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An organization requires a range of IP addresses to assign one to each of its 1500 computers. The organization has approached on Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space 202.61.0.0/17. The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP’s router using route aggregation. Which of the following address spaces are potential candidates from which the ISP can allot any one to the organization? I. 202.61.84.0/21 II. 202.61.104.0/21 III. 202.61.64.0/21 IV. 202.61.144.0/21

An organization requires a range of IP addresses to assign one to each of its 1500 computers. The organization has approached on Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space 202.61.0.0/17. The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP’s router using route aggregation. Which of the following address spaces are potential candidates from which the ISP can allot any one to the organization? I. 202.61.84.0/21 II. 202.61.104.0/21 III. 202.61.64.0/21 IV. 202.61.144.0/21 Correct Answer II and III only

Concept:

CIDR stands for Classless Inter Domain Routing. CIDR denotes the number of bits in the Netid.  Upon subnetting, any IP address consist of three subparts: Netid (bits for identifying the network), subnetid (bits for subnetting), and Hostid (bits for address allocation to hosts)

Calculation:

Given CIDR is 17, that means Netid=17.

There are 1500 devices, so the number of bits in Hostid= 11 (because 211 ≥ 1500)

Therefore, subnet bits= 32- (17+11) = 4

Hence, in any IP address in the given network, first 17 bits be part of network, next 4 of subnet ans last 11 bits would be for host address. Host bits should be all 0s

Address I: Cannot be allotted

202.61.84.0/21= 202.61. 0 1010 100.00000000. This address is not possible because the all the 11 host bits should be zero, but it contains a 1.

Address II: Can be allotted

202.61.104.0/21= 202.61. 0 1101 000.00000000. This address is possible as all the host bits are 0 and network bits are also satisfied.

Address III: Can be allotted

202.61.64.0/21= 202.61.0 1000 000.0000000. This address is possible as all the host bits are 0 and network bits are also satisfied

Address IV: Cannot be allotted

202.61.144.0/21= 202.61. 1 0010 000. 00000000. This address is not possible because bit in Netid has become 1.

Related Questions

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