A local company (let’s call it “Lord and Master Ltd.”) is using an ISP (Internet

Question

A local company (let’s call it “Lord and Master Ltd.”) is using an ISP (Internet Service Provider) to connect their internal TCP/IP network to the Internet. The ISP has allocated one Class-B subnet to L&M. The network and subnet-mask which have been allocated are: network – 150.47.0.0/20. L&M has been allocated the subnet number 3 (out of the possible 14 from the ISP, since ISP does not use all "0" and all "1" addresses for hosts) for its own exclusive use. L&M is a mid-size company but still with no knowledge about NAT.

1- How many Internet-connected hosts can L&M have?

2- What is the subnet IP address for Lord and Master Ltd?

3- Currently, L&M has 8 departments with no more than 254 nodes needed in each department for immediate growth. If subneting is suggested to accommodate this situation so that there is a subnet for each department, what subnet mask should be used within L&M in dotted decimal representation?

4- What is the broadcast address to cover all hosts in all the L&M subnets in dotted-decimal representation?

Explanation / Answer

Answer to 1:

There is no NAT. So number of hosts which may connect to Internet is the the number of IP addresses allocated to L&M Ltd.

The Class-B subnet allocated to L&M Ltd is 150.47.0.0/20 i.e. having 20 bits network id. Thus there are (32-20)=12 bits host id.

Therefore 212 = 4096 hosts of L&M ltd can connect to Internet.

Answer to 2:

Subnet IP address = 150.47.0.0 & 255.255.240.0 = 150.47.0.0

Answer to 3:

Since each department requires atmost 254 hosts which is less than 255 we can assign the following subnet to 8 departments:

Department 1 : 150.47.1.0/24

Department 2 : 150.47.2.0/24

Department 3 : 150.47.3.0/24

Department 4 : 150.47.4.0/24

Department 5 : 150.47.5.0/24

Department 6 : 150.47.6.0/24

Department 7 : 150.47.7.0/24

Department 8 : 150.47.8.0/24

Thus the subnet mask within L&M ltd for these departments is 255.255.255.0

Answer to 4:

For L&M Ltd,

Subnet address = 150.47.0.0

Subnet mask = 255.255.240.0

Complement of Subnet mask = 0.0.15.255

So broadcast address = 150.47.0.0 & 0.0.15.255 = 150.47.15.255

NB: Hope it helps to get you answer. Let me know any concern to help you better.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.