Computer Networks · TEE · mocks

Q1D · IPv4 address block

TEE mock A · full paper

TEE · 100 attempted marks · solutions for every choice · current syllabus 702AI0C023

TEE practice mock · Q1D. One address in a block is 192.168.8.173/27. Find its subnet mask, first/last address, total address count and ordinary usable-host range. [5 mock marks]

Answer

Given

IP/prefix = 192.168.8.173/27.

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=32−27=5; N=25=32.

Step 2. Construct the mask

Mask=255.255.255.224; binary=11111111.11111111.11111111.11100000.

Step 3. Find the changing-octet interval

The changing octet is number 4. Its mask value is 224, so step=256−224=32. The IP value 173 lies in 160…191.

Step 4. Form the first/network address

Keep earlier octets; place 160 in octet 4; set any later octets to zero. Network=192.168.8.160.

Step 5. Show the AND operation

OctetIP bitsMask bitsAND resultDecimal
1110000001111111111000000192
2101010001111111110101000168
30000100011111111000010008
4101011011110000010100000160

Step 6. Form the last/broadcast address

Use 191 in the changing octet and 255 in any later octets. Broadcast=192.168.8.191.

Step 7. State usable hosts

First usable=192.168.8.161; last usable=192.168.8.190; count=32−2=30.

Final answer: Network 192.168.8.160; broadcast 192.168.8.191; 32 total, 30 ordinary usable; range 192.168.8.161–192.168.8.190.