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 subnet uses the mask 255.255.255.240. How many usable host addresses H does it have?
255.255.255.240
H =
2.IPv4 subnettingHard
A host has the address 172.20.14.135/21. What is the third octet x of its subnet's broadcast address?
172.20.14.135/21
x =
3.Computer networksMedium
What is the main difference between the HTTP methods GET and POST?
- AGET deletes a resource, POST reads it from the server's database
- BGET retrieves a resource, POST submits data in the request body
- CGET is encrypted, POST is not
- DGET uses UDP, POST uses TCP
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
- CNeither ever resends lost frames
- DGo-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
5.Number systemsMedium
Set the bits so they store −63 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −638 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 1100 << 2. What is the result's value R in decimal?
1011 1100<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: H = 14
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).
Prefix length
255.255.255.240 = /28
Usable hosts
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: x = 15
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
23 = 8
Subnet of this host
network: 172.20.8.0
broadcast: 172.20.15.255
Answer
x = 15
3.Computer networks
Answer: B · GET retrieves a resource, POST submits data in the request body
Why this is the answer
GET asks for a resource and should not change server state. POST sends data, such as a form, in the body for the server to process.
Answer
B · GET retrieves a resource, POST submits data in the request body
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: U = 193
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
63 = 0011 1111
inverted: 1100 0000
+1 → 1100 0001
Read it as unsigned
1100 0001₂ = 256 − 63 = 193
Answer
U = 193
6.Number systems
Answer: R = 240
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
Shifting left by 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1011 1100 << 2
= 1111 0000
Value of the result
1111 0000₂ = 240
Answer
R = 240
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