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
A department needs 1068 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1068 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.75.136.0/24, 10.75.137.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.75.136.0/24 … 10.75.137.0/24
p =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A12 ms
- B8 ms
- C0.125 ms
- D1 ms
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?
- ATo let the routers along the path update their routing tables for the next connection
- BSo delayed segments die out and the final ACK can be resent if it was lost
- CTo wait for the DNS cache to expire
- DTo keep the encryption keys for later reuse
5.Number systemsMedium
The 8-bit pattern 1000 0010 stores a signed integer in two's complement. What is its value N in decimal?
1000 00108 bit
N =
6.Number systemsHard
A system must give every one of 2664 items its own binary code of the same length. What is the smallest number of bits b needed?
2664
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 21
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).
Host bits needed
210 − 2 = 1022 < 1068
211 − 2 = 2046 ≥ 1068 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: p = 23
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
136 = 10001000
137 = 10001001
common leading bits: 7
Prefix length
p = 16 + 7 = 23
Answer
p = /23
Summary route: 10.75.136.0/23
3.Computer networks
Answer: D · 1 ms
Why this is the answer
1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.
Answer
D · 1 ms
4.Computer networks
Answer: B · 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
B · So delayed segments die out and the final ACK can be resent if it was lost
5.Number systems
Answer: N = −126
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.
Add the place values
The other bits keep their normal positive values.
−128 + 2 = −126
Answer
N = −126
6.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 2664
212 = 4096 ≥ 2664
Answer
b = 12
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