Computer Networks

A company has been allocated the network 192.168.10.0/24. They need to create 4 subnets with at least 50 hosts each. What subnet mask should they use?

Step 1: Determine how many bits to borrow for subnets.
Need 4 subnets -> 2^n >= 4 -> n = 2 bits

Step 2: Check remaining host bits can support 50 hosts.
Original: /24 means 8 host bits
After borrowing 2 bits: 6 host bits remain
Hosts per subnet = 2^6 - 2 = 64 - 2 = 62 hosts
62 >= 50, so this works.

Step 3: Calculate new subnet mask.
Original /24 = 255.255.255.0
Borrowing 2 bits: /24 + 2 = /26
/26 in binary: 11111111.11111111.11111111.11000000
Decimal: 255.255.255.192

Step 4: Verify subnets.
Subnet 1: 192.168.10.0 - 192.168.10.63 (usable: .1 - .62)
Subnet 2: 192.168.10.64 - 192.168.10.127
Subnet 3: 192.168.10.128 - 192.168.10.191
Subnet 4: 192.168.10.192 - 192.168.10.255

Answer: /26 or 255.255.255.192

In TCP, a sender has a window size of 8000 bytes and RTT of 100ms. What is the theoretical maximum throughput? If the packet size is 1000 bytes, how many packets can be in flight simultaneously?

Step 1: Calculate throughput using Bandwidth-Delay concept.
Throughput = Window Size / RTT
Window Size = 8000 bytes = 64000 bits
RTT = 100ms = 0.1 seconds

Step 2: Calculate throughput.
Throughput = 64000 bits / 0.1 s = 640,000 bps = 640 Kbps

Step 3: Calculate packets in flight.
Packet size = 1000 bytes
Window size = 8000 bytes
Packets in flight = 8000 / 1000 = 8 packets

Step 4: Verify interpretation.
With window size of 8000 bytes, sender can send 8 full packets (1000 bytes each) before needing acknowledgment. This maximizes pipeline utilization.

Answer: Maximum throughput = 640 Kbps, Packets in flight = 8 packets

A network uses CSMA/CD with a bandwidth of 1 Gbps. The maximum distance between any two stations is 2 km, and the signal propagation speed is 2 x 10^8 m/s. What is the minimum frame size required?

Step 1: Understand CSMA/CD timing requirement.
For CSMA/CD to work, transmission time must be at least twice the propagation time.
Minimum frame size condition: Transmission Time >= 2 x Propagation Time

Step 2: Calculate propagation time.
Distance = 2 km = 2000 meters
Propagation speed = 2 x 10^8 m/s
Propagation Time = Distance / Speed = 2000 / (2 x 10^8) = 10^-5 seconds = 10 microseconds

Step 3: Calculate required transmission time.
Transmission Time >= 2 x 10^-5 seconds = 20 microseconds

Step 4: Calculate minimum frame size.
Bandwidth = 1 Gbps = 10^9 bits/second
Minimum Frame Size = Bandwidth x Transmission Time
= 10^9 bits/s x 20 x 10^-6 s
= 10^9 x 20 x 10^-6 bits
= 20,000 bits
= 2,500 bytes

Answer: Minimum frame size = 2500 bytes (or 20,000 bits)

In a network using pure Aloha, stations transmit at 1000 frames/second. If the frame transmission time is 1ms, what is the throughput?

Which layer of the OSI model is responsible for routing packets between different networks?

Step 1: The OSI model has 7 layers.
Step 2: The Network layer (Layer 3) handles logical addressing and routing.
Step 3: It determines the best path for data packets to travel from source to destination across multiple networks.
Step 4: Routers operate at this layer using IP addresses.
Answer: Network layer

What is the default port number for HTTP?

Step 1: HTTP (Hypertext Transfer Protocol) uses well-known port numbers.
Step 2: Port 80 is the standard/default port for HTTP traffic.
Step 3: Port 443 is for HTTPS (secure HTTP).
Step 4: Port 21 is for FTP, and 8080 is commonly used as an alternative HTTP port.
Answer: 80

Which protocol is connectionless and provides best-effort delivery?

Step 1: TCP is connection-oriented with guaranteed delivery.
Step 2: UDP (User Datagram Protocol) is connectionless.
Step 3: UDP sends data without establishing a connection or confirming receipt.
Step 4: It provides faster transmission but no reliability guarantees.
Answer: UDP