Tag: Computer Networks Course

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 1st transmission = 1 MSS
  • Window size at the start of 2nd transmission = 2 MSS
  • Window size at the start of 3rd transmission = 4 MSS
  • Window size at the start of 4th 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 5th transmission = 7 MSS
  • Window size at the start of 6th transmission = 8 MSS
  • Window size at the start of 7th transmission = 9 MSS
  • Window size at the start of 8th transmission = 10 MSS
  • Window size at the start of 9th transmission = 11 MSS
  • Window size at the start of 10th transmission = 12 MSS

 

From here,

  • Window size at the end of 9th transmission or at the start of 10th 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 1st transmission = 2 MSS
  • Window size at the start of 2nd transmission = 4 MSS
  • Window size at the start of 3rd 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 4th transmission = 9 MSS
  • Window size at the start of 5th 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 6th transmission = 2 MSS
  • Window size at the start of 7th transmission = 4 MSS
  • Window size at the start of 8th 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 9th transmission = 6 MSS
  • Window size at the start of 10th transmission = 7 MSS
  • Window size at the start of 11th transmission = 8 MSS

 

From here,

Window size at the end of 10th transmission

= Window size at the start of 11th 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 1st transmission = 1 MSS
  • Window size at the start of 2nd transmission = 2 MSS
  • Window size at the start of 3rd transmission = 4 MSS
  • Window size at the start of 4th transmission = 8 MSS
  • Window size at the start of 5th 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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How Digital Signature Works | Algorithm

Digital Signatures-

 

  • The signature on a document is the proof to the receiver that the document is coming from the correct entity.
  • A digital signature guarantees the authenticity of an electronic document in digital communication.

 

How Digital Signature Works?

 

  • The sender of the document digitally signs the document.
  • The receiver of the document verifies the signature.

 

The steps involved in the digital signature algorithm are-

 

At Sender Side-

 

At sender side,

  • Using a hash function, sender converts the message to be sent into a digested form.
  • There are various hash functions that may be used like SHA-1, MD5 etc.
  • The message in digested form is called as message digest.
  • Sender encrypts the message digest using his private key.
  • The encrypted message digest is called as signed digest or signature of the sender.
  • Sender sends the signed digest along with the original message to the receiver.

 

 

At Receiver Side-

 

At receiver side,

  • Receiver receives the original message and the signed digest.
  • Using a hash function, receiver converts the original message into a message digest.
  • Also, receiver decrypts the received signed digest using the sender’s public key.
  • On decryption, receiver obtains the message digest.
  • Now, receiver compares both the message digests.
  • If they are same, then it is proved that the document is coming from the correct entity.

 

 

Also Read- RSA Algorithm

 

Important Points-

 

Point-01:

 

After digitally signing the document, sender sends the following two things to the receiver-

  • Signed digest or signature
  • Original message

 

Point-02:

 

  • Sender uses his private key to digitally sign the document.
  • Receiver uses the sender’s public key to verify the signature.

 

Point-03:

 

  • Digital signature of a person varies from document to document.
  • This ensures authenticity of the document.

 

Point-04:

 

In digital signature,

  • There is one to one relationship between a message and a signature.
  • Each message has its own signature.

 

Point-05:

 

Digital signature verifies-

  • Authenticity
  • Integrity
  • Non-repudiation

 

Also Read- Diffie Hellman Key Exchange Algorithm

 

PRACTICE PROBLEMS BASED ON DIGITAL SIGNATURES-

 

Problem-01:

 

Anarkali digitally signs a message and sends it to Salim. Verification of the signature by Salim requires-

  1. Anarkali’s public key
  2. Salim’s public key
  3. Salim’s private key
  4. Anarkali’s private key

 

Solution-

 

Clearly, Option (A) is correct.

 

Problem-02:

 

Consider that B wants to send a message m that is digitally signed to A. Let the pair of private and public keys for A and B be denoted by Kx and Kx+ for x = A, B respectively. Let Kx(m) represent the operation of encrypting m with a key Kx and H(m) represent the message digest. Which one of the following indicates the correct way of sending the message m along with the digital signature to A?

  1. {m, KB+(H(m))}
  2. {m, KB(H(m))}
  3. {m, KA(H(m))}
  4. {m, KA+(H(m))}

 

Solution-

 

Clearly, Option (B) is correct.

 

To gain better understanding about Digital Signatures,

Watch this Video Lecture

 

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Diffie Hellman Key Exchange | Asymmetric Encryption

Asymmetric Encryption-

 

Before you go through this article, make sure that you have gone through the previous article on Asymmetric Key Cryptography.

 

In asymmetric encryption,

  • Sender and receiver use different keys to encrypt and decrypt the message.
  • The famous asymmetric encryption algorithms are-

 

 

In this article, we will discuss about Diffie Hellman Key Exchange Algorithm.

 

Symmetric Key Cryptography-

 

In symmetric key cryptography,

  • Both sender and receiver use a common secret key to encrypt and decrypt the message.
  • The major issue is exchanging the secret key between the sender and the receiver.
  • Attackers might intrude and know the secret key while exchanging it.

 

Read More- Symmetric Key Cryptography

 

Diffie Hellman Key Exchange-

 

As the name suggests,

  • This algorithm is used to exchange the secret key between the sender and the receiver.
  • This algorithm facilitates the exchange of secret key without actually transmitting it.

 

Diffie Hellman Key Exchange Algorithm-

 

Let-

  • Private key of the sender = Xs
  • Public key of the sender = Ys
  • Private key of the receiver = Xr
  • Public key of the receiver = Yr

 

Using Diffie Hellman Algorithm, the key is exchanged in the following steps-

 

Step-01:

 

  • One of the parties choose two numbers ‘a’ and ‘n’ and exchange with the other party.
  • ‘a’ is the primitive root of prime number ‘n’.
  • After this exchange, both the parties know the value of ‘a’ and ‘n’.

 

Step-02:

 

  • Both the parties already know their own private key.
  • Both the parties calculate the value of their public key and exchange with each other.

 

Sender calculate its public key as-

Ys = aXs mod n

Receiver calculate its public key as-

Yr = aXr mod n

 

Step-03:

 

  • Both the parties receive public key of each other.
  • Now, both the parties calculate the value of secret key.

 

Sender calculates secret key as-

Secret key = (Yr)Xs mod n

Receiver calculates secret key as-

Secret key = (Ys)Xr mod n

 

Finally, both the parties obtain the same value of secret key.

 

PRACTICE PROBLEMS BASED ON DIFFIE HELLMAN KEY EXCHANGE-

 

Problem-01:

 

Suppose that two parties A and B wish to set up a common secret key (D-H key) between themselves using the Diffie Hellman key exchange technique. They agree on 7 as the modulus and 3 as the primitive root. Party A chooses 2 and party B chooses 5 as their respective secrets. Their D-H key is-

  1. 3
  2. 4
  3. 5
  4. 6

 

Solution-

 

Given-

  • n = 7
  • a = 3
  • Private key of A = 2
  • Private key of B = 5

 

Step-01:

 

Both the parties calculate the value of their public key and exchange with each other.

 

Public key of A

= 3private key of A mod 7

= 32 mod 7

= 2

 

Public key of B

= 3private key of B mod 7

= 35 mod 7

= 5

 

Step-02:

 

Both the parties calculate the value of secret key at their respective side.

 

Secret key obtained by A

= 5private key of A mod 7

= 52 mod 7

= 4

 

Secret key obtained by B

= 2private key of B mod 7

= 25 mod 7

= 4

 

Finally, both the parties obtain the same value of secret key.

The value of common secret key = 4.

Thus, Option (B) is correct.

 

Problem-02:

 

In a Diffie-Hellman Key Exchange, Alice and Bob have chosen prime value q = 17 and primitive root = 5. If Alice’s secret key is 4 and Bob’s secret key is 6, what is the secret key they exchanged?

  1. 16
  2. 17
  3. 18
  4. 19

 

Solution-

 

Given-

  • n = 17
  • a = 5
  • Private key of Alice = 4
  • Private key of Bob = 6

 

Step-01:

 

Both Alice and Bob calculate the value of their public key and exchange with each other.

 

Public key of Alice

= 5private key of Alice mod 17

= 54 mod 17

= 13

 

Public key of Bob

= 5private key of Bob mod 17

= 56 mod 17

= 2

 

Step-02:

 

Both the parties calculate the value of secret key at their respective side.

 

Secret key obtained by Alice

= 2private key of Alice mod 7

= 24 mod 17

= 16

 

Secret key obtained by Bob

= 13private key of Bob mod 7

= 136 mod 17

= 16

 

Finally, both the parties obtain the same value of secret key.

The value of common secret key = 16.

Thus, Option (A) is correct.

 

To gain better understanding about Diffie Hellman Key Exchange Algorithm,

Watch this Video Lecture

 

Next Article- Digital Signatures

 

Get more notes and other study material of Computer Networks.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Public Key Cryptography | RSA Algorithm Example

Cryptography in Network Security-

 

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

 

We have discussed-

  • Cryptography is a method of storing and transmitting data in a particular form.
  • Cryptography techniques are-

 

 

In this article, we will discuss about Asymmetric Key Cryptography.

 

Asymmetric Key Cryptography-

 

In this technique,

  • Sender and receiver use different keys to encrypt and decrypt the message.
  • It is called so because sender and receiver use different keys.
  • It is also called as public key cryptography.

 

Working-

 

The message exchange using public key cryptography involves the following steps-

 

 

Step-01:

 

At sender side,

  • Sender encrypts the message using receiver’s public key.
  • The public key of receiver is publicly available and known to everyone.
  • Encryption converts the message into a cipher text.
  • This cipher text can be decrypted only using the receiver’s private key.

 

Step-02:

 

  • The cipher text is sent to the receiver over the communication channel.

 

Step-03:

 

At receiver side,

  • Receiver decrypts the cipher text using his private key.
  • The private key of the receiver is known only to the receiver.
  • Using the public key, it is not possible for anyone to determine the receiver’s private key.
  • After decryption, cipher text converts back into a readable format.

 

Advantages-

 

The advantages of public key cryptography are-

  • It is more robust.
  • It is less susceptible to third-party security breach attempts.

 

Disadvantages-

 

The disadvantages of public key cryptography are-

  • It involves high computational requirements.
  • It is slower than symmetric key cryptography.

 

Number of Keys Required-

 

To use public key cryptography,

  • Each individual requires two keys- one public key and one private key.
  • For n individuals to communicate, number of keys required = 2 x n = 2n keys.

 

Asymmetric Encryption Algorithms-

 

The famous asymmetric encryption algorithms are-

 

 

  1. RSA Algorithm
  2. Diffie-Hellman Key Exchange

 

In this article, we will discuss about RSA Algorithm.

 

RSA Algorithm-

 

Let-

  • Public key of the receiver = (e , n)
  • Private key of the receiver = (d , n)

 

Then, RSA Algorithm works in the following steps-

 

Step-01:

 

At sender side,

  • Sender represents the message to be sent as an integer between 0 and n-1.
  • Sender encrypts the message using the public key of receiver.
  • It raises the plain text message ‘P’ to the eth power modulo n.
  • This converts the message into cipher text ‘C’.

 

C = Pe mod n

 

Step-02:

 

  • The cipher text ‘C’ is sent to the receiver over the communication channel.

 

Step-03:

 

At receiver side,

  • Receiver decrypts the cipher text using his private key.
  • It raises the cipher text ‘C’ to the dth power modulo n.
  • This converts the cipher text back into the plain text ‘P’.

 

P = Cd mod n

 

NOTE-

 

‘e’ and ‘d’ must be multiplicative inverses modulo Ø(n)

 

After decryption, receiver must have-

P = Cd mod n

P = (Pe mod n)d mod n

P = Ped mod n

For this equation to be true, by Euler’s Theorem, we must have-

ed = 1 mod Ø(n)

OR

ed = kØ(n) + 1

Thus, e and d must be multiplicative inverses modulo Ø(n).

 

Steps to Generate Public Key And Private Key-

 

An individual can generate his public key and private key using the following steps-

 

Step-01:

 

Choose any two prime numbers p and q such that-

  • They are different.
  • They are very large.

 

Step-02:

 

Calculate ‘n’ and toilent function Ø(n) where-

  • n = p x q
  • Ø(n) = (p-1) x (q-1)

 

Step-03:

 

Choose any value of ‘e’ such that-

  • 1 < e < Ø(n)
  • gcd (e, Ø(n)) = 1

 

Step-04:

 

Determine ‘d’ such that-

 

 

  • You already know the value of ‘e’ and Ø(n).
  • Choose the least positive integer value of ‘k’ which gives the integer value of ‘d’ as a result.
  • Use trial and error method.
  • Start substituting different values of ‘k’ from 0.

 

PRACTICE PROBLEMS BASED ON RSA ALGORITHM-

 

Problem-01:

 

In a RSA cryptosystem, a participant A uses two prime numbers p = 13 and q = 17 to generate her public and private keys. If the public key of A is 35, then the private key of A is _______.

 

Solution-

 

Given-

  • Prime numbers p = 13 and q = 17
  • Public key = 35

 

Step-01:

 

Calculate ‘n’ and toilent function Ø(n).

 

Value of n,

n = p x q

n = 13 x 17

∴ n = 221

 

Toilent function,

Ø(n) = (p-1) x (q-1)

Ø(n) = (13-1) x (17-1)

∴ Ø(n) = 192

 

Step-02:

 

  • We are already given the value of e = 35.
  • Thus, public key = (e , n) = (35 , 221)

 

Step-03:

 

Determine ‘d’ such that-

 

 

Here,

  • The least value of ‘k’ which gives the integer value of ‘d’ is k = 2.
  • On substituting k = 2, we get d = 11.

 

Thus, private key of participant A = (d , n) = (11, 221).

 

Problem-02:

 

In the RSA public key cryptosystem, the private and public keys are (e, n) and (d, n) respectively, where n = p x q and p and q are large primes. Besides, n is public and p and q are private. Let M be an integer such that 0 < M < n and f(n) = (p-1)(q-1).

 

Now consider the following equations-

I. M’ = Me mod n and M = (M’)d mod n

II. ed ≡ 1 mod n

III. ed = 1 mod f(n)

IV. M’ = Me mod f(n) and M = (M’)d mod f(n)

 

Which of the above equations correctly represent RSA cryptosystem?

  1. I and II
  2. I and III
  3. II and IV
  4. III and IV

 

Solution-

 

Clearly, Option (B) is correct.

 

To gain better understanding about RSA Algorithm,

Watch this Video Lecture

 

Next Article- Diffie Hellman Key Exchange Algorithm

 

Get more notes and other study material of Computer Networks.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Symmetric Key Cryptography | Cryptography Techniques

Cryptography in Network Security-

 

In network security,

  • Cryptography is a method of storing and transmitting data in a particular form.
  • It ensures that only the person for whom the message is intended can read the message.

 

The message exchange using cryptography involves the following steps-

 

 

Step-01:

 

At sender side,

  • Using an encryption algorithm, the message is converted into an unreadable form.
  • The message in unreadable form is called as cipher text.

 

Step-02:

 

  • The cipher text is sent to the receiver over the communication channel.
  • Since the message is encrypted, the attackers can not read the message.

 

Step-03:

 

At receiver side,

  • Using a decryption algorithm, the message is again converted into the readable form.
  • Then, receiver can read the message.

 

Cryptography Techniques-

 

Cryptography techniques may be classified as-

 

 

  1. Symmetric Key Cryptography
  2. Asymmetric Key Cryptography

 

In this article, we will discuss about symmetric key cryptography.

 

Symmetric Key Cryptography-

 

In this technique,

  • Both sender and receiver uses a common key to encrypt and decrypt the message.
  • This secret key is known only to the sender and to the receiver.
  • It is also called as secret key cryptography.

 

Working-

 

The message exchange using symmetric key cryptography involves the following steps-

 

 

 

  • Before starting the communication, sender and receiver shares the secret key.
  • This secret key is shared through some external means.
  • At sender side, sender encrypts the message using his copy of the key.
  • The cipher text is then sent to the receiver over the communication channel.
  • At receiver side, receiver decrypts the cipher text using his copy of the key.
  • After decryption, the message converts back into readable format.

 

Symmetric Encryption Algorithms-

 

Some of the encryption algorithms that use symmetric key are-

  • Advanced Encryption Standard (AES)
  • Data Encryption Standard (DES)

 

Advantages-

 

The advantages of symmetric key algorithms are-

  • They are efficient.
  • They take less time to encrypt and decrypt the message.

 

Disadvantages-

 

Point-01:

 

The number of keys required is very large.

 

In symmetric key cryptography,

  • Each pair of users require a unique secret key.
  • If N people in the world wants to use this technique, then there needs to be N(N-1) / 2 secret keys.
  • For 1 million people to communicate, a half billion secret keys would be needed.

 

How N(N-1)/2 Keys Will Be Required?

 

  • Consider a complete graph with N nodes.
  • Consider each node represents one person.
  • Then, each person will require (N-1) keys to communicate with other (N-1) people.
  • Thus, each edge must have a unique key for communication.
  • Thus, Number of keys required = Number of edges = nC2 = n(n-1)/2.

 

Point-02:

 

  • Sharing the secret key between the sender and receiver is an important issue.
  • While sharing the key, attackers might intrude.

 

To overcome this disadvantage,

Diffie Hellman Key Exchange Algorithm is used for exchanging the secret key.

 

Important Points-

 

Point-01:

 

In symmetric key cryptography,

  • Both sender and receiver uses the same key.
  • Sender encrypts the message using his copy of the key.
  • Receiver decrypts the message using his copy of the key.
  • The key must not be known to anyone else other than sender and receiver.
  • If the secret key is known to any intruder, he could decrypt the message.

 

Point-02:

 

  • This cryptography technique is called as symmetric key cryptography.
  • It is because both sender and receiver use the same key on their sides.

 

Point-03:

 

  • This cryptography technique is called as secret key cryptography.
  • It is because the key has to be kept secret between the sender and receiver.

 

To gain better understanding about Symmetric Key Cryptography,

Watch this Video Lecture

 

Next Article- Asymmetric Key Cryptography | RSA Algorithm

 

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Watch video lectures by visiting our YouTube channel LearnVidFun.