Computer Networks · TEE · papers

Q1D · IPv4 address block

Final · 5 December 2025

TEE · Actual TEE paper · all questions · current syllabus 702AI0C023

Actual PYQ · Q1D. A block of address is granted to a small organization and one of the address is 200.140.64.48/26. Determine the first address and last address in the block. Determine the total number of address in the block. [5 marks]

Answer

Given

IP/prefix = 200.140.64.48/26.

To find

The complete containing address block and its usable host range.

Formula and why

h=32-p,\qquad N=2^hLaTeX: h=32-p,\qquad N=2^h
\text{step}=256-\text{changing mask octet}LaTeX: \text{step}=256-\text{changing mask octet}

Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.

Step-by-step solution

Step 1. Count host bits and addresses

h=32−26=6; N=26=64.

Step 2. Construct the mask

Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.

Step 3. Find the changing-octet interval

The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 48 lies in 0…63.

Step 4. Form the first/network address

Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=200.140.64.0.

Step 5. Show the AND operation

OctetIP bitsMask bitsAND resultDecimal
1110010001111111111001000200
2100011001111111110001100140
301000000111111110100000064
40011000011000000000000000

Step 6. Form the last/broadcast address

Use 63 in the changing octet and 255 in any later octets. Broadcast=200.140.64.63.

Step 7. State usable hosts

First usable=200.140.64.1; last usable=200.140.64.62; count=64−2=62.

Final answer: Network 200.140.64.0; broadcast 200.140.64.63; 64 total, 62 ordinary usable; range 200.140.64.1–200.140.64.62.

Source: Final · 5 December 2025, original question paper page 1. Original marks retained; solution independently authored.