3. (6) Given a network of 138.226.178.128/26, how many subnets of 12 addresses e
ID: 3634205 • Letter: 3
Question
3. (6) Given a network of 138.226.178.128/26, how many subnets of 12 addresses each can be
created in this network? Choose the maximum possible number.
a. 1
b. 2
c. 4
d. 8 answer: ______
Following that, specify the sub-network addresses corresponding to the answer you
provided above. For example, suppose your answer is b. (2), then you should provide
the network addresses for the 2 subnets individually in the CIDR notation.
answer: ____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________
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15. (2) A DHCP request from Host A is relayed by Router 128.132.36.162/28. When the DHCP
server receives the request, what would be the correct IP address that can be assigned to A?
a. 128.132.36.158
b. 128.132.36.173
c. 128.132.36.192
d. none of the above answer: ______
19. (6) Given Host A's address 122.219.36.189/28, and Host B's address 122.219.36.165/28.
What is the network mask in dotted decimal form?
answer: ____________________________________________________________
What is the prefix (i.e., network address) of Host A in dotted decimal form?
answer: ____________________________________________________________
What is the prefix (i.e., network address) of Host B in dotted decimal form?
answer: ____________________________________________________________
What is the network relation of Host A and B?
a. A and B are in the same network.
b. A and B are in the different networks.
c. Not enough information to determine. answer: _________
What is the suffix of Host A? answer: _________
What is the suffix of Host B? answer: _________
20. (10) Routing questions.
Given a router with 4 interfaces:
Interface Address
a. Ethernet0 128.192.160.1/22
b. Ethernet1 128.192.164.1/22
c. Ethernet2 128.192.168.1/21
d. Ethernet3 128.192.176.1/20
And its routing table of
Network netmask nexthop
128.192.0.0 255.255.128.0 (/17) 128.192.160.2
128.192.128.0 255.255.224.0 (/19) 128.192.191.254
128.192.192.0 255.255.192.0 (/18) 128.192.166.1
defaultroute 128.192.168.2
Assume the ARP cache of the router is empty. So after figuring out the next hop address,
the router needs to broadcast an ARP request to resolve the physical address of the target.
20.1 Which interface would a packet destined for 128.192.18.1 go out?
answer: _________
What is the target IP address for the ARP request? _________
20.2 Which interface would a packet destined for 128.192.200.1 go out?
answer: _________
What is the target IP address for the ARP request? _________
20.3 Which interface would a packet destined for 128.192.190.1 go out?
answer: _________
What is the target IP address for the ARP request? _________
20.4 Which interface would a packet destined for 131.94.133.12 go out?
answer: _________
What is the target IP address for the ARP request? _________
20.5 Which interface would a packet destined for 128.192.130.1 go out?
answer: _________
What is the target IP address for the ARP request? _________
I appreciate your help
Explanation / Answer
for 3. (6) from 138.226.178.128/26 we can say we have 2^6 addresses with us it can be divided into four sixteens so max=4 option is C subnetworks are 138.226.178.128/28, 138.226.178.144/28, 138.226.178.160/28, 138.226.178.176/28 we are using 2 bits out of 6 bits given for host address so those bits are 00,01,10,11 respectively. 19. (6) network mask in dotted decimal format is 11111111.11111111.11111111.11110000 for both Given Host A's address 122.219.36.189/28, and Host B's address 122.219.36.165/28 A's network address is 122.219.36.176 by taking first 28 bits and making last 4 as 0's(you can change this to dotted decimal format urself) B's network address is 122.219.36.160 by taking first 28 bits and making last 4 as 0's(you can change this to dotted decimal format urself) A & B are having different network addresses it can be infered they belong to different networks..
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