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 
Discuss it
Question 1 Explanation:
This scheme is used for the byte synchronization and the encode/decode scheme which transmits 8 bits as a 10bit code group. The features of this scheme are lowcost component design and good transition density for easy clock recovery. It is used by Gigaethernet technology over the fibre optic medium.
Question 2 
Which of the following is an unsupervised neural network?
RBS  
Hopfield  
Back propagation  
Kohonen 
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 
Discuss it
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.
Process Time P1 10 P2 5 P3 20 P4 8 P5 15If roundrobin scheduling with 5 ms is used what is the average waiting time of the processes in the queue?
27 ms  
26.2 ms  
27.5 ms  
27.2 ms 
Discuss it
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 
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 
Discuss it
Question 6 Explanation:
By taking a truth table, we can see the output:
X Y X xor Y (X xor Y) xor Y 0 0 0 0 0 1 1 0 1 0 1 1 1 1 0 1Clearly the output depends on the value of X. Alternate approach: Through simplification of the expression, we can find the solution.
(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 zbuffer algorithm?
It is a depth sort algorithm  
No limitation on total number of objects  
. Comparisons of objects is done  
zbuffer is initialized to background colour at start of algorithm

Discuss it
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 floatingpoint number C1D00000 (hexadecimal notation)? (Assume 32bit, single precision floating point IEEE representation)
28  
15  
26  
28 
Discuss it
Question 8 Explanation:
Floating Point number in Hexadecimal = C1D00000
Floating Point number in Binary = 1100 0001 1101 0000 0000 0000 0000 0000
In 32bit, 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 = 131127 = 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 
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% 
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.