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 172.19.0.0/16 is divided into equal subnets with prefix /18. How many subnets N are there?
172.19.0.0/16/18
N =
2.IPv4 subnettingHard
A host has the address 10.20.198.93/27. What is the fourth octet x of its network address?
10.20.198.93/27
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ACSMA/CA
- BCSMA/CD
- CTDMA
- DToken passing
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
- BGo-Back-N resends only the lost frame; Selective Repeat resends that frame and all later ones
- CGo-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
- DNeither ever resends lost frames
5.Number systemsMedium
Set the bits so they store −64 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −648 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0001 1110 OR 0101 0100. What is the result's value R in decimal?
0001 1110OR0101 0100
R =
3Solutions
1.IPv4 subnetting
Answer: N = 4
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).
Borrowed bits
18 − 16 = 2
Number of subnets
N = 22 = 4
Answer
N = 4
Each /18 subnet has 214 − 2 = 16382 usable hosts.
2.IPv4 subnetting
Answer: x = 64
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
25 = 32
Subnet of this host
network: ⌊93/32⌋·32 = 64
broadcast: 64 + 32 − 1 = 95
usable hosts: 65 … 94
Answer
x = 64
3.Computer networks
Answer: B · CSMA/CD
Why this is the answer
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer
B · CSMA/CD
4.Computer networks
Answer: C · 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
C · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
5.Number systems
Answer: U = 192
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
64 = 0100 0000
inverted: 1011 1111
+1 → 1100 0000
Read it as unsigned
1100 0000₂ = 256 − 64 = 192
Answer
U = 192
6.Number systems
Answer: R = 94
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.
Bit by bit
OR gives 1 where at least one bit is 1.
0001 1110
OR 0101 0100
= 0101 1110
Value of the result
0101 1110₂ = 94
Answer
R = 94
ExamProblems · examproblems.com/courses/computer-networks