Help: Assume a host computer has the following configuration: IP Address: 72.240
ID: 3550951 • Letter: H
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Assume a host computer has the following configuration: IP Address: 72.240.84.176 Subnet Mask: 255.255.248.0 Default Gateway: 72.240.80.1 What is the Class of this network? Has this network been sub-netted? How do you know? How many possible hosts would there be on the above network if all usable addresses were assigned (show your work)? How would this IP address be expressed using CIDR notation? What is the range of the block of addresses this Host belongs to (show your work)? Give the network address using CIDR notation. Give the broadcast address using the CIDR notation.Explanation / Answer
a.) Class - A . It depends on the value of the first byte Class A (0-127) , Class B(128-191), Class C(192-223), Class D(225-239), Class E(240-255). Since 0 < 74 < 127 this is Class A.
b.) Yes , network is sub-netted. If we notice the subnet mask its 11111111 11111111 11111000 00000000, so it is divided as 21 bits and 11 bits part. There can be 2^11 that is 2048 atmost peers in the sub-net. And since the defalut gate-way is given , which says all the request's out of the subnet will be from the gateway ip, So obviously subnetted
c.)As indicated in part b there can be 2048 addresses.
d.) CDRI == 21 . CIDR specifies an IP address range,and since subnet mask is 11111111 11111111 11111000 00000000
count number of ones for CIDR.
e.) to get it do a logical "AND" operation between given ip and sub-net mask and we get 72.240.80.0
11111111 11111111 11111000 00000000 (subnet mask)
+01001000 11110000 01010100 10110000 (ip)
= 01001000 11110000 01010000 00000000 (72.240.80.0)
so range from 72.240.80.0 to 74.240.87.255
f.) CDRI for the network is 72.240.80.0/21 .
g.) Broadcast address is 74.240.87.255 which is the last address of the range.
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