CompTIA CloudNetX

1. 1. Write the place valuesAn 8-bit number has fixed place values, left to right: 128, 64, 32, 16, 8, 4, 2, 1.
2. 2. Line the bits up under the place values128→1, 64→1, 32→0, 16→0, 8→0, 4→0, 2→0, 1→0.
3. 3. Keep only the place values above a 1The 128 column and the 64 column hold a 1. Every other column holds 0 and contributes nothing.
4. 4. Add them128 + 64 = 192.

Answer: 11000000 = 192. This is why a /24 subnet mask octet of 11000000 reads as 192 in dotted decimal.

1. 1. Start at the largest place valueAsk: does 128 fit into 172? Yes. Write a 1 and subtract: 172 − 128 = 44.
2. 2. Next column, 6464 does not fit into 44. Write 0. Remainder stays 44.
3. 3. Next column, 3232 fits into 44. Write 1. 44 − 32 = 12.
4. 4. Continue down16 into 12? No → 0. 8 into 12? Yes → 1, remainder 4. 4 into 4? Yes → 1, remainder 0. 2 → 0. 1 → 0.
5. 5. Read the bits in order1, 0, 1, 0, 1, 1, 0, 0.

Answer: 172 = 10101100. Check it by adding the on-bits back: 128 + 32 + 8 + 4 = 172.

1. 1. Split into 4-bit groups11011110 becomes 1101 and 1110.
2. 2. Convert each group with place values 8, 4, 2, 11101 = 8 + 4 + 1 = 13. 1110 = 8 + 4 + 2 = 14.
3. 3. Replace values over 9 with letters10=A, 11=B, 12=C, 13=D, 14=E, 15=F. So 13 = D and 14 = E.
4. 4. Join the digitsD followed by E.

Answer: 11011110 = DE in hex, which is 222 in decimal. Each pair of hex digits is exactly one byte, which is why a MAC address is six hex pairs.

1. 1. Find the interesting octetThe first three octets of the mask are 255, so only the fourth octet matters. 192 in binary is 11000000, so 2 host-network bits are borrowed.
2. 2. Find the block size256 − 192 = 64. Subnets step in blocks of 64: 0, 64, 128, 192.
3. 3. Place the host77 falls between 64 and 127, so the host sits in the 192.168.10.64 subnet.
4. 4. Name the boundariesNetwork address = 192.168.10.64. Broadcast = one below the next block = 192.168.10.127.
5. 5. Usable hostsEverything between the two boundaries: .65 through .126.

Answer: Network 192.168.10.64, broadcast 192.168.10.127, usable range 192.168.10.65–192.168.10.126 (62 usable addresses).

1. 1. Count host bitsAn IPv4 address is 32 bits. 32 − 22 = 10 host bits.
2. 2. Raise 2 to that power2^10 = 1024 total addresses in the block.
3. 3. Remove the two reserved addressesThe lowest address is the network address and the highest is the broadcast address: 1024 − 2.

Answer: 1022 usable host addresses. The general formula is 2^(32 − prefix) − 2.

1. 1. Local checksThe resolver checks its own cache, then the hosts file. A stale entry here explains a machine that reaches the wrong server while everyone else is fine.
2. 2. Recursive resolverThe query goes to the configured DNS server, usually learned by DHCP. If that server has a cached answer within its TTL, it replies immediately.
3. 3. Root and TLDWith no cache, the resolver asks a root server, which refers it to the .com name servers, which refer it to the authoritative servers for example.com.
4. 4. Authoritative answerThe authoritative server returns the A record (IPv4) or AAAA record (IPv6) for www.example.com.
5. 5. Cache and connectThe answer is cached for its TTL, then the workstation opens a TCP connection to that IP on port 443.

Answer: Name resolution is a chain: cache → hosts file → recursive resolver → root → TLD → authoritative. Test the chain with nslookup or dig; if the IP is correct but the page fails, the fault is below DNS.

1. 1. Reserve for the hostThe host operating system and hypervisor need their own memory — reserve about 4 GB.
2. 2. Divide what remains16 − 4 = 12 GB available. 12 ÷ 4 = 3 VMs.
3. 3. Check CPUCores can be oversubscribed; memory generally cannot. Two virtual CPUs each across 3 VMs is 6 of 8 cores — comfortable.
4. 4. Check diskDynamically expanding disks grow over time; confirm free space on the host volume before you build.

Answer: Three 4 GB VMs. Memory is the hard limit: overcommit it and the host swaps to disk and everything crawls.