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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

Dialog Control and Synchronization are function of which layer?(a) Presentation Layer(b) Application Layer(c) Session Layer(d) Data Link LayerThe question was asked in an interview for internship.This intriguing question originated from TCP/IP and OSI Reference Model topic in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» CORRECT answer is (c) Session Layer

The explanation: DIALOG Control and Synchronization are function of the Session Layer.
2.

UDP stands for ________(a) User Datagram Protocol(b) Used Data Protocol(c) Unified Definition Protocol(d) Undefined Diagnostic ProtocolThis question was posed to me in an interview for internship.My question comes from TCP/IP and OSI Reference Model in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The CORRECT option is (a) USER Datagram PROTOCOL

Best EXPLANATION: UDP stands for User Datagram Protocol.

3.

A digitized voice channel is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (two samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate?(a) 64 kbps(b) 32 kbps(c) 128 kbps(d) 16 kbpsI got this question in an internship interview.I would like to ask this question from Layers topic in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The CORRECT OPTION is (a) 64 kbps

The EXPLANATION is: 2 × 4000 × 8 = 64,000 = 64kbps.

4.

It is desirable to revoke a certificate before it expires because –(a) the user is no longer certified by this CA(b) the CA’s certificate is assumed to be compromised(c) the user’s private key is assumed to be compromised(d) all of the mentionedThis question was posed to me during an internship interview.I want to ask this question from Overview topic in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The CORRECT answer is (d) all of the mentioned

The BEST explanation: All of the options are TRUE with regard to REVOCATION of a CERTIFICATE.

5.

What is the hamming distance between these 2 codes: 10010010 and 11011001?(a) 3(b) 4(c) 6(d) 2This question was posed to me in quiz.Asked question is from Layers topic in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» CORRECT choice is (b) 4

Explanation: HAMMING distance is number of dissimilar BITS between 2 streams.
6.

Generate the CRC codeword for the message x^3+1 using the generator polynomial x^3+x+1.(a) 1001101(b) 1001001(c) 1001110(d) 1001111The question was posed to me in an interview for job.My question comes from Layers in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» CORRECT option is (C) 1001110

Easy explanation: M(x) = 1001. G(x) = 1011.

x^3*M(x) = 1001000.

On DIVIDING we get remainder as 110. THEREFORE, Codeword is 1001110.
7.

The Index of Coincidence is(a) 0.065(b) 0.048(c) 0.067(d) 0.044I got this question in class test.This interesting question is from Overview topic in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT choice is (d) 0.044

To ELABORATE: NUMBER of letters = 145.From this, IC=0.0438697 .This is very strong evidence thatthe MESSAGE CAME from a polyalphabetic ciphering scheme.
8.

How many rounds does the AES-192 perform?(a) 10(b) 12(c) 14(d) 16I have been asked this question during an interview.This intriguing question comes from Overview topic in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct CHOICE is (b) 12

Explanation: AES 192 PERFORMS 12 ROUNDS.

9.

How many keys are used in the ANSI X9.17 PRNG?(a) 3(b) 2(c) 4(d) 6I have been asked this question in quiz.This intriguing question originated from Overview topic in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT choice is (a) 3

Best EXPLANATION: 2 keys are used for the 3DES Encrypt DECRYPT Encrypt METHOD.
10.

DSL stands for ________(a) Data Storage Line(b) Digital Subscriber Line(c) Data Service Language(d) Data Secure LanguageI had been asked this question during an interview for a job.This key question is from TCP/IP and OSI Reference Model in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT option is (B) DIGITAL Subscriber LINE

Explanation: DSL STANDS for Digital Subscriber Line.
11.

HTTP stands for ________(a) Hash Text Transfer Protocol(b) Hyper Text Transfer Protocol(c) Hash Transfer Text Protocol(d) none of the mentionedThe question was posed to me in final exam.This intriguing question originated from TCP/IP and OSI Reference Model topic in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT CHOICE is (B) Hyper TEXT Transfer Protocol

The explanation: HTTP stands for Hyper Text Transfer Protocol.
12.

A device that connects networks with different protocols –(a) Switch(b) Hub(c) Gateway(d) Proxy ServerI got this question in semester exam.The origin of the question is Network Hardware topic in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The CORRECT CHOICE is (c) GATEWAY

Explanation: Such a DEVICE is KNOWN as a Gateway.

13.

What is the function of Network Interface Cards?(a) connects the clients, servers and peripherals to the network through a port(b) allows you to segment a large network into smaller, efficient networks(c) connects networks with different protocols like TCP/IP(d) boost the signal between two cable segments or wireless access pointsThis question was addressed to me in unit test.Enquiry is from Network Hardware topic in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct OPTION is (a) connects the clients, servers and peripherals to the network through a PORT

For explanation I would SAY: A Network Interface CARDS connects the clients, servers and peripherals to the network through a port.

14.

Which topology requires the most amount of wiring?(a) Mesh(b) Star(c) Bus(d) RingI got this question in quiz.This interesting question is from Topologies topic in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» CORRECT OPTION is (a) MESH

Explanation: This is the DISADVANTAGE of mesh TOPOLOGY.
15.

2 Half duplex systems can make a Full-Duplex.(a) True(b) FalseThe question was asked during a job interview.I would like to ask this question from Topologies topic in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» CORRECT ANSWER is (a) TRUE

To EXPLAIN: The STATEMENT is true.
16.

The property that any extracted subsequence should pass the test for randomness is(a) Scalability(b) Uniformity(c) Stability(d) ConsistencyI have been asked this question in a national level competition.I want to ask this question from Overview in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT option is (a) SCALABILITY

Best explanation: Scalability is the PROPERTY where any extracted SUBSEQUENCE should pass the test for randomness.
17.

The last two blocks of the XTS-AES mode are(a) padded as 10*(b) encrypted/ decrypted using ciphertext-stealing(c) padded as 10*1(d) padded and then swapped after encryption/ decryptionThe question was asked in a job interview.Question is taken from Overview in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct CHOICE is (b) encrypted/ decrypted using ciphertext-stealing

Easiest explanation: The correct term used to ENCRYPT/ decrypt the last 2 blocks is ‘cipher-text stealing’ where C(m) and C(m-1) are INTERCHANGED with each other.

18.

A total of seven messages are required in the Public-Key distribution scenario. However, the initial five messages need to be used only infrequently because both A and B can save the other’s public key for future – a technique known as ____________(a) time stamping(b) polling(c) caching(d) squeezingThe question was posed to me during an interview.My doubt is from Overview in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct CHOICE is (c) caching

To EXPLAIN: This technique is known as caching.

19.

Which one of the following modes of operation in DES is used for operating short data?(a) Cipher Feedback Mode (CFB)(b) Cipher Block chaining (CBC)(c) Electronic code book (ECB)(d) Output Feedback Modes (OFB)I got this question at a job interview.My question is based upon Overview in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct answer is (C) Electronic code book (ECB)

Easy explanation: The Electronic code book mode is used for OPERATING on short data as the same key is used for each block. THUS repetitions in PLAIN Text lead to repetitions in Cipher Text.

20.

Find the solution of x^2≡3 mod 23(a) x≡±16 mod 23(b) x≡±13 mod 23(c) x≡±22 mod 23(d) x≡±7 mod 23The question was posed to me in exam.Asked question is from Overview in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct choice is (a) x≡±16 mod 23

To EXPLAIN I would say: a=33^((23+1)/4)≡3^6≡1(QR and there is SOLUTION)

x ≡ ±3(23 + 1)/4 (mod 23) ≡±16i.e. x = 7 and 16

21.

File, Transfer, Access and Management (FTAM) is a function of which layer ?(a) Presentation Layer(b) Application Layer(c) Session Layer(d) Data Link LayerI have been asked this question in exam.Asked question is from TCP/IP and OSI Reference Model in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Right option is (b) Application LAYER

To EXPLAIN I would say: FTAM is an Application Layer FUNCTION.

22.

ICMP stands for ________(a) Internal Control Message Protocol(b) Internet Cipher Mail Protocol(c) Internal Cipher Mail Protocol(d) Internet Control Message ProtocolI have been asked this question in an online interview.I'm obligated to ask this question of TCP/IP and OSI Reference Model topic in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct answer is (d) INTERNET Control MESSAGE Protocol

For EXPLANATION I WOULD say: ICMP stands for Internet Control Message Protocol.

23.

A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the requirement to make this frame collision-free?(a) 2msec(b) 4msec(c) 2sec(d) 4secI have been asked this question in an international level competition.My query is from Layers topic in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The CORRECT choice is (a) 2msec

To explain: AVERAGE frame TRANSMISSION time

Tfr = 200 bits/200 kbps or 1 MS

Vulnerable time = 2x Tfr = 2 × 1 ms = 2 ms.

24.

Five channels, each with a 100-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 kHz between the channels to prevent interference?(a) 4600 kHz(b) 500 kHz(c) 540 kHz(d) 580 kHzI got this question in quiz.Question is from Layers in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct choice is (c) 540 KHZ

To explain: For five CHANNELS, we need at least four guard bands. This means that the required bandwidth is at least 5 × 100 + 4 × 10 = 540 kHz.

25.

What are the bits transmitted for the Unipolar system?(a) 01101(b) 11010(c) 10110(d) 01001I had been asked this question during an interview.Asked question is from Layers in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT choice is (C) 10110

The BEST I can explain: High – 1, Low – 0.
26.

A channel has a 1-MHz bandwidth. The SNR for this channel is 63. What is the appropriate bit rate?(a) 4 Mbps(b) 6 Mbps(c) 8 Mbps(d) 12 MbpsThis question was posed to me in unit test.I want to ask this question from Layers topic in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Right choice is (b) 6 Mbps

To explain I would SAY: C = Blog_2(1 + SNR) = 10^6 log_2(1 + 63) = 6Mbps.

27.

We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?(a) 1024(b) 512(c) 256(d) 128The question was posed to me in my homework.My query is from Layers in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Right CHOICE is (d) 128

Easy explanation: 256000= 2 × 20000 × log L = 6.625.

L = 98.7 Levels.

Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit RATE. If we have 128 levels, the bit rate is 280 KBPS.

28.

Which function can be used in AES multiplication –(a) m(x)=x^7+x^4+x^3(b) m(x)=x^8+x^4+x^3+x+1(c) m(x)=x^8+x^3+x^2+x+1(d) m(x)=x^8+x^5+x^3+xThis question was addressed to me in an interview for job.My question is taken from Overview in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The CORRECT answer is (b) m(X)=x^8+x^4+x^3+x+1

Explanation: m(x)=x^8+x^4+x^3+x+1 STANDS for 100011011 which is an IRREDUCIBLE polynomial. Others are not irreducible polynomials.

29.

What is the hamming code for the data: 1001101?(a) 10011100101(b) 11010000101(c) 10001100101(d) 11111001011This question was addressed to me during an interview for a job.This key question is from Layers topic in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct choice is (a) 10011100101

To ELABORATE: Find the 1st 2ND 4th and 8th BITS using the hamming ALGORITHM and thus proceed to get the hamming code.

30.

A device that provides a central connection point for cables is –(a) Switch(b) Hub(c) Gateway(d) Proxy ServerThis question was posed to me in unit test.My doubt stems from Network Hardware topic in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct option is (C) Gateway

Explanation: Such a device is known as a Hub and is USUALLY USED in STAR TOPOLOGIES.

31.

What is the block size in the Simplified AES algorithm?(a) 8 bits(b) 40 bits(c) 16 bits(d) 36 bitsThe question was asked by my college professor while I was bunking the class.My enquiry is from Overview in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Right answer is (B) 40 BITS

To explain: The block size for the AES ALGORITHM is 16 bits.

32.

RC5 uses 2 magic constants to define their subkeys. These are(a) Base of natural Logarithm and Golden ratio(b) Base of natural Logarithm and Pi(c) Golden Ratio andPi(d) Pi and Golden RationThe question was asked by my school principal while I was bunking the class.Origin of the question is Overview in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT option is (a) Base of natural Logarithm and GOLDEN ratio

Explanation: The initialize operations makes use of MAGIC constants defined as follows:

P_w=Odd[(e-2) 2^w]

Q_w=Odd[(φ-1) 2^w].
33.

Flow Control and Error Control are functions of which layer?(a) Physical Layer(b) Application Layer(c) Data Link Layer(d) Network LayerThis question was posed to me in an interview.My doubt is from TCP/IP and OSI Reference Model in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct option is (c) Data Link Layer

The best EXPLANATION: FLOW CONTROL and ERROR Control are functions of the Data Link Layer.

34.

Division of(131B6C3) base 16 by (lA2F) base 16 yeilds(a) 1AD(b) DAD(c) BAD(d) 9ADThis question was posed to me by my school teacher while I was bunking the class.This is a very interesting question from Overview topic in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT answer is (d) 9AD

The best I can explain: Base 16 division to be FOLLOWED where A-F STAND for 10-15.
35.

Which of these is a connection oriented service?(a) X.25(b) Frame Relay(c) ATM(d) All of the mentionedI have been asked this question by my college director while I was bunking the class.Question is taken from TCP/IP and OSI Reference Model in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct OPTION is (d) All of the MENTIONED

Easy explanation: All the mentioned are connection oriented SERVICES.

36.

SMTP stands for ________(a) Service Message Transmission Permission(b) Secure Message Transfer Protocol(c) Simple Mail Transfer Protocol(d) Simple Message Transfer ProtocolThis question was addressed to me in an interview.My doubt stems from TCP/IP and OSI Reference Model in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct choice is (C) Simple Mail TRANSFER Protocol

The explanation: SMTP STANDS for Simple Mail Transfer Protocol.

37.

A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (including the delays in the devices and ignoring the time needed to send a jamming signal, as we see later) is 25.6 μs, what is the minimum size of the frame?(a) 128 bytes(b) 32 bytes(c) 16 bytes(d) 64 bytesI had been asked this question during an online exam.I would like to ask this question from Layers in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct CHOICE is (d) 64 bytes

To EXPLAIN: The MINIMUM frame transmission time is Tfr = 2 × Tp = 51.2 μs. This means, in the WORST case, a station needs to transmit for a period of 51.2 μs to detect the collision.

The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for STANDARD Ethernet.

38.

Find the first 8 bits for Blum Blum Shub Bit Generator when seed = 101355 and n = 192649.(a) 10101010(b) 11100010(c) 11001011(d) 11001110The question was asked in quiz.My question comes from Overview in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT answer is (d) 11001110

For explanation: The BLUM blum SHUB algorithm is as follows

Xo = s^2 mod n

for i=1 to 8

Xi = X(i-1)^2 mod n

Bi = Xi mod 2

Using this we compute the bits as – 11001110.
39.

Multiply the polynomials P1 = (x^5 + x^2 + x) by P2 = (x^7 + x^4 + x^3 + x^2 + x) in GF(28) with irreducible polynomial (x^8 + x^4 + x^3 + x + 1). The result is(a) x^4+ x^3+ x+1(b) x^5+ x^3+ x^2+x+1(c) x^5+ x^4+ x^3+x+1(d) x^5+ x^3+ x^2+xThe question was posed to me during an internship interview.The query is from Overview topic in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» CORRECT option is (b) X^5+ x^3+ x^2+x+1

The best I can EXPLAIN: On performing polynomial multiplication we get with RESPECT to MODULUS (x^8 + x^4 + x^3 + x + 1) we get x^5+ x^3+ x^2+x+1.
40.

Logical Addressing and Routing are functions of which layer?(a) Physical Layer(b) Transport Layer(c) Data Link Layer(d) Network LayerI got this question by my school teacher while I was bunking the class.I'd like to ask this question from TCP/IP and OSI Reference Model in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct CHOICE is (d) Network Layer

The explanation is: . Logical Addressing and ROUTING are FUNCTIONS of the Network Layer.

41.

Calculate the theoretical highest bit rate of a regular telephone line. The signal-to-noise ratio is usually 3162.(a) 34.860 kbps(b) 17.40 kbps(c) 11.62 kbps(d) none of the MentionedThe question was asked in semester exam.This intriguing question comes from Layers in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The CORRECT CHOICE is (a) 34.860 kbps

The best I can explain: C = B log (1+SNR) = 3000 log (1+3162) = 34,860 bps.

Logarithm here is in terms of base 2.

42.

Which topology uses the token passing algorithm?(a) Mesh(b) Star(c) Bus(d) RingThis question was posed to me by my college director while I was bunking the class.My doubt stems from Topologies in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The CORRECT choice is (d) Ring

Easiest explanation: The ring topology uses the TOKEN PASSING ALGORITHM. The token holder gets its turn at communication.

43.

How many links are there for N nodes in the mesh topology?(a) 1(b) N(N+1)/2(c) N(d) N(N-1)/2The question was asked in unit test.This is a very interesting question from Topologies in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT option is (d) N(N-1)/2

The EXPLANATION: There are N(N-1)/2 LINKS for N nodes.
44.

Ethernet LANs use which topology?(a) Mesh(b) Star(c) Bus(d) RingThis question was addressed to me in an interview for job.I'm obligated to ask this question of Topologies topic in section Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» CORRECT option is (c) BUS

Easy EXPLANATION: ETHERNET LANs use bus topology,
45.

12. For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where Cipher message=11 and thus find the plain text.(a) 88(b) 122(c) 143(d) 111I have been asked this question in a job interview.My question is from Overview in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Right ANSWER is (a) 88

The best I can explain: N = PQ = 11 × 19 = 187.

C=M^e mod n ;C=11^23 mod 187 ; C = 88 mod 187.

46.

‘Jitter’ refers to-(a) errorless delivery(b) variation in the packet arrival time(c) timely delivery of message(d) none of the mentionedThis question was posed to me in an internship interview.I would like to ask this question from Topologies in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct choice is (B) variation in the packet ARRIVAL time

To explain: ‘JITTER’ refers to variation in the packet arrival time.

47.

What is the bit rate for high-definition TV (HDTV)?(a) 1.4 Gbps(b) 2 Gbps(c) 1.5 Gbps(d) 1.8 GbpsThe question was posed to me in an international level competition.I'd like to ask this question from Layers in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct choice is (c) 1.5 Gbps

Explanation: The HDTV screen is normally a ratio of 16: 9 (in contrast to 4: 3 for regular TV)

There are 1920 by 1080 PIXELS PER screen, and

Screen is renewed 30 TIMES per second.

24-bits represent one color pixel.

Therefore Bit Rate = 1920 × 1080 × 30 × 24 = 1,492992,000 = 1.5 Gbps.

48.

Which topology has the toughest fault identification?(a) Mesh(b) Star(c) Bus(d) RingI got this question in a national level competition.The question is from Topologies in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct choice is (c) Bus

The EXPLANATION is: In the bus topology, FAULT identification is TOUGHER.

49.

A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces 250 frames per second?(a) 38 frames(b) 48 frames(c) 96 frames(d) 126 framesThis question was addressed to me in examination.My doubt stems from Layers in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

Correct choice is (a) 38 frames

For explanation I would say: If the system creates 250 frames PER second, or 1/4 frames per millisecond, then G = ¼.

S = G× e−2G = 0.152 (15.2 percent).

This MEANS that

Throughput = 250 × 0.152 = 38.

Only 38 frames out of 250 will probably SURVIVE.

50.

Find the checksum byte for the fallowing data words: 10110011, 10101011, 01011010, 11010101(a) 10001101(b) 01110010(c) 10110101(d) 01110010This question was addressed to me during an interview for a job.This key question is from Layers topic in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer» RIGHT choice is (a) 10001101

Best explanation: Do BINARY addition to COMPUTE the RESULT.