Workstation A has been assigned an IP address of 192.0.2.24/28. Workstation B has been assigned an IP address of 192.0.2.100/28. The two workstations are connected with a straight-through cable…

Question

Options

• AReplace the straight-through cable with a crossover cable.
• BChange the subnet mask of the hosts to /26.
• CChange the subnet mask of the hosts to /26.
• DChange the address of Workstation A to 192.0.2.15.
• EChange the address of Workstation B to 192.0.2.111.

• A
  70% (33)
• B
  2% (1)
• C
  15% (7)
• D
  9% (4)
• E
  4% (2)

Explanation

There are two separate problems here. First, a straight-through cable is used for unlike devices (PC to switch/router); connecting two PCs directly requires a crossover cable - Answer A fixes this physical-layer issue. Second, the hosts are on different subnets: 192.0.2.24/28 falls in the 192.0.2.16–30 subnet, while 192.0.2.100/28 falls in the 192.0.2.96–110 subnet. Because /28 gives a block size of 16, the two IPs land in non-overlapping ranges and cannot communicate at Layer 3 without a router. Changing the mask to /25 (block size 128) would place both .24 and .100 in the single 192.0.2.0–127 range, fixing the subnet mismatch. Changing to /26 (block size 64) does NOT help - .24 lands in the .0–63 block and .100 lands in the .64–127 block, so they remain on separate subnets. Options D and E are traps: 192.0.2.15 is the broadcast address of the .0/28 subnet, and 192.0.2.111 is the broadcast address of the .96/28 subnet - neither is a valid host address.

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