IPv4 subnetting
- Hosts per subnet
- 2(32 − prefix) − 2 (/31: 2, /32: 1)
- Block size
- 256 − (mask octet)
- Network address
- IP AND mask
- Broadcast address
- IP OR (NOT mask): all host bits set to 1
- Subnets from n borrowed bits
- 2n
Computer Networks: cram pack
Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.
1Formulas and methods
Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- From decimal
- divide by the base, read the remainders bottom-up
- Hex ↔ binary
- one hex digit = 4 bits
- Two’s complement
- invert the bits and add 1
2Practice problems
1.IPv4 subnettingMedium
The network 172.16.0.0/16 is divided into equal subnets with prefix /19. How many subnets N are there?
172.16.0.0/16/19
N =
2.IPv4 subnettingHard
The network 192.168.223.0/24 is split with VLSM for these LANs: A: 25 hosts, B: 46 hosts, C: 9 hosts, D: 16 hosts, E: 3 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN E?
A: 25B: 46C: 9D: 16E: 3
x =
3.Computer networksMedium
Which range of TCP/UDP port numbers is reserved as "well-known" ports?
- A49152–65535
- B0–65535
- C1024–49151
- D0–1023
4.Computer networksHard
Why does the side that closes a TCP connection first stay in the TIME_WAIT state for twice the maximum segment lifetime?
- ASo delayed segments die out and the final ACK can be resent if it was lost
- BTo wait for the DNS cache to expire
- CTo let the routers along the path update their routing tables for the next connection
- DTo keep the encryption keys for later reuse
5.Number systemsMedium
Set the bits so they store −83 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −838 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 1000 OR 0101 0111. What is the result's value R in decimal?
1011 1000OR0101 0111
R =
3Solutions
1.IPv4 subnetting
Answer: N = 8
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
Borrowed bits
19 − 16 = 3
Number of subnets
N = 23 = 8
Answer
N = 8
Each /19 subnet has 213 − 2 = 8190 usable hosts.
2.IPv4 subnetting
Answer: x = 144
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
B: 46 hosts → 26 = 64 (/26)
A: 25 hosts → 25 = 32 (/27)
D: 16 hosts → 25 = 32 (/27)
C: 9 hosts → 24 = 16 (/28)
E: 3 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .63
A: .64 – .95
D: .96 – .127
C: .128 – .143
E: .144 – .151
Answer
x = 144
LAN E: 192.168.223.144/29.
3.Computer networks
Answer: D · 0–1023
Why this is the answer
Ports 0–1023 are well-known ports for standard services. 1024–49151 are registered ports and 49152–65535 are dynamic or private ports.
Answer
D · 0–1023
4.Computer networks
Answer: A · So delayed segments die out and the final ACK can be resent if it was lost
Why this is the answer
TIME_WAIT ensures a lost final ACK can be repeated and old duplicate segments expire before the same address and port pair is reused.
Answer
A · So delayed segments die out and the final ACK can be resent if it was lost
5.Number systems
Answer: U = 173
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Invert and add one
83 = 0101 0011
inverted: 1010 1100
+1 → 1010 1101
Read it as unsigned
1010 1101₂ = 256 − 83 = 173
Answer
U = 173
6.Number systems
Answer: R = 255
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Bit by bit
OR gives 1 where at least one bit is 1.
1011 1000
OR 0101 0111
= 1111 1111
Value of the result
1111 1111₂ = 255
Answer
R = 255
ExamProblems · examproblems.com/courses/computer-networks