are the network bits)
bits remaining for the host = 32-25 = 7
number of subnets = 21 = 2
now host per subnets = we have only 7 bits left in the last octet hence
hosts = 27 = 128
we know that two host are reserved
1) network address
2) broadcast address
hence valid hosts per subnets = 128-2 = 126
block size = 0, 128, 255
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we need only one subnet hence..
Subnet mask: 255.255.255.128
Network Address = 192.180.176.0/25
Broadcast Address = 192.180.176.127
First Host = 192.180.176.1
Last Host = 192.180.176.126
Hosts/Network = 126
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Kindly Check and Verify Thanks...!!!
Subnet No: Network IP Hosts range Broadcast address 1201.89.5.0/27
201.89.5.1 -
201.89.5.30
201.89.5.312
201.89.5.32/27
201.89.5.33 -
201.89.5.62
201.89.5.633
201.89.5.64/27
201.89.5.65 -
201.89.5.94
201.89.5.954
201.89.5.96/27
201.89.5.97 -
201.89.5.126
201.89.5.1275
201.89.5.128/27
201.89.5.129 -
201.89.5.158
201.89.5.1596
201.89.5.160/27
201.89.5.161-
201.89.5.190
201.89.5.1917
201.89.5.192/27
201.89.5.193-
201.89.5.222
201.89.5.223
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