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IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
IP Subnetting
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IP Subnetting

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Basic IP Subnetting

Basic IP Subnetting

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  • 1. IP Subnetting Shahzad Rashid NOC Executive Engineer (TXN) Telenor Pakistan
  • 2. Contents
    • IP Classes
    • Why Subnetting
    • Subnetting for Networks
    • Subnetting for Hosts
  • 3.
      • IP Classes:
        • Class A: 1-126 (16 Million Hosts on 127 Networks)
        • Class B: 128-191 (65000 Hosts on 16000 Networks)
        • Class C: 192-223 (254 Hosts on 2 Million Networks)
        • Class D: 224-239 (Reserved for Multicast Groups)
        • Class E: 240-254 (For Future use and R & D)
        • Loop Back Address: 127.x.x.x (Reserved for Loopback Tests)
        • .1/.2 are usually used as Gateway Addresses
        • The last address e.g. .255 is used as Broadcast Address
      • Why Subnetting:
        • To preserve Public IPv4 Addresses
  • 4. Subnetting for Networks
    • .
      • Step 1: Convert the Number of Networks to Binary
      • Step 2: Reserve bits in Subnet Mask according to the Number of Networks Converted to Binary (From Left to Right)
      • Step 3: Use the Last Reserved bit to find your Network Ranges
  • 5. Practice Scenario
    • .
      • Subnetting for 5 Networks using Class C IP Address: 192.168.1.0
      • Default Class C Mask: 255.255.255.0
      • Step 1: Convert required Networks to Binary
      • 128 64 32 16 8 4 2 1
      • 0 0 0 0 0 1 0 1
      • Bit 1 and 3 are Turned ON. That makes a total of 5
      • Now Count from Right to Left and see how far is the last bit Turned ON
      • 3 is the count from Right to Left
  • 6. Practice Scenario (Contd…)
    • .
      • Step 2:
      • Subnet mask: 255.255.255.0
      • Reserve 3 bits in the Fourth Octet of Subnet Mask from Left to Right (Turn them ON and turn OFF the remaining bits)
      • 255 255 255 0
      • 11111111 11111111 11111111 000000000
      • 11111111 11111111 11111111 111 000000
      • First 3 Bits of the Last (Fourth) Octet Reserved
      • Subnet Mask comes out to be : 255.255.255.224
      • This is the Subnet Mask for 5 Networks
  • 7. Practice Scenario (Contd…)
    • .
      • Step 3:
      • 255 255 255 0
      • 11111111 11111111 11111111 000000000
      • 11111111 11111111 11111111 111 000000
      • Refer to the Last Octet
      • 128 64 32 16 8 4 2 1
      • 1 1 1 0 0 0 0 0
      • Starting from Left to Right notice the last bit Turned ON
      • Last bit Turned ON is 32
      • Network Ranges will be determined using 32
  • 8. Practice Scenario (Contd…)
    • .
      • Step 3 (Contd…):
      • Network Ranges (The best possible utilization of IP for 5 Networks)
      • 1. 192.168.1.1 ----- 192.168.1.31
      • 2. 192.168.1.32 ----- 192.168.1.63
      • 3. 192.168.1.64 ----- 192.168.1.95
      • 4. 192.168.1.96 ----- 192.168.1.127
      • 5. 192.168.1.128 ----- 192.168.1.159
      • 6. 192.168.1.160 ----- 192.168.1.191
      • 7. 192.168.1.192 ----- 192.168.1.223
      • These are the 7 Seven Network Ranges which is the most appropriate Subnetting for 5 Networks without misusing the IP’s
  • 9. Subnetting for Hosts
    • .
      • Step 1: Convert the Number of Hosts to Binary
      • Step 2: Reserve bits in Subnet Mask according to the Number of Hosts Converted to Binary (From Right to Left)
      • Step 3: Use Last Reserved bit to find your Network Ranges
  • 10. Practice Scenario
    • .
      • Subnetting for 50 Hosts using Class C IP Address: 192.168.1.0
      • Default Class C Mask: 255.255.255.0
      • Step 1: Convert required Networks to Binary
      • 128 64 32 16 8 4 2 1
      • 0 0 1 1 0 0 1 0
      • Bit 2, 5 and 6 are Turned ON. That makes a total of 50
      • Now Count from Left to Right and see how far is the last bit Turned ON
      • 6 is the count from Right to Left
  • 11. Practice Scenario (Contd…)
    • .
      • Step 2:
      • Subnet mask: 255.255.255.0
      • Reserve 6 bits in the Fourth Octet of Subnet Mask from Right to Left (Turn them OFF and turn ON the remaining bits)
      • 255 255 255 0
      • 11111111 11111111 11111111 000000000
      • 11111111 11111111 11111111 11 0000000
      • Last 6 Bits of the Last (Fourth) Octet Reserved
      • Subnet Mask comes out to be : 255.255.255.192
      • This is the Subnet Mask for 50 Hosts
  • 12. Practice Scenario (Contd…)
    • .
      • Step 3:
      • 255 255 255 0
      • 11111111 11111111 11111111 000000000
      • 11111111 11111111 11111111 11 0000000
      • Refer to the Last Octet
      • 128 64 32 16 8 4 2 1
      • 1 1 0 0 0 0 0 0
      • Starting from Left to Right notice the last bit Turned ON
      • Last bit Turned ON is 64
      • Network Ranges for 50 Hosts will be determined using 64
  • 13. Practice Scenario (Contd…)
    • .
      • Step 3 (Contd…):
      • Network Ranges (The best possible utilization of IP for 50 Hosts)
      • 1. 192.168.1.1 ----- 192.168.1.63
      • 2. 192.168.1.64 ----- 192.168.1.127
      • 3. 192.168.1.128 ----- 192.168.1.191
      • 4. 192.168.1.192 ----- 192.168.1.255
      • Each Network Range can have around 60-61 Hosts each, which is the best possible Subnetting for 50 Hosts
  • 14. Thanks for Viewing
    • .

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