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 192.168.93.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 2?
192.168.93.0/24/26subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.163.90.0/24, 10.163.91.0/24, 10.163.92.0/24, 10.163.93.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.163.90.0/24 … 10.163.93.0/24
p =
3.Computer networksMedium
What does an MX record in DNS specify?
- AThe IPv4 address of a web server
- BAn alias name for another host
- CThe mail servers that accept email for a domain
- DThe name servers for a zone
4.Computer networksHard
A TCP connection has a 64 KB window and a 100 ms round-trip time, and no window scaling. What is its approximate maximum throughput?
- AAbout 50 Mbit/s
- BAbout 5 Mbit/s
- CAbout 640 Mbit/s
- DAbout 0.5 Mbit/s
5.Number systemsMedium
Set the bits so they store −102 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1028 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 0001 OR 0011 1010. What is the result's value R in decimal?
1111 0001OR0011 1010
R =
3Solutions
1.IPv4 subnetting
Answer: x = 127
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).
Block size
26 = 64
Subnet number 2
network: (2 − 1)·64 = 64
broadcast: 64 + 64 − 1 = 127
Answer
x = 127
2.IPv4 subnetting
Answer: p = 21
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
90 = 01011010
93 = 01011101
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.163.88.0/21 (it also covers 4 network(s) that are not in the list).
3.Computer networks
Answer: C · The mail servers that accept email for a domain
Why this is the answer
MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.
Answer
C · The mail servers that accept email for a domain
4.Computer networks
Answer: B · About 5 Mbit/s
Why this is the answer
Throughput ≤ window / RTT = 65,536 × 8 bits / 0.1 s ≈ 5.2 Mbit/s, however fast the link is.
Answer
B · About 5 Mbit/s
5.Number systems
Answer: U = 154
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
102 = 0110 0110
inverted: 1001 1001
+1 → 1001 1010
Read it as unsigned
1001 1010₂ = 256 − 102 = 154
Answer
U = 154
6.Number systems
Answer: R = 251
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.
1111 0001
OR 0011 1010
= 1111 1011
Value of the result
1111 1011₂ = 251
Answer
R = 251
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