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 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/243 subnets
H =
2.IPv4 subnettingHard
The network 192.168.161.0/24 is split with VLSM for these LANs: A: 8 hosts, B: 51 hosts, C: 67 hosts, D: 4 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: 8B: 51C: 67D: 4
x =
3.Computer networksMedium
A signal travels 2000 km through fibre at about 2 × 10⁸ m/s. What is the propagation delay?
- A1 ms
- B0.1 ms
- C10 ms
- D100 ms
4.Computer networksHard
With sliding-window protocols, how does Go-Back-N differ from Selective Repeat after one frame is lost?
- ABoth resend the whole window from the start
- BNeither ever resends lost frames
- CGo-Back-N resends only the lost frame; Selective Repeat resends that frame and all later ones
- DGo-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
5.Number systemsMedium
The 8-bit pattern 1100 1101 stores a signed integer in two's complement. What is its value N in decimal?
1100 11018 bit
N =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 3 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 208
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
C: 67 hosts → 27 = 128 (/25)
B: 51 hosts → 26 = 64 (/26)
A: 8 hosts → 24 = 16 (/28)
D: 4 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .127
B: .128 – .191
A: .192 – .207
D: .208 – .215
Answer
x = 208
LAN D: 192.168.161.208/29.
3.Computer networks
Answer: C · 10 ms
Why this is the answer
2 × 10⁶ m / 2 × 10⁸ m/s = 0.01 s = 10 ms. Propagation delay depends on distance and medium, not packet size.
Answer
C · 10 ms
4.Computer networks
Answer: D · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
Why this is the answer
A Go-Back-N receiver discards out-of-order frames, so the sender must repeat everything after the loss. Selective Repeat buffers them and asks only for the missing one.
Answer
D · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
5.Number systems
Answer: N = −51
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 + 64 + 8 + 4 + 1 = −51
Answer
N = −51
6.Number systems
Answer: M = 768 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
512·512·24 = 6291456 bit = 786432 B
Convert to KiB
786432/1024 = 768 KiB
Answer
M = 768 KiB
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