TCP Congestion Control | Practice Problems

TCP Congestion Control-

 

Before you go through this article, make sure that you have gone through the previous article on TCP Congestion Control.

 

TCP congestion control policy consists of following three phases-

 

 

1. Slow Start
2. Congestion Avoidance
3. Congestion Detection

 

In this article, we will discuss practice problems on TCP Congestion Control.

 

PRACTICE PROBLEMS BASED ON TCP CONGESTION CONTROL-

 

Problem-01:

 

The growth of congestion window takes place-

1. Infinitely
2. Up to Threshold
3. Up to the size of receiver’s window
4. Up to timeout

 

Solution-

 

Option (C) is correct.

 

Problem-02:

 

Consider the effect of using slow start on a line with a 10 msec RTT and no congestion. The receiver window is 24 KB and the maximum segment size is 2 KB. How long does it take before the first full window can be sent?

 

Solution-

 

Given-

• Receiver window size = 24 KB
• Maximum Segment Size = 2 KB
• RTT = 10 msec

 

Receiver Window Size-

 

Receiver window size in terms of MSS

= Receiver window size / Size of 1 MSS

= 24 KB / 2 KB

= 12 MSS

 

Slow Start Threshold-

 

Slow start Threshold

= Receiver window size / 2

= 12 MSS / 2

= 6 MSS

 

Slow Start Phase-

 

• Window size at the start of 1^st transmission = 1 MSS
• Window size at the start of 2^nd transmission = 2 MSS
• Window size at the start of 3^rd transmission = 4 MSS
• Window size at the start of 4^th transmission = 6 MSS

 

Since the threshold is reached, so it marks the end of slow start phase.

Now, congestion avoidance phase begins.

 

Congestion Avoidance Phase-

 

• Window size at the start of 5^th transmission = 7 MSS
• Window size at the start of 6^th transmission = 8 MSS
• Window size at the start of 7^th transmission = 9 MSS
• Window size at the start of 8^th transmission = 10 MSS
• Window size at the start of 9^th transmission = 11 MSS
• Window size at the start of 10^th transmission = 12 MSS

 

From here,

• Window size at the end of 9^th transmission or at the start of 10^th transmission is 12 MSS.
• Thus, 9 RTT’s will be taken before the first full window can be sent.

 

So,

Time taken before the first full window is sent

= 9 RTT’s

= 9 x 10 msec

= 90 msec

 

Problem-03:

 

Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of slow start phase is 2 MSS and the threshold at the start of first transmission is 8 MSS. Assume that a time out occurs during the fifth transmission. Find the congestion window size at the end of tenth transmission.

1. 8 MSS
2. 14 MSS
3. 7 MSS
4. 12 MSS

 

Solution-

 

Given-

• Window size at the start of slow start phase = 2 MSS
• Threshold at the start of first transmission = 8 MSS
• Time out occurs during 5th transmission

 

Slow Start Phase-

 

• Window size at the start of 1^st transmission = 2 MSS
• Window size at the start of 2^nd transmission = 4 MSS
• Window size at the start of 3^rd transmission = 8 MSS

 

Since the threshold is reached, so it marks the end of slow start phase.

Now, congestion avoidance phase begins.

 

Congestion Avoidance Phase-

 

• Window size at the start of 4^th transmission = 9 MSS
• Window size at the start of 5^th transmission = 10 MSS

 

It is given that time out occurs during 5th transmission.

 

TCP reacts by-

• Setting the slow start threshold to half of the current congestion window size.
• Decreasing the congestion window size to 2 MSS (Given value is used).
• Resuming the slow start phase.

 

So now,

• Slow start threshold = 10 MSS / 2 = 5 MSS
• Congestion window size = 2 MSS

 

Slow Start Phase-

 

• Window size at the start of 6^th transmission = 2 MSS
• Window size at the start of 7^th transmission = 4 MSS
• Window size at the start of 8^th transmission = 5 MSS

 

Since the threshold is reached, so it marks the end of slow start phase.

Now, congestion avoidance phase begins.

 

Congestion Avoidance Phase-

 

• Window size at the start of 9^th transmission = 6 MSS
• Window size at the start of 10^th transmission = 7 MSS
• Window size at the start of 11^th transmission = 8 MSS

 

From here,

Window size at the end of 10^th transmission

= Window size at the start of 11^th transmission

= 8 MSS

 

Thus, Option (A) is correct.

 

Problem-04:

 

Suppose that the TCP congestion window is set to 18 KB and a time out occurs. How big will the window be if the next four transmission bursts are all successful? Assume that the MSS is 1 KB.

 

Solution-

 

Congestion Window Size-

 

Congestion window size in terms of MSS

= 18 KB / Size of 1 MSS

= 18 KB / 1 KB

= 18 MSS

 

Reaction Of TCP On Time Out-

 

TCP reacts by-

• Setting the slow start threshold to half of the current congestion window size.
• Decreasing the congestion window size to 1 MSS.
• Resuming the slow start phase.

 

So now,

• Slow start threshold = 18 MSS / 2 = 9 MSS
• Congestion window size = 1 MSS

 

Slow Start Phase-

 

• Window size at the start of 1^st transmission = 1 MSS
• Window size at the start of 2^nd transmission = 2 MSS
• Window size at the start of 3^rd transmission = 4 MSS
• Window size at the start of 4^th transmission = 8 MSS
• Window size at the start of 5^th transmission = 9 MSS

 

Thus, after 4 successful transmissions, window size will be 9 MSS or 9 KB.

 

Problem-05:

 

On a TCP connection, current congestion window size is 4 KB. The window advertised by the receiver is 6 KB. The last byte sent by the sender is 10240 and the last byte acknowledged by the receiver is 8192.

 

Part-01:

 

The current window size at the sender is ____.

1. 2048 B
2. 4096 B
3. 6144 B
4. 8192 B

 

Part-02:

 

The amount of free space in the sender window is ____.

1. 2048 B
2. 4096 B
3. 6144 B
4. 8192 B

 

Solution-

 

Part-01:

 

Sender window size

= min (Congestion window size, Receiver window size)

= min(4KB , 6KB)

= 4 KB

= 4096 B

Thus, Option (B) is correct.

 

Part-02:

 

Given-

• Last byte acknowledged by the receiver = 8192
• Last byte sent by the sender = 10240

 

From here,

• It means bytes from 8193 to 10240 are still present in the sender’s window.
• These bytes are waiting for their acknowledgement.
• Total bytes present in sender’s window = 10240 – 8193 + 1 = 2048 bytes.

 

From here,

Amount of free space in sender’s window currently

= 4096 bytes – 2048 bytes

= 2048 bytes

 

This indicates that half of the sender’s window is currently empty.

Thus, Option (A) is correct.

 

Next Article- TCP Timers

 

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