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01-02: Exercises

Question

A Class B network with network ID 172.20.0.0 is given.
The network needs to be divided into at least 15 subnets.

Based on this requirement, answer the following:

  1. How many bits have to be borrowed from the host portion?
  2. How many total bits will be used to identify the subnet?
  3. What is the new subnet mask in dotted decimal notation?
  4. What is the magic number (block size)?
  5. What are the network IDs of:
  6. the first subnet
  7. the second subnet
  8. the last required (15th) subnet
  9. the last created (16th) subnet
  10. How many usable host addresses are available per subnet?
  11. What are the usable host ranges of:
  12. the first subnet
  13. the second subnet
  14. the last required (15th) subnet

Solution

Step 1: Number of bits to be borrowed

Given: - Class B network → default prefix = /16
- Required subnets = 15

Checking powers of 2:

  • 2¹ = 2
  • 2² = 4
  • 2³ = 8
  • 2⁴ = 16

Thus, 4 bits have to be borrowed (since 2⁴ = 16 ≥ 15).

👉 (1) Bits borrowed = 4

Step 2: Total subnet bits

👉 (2) Total subnet bits = 16 + 4 = 20 bits (/20)

Step 3: New subnet mask

Default mask (binary):
11111111.11111111.00000000.00000000

After borrowing 4 bits:
11111111.11111111.11110000.00000000

Convert to decimal:

👉 (3) Subnet mask = 255.255.240.0 (/20)

Step 4: Magic number (Block Size)

The change occurs in the 3rd octet.

Magic number = 256 − 240 = 16

👉 (4) Magic number = 16

(Subnet IDs increase by 16 in the 3rd octet)

Step 5: Subnet network IDs

Using block size 16:

  • 1st → 172.20.0.0/20
  • 2nd → 172.20.16.0/20
  • 3rd → 172.20.32.0/20
  • ...
  • 15th → 172.20.224.0/20
  • 16th → 172.20.240.0/20

👉 (5) Network IDs: - First subnet → 172.20.0.0
- Second subnet → 172.20.16.0
- Last required (15th) → 172.20.224.0
- Last created (16th) → 172.20.240.0

Step 6: Number of hosts per subnet

Remaining host bits: 32 − 20 = 12

Number of usable hosts: 2¹² − 2 = 4096 − 2 = 4094

👉 (6) Usable hosts per subnet = 4094

Step 7: Host ranges

First subnet (172.20.0.0/20)

  • Network ID = 172.20.0.0
  • Broadcast ID = 172.20.15.255
  • Usable host range = 172.20.0.1 to 172.20.15.254

Second subnet (172.20.16.0/20)

  • Network ID = 172.20.16.0
  • Broadcast ID = 172.20.31.255
  • Usable host range = 172.20.16.1 to 172.20.31.254

Last required (15th) subnet (172.20.224.0/20)

  • Network ID = 172.20.224.0
  • Broadcast ID = 172.20.239.255
  • Usable host range = 172.20.224.1 to 172.20.239.254

👉 (7) Host ranges: - First subnet → 172.20.0.1 to 172.20.15.254
- Second subnet → 172.20.16.1 to 172.20.31.254
- Last required subnet → 172.20.224.1 to 172.20.239.254

Final Summary

  1. Bits borrowed from host portion = 4

  2. Total subnet bits = 20 bits (/20)

  3. Subnet mask = 255.255.240.0

  4. Magic number (block size) = 16

  5. Subnet network IDs:

  6. First subnet → 172.20.0.0
  7. Second subnet → 172.20.16.0
  8. Last required (15th) → 172.20.224.0

  9. Last created (16th) → 172.20.240.0

  10. Usable hosts per subnet = 4094

  11. Host ranges:

  12. First subnet → 172.20.0.1 to 172.20.15.254
  13. Second subnet → 172.20.16.1 to 172.20.31.254
  14. Last required subnet → 172.20.224.1 to 172.20.239.254