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 606 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
606 hosts
p =
2.IPv4 subnettingHard
The network 192.168.24.0/24 is split with VLSM for these LANs: A: 84 hosts, B: 5 hosts, C: 11 hosts, D: 36 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 C?
A: 84B: 5C: 11D: 36
x =
3.Computer networksMedium
A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?
- ABoth routes, splitting the traffic equally
- B10.1.0.0/16, because it is the longest matching prefix
- C10.0.0.0/8, because it was added first
- D10.0.0.0/8, because it covers more addresses than the /16
4.Computer networksHard
Which routing protocol is a link-state protocol?
- ARIP
- BBGP
- COSPF
- DARP
5.Number systemsMedium
The 8-bit pattern 1010 0001 stores a signed integer in two's complement. What is its value N in decimal?
1010 00018 bit
N =
6.Number systemsHard
What is the most negative integer N that can be stored in 12-bit two's complement?
12 bit
N =
3Solutions
1.IPv4 subnetting
Answer: p = 22
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
29 − 2 = 510 < 606
210 − 2 = 1022 ≥ 606 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
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: 84 hosts → 27 = 128 (/25)
D: 36 hosts → 26 = 64 (/26)
C: 11 hosts → 24 = 16 (/28)
B: 5 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .127
D: .128 – .191
C: .192 – .207
B: .208 – .215
Answer
x = 192
LAN C: 192.168.24.192/28.
3.Computer networks
Answer: B · 10.1.0.0/16, because it is the longest matching prefix
Why this is the answer
IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.
Answer
B · 10.1.0.0/16, because it is the longest matching prefix
4.Computer networks
Answer: C · OSPF
Why this is the answer
OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.
Answer
C · OSPF
5.Number systems
Answer: N = −95
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 + 32 + 1 = −95
Answer
N = −95
6.Number systems
Answer: N = −2048
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.
Most negative value
−211 = −2048
Answer
N = −2048
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