Thursday, 21 January 2016

Subnetting/supernetting Examples

Q1.      IP address: 130.45.34.58/20 What is the subnet address?
Q2.      IP address: 19.30.80.5/18 What is the subnet address?
Q3.      A company is granted the site address 201.70.64.0. The company needs six subnets.Design the subnets.
Q4.      A company needs 1000 addresses. Which of the following set of class C blocks can be used to form a supernet for this company?
            A.            198.47.32.0         198.47.33.0         198.47.34.0
B.            198.47.32.0         198.47.42.0         198.47.52.0         198.47.62.0
C.            198.47.31.0         198.47.32.0         198.47.33.0         198.47.52.0
D.            198.47.32.0         198.47.33.0         198.47.34.0         198.47.35.0
Q5.      We need to make supernetwork out of 16 class C blocks. What is the supernet mask?
Q6.      A supernet has a first address of 205.16.32.0/21. A router receives three packets with the          following destination addresses:
   205.16.37.44
   205.16.42.56
   205.17.33.76
Q7.      A supernet has a first address of 205.16.32.0/21. How many blocks are in this supernet and      what is range of addresses?
Q8.       Which of the following can be beginning address of a block that contains 16 addresses?
                A.            205.16.37.32
                B.            190.16.42.44
                C.            17.17.33.80

                D.            123.45.24.52
Q9.       Which of the following can be the beginning address of a block that contains 1024                    addresses?
                A.            205.16.37.32
                B.            190.16.42.0
                C.            17.17.32.0

                D.            123.45.24.52
Q10.    A small firm is given a block with the beginning address and the prefix length                         205.16.37.24/29.
            What is the range of the block?
Q11.    What is the network address if one of the addresses is 167.199.170.82/27?
Q12.    An organization is granted the block 130.34.12.64/26. The organization needs to have four     subnets. What are the subnet addresses and the range of addresses for each subnet?
Q13.    An ISP granted a block of addresses starting with 190.100.0.0/16. The ISP needs to                 distribute these addresses to three groups of customers as follows:
   1.       The first group has 64 customers; each needs 256 addresses.
   2.       The second group has 128 customers; each needs 128 addresses.
   3.       The third group has 128 customers; each needs 64 addresses.

   Design the subblocks and give the slash notation for each subblock. Find out how many          addresses are still available after these allocations.

Best of Luck !
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2 comments:

  1. Unknown23 January 2016 at 12:12

    Kindly share the solutions for above questions for verification purpose

    ReplyDelete
  2. Tanuj Aggarwal23 January 2016 at 13:58

    Answers

    Ans.1 The subnetwork address is 130.45.32.0
    Ans.2 The subnetwork address is 19.30.64.0
    Ans.3 Company can have 8 subnets (not 6)
    201.70.64.0
    201.70.64.32
    201.70.64.64
    201.70.64.96
    201.70.64.128
    201.70.64.160
    201.70.64.192
    201.70.64.224
    Subnet mask = 255.255.255.224
    Ans.4 D is correct
    Ans.5 We need 16 blocks. For 16 blocks we need to change four 1s to 0s in the default mask. So the mask is
    11111111.11111111.11110000.00000000
    OR
    255.255.240.0
    Ans.6 First is correct
    Ans.7 205.16.32.0 to 205.16.39.255
    Ans.8 The address 205.16.37.32 & 17.17.33.80 is correct.
    Ans.9 17.17.32.0 is correct
    Ans10 205.16.37.24 to 205.16.37.31 (There are only 8 addresses in this block.)
    Ans.11 The network address is 167.199.170.64/27.
    Ans.12 130.34.12.64/28 to 130.34.12.79/28
    130.34.12.80/28 to 130.34.12.95/28
    130.34.12.96/28 to 130.34.12.111/28
    130.34.12.112/28 to 130.34.12.127/28
    Ans.13 Group 1.
    For this group, each customer needs 256 addresses. This means the suffix length is 8 (2^8 =256). The prefix length is 38-8=24.
    190.100.0.0/24 to 190.100.0.255/24
    190.100.1.0/24 to 190.100.1.255/24
    ………………….and so on…………….......
    190.100.63.0/24 to 190.100.63.255/24
    Total= 64x256=16,384
    Group 2.
    For this group, each customer needs 128 addresses. This means the suffix length is 7 (2^7 =128). The prefix length is 32-7=25.
    190.100.64.0/25 to 190.100.64.127/25
    190.100.64.128/25 to 190.100.64.255/25
    ………………….and so on…………….......
    190.100.127.128/25 to 190.100.127.255/25
    Total=128x128=16,384
    Group 3.
    For this group, each customer needs 64 addresses. This means the suffix length is 6 (2^6 =64). The prefix length is 32-6=26.
    190.100.128.0/26 to 190.100.128.63/26
    190.100.128.64/26 to 190.100.128.127/26
    ………………….and so on…………….......
    190.100.159.192/26 to 190.100.159.255/26
    Total=128x64=8,192

    Number of granted addresses =65,536
    Number of allocated addresses =40,960
    Number of available addresses =24,576

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