# ISRO CS 2011

Question 1 |

The encoding technique used to transmit the signal in giga ethernet technology over fiber optic medium is

Differential Manchester encoding | |

Non return to zero | |

4B/5B encoding | |

8B/10B encoding |

**Misc Topics in Computer Networks**

**ISRO CS 2011**

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Question 1 Explanation:

This scheme is used for the byte synchronization and the encode/decode scheme which transmits 8 bits as a 10-bit code group. The features of this scheme are low-cost component design and good transition density for easy clock recovery. It is used by Giga-ethernet technology over the fibre optic medium.

Question 2 |

Which of the following is an unsupervised neural network?

RBS | |

Hopfield | |

Back propagation | |

Kohonen |

**ISRO CS 2011**

**Discuss it**

Question 3 |

In compiler terminology reduction in strength means

Replacing run time computation by compile time computation | |

Removing loop invariant computation | |

Removing common subexpressions | |

replacing a costly operation by a relatively cheaper one |

**Code Generation and Optimization**

**ISRO CS 2011**

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Question 3 Explanation:

Strength Reduction is a compiler optimization in which costly operations are replaced by cheaper ones. Example: Exponentiation is replaced by multiplication and multiplication is in return replaced by addition.
The following code is having multiplication operator:

a = 10; for (i = 0; i < X; i++) { Z[i] = a * i; }This code can be replaced by the following code by replacing the multiplication with addition.

a = 10; k = 0; for (i = 0; i < X; i++) { Z[i] = k; k = k + a; }So, option (D) is correct.

Question 4 |

The following table shows the processes in the ready queue and time required for each process for completing its job.

If round-robin scheduling with 5 ms is used what is the average waiting time of the processes in the queue?Process TimeP1 10 P2 5 P3 20 P4 8 P5 15

27 ms | |

26.2 ms | |

27.5 ms | |

27.2 ms |

**OS Process Management**

**ISRO CS 2011**

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Question 4 Explanation:

The Gantt chart for the processes is shown:
Waiting time of a process = Sum of the periods spent waiting in the ready queue.

Waiting time = completion time - burst time Waiting time of P1 = 30 -10 = 20 Waiting time of P2 = 10 - 5 = 5 Waiting time of P3 = 58 - 20 = 38 Waiting time of P4 = 38 - 8 = 30 Waiting time of P5 = 53 - 15 = 38 Average waiting time = 20 + 5 + 38 + 30 + 38 =131/5 = 26.2

Question 5 |

MOV [BX], AL type of data addressing is called ?

register | |

immediate | |

register indirect | |

register relative |

**Computer Organization and Architecture**

**ISRO CS 2011**

**Pipelining and Addressing modes**

**Discuss it**

Question 5 Explanation:

As the contents of AL are copied to the address equal to the value of BX, most suitably it is register indirect mode.

Question 6 |

Evaluate (X xor Y) xor Y?

All 1's | |

All 0's | |

X | |

Y |

**Digital Logic & Number representation**

**ISRO CS 2011**

**Logic functions and Minimization**

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Question 6 Explanation:

By taking a truth table, we can see the output:

Clearly the output depends on the value of X. Alternate approach: Through simplification of the expression, we can find the solution.X Y X xor Y (X xor Y) xor Y0 0 0 0 0 1 1 0 1 0 1 1 1 1 0 1

(X xor Y) xor Y (XY' + X'Y) xor Y (XY' + X'Y)'Y + (XY' + X'Y)Y' (X' + Y)(X + Y') + XY' XY + XY' = X(Y + Y') = X

Question 7 |

Which of the following is true about z-buffer algorithm?

It is a depth sort algorithm | |

No limitation on total number of objects | |

. Comparisons of objects is done | |

z-buffer is initialized to background colour at start of algorithm |

**ISRO CS 2011**

**Computer Graphics**

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Question 7 Explanation:

Option 1: False, as it is a depth buffer and not depth sort algorithm.
Option 2: True, as the size of objects can be large.
Option 3: False, as only the depth of object is compared and not the entire object.
Option 4: False, no background colour is initiated even at the start of algorithm.
Option (B) is correct.

Question 8 |

What is the decimal value of the floating-point number C1D00000 (hexadecimal notation)? (Assume 32-bit, single precision floating point IEEE representation)

28 | |

-15 | |

-26 | |

-28 |

**Digital Logic & Number representation**

**Number Representation**

**ISRO CS 2011**

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Question 8 Explanation:

Floating Point number in Hexadecimal = C1D00000
Floating Point number in Binary = 1100 0001 1101 0000 0000 0000 0000 0000
In 32-bit, single precision floating point IEEE representation, first MSB represents sign of mantissa: 1 is used to represent a negative mantissa and 0 for a positive value of mantissa, next 8 bits are for exponent value and then 23 bits represents mantissa.

Value of exponent = 131-127 = 4 Mantissa = -1.101000000...0 Floating point number = -1.10100...0000 x 2^{4}= -11010 = -26

Question 9 |

What is the raw throughput of USB 2.0 technology?

480 Mbps | |

400 Mbps | |

200 Mbps | |

12 Mbps |

**Computer Organization and Architecture**

**ISRO CS 2011**

**Pipelining and Addressing modes**

**Discuss it**

Question 9 Explanation:

USB 2.0 has a maximum signaling rate of 480 Mbit/s (High Speed or High Bandwidth).
Option (A) is correct.

Question 10 |

Below is the precedence graph for a set of tasks to be executed on a parallel processing system S.
What is the efficiency of this precedence graph on S if each of the tasks T1, T2, T3,....T8 takes the same time and the system S has five processors?

25% | |

40% | |

50% | |

90% |

**OS CPU Scheduling**

**ISRO CS 2011**

**Discuss it**

Question 10 Explanation:

It can be seen that along with the sequential execution of T1 and T2, (T3, T6), (T4, T7) and (T5, T8), all these three processes can be executed in parallel.
So total number of processes that can be executed in 4 units time using 5 available processors = 5*4 = 20
But here processes that are executing in 4 units time = 8
Throughput = 8/20 * 100 = 40%
So, option (B) is correct.

There are 80 questions to complete.