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 192.168.10.0/24 must be divided into at least 10 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2410 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.222.155.0/24, 10.222.156.0/24, 10.222.157.0/24, 10.222.158.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.222.155.0/24 … 10.222.158.0/24
p =
3.Computer networksMedium
Which single prefix summarises both 192.168.0.0/24 and 192.168.1.0/24?
- A192.168.0.0/25
- B192.168.1.0/23
- C192.168.0.0/22
- D192.168.0.0/23
4.Computer networksHard
Why was the TCP window scale option introduced?
- AThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
- BThe 32-bit window field was too large and wasted header space in every TCP segment sent
- CWithout it, TCP could not detect lost segments on long-distance links
- DUDP needed a way to control its sending rate on congested paths
5.Number systemsMedium
Set the bits so they store −112 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1128 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 20 bits?
20 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 14
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).
Bits to borrow
24 = 16 ≥ 10 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
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
155 = 10011011
158 = 10011110
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.222.152.0/21 (it also covers 4 network(s) that are not in the list).
3.Computer networks
Answer: D · 192.168.0.0/23
Why this is the answer
0 and 1 share all bits of the third octet except the last, so one bit less of prefix (/23) starting at 192.168.0.0 covers both exactly.
Answer
D · 192.168.0.0/23
4.Computer networks
Answer: A · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
Why this is the answer
The original header allows at most 65,535 bytes of unacknowledged data. Window scaling multiplies this by a power of two so high bandwidth-delay paths can be filled.
Answer
A · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: U = 144
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
112 = 0111 0000
inverted: 1000 1111
+1 → 1001 0000
Read it as unsigned
1001 0000₂ = 256 − 112 = 144
Answer
U = 144
6.Number systems
Answer: N = 1.049 × 10⁶
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Largest value
220 − 1 = 1048575
Answer
N = 1048575
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