Question

Please explain answer and show steps if possible

A new network is being designed for your company, Acme, Inc. If you use a Class C IP network, which subnet masks will provide one usable subnet per department while allowing enough usable host addresses for each department specified in the table? (Choose three)

Department	Number of Users
Corporate	60
Customer Support	62
Financial	25
HR	5
Engineering	4

1. 255.255.255.0
2. 255.255.255.128
3. 255.255.255.192
4. 255.255.255.224
5. 255.255.255.240
6. 255.255.255.248
7. 255.255.255.252

The network address of 172.16.0.0/20 provides how many subnets and hosts?

1. 7 subnets, 30 hosts each
2. 16 subnets, 8,190 hosts each
3. 16 subnets, 4,094 hosts each
4. 7 subnets, 2,046 hosts each

You need to configure a server that is on the subnet 192.168.19.24/29. The router has the last available host address. Which of the following should you assign to the server?

1. 192.168.19.0 255.255.255.0
2. 192.168.19.33 255.255.255.240
3. 192.168.19.26 255.255.255.248
4. 192.168.19.30 255.255.255.248

On a VLSM network, which mask should you use on point-to-point WAN links to reduce the waste of IP addresses?

1. /27
2. /30
3. /28
4. /29

Answer 1

1)

Department	Number of Users	Subnet Mask
Corporate	60	255.255.255.192 (62 hosts)
Customer Support	62	255.255.255.192 (62 hosts)
Financial	25	255.255.255.224 (30 hosts)
HR	5	255.255.255.248 (6 hosts)
Engineering	5	255.255.255.248 (6 hosts)

Therefore subnet masks required are : c) 255.255.255.192 d) 255.255.255.224 f) 255.255.255.248

2)

Given network address 172.16.0.0/20

Note that it is class B network as it ranges (128 - 191) in the first octet.

/20 indicates subnet mask 255.255.240.0 so it has 4 subnet bits.

Therefore, Number of subnets = 2⁴ = 16 subnets.

Host id = 32 - 20 = 12 bits.

Therefore, Hosts per subnet = 2¹² = 4096/ subnet.

Usable hosts = 4096 - 2 = 4094/ subnet.

Therefore, option c) is correct 16 subnets, 4,094 hosts each.

3)

Given subnet 192.168.19.24/29

/29 has a subnet mask of 255.255.255.248.

Host ID bits = 32 - 29 = 3

Hosts per subnet = 2³ = 8.

Each subnet is a size of 8 in the fourth octet. So subnet in the fourth octets are 0, 8 ,26, 24, 32, 40...etc

Broadcast address of subnet  192.168.19.24 is 192.168.19.31

Last available host address o s(Broadcast address - 1) which is 192.168.19.30

Therefore, option D which 192.168.19.30 , 255.255.255.248 is the correct option.

4)

Point to point wan links has only two hosts. So it just needs two host address.

/30 indicates subnet mask 255.255.255.252

/Host ID bits = 32 - 30 = 2

Number of hosts = 2² = 4 but number of usable hosts = 4 - 2 = 2

/30 provides exactly 2 hosts per subnet. So there is no wastage og host addresses.

Therefore, Option b is correct.