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 10.106.148.0/22 is divided into equal subnets with prefix /24. How many subnets N are there?
10.106.148.0/22/24
N =
2.IPv4 subnettingHard
A host has the address 10.20.34.39/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.34.39/28
x =
3.Computer networksMedium
What uniquely identifies one end of a TCP connection on the internet?
- AAn IP address together with a port number
- BA MAC address alone
- CA domain name alone
- DA subnet mask and gateway
4.Computer networksHard
Why was the TCP window scale option introduced?
- AWithout it, TCP could not detect lost segments on long-distance links
- BThe 32-bit window field was too large and wasted header space in every TCP segment sent
- CThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
- DUDP needed a way to control its sending rate on congested paths
5.Number systemsMedium
Set the bits so they store −8 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −88 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1000 1001 << 1. What is the result's value R in decimal?
1000 1001<< 1
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
24 − 22 = 2
Number of subnets
N = 22 = 4
Answer
N = 4
Each /24 subnet has 28 − 2 = 254 usable hosts.
2.IPv4 subnetting
Answer: x = 47
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
24 = 16
Subnet of this host
network: ⌊39/16⌋·16 = 32
broadcast: 32 + 16 − 1 = 47
usable hosts: 33 … 46
Answer
x = 47
3.Computer networks
Answer: A · An IP address together with a port number
Why this is the answer
A TCP connection is identified by source IP and port plus destination IP and port. Ports separate the many applications on one host.
Answer
A · An IP address together with a port number
4.Computer networks
Answer: C · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
Why this is the answer
The original header allows at most 65,535 bytes of unacknowledged data. Window scaling multiplies this by a power of two so high bandwidth-delay paths can be filled.
Answer
C · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: U = 248
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
8 = 0000 1000
inverted: 1111 0111
+1 → 1111 1000
Read it as unsigned
1111 1000₂ = 256 − 8 = 248
Answer
U = 248
6.Number systems
Answer: R = 18
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 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1000 1001 << 1
= 0001 0010
Value of the result
0001 0010₂ = 18
Answer
R = 18
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