Quiz Electronics Engineering 13 April 2020

Quiz Electronics Engineering

Exam: NIC

Topic: Miscellaneous

Date: 13 /04/2020

 

Each Question carries 1 Mark

Negative Marking:  1/4

Time: 10 Minutes

 

Q1. A pipeline P operating at 400 MHz has a speedup factor of 6 and operating at 70% efficiency. How many stages are there in pipeline?

(a) 5

(b) 6

(c) 8

(d) 9

L1 Difficulty 4

QTags Operating System

QCreator Vikram Kumar

 

 

Q2. How much speed do we gain by using the cache, when cache is used 80% of the time? Assume cache is faster than main memory.

(a) 5.27

(b) 2.00

(c) 4.16

(d) 6.09

L1 Difficulty 4

QTags Computer Organization & Microprocessor

QCreator Vikram Kumar

 

 

Q3. In 8086, the jump condition for the instruction JNBE is?

(a) CF=0 or ZF =0

(b) ZF = 0 or SF = 1

(c) CF =0 or ZF=0

(d) CF = 0

L1 Difficulty 3

QTags Computer Organization & Microprocessor

QCreator Vikram Kumar

 

 

 

Q4. How many number of times the instruction sequence below will loop before coming out of the loop?

MOV AL, 00H

A1:      INC AL

JNZ A1

(a) 1

(b) 255

(c) 256

(d) will not come out of the loop

L1 Difficulty 4

QTags Computer Organization & Microprocessor

QCreator Vikram Kumar

 

 

Q5. Ethernet layer-2 switch is a network element type which gives.

(a) Different collision domain and same broadcast domain

(b) Different collision domain and different broadcast domain.

(c) Same collision domain and same broadcast domain.

(d) Same collision domain and different broadcast domain.

L1 Difficulty 4

QTags Networking

QCreator Vikram Kumar

 

 

Q6. What will be the efficiency of a Stop and Wait protocol, if the transmission time for a frame is 20ns and the propagation time is 30 ns?

(a) 20%

(b) 25%

(c) 40%

(d) 66%

L1 Difficulty 3

QTags Networking

QCreator Vikram Kumar

 

 

Q7. What is IP class and number of sub networks if the subnet mask is 255.224.0.0 ?

(a) Class A, 3

(b) Class A, 8

(c) Class B, 3

(d) Class B, 32

L1 Difficulty 2

QTags Networking

QCreator Vikram Kumar

 

 

Q8. Which algorithm is used to shape the bursty traffic into a fixed rate traffic by averaging the data rate?

(a) Solid bucket algorithm

(b) Spanning tree algorithm

(c) Hocken helm algorithm

(d) Leaky bucket algorithm

L1 Difficulty 4

QTags Networking

QCreator Vikram Kumar

 

 

Q9. A packet filtering firewall can

(a) Deny certain users from accessing a service

(b) Block worms and viruses from entering the network

(c) Disallow some files from being accessed through FTP

(d) Block some hosts from accessing the network

L1 Difficulty 3

QTags Networking

QCreator Vikram Kumar

 

 

Q10. The state of a process after it encounters an I/O instruction is?

(a) Ready

(b) Blocked

(c) Idle

(d) Running

L1 Difficulty 3

QTags Operating System

QCreator Vikram Kumar

 

 

SOLUTIONS

 

S1. Ans.(d)

Sol. Let number of stages be K.

Given:

  • Efficiency of pipeline = 70% = 0.7
  • Speed-up factor = 6

We know that, Efficiency of K-stage pipeline  =

0.7 =

K =    = 8.57 = 9(approx.)

 

 

S2. Ans.(c)

Sol. Assuming that memory is accessed only on a cache miss and when cache miss occurs at first cache is accessed and then main memory. So, both times are added in that case.

We know that, Speed gain  =

=   =

= x(let)  where, 0 ≤ x ≤  0.1 Since cache is faster than memory by at least 10 times.

Speed gain =

For minimum x = 0, speed gain maximum = 5

For maximum x = 0.1, speed gain minimum = 3.33

Hence, speed gained by using the cache is 4.16.

 

 

S3. Ans.(c)

Sol. JNBE instruction means Jump short if neither below nor equal to CF = 0 and ZF = 0.

 

 

S4. Ans.(c)

Sol. JNZ instruction means Jump if ZF = 0 i.e. if accumulator is not empty. AL will contain 0000 0000, on incrementing it will be 0000 0001 and so on. It is a 8-bit register so maximum it can hold value = 28 = 256. After this accumulator will get overflow and it becomes 0. So, Loop will continue for 256 times.

 

 

S5. Ans.(a)

Sol.

  • A Collision Domain is the part of a network where packet collisions can occur. A collision occurs when two devices sharing the same medium or through repeaters send a packet at the same time. It consists of nodes on the same set of inter-connected repeaters, divided by switches and learning bridges.
  • A broadcast domain is a logical division of a computer network, in which all nodes can reach each other by broadcast at the data link layer. A broadcast domain can be within the same LAN segment or bridged to other LAN segments. Routers form boundaries between broadcast domains.

Ethernet layer-2 switch is a network element type which gives different collision domain and same broadcast domain. A Wireless router creates multiple broadcast and collision domain.

 

 

S6. Ans.(b)

Sol. Given:

  • Transmission time (Tt) = 20 ns
  • Propagation time (Tp) = 30 ns

Efficiency =    =    =    =    =  25%

 

 

S7. Ans.(b)

Sol. Subnet mask = 255.224.0.0

It can be written as, 1111 1111.1110 0000.0.0 .

For finding IP class we check which octets in the subnet mask are totally filled with 1’s.

Since in the first octet only all the bits are 1 so IP class is A.

Since there are three 1’s in the second octet that means there are 3 subnet ID bits.

So, total number of subnetworks = 23 =8

 

 

S8. Ans.(d)

Sol. A leaky bucket algorithm that may be used to determine whether some sequence of discrete events conforms to defined limits on their average and peak rate of frequency.

Traffic Shaping:- It is a method of congestion control by regulating average rate of data flow before entering the packet into the network. At connection set-up time, the sender and carrier share a common traffic pattern.

There are two types of Traffic shaping algorithm :-

  1. Leaky Bucket Algorithm
  2. Token Bucket Algorithm
  3. Leaky Bucket Algorithm:- It is implemented as a single-server queue with constant service time to control data rate in a network. It temporarily stores a variable number of requests and organize them into a set-rate output of packets in an Asynchronous Transfer Mode network. When buffer is overflow then packets are discarded. In this algorithm the input rate varies but the output rate remains constant. This saves busty traffic into fixed rate traffic by averaging the data rate.
  4. Token Bucket Algorithm:- It is used in packet switched computer networks and telecommunications networks. It can be used to check that data transmissions matches to defined limits on bandwidth and burstiness. It throws away tokens when the bucket is full but never discards packets.

 

 

S9. Ans.(d)

Sol. Packet filtering is a firewall technique. It monitors outgoing and incoming packets and allow them to pass or halt based on source and destination IP addresses, protocols and ports. It prevents worms and viruses from entering the network.

 

 

S10. Ans.(b)

Sol.

  • Blocked or wait state: Whenever the process requests access to I/O ,it enters the blocked or wait state. It continues to wait in the main memory and once the I/O operation is completed the process goes to the ready state.
  • Ready: The process is loaded into the main memory after creation. The process here is ready to run and is waiting for CPU time for its execution. Processes that are ready for execution by the CPU are maintained in a queue for ready processes.
  • Run – The instruction within the process are executed.

 

 

 

 

 

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