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 10.187.240.0/20 is divided into equal subnets with prefix /27. How many subnets N are there?
10.187.240.0/20/27
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.210.17.0/24, 10.210.18.0/24, 10.210.19.0/24, 10.210.20.0/24, 10.210.21.0/24, 10.210.22.0/24, 10.210.23.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.210.17.0/24 … 10.210.23.0/24
p =
3.Computer networksMedium
How does TLS typically use asymmetric and symmetric cryptography?
- AOnly asymmetric cryptography is used throughout
- BAsymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
- CSymmetric methods agree keys and authenticate; asymmetric ciphers encrypt all the data
- DOnly symmetric cryptography is used, with a fixed shared password
4.Computer networksHard
TCP congestion control's "slow start" phase:
- AHalves the congestion window every round trip until a packet loss is detected
- BIncreases the congestion window exponentially until a threshold or loss
- CIs used only by UDP
- DKeeps the window fixed at one segment
5.Number systemsMedium
Set the bits so they store −105 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1058 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1000 1101 << 3. What is the result's value R in decimal?
1000 1101<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: N = 128
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
27 − 20 = 7
Number of subnets
N = 27 = 128
Answer
N = 128
Each /27 subnet has 25 − 2 = 30 usable hosts.
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
17 = 00010001
23 = 00010111
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.210.16.0/21 (it also covers 1 network(s) that are not in the list).
3.Computer networks
Answer: B · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
Why this is the answer
Public-key methods and certificates set up a shared secret and prove identity. Fast symmetric ciphers such as AES then protect the bulk data.
Answer
B · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
4.Computer networks
Answer: B · Increases the congestion window exponentially until a threshold or loss
Why this is the answer
Slow start grows the congestion window exponentially; after the threshold, congestion avoidance grows it linearly, and loss reduces it.
Answer
B · Increases the congestion window exponentially until a threshold or loss
5.Number systems
Answer: U = 151
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
105 = 0110 1001
inverted: 1001 0110
+1 → 1001 0111
Read it as unsigned
1001 0111₂ = 256 − 105 = 151
Answer
U = 151
6.Number systems
Answer: R = 104
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
Shifting left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1000 1101 << 3
= 0110 1000
Value of the result
0110 1000₂ = 104
Answer
R = 104
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