With reference to the given exhibit, determine the subnet mask assigned to LAN Semenyih?

Question

BlackTom AI · Accepted Answer

Accommodating 70 hosts requires at least 70 usable IP addresses. For IPv4 subnetting the usable host count in a subnet is 2^h − 2, where h is the number of host bits (the subtraction of 2 accounts for the network and broadcast addresses).

Step-by-step reasoning and calculations:
1. Determine minimum host bits h so that 2^h − 2 ≥ 70.
   - If h = 6: 2^6 − 2 = 64 − 2 = 62 usable hosts → too few.
   - If h = 7: 2^7 − 2 = 128 − 2 = 126 usable hosts → sufficient.

2. With h = 7 host bits, the subnet mask length is 32 − 7 = 25 bits, i.e., a /25 mask.

3. Convert /25 to dotted-decimal: the first three octets are all 255 (24 bits). The 25th bit is the highest bit of the fourth octet: binary for the fourth octet mask = 10000000 which is 128 in decimal. So the mask is 255.255.255.128.

Why this answer is correct:
- A /25 provides 126 usable addresses, which covers the 70 hosts with room for growth. It is the smallest mask that meets the requirement (minimizes wasted address space while providing enough hosts).

Why the other options are incorrect:
- a. 255.255.192.0 is a /18 (14 host bits) and would provide a very large number of hosts (2^14 − 2 = 16382 usable). It is over-provisioned and not the minimal correct mask for 70 hosts.
- c. 255.255.255.224 is a /27 (5 host bits) → 2^5 − 2 = 30 usable hosts, insufficient for 70.
- d. 255.255.255.192 is a /26 (6 host bits) → 2^6 − 2 = 62 usable hosts, still insufficient for 70.

Conclusion: The correct subnet mask assigned to LAN Semenyih is 255.255.255.128, which matches the provided answer (b).

With reference to the given exhibit, determine the subnet mask assigned to LAN Semenyih?单项选择题

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