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Subnetting Work Sheet No Binary

Feb 25th, 2025 (edited)
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  1. John's Quick Subnetting Worksheet
  2. (No Binary Required!)
  3. Practice Subnetting IPv4 https://subnetipv4.com/
  4.  
  5. IP Address: OCTET . OCTET . OCTET . OCTET
  6. πŸ“Œ Step 1: Write Down the IP Address
  7. πŸ–Š IP Address: ____ . ____ . ____ . ____ / ______ (CIDR Notation)
  8.  
  9. πŸ“Œ Step 2: Identify the Default Subnet Mask
  10. βœ” Class A: 1 – 127 β†’ Subnet Mask: 255.0.0.0
  11. βœ” Class B: 128 – 191 β†’ Subnet Mask: 255.255.0.0
  12. βœ” Class C: 192 – 223 β†’ Subnet Mask: 255.255.255.0
  13. πŸ‘‰ If no bits are borrowed β€œGO TO STEP 8” the default subnet mask remains unchanged, and the network address is determined by the default class of the IP address:
  14. β€’ Class A (1-127): Network Address = X.0.0.0 β†’ Subnet Mask: 255.0.0.0
  15. β—¦ Example: If the IP address is 10.45.78.120, the adjusted network address would be 10.0.0.0.
  16. β€’ Class B (128-191): Network Address = X.Y.0.0 β†’ Subnet Mask: 255.255.0.0
  17. β—¦ Example: If the IP address is 172.16.45.200, the adjusted network address would be 172.16.0.0.
  18. β€’ Class C (192-223): Network Address = X.Y.Z.0 β†’ Subnet Mask: 255.255.255.0
  19. β—¦ Example: If the IP address is 192.168.1.56, the adjusted network address would be 192.168.1.0.
  20. Since no additional subnetting occurs, the network remains a single large network, maximizing the number of available host addresses.
  21.  
  22.  
  23. πŸ“Œ Step 3: Determine if there is an "Interesting Octet"
  24. πŸ›  The "Interesting Octet" is the Octet where subnetting (borrowing bits) occurs.
  25. βœ” Write 255 in all Octets in the subset mask before this point.
  26.  
  27. πŸ“Œ Step 4: Determine Borrowed Bits (If Any) Borrowed Bits ______?
  28.  
  29. βœ” If subnetting is needed, borrow bits from the β€œHost Portion”
  30. βœ” Use the table below to calculate the Subnet Mask Value:
  31. Borrowed Bits
  32. 1 2 3 4 5 6 7 8
  33. Subnet Value
  34. 128 64 32 16 8 4 2 1
  35. Example: Borrowed bits in a /28 the first 3 octets add up to 24 so 28 would be β€œ4” borrowed bits 24 + 4 = 28, which would add up to β€œ240”
  36. βœ” Add up the values below the β€œBorrowed Bits” in the β€œTable Shown Above” to determine the "Interesting Octet."
  37. 128 + 64 + 32 + 16 = 240. So now β€œ240” is the β€œInteresting Octet”
  38. βœ” All β€œempty Octets” to the right, (if Any), should be replaced with ZERO
  39. πŸ“Œ IP Address: ____ . ____ . ____ . ____
  40. πŸ“Œ Subnet Mask: ____ . ____ . ____ . ____
  41.  
  42. πŸ“Œ Interesting Octet in the Subnet Mask: ______ (Circle β€œNONE” if no bits are borrowed)
  43.  
  44. πŸ“Œ Step 5: Identify the Corresponding IP Address Octet
  45. βœ” Find the Octet in the IP Address that aligns with the Interesting Octet from Step 3.
  46. πŸ“Œ Corresponding IP Address Octet: ______ (Circle β€œNONE” if no bits are borrowed)
  47.  
  48. πŸ“Œ Step 6: Calculate the Magic Number
  49. πŸ”’ Formula:
  50. ο‚§ Magic Number = 256 βˆ’ ______ (The Subnet Mask’s "Interesting Octet" first non 255 Octet)
  51. βœ” If no bits are borrowed, circle "NONE" as the magic number does not apply.
  52. βœ” If subnetting is applied, use the formula above.
  53. πŸ“Œ Magic Number: ______
  54. Example Calculation:
  55. βœ” Subnet Mask: 255.255.240.0
  56. βœ” Interesting Octet: 240
  57. βœ” Magic Number: 256 - 240 = 16
  58.  
  59. πŸ“Œ Step 7: Find the Closest Multiple
  60. πŸ“Œ Corresponding IP Address Octet: ______
  61. πŸ“Œ Magic Number: ______
  62. πŸ‘‰ To find the closest multiple of the magic number that does not exceed the Corresponding Octet:
  63. 1️⃣ Divide the Corresponding IP Address Octet by the Magic Number
  64. βœ” Example: If the Coresponding IP Octet = 149 and the Magic Number = 2
  65. 149 Γ· 2 = 74.5
  66. 2️⃣ Ignore the decimal (Take only the integer part)
  67. βœ” 74
  68. 3️⃣ Multiply the result by the Magic Number
  69. βœ” 74 Γ— 2 = 148
  70. ATTENTION! Key Rules to Follow: πŸ“Œ Calculated Closest Multiple: ______
  71. βœ” The closest multiple must be β€œLESS THAN OR EQUAL TO” the corresponding IP octet.
  72. βœ” If bits are borrowed, always choose the closest valid multiple that does β€œNOT Exceed” the β€œCorresponding IP Address Octet”
  73. βœ” If the closest multiple equals the corresponding IP octet, keep it as is.
  74.  
  75. βœ” If the calculated multiple exceeds the corresponding IP octet, use the previous multiple.
  76.  
  77. βœ” If no valid multiple exists, replace the corresponding IP octet with ZERO.
  78. πŸ“Œ Closest Multiple: ______ (Circle "NONE" if no bits are borrowed)
  79.  
  80. πŸ“Œ Step 8: Adjust the Original Subnet Number (Network Address)
  81. Original Network Address: ____ . ____ . ____ . ____
  82. NO "BORROWED BITS" in the Subnet Mask, use the DEFAULT SUBNET MASK for that IP ADDRESS CLASS. Then, replace all the numbers, (OCTETS), in the IP Address that correspond to the "0s" in the Subnet Mask, with ZEROS.
  83.  
  84. IF THERE ARE BORROWED BITS β€œClosest Multiple”
  85. THIS ADJUSTED NETWORK ADDRESS BELOW IS NOW YOUR NEW CALCULATED NETWORK ADDRESS
  86. Adjusted Network Address with the Closest Multiple: ____ . ____ . ____ . ____
  87.  
  88. πŸ“Œ Step 9: Calculate the Broadcast Address
  89. πŸ“Œ Adjusted Network Address with the closest multiple: ____ . ____ . ____ . ____
  90. πŸ“Œ Magic Number: ______
  91. πŸ“Œ Replaced Adjusted Corresponding IP Octet: ______
  92. πŸ‘‰ Formula:
  93. 1️⃣ Carry down all the Adjusted Network Address Octets to the corresponding positions in the Broadcast Address, up to (but not including) the Replaced Adjusted Corresponding IP Octet.
  94. 2️⃣ Calculate the Broadcast Octet:
  95. Broadcast Octet = (Adjusted Corresponding IP Address Octet) + (Magic Number βˆ’ 1)
  96. 3️⃣ Bring down the calculated Broadcast Octet and insert it into its correct position.
  97. 4️⃣ Replace all Octets to the right of this Broadcast Octet with 255 to complete the Broadcast Address.
  98. πŸ“Œ Broadcast Address: ____ . ____ . ____ . ____
  99.  
  100. πŸ“Œ Step 10: Determine the Host Range
  101. πŸ“Œ Adjusted Network Address with the closest multiple: ____ . ____ . ____ . ____
  102. βœ” First Host: Network Address + 1 β†’ ____ . ____ . ____ . ____
  103. βœ” Last Host: Broadcast Address Last Octet - 1 β†’ ____ . ____ . ____ . ____
  104.  
  105. πŸ“Œ Step 11: Identify the Next Subnet
  106. πŸ“Œ Broadcast Address: ____ . ____ . ____ . ____
  107. πŸ“Œ Next Subnet = Broadcast Address + 1
  108. βœ” Rules:
  109. πŸ”Ή If the last Octet reaches 254 or 255, increment the previous Octet and reset the last Octet
  110. to ZERO
  111. πŸ”Ή If all three Octets reach 254 or 255, reset them ALL to ZERO
  112. πŸ“Œ Next Subnet: ____ . ____ . ____ . ____
  113.  
  114. πŸ“Œ Step 12: Final Subnetting Results
  115. βœ” Write down all network details based on your calculations:
  116. Network Component
  117. Calculated Address
  118. Network Address
  119. ____ . ____ . ____ . ____
  120. First Host
  121. ____ . ____ . ____ . ____
  122. Last Host
  123. ____ . ____ . ____ . ____
  124. Broadcast Address
  125. ____ . ____ . ____ . ____
  126. Next Subnet
  127. ____ . ____ . ____ . ____
  128.  
Tags: Subnetting
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