255.255.255.248 is a legal Class C netmask True &nbs

255.255.255.248 is a legal Class C netmask True False
192.168.2.255 is a likely IP address for a host True False
If we use 4 bits from the last octet for the network, we would have 16 networks, with 16 hosts each True False

[Show your work – no calculators of any sorts]

We have a subnetted network with netmask 255.255.255.128 - What is the broadcast address for the first network? (Just the last octet is fine.)

Expert Solution

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1) 255.255.255.248 is a legal Class C netmask                                  True                False

- Range of class C IP address is 192.0. 0.0 to 223.255. 255.0.
- For subnet having 255.255.255.248 in class C, 3 bits are reserved for host ()
- Thus, the answer is TRUE

2.) 192.168.2.255 is a likely IP address for a host                              True                False

- This is a class C IP address.
- 24 bits are reserved for network bits and 8 are reserved for host bits.
- For each network, first and last address are reserved
- First address is reserved for network address and last is reserved for broadcast address and cannot be assigned to any host.
- The given IP address uses the last adress with a network and cannot be a host address
- Thus, answer is FALSE

3.) If we use 4 bits from the last octet for the network, we would have 16 networks, with 16 hosts each                        True                False

- If 4 bits are reserved for network from last octet, then number of network would be 2^4 = 16 is right.
- But, for host in each network, first and last address is reserved as stated in 2).
- First is reserved for network address and last address in a network is reserved for broadcase address.
- Thus, number of host in each network would be 2^4 - 2 = 14 hosts per network.
- Thus, the answer is FALSE

4) We have a subnetted network with netmask 255.255.255.128 - What is the broadcast address for the first network? (Just the last octet is fine.)

- According to this subnet, 25 bits are reserved for network bits.
- 11111111.11111111.11111111.10000000
- Last 7 bits are host bits.
- Therefore, for network address, convert the host bits to all 0
- So ,for network address, the last octed is 10000000
- But, for broadcast, we consider the last address.
- Convert all host bits to 1 for broadcast address.
- 1 1111111

Last octed of broadcast adress of first network - 11111111 (first 1 is a part of network bit and last 7 host bits)

venereology answered 1 year ago

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