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 1134 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1134 hosts
p =
2.IPv4 subnettingHard
The network 192.168.149.0/24 is split with VLSM for these LANs: A: 95 hosts, B: 23 hosts, C: 22 hosts, D: 5 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?
A: 95B: 23C: 22D: 5
x =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A12 ms
- B1 ms
- C0.125 ms
- D8 ms
4.Computer networksHard
TCP congestion control's "slow start" phase:
- AKeeps the window fixed at one segment
- BIs used only by UDP
- CIncreases the congestion window exponentially until a threshold or loss
- DHalves the congestion window every round trip until a packet loss is detected
5.Number systemsMedium
The 8-bit pattern 1000 1011 stores a signed integer in two's complement. What is its value N in decimal?
1000 10118 bit
N =
6.Number systemsHard
An uncompressed image is 256 × 256 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
256 × 256 px8 bit/px1 KiB = 1024 B
M = KiB
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 < 1134
211 − 2 = 2046 ≥ 1134 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 192
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
A: 95 hosts → 27 = 128 (/25)
B: 23 hosts → 25 = 32 (/27)
C: 22 hosts → 25 = 32 (/27)
D: 5 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .127
B: .128 – .159
C: .160 – .191
D: .192 – .199
Answer
x = 192
LAN D: 192.168.149.192/29.
3.Computer networks
Answer: B · 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
B · 1 ms
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: N = −117
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 + 8 + 2 + 1 = −117
Answer
N = −117
6.Number systems
Answer: M = 64 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
256·256·8 = 524288 bit = 65536 B
Convert to KiB
65536/1024 = 64 KiB
Answer
M = 64 KiB
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