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 1259 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1259 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.218.110/29. What is the fourth octet x of its network address?
10.20.218.110/29
x =
3.Computer networksMedium
What is the usual order of messages when a client obtains an address from DHCP?
- ADiscover, Request, Offer, Acknowledge
- BRequest, Offer, Discover, Acknowledge
- CDiscover, Offer, Request, Acknowledge
- DOffer, Discover, Acknowledge, Request
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 keep the encryption keys for later reuse
- BTo wait for the DNS cache to expire
- CTo let the routers along the path update their routing tables for the next connection
- DSo delayed segments die out and the final ACK can be resent if it was lost
5.Number systemsMedium
Write −55 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −558 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 10-bit two's complement?
10 bit
N =
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 < 1259
211 − 2 = 2046 ≥ 1259 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 104
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
23 = 8
Subnet of this host
network: ⌊110/8⌋·8 = 104
broadcast: 104 + 8 − 1 = 111
usable hosts: 105 … 110
Answer
x = 104
3.Computer networks
Answer: C · Discover, Offer, Request, Acknowledge
Why this is the answer
The client broadcasts Discover, servers reply with Offers, the client Requests one offer and the chosen server confirms with an Acknowledge.
Answer
C · Discover, Offer, Request, Acknowledge
4.Computer networks
Answer: D · 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
D · So delayed segments die out and the final ACK can be resent if it was lost
5.Number systems
Answer: U = 201
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.
Invert and add one
55 = 0011 0111
inverted: 1100 1000
+1 → 1100 1001
Read it as unsigned
1100 1001₂ = 256 − 55 = 201
Answer
U = 201
6.Number systems
Answer: N = −512
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.
Most negative value
−29 = −512
Answer
N = −512
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