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 682 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
682 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.111.115/22. What is the third octet x of its subnet's broadcast address?
172.20.111.115/22
x =
3.Computer networksMedium
Which metric does RIP use to choose routes?
- AMeasured delay
- BLink bandwidth
- CHop count
- DBusiness policy
4.Computer networksHard
TCP congestion control's "slow start" phase:
- AHalves the congestion window every round trip until a packet loss is detected
- BIs used only by UDP
- CIncreases the congestion window exponentially until a threshold or loss
- DKeeps the window fixed at one segment
5.Number systemsMedium
Set the bits so they store −98 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −988 bit
U =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: p = 22
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
29 − 2 = 510 < 682
210 − 2 = 1022 ≥ 682 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 111
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 in the third octet
22 = 4
Subnet of this host
network: 172.20.108.0
broadcast: 172.20.111.255
Answer
x = 111
3.Computer networks
Answer: C · Hop count
Why this is the answer
RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.
Answer
C · Hop count
4.Computer networks
Answer: C · 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
C · Increases the congestion window exponentially until a threshold or loss
5.Number systems
Answer: U = 158
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
98 = 0110 0010
inverted: 1001 1101
+1 → 1001 1110
Read it as unsigned
1001 1110₂ = 256 − 98 = 158
Answer
U = 158
6.Number systems
Answer: M = 1536 KiB
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.
Bits and bytes
1024·768·16 = 12582912 bit = 1572864 B
Convert to KiB
1572864/1024 = 1536 KiB
Answer
M = 1536 KiB
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