Transmission Control Protocol | Practice Problems

PRACTICE PROBLEMS BASED ON TRANSMISSION CONTROL PROTOCOL-

 

Problem-01:

 

How many TCP connections can be opened between two ports?

1. Multiple
2. Single
3. Zero
4. None

 

Solution-

 

Option (B) is correct.

 

Problem-02:

 

TCP protects itself from miss delivery by IP with the help of-

1. Source IP Address in IP header
2. Destination IP Address in IP header
3. Pseudo header
4. Source port and Destination port

 

Solution-

 

Option (C) is correct.

 

Problem-03:

 

What addressing system has topological significance?

1. Logical or Network Address
2. LAN or Physical Address
3. Port Addressing System
4. Multicast Addressing System

 

Solution-

 

Option (A) is correct.

 

Problem-04:

 

If WAN link is 2 Mbps and RTT between source and destination is 300 msec, what would be the optimal TCP window size needed to fully utilize the line?

1. 60,000 bits
2. 75,000 bytes
3. 75,000 bits
4. 60,000 bytes

 

Solution-

 

Given-

• Bandwidth = 2 Mbps
• RTT = 300 msec

 

Optimal TCP Window Size-

 

Optimal TCP window size

= Maximum amount of data that can be sent in 1 RTT

= 2 Mbps x 300 msec

= 600 x 10³ bits

= 60,0000 bits

= 75,000 bytes

 

Thus, Option (B) is correct.

 

Problem-05:

 

Suppose host A is sending a large file to host B over a TCP connection. The two end hosts are 10 msec apart (20 msec RTT) connected by a 1 Gbps link. Assume that they are using a packet size of 1000 bytes to transmit the file. For simplicity, ignore ack packets. At least how big would the window size (in packets) have to be for the channel utilization to be greater than 80%?

1. 1000
2. 1500
3. 2000
4. 2500

 

Solution-

 

Given-

• RTT = 20 msec
• Bandwidth = 1 Gbps
• Packet size = 1000 bytes
• Efficiency >= 80%

 

Window Size For 100% Efficiency-

 

For 100% efficiency,

Window size

= Maximum number of bits that can be transmitted in 1 RTT

= 1 Gbps x 20 msec

= (10⁹ bits per sec) x 20 x 10^-3 sec

= 20 x 10⁶ bits

= 2 x 10⁷ bits

 

Window Size For 80% Efficiency-

 

For 80% efficiency,

Window size

= 0.8 x 2 x 10⁷ bits

= 1.6 x 10⁷ bits

 

In terms of packets,

Window size

= 1.6 x 10⁷ bits / Packet size

= 1.6 x 10⁷ bits / (1000 x 8 bits)

= 0.2 x 10⁴ packets

= 2000 packets

 

Thus, Option (C) is correct.

 

Problem-06:

 

A TCP machine is sending windows of 65535 B over a 1 Gbps channel that has a 10 msec one way delay.

1. What is the maximum throughput achievable?
2. What is the line efficiency?

 

Solution-

 

Given-

• Window size = 65535 bytes
• Bandwidth = 1 Gbps
• One way delay = 10 msec

 

Method-01:

 

Maximum amount of data that can be sent in 1 RTT

= 1 Gbps x (2 x 10 msec)

= (10⁹ bits per sec) x 20 x 10^-3 sec

= 20 x 10⁶ bits

= 25 x 10⁵ bytes

 

Amount of data that is actually being sent in 1 RTT = 65535 bytes

 

Thus,

Line Efficiency(η)

= Amount of data being sent in 1 RTT / Maximum amount of data that can be sent in 1 RTT

= 65535 bytes / 25 x 10⁵ bytes

= 0.026214

= 2.62%

 

Now,

Maximum Achievable Throughput

= Efficiency x Bandwidth

= 0.0262 x 1 Gbps

= 26.214 Mbps

 

Method-02:

 

Maximum Achievable Throughput

= Number of bits sent per second

= 65535 B / 20 msec

= (65535 x 8 bits) / (20 x 10^-3 sec)

= 26.214 Mbps

 

Now,

Line Efficiency

= Throughput / Bandwidth

= 26.214 Mbps / 1 Gbps

= 26.214 x 10^-3

= 0.026214

= 2.62%

 

Next Article- TCP Congestion Control

 

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