GATE 2015 (Set-01)

Question 1
Didn’t you buy _________ when you went shopping?
A
any paper
B
much paper
C
no paper
D
a few paper
Question 1 Explanation: 
In negative sentence we can use "any" with plural countable nouns.
Question 2
Which of the following combinations is incorrect?
A
Acquiescence – Submission
B
Wheedle – Roundabout
C
Flippancy – Lightness
D
Profligate – Extravagant
Question 2 Explanation: 
Acceptance = The reluctant acceptance of something without protest
Wheedle = Use endearments (or) flattery to persuade some to do something
Flippancy = Lack of respect (or) seriousness
Profligate = Recklessly extravagant
Question 3
Given set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is probability that the sum of the two numbers equals 16?
A
0.20
B
0.25
C
0.30
D
0.33
Question 3 Explanation: 
Favourable outcomes = {2,14}, {3,13}, {4,12}, {5,11} = 4
Total outcomes = 4C1 × 5C1 = 20
Probability = Favourable outcomes/ Total outcomes = 4/20 = 0.20
Question 4
A
She had a terrible time at the party.
B
She had a horrible time at the party.
C
She had a terrific time at the party
D
She had a terrifying time at the party
Question 4 Explanation: 
The options A, B and D means that she didn't enjoy the party.
→ Horrible and terrible means fearful.
→ Terrific = Wonderful
→ So, C is the Answer.
Question 5
       
A
Statement I alone is sufficient, but statement II alone is not sufficient.
B
Statement II alone is sufficient, but statement I alone is not sufficient.
C
Both statements together are sufficient, but neither statement alone is sufficient.
D
Statement I and II together are not sufficient.
Question 5 Explanation: 
When you climbing, only height matters not the width.
Statement I:
Each step is 3/4 foot high.
No. of steps = 9/(3/4) = 12 steps
Statement I alone is sufficient.
Statement II:
Each step is 1 foot wide.
Statement II alone is not sufficient.
Question 6

     
A
32
B
33
C
34
D
35
Question 6 Explanation: 
Male to female students in each department = 5 : 4
There are 40 males in Electrical Engineering.
⇒ Number of females in Electrical Engineering = 4/5 × 40 = 32
Total number of students in Electrical Engineering = 40+32 = 72
This constitutes 20% of the strength of college.
Number of students in Civil department = 30/100 × 72 = 108
Number of female students in Civil department = 4/(4+5) × 108 = 48
Number of students in Mechanical department = 10/20 × 72 = 36
Number of female students in Mechanical department = 4/(4+5) × 36 = 16
Required Difference = 48 - 16 = 32
Question 7
Select the alternative meaning of the underlined part of the sentence. The chain snatchers took to their heels when the police party arrived.
A
took shelter in a thick jungle
B
open indiscriminate fire
C
took to flight
D
unconditionally surrendered
Question 7 Explanation: 
The words took to their heels and took to flight means that runaway.
Question 8
     
A
Only relation I is true
B
Only relation II is true
C
Relations II and III are true.
D
Relations I and III are true.
Question 8 Explanation: 
75% of students have a chance of passing atleast one subject = 1 - (1 - m) (1 - p) (1 - c) = 0.75 ----(i)
50% of students have a chance pass in atleast two
(1 - m)pc + (1 - p)mc + (1 - c) mp + mpc = 0.5 ----(ii)
40% of students have a chance of passing exactly two
(1 - m)pc + (1 - p)mc + (1 - c)mp = 0.4 ----(iii)
From equation (ii) and (iii) we can get
mpc = 0.1
⇒ m*p*c = 1/10
Statement (III) is correct.
→ Simplify eq (i), we get ⇒ p+c+m - (mp+mc+pc) + mpc = 0.75
⇒ p+c+m - (mp+mc+pc) = 0.65 ----(iv) → Simplify equation (iii), we get
⇒ pc + mc + mp - 3mpc = 0.4
From (iv) and (v)
p + c +m - 0.7 = 0.65
⇒ p + c + m = 1.35 = 27/20
Statement (I) is correct.
Question 9
A
Statements I and II follow
B
Statements I and III follow
C
Statements II and III follow
D
All statements follow
Question 9 Explanation: 
The statement III is not correct, banning of water in lower areas is not the solution of the problem.
Question 10
A
2.290
B
2.970
C
6.795
D
8.795
Question 10 Explanation: 
Total number of students = 44 (Answered- correctly+Answered-wrongly+Not-attempted)
Total marks obtained = 21×2+15×3+11×1+23×2+31×5=299
Average marks = 299/44 = 6.795
Question 11
Which one of the following is True at any valid state in shift-reduce parsing?  
A
Viable prefixes appear only at the bottom of the stack and not inside
B
Viable prefixes appear only at the top of the stack and not inside
C
The stack contains only a set of viable prefixes
D
The stack never contains viable prefixes
Question 11 Explanation: 
A handle is actually on the top of the stack. A viable prefixes is prefix of the handle and so can never extend to the right of handle, i.e., top of stack. So set of viable prefixes is in stack.
Question 12
   
A
2 3 1 4
B
3 4 2 1
C
3 1 4 2
D
3 1 2 4
Question 12 Explanation: 
Condition coverage is also known as predicate coverage in which each of the Boolean expression evaluated to both true and false, which is nothing but white-box testing, which tests internal structures of an application. Hence A-3.
Equivalence class partitioning ⇒ is a software testing technique that divides the input data of a software unit into partitions of equivalent data from which test cases can be derived, which is nothing but black box testing Hence B-1.
Volume testing ⇒ volume testing refers to testing a software application with certain amount of data which is nothing but system testing C-2.
Alpha testing ⇒ Alpha testing is simulated or actual operation testing by potential user/customers, which is nothing but performance testing. D-4.
Question 13
 
A
Either S1 or S2
B
Either S2 or S3
C
Only S2 and S3
D
All of S1, S2 and S3
Question 13 Explanation: 
Any implied register is never explicitly mentioned as an operand in an operation. So as the question asks what can be specified using the address fields, implied accumulator register can't be represented in address field. So S3 is wrong. Hence Option A is the correct answer.
Question 14
Which one of the following is the recurrence equation for the worst case time complexity of the Quicksort algorithm for sorting n(≥ 2) numbers? In the recurrence equations given in the options below, c is a constant.
A
T(n) = 2T (n/2) + cn
B
T(n) = T(n – 1) + T(1) + cn
C
T(n) = 2T (n – 2) + cn
D
T(n) = T(n/2) + cn
Question 14 Explanation: 
When the pivot is the smallest (or largest) element at partitioning on a block of size n the result yields one empty sub-block, one element (pivot) in the correct place and sub block of size n-1. Hence recurrence relation T(n) = T(n-1)+T(1) + cn.
Question 15
A
I only
B
III only
C
III and IV only
D
I and IV only
Question 15 Explanation: 
1 ⇒ is recursive,
This one is true, because L1 is context free which is nothing but recursive, recursive language is closed under complement hence true.
2 ⇒ (complement of L2) is recursive
If L2 and both are recursive enumerable then is recursive
Hence option 2 is false
3 ⇒ is context free
Which is false because context free language does not closed under complement
4 ⇒ ∪L2 is recursive enumerable
⇒ recursive
Every recursive language is also recursive enumerable
L2 ⇒ recursive enumerable
∪ L2 ⇒ recursive enumerable
Because recursive enumerable language closed under union.
Question 16
   
A
III only
B
I and III only
C
I and IV only
D
II and IV only
Question 16 Explanation: 
S1: FALSE
The sequence number of the subsequent segment depends on the number of 8-byte characters in the current segment.
S2: TRUE
Depending on the value of α or Estimated RTT it may or may not be greater than 1.
S3: FALSE
It is the size of the receiver's buffer that's never changed. Receive Window is the part of the receiver's buffer that's changing all the time depending on the processing capability at the receiver's side and the network traffic.
S4: TRUE
The number of unacknowledged bytes that A sends cannot exceed the size of the receiver's window. But if it can't exceed the receiver's window, then it surely has no way to exceed the receiver's buffer as the window size is always less than or equal to the buffer size. On the other hand, for urgent messages, the sender CAN send it in even though the receiver's buffer is full.
Question 17
     
A
3
B
4
C
5
D
6
Question 17 Explanation: 
If we execute P2 process after P1 process, then B = 3
If we execute P1 process after P2 process, then B = 4
If we did preemption between P1 & P2 processes, then B = 2 (Preemption have done from P1 to P2) or B = 3 (Preemption have done from P2 to P1). So, among 2 & 3 values, only one value will be saved in B. So, total no. of distinct values that B can possibly take after the execution is 3.
Question 18
Consider a 4-bit Johnson counter with an initial value of 0000. The counting sequence of this counter is
A
0, 1, 3, 7, 15, 14, 12, 8, 0
B
0, 1, 3, 5, 7, 9, 11, 13, 15, 0
C
0, 2, 4, 6, 8, 10, 12, 14, 0
D
0, 8, 12, 14, 15, 7, 3, 1, 0
Question 18 Explanation: 
In a Johnson’s counter LSB is complemented and a circular right shift operation has to be done to get the next state.

The state sequence is 0,8,12,14,15,7,3,1,0.
Question 19
Which one of the following fields of an IP header is NOT modified by a typical IP router?
A
Checksum
B
Source address
C
Time to Live (TTL)
D
Length
Question 19 Explanation: 
Option C (TTL) is decremented by each visited router. When it reaches to Zero, then Packet will be discarded.
Option A (Checksum) needs to be updated by each visited Router since TTL Value is modified.
Option D (Length) also modified whenever there is a need of performing the fragmentation process.
Option B (Source Address) can’t be modified by an IP router. Only NAT can modify it.
Question 20
SELECT operation in SQL is equivalent to
A
the selection operation in relational algebra
B
the selection operation in relational algebra, except that SELECT in SQL retains duplicates
C
the projection operation in relational algebra
D
the projection operation in relational algebra, except that SELECT in SQL retains duplicates
Question 20 Explanation: 
SELECT operation in SQL perform vertical partitioning which is performed by projection operation in relational calculus but SQL is multi sets; hence (D).
Question 21
 
A
5
B
6
C
7
D
8
Question 21 Explanation: 
A = LU

l11 = 2 -----(1)
l11u12 = 2
u12 = 2/2
u12 = 1 ----- (2)
l21 = 4 ----- (3)
l21u12+l22 = 9
l22 = 9 - l21u12 = 9 - 4 × 1 = 5
Question 22
A
P-iii, Q-ii, R-iv, S-i
B
P-i, Q-ii, R-iv, S-iii
C
P-ii, Q-iii, R-iv, S-i
D
P-ii, Q-i, R-iii, S-iv
Question 22 Explanation: 
Prim’s algorithm always select minimum distance between two of its sets which is nothing but greedy method.
Floyd-warshall always changes it distance at each iteration which is nothing but dynamic programming.
Merge sort in merge sort first we always divide and merge to perform sorting hence divide and conquer.
Hamiltonian circuit used to reach the entire vertex once, if some vertex is repeating in its path it will backtrack.
Question 23
A
-5
B
6
C
7
D
8
Question 23 Explanation: 
Function f1 will not swap the value of 'a' and 'b' because f1 is call by value.
But f2 will swap the value of 'b' and 'c' because f2 is call by reference. So finally the value of
a=4
b=6
c=5
So, answer will be
c - a - b
5 - 4 - 6 = -5
Question 24
A
h(x)/g(x)
B
-1/x
C
g(x)/h(x)
D
x/(1-x)2
Question 24 Explanation: 
g(x)= 1 – x, h(x)=x/x-1 -------- (2)
Replace x by h(x) in (1), replacing x by g(x) in (2),
g(h(x))=1-h(x)=1-x/x-1=-1/x-1
h(g(x))=g(x)/g(x)-1=1-x/-x
⇒ g(h(x))/h(g(x))=x/(x-1)(1-x)=(x/x-1)/1-x=h(x)/g(x)
Question 25
The height of a tree is the length of the longest root-to-leaf path in it. The maximum and minimum number of nodes in a binary tree of height 5 are
A
63 and 6, respectively
B
64 and 5, respectively
C
32 and 6, respectively
D
31 and 5, respectively
Question 25 Explanation: 
Maximum number of nodes in a binary tree of height h is,
2h+1 - 1 = 25+1 - 1 = 63
Minimum number of nodes in a binary tree of height h is
h + 1 = 5 + 1 = 6
Question 26
 
A
2036, 2036, 2036
B
2012, 4, 2204
C
2036, 10, 10
D
2012, 4, 6
Question 26 Explanation: 
⇒ Address of x = 2000
⇒ x [4] [3] can represents that x is a 2-dimensional array.
⇒ x+3 = (Address of x) + 3 * 4 * 3 [3×4×3 is inner dimention]
= 2000 + 36
= 2036
⇒ *(x+3) also returns the address i.e., 2036.
The '*' represents 1 - D but x is starting at 2036.
⇒ *(x+3)+3 = *(Address of x + 2 * 3 * 4) + 3
= *(2000 + 24) +3
= *(2024) + 3 ['*' will change from 2D to 1D]
= 2024 + 3 * 4
= 2024 + 12
= 2036
Question 27
A
1
B
2
C
3
D
4
Question 27 Explanation: 
M accepts the strings which end with a and N accepts the strings which end with B. Their intersection should accept empty language.
Question 28
Consider a non-pipelined processor with a clock rate of 2.5 gigahertz and average cycles per instruction of four. The same processor is upgraded to a pipelined processor with five stages; but due to the internal pipeline delay, the clock speed is reduced to 2 gigahertz. Assume that there are no stalls in the pipeline. The speed up achieved in this pipelined processor is __________.
A
3.2
B
3.3
C
3.4
D
3.5
Question 28 Explanation: 
Given that the processor clock rate = 2.5 GHz, the processor takes 2.5 G cycles in one second.
Time taken to complete one cycle = (1 / 2.5 G) seconds
Since it is given that average number of cycles per instruction = 4, the time taken for completing one instruction = (4 / 2.5 G) = 1.6 ns
In the pipelined case we know in the ideal case CPI = 1, and the clock speed = 2 GHz.
Time taken for one instruction in the pipelined case = (1 / 2 G) = 0.5 ns
Speedup = 1.6/0.5 = 3.2
Question 29
The least number of temporary variables required to create a three-address code in static single assignment form for the expression q + r/3 + s – t * 5 + u * v/w is _________.
A
8
B
9
C
10
D
11
Question 29 Explanation: 
We will need one temporary variable for storing the result of every binary operation as Static Single Assignment implies the variable cannot be repeated on left hand side of assignment.
The given expression:
q+r/3+s−t∗5+u∗v/w
t1=r/3;
t2=t∗5;
t3=u∗v;
t4=t3/w;
t5=q+t1;
t6=t5+s;
t7=t6−t2;
t8=t7+t4;
So in total we need 8 temporary variables. If it was not mentioned as static single assignment then answer would have been 3 because we can re-use the same temporary variable several times.
Question 30
A
pr = 0
B
pr = 1
C
0 < pr ≤ 1/5
D
1/5 < pr < 1
Question 30 Explanation: 
Total number of elements i.e., ordered triplets (x,y,z) in L3 are 5×5×5 i.e., 125 n(s) = 125
Let A be the event that an element (x,y,z)∈ L3 satisfies x ∨(y∧z) = (x∨y) ∧ (x∨z) Since q∨(r∧s) = q∨p = q
and (q∨r)∧(q∨s)=t∧t=t q∨(r∧s)≠(q∨r)∧(q∨s)
Therefore, (x,y,z) = (q,r,s),(q,s,r),(r,q,s),(r,s,q),(s,r,q),(s,q,r)
i.e., 3! = 6 elements will not satisfy distributive law and all other (x,y,z) choices satisfy distributive law
n(A) = 125-6 = 119
∴ required probability is 119/125
⇒ 1/5
Question 31
A
10110
B
10010
C
01010
D
01001
Question 31 Explanation: 
In q0 state for '1', a '1' is pushed and for a '0', a '0' is pushed. In q1 state, for a '0' a '1' is popped, and for '1' a '0' is popped. So the given PDA is accepting all strings of form x0(xr)' or x1(xr)' or x(xr)' , where (xr)' is the complement of reverse of x.
Question 32
 
A
{r = qx+y ∧ r
B
{x = qy+r ∧ r
C
{y = qx+r ∧ 0
D
{q+10}
Question 32 Explanation: 
The loop terminates when r In each iteration q is incremented by 1 and y is subtracted from 'r'. Initial value of 'r' is 'x'. So, loop iterates x/y times and q will be equal to x/y and r = x%y.
⇒ x = qy + r
Question 33

An algorithm performs (log N)1/2 find operations, N insert operations, (log N)1/2 delete operations, and (log N)1/2 decrease-key operations on a set of data items with keys drawn from a linearly ordered set. For a delete operation, a pointer is provided to the record that must be deleted. For the decrease-key operation, a pointer is provided to the record that has its key decreased. Which one of the following data structures is the most suited for the algorithm to use, if the goal is to achieve the best total asymptotic complexity considering all the operations?

A
Unsorted array
B
Min-heap
C
Sorted array
D
Sorted doubly linked list
Question 33 Explanation: 
Question 34
A
40, 30, 20, 10, 15, 16, 17, 8, 4, 35
B
40, 35, 20, 10, 30, 16, 17, 8, 4, 15
C
40, 30, 20, 10, 35, 16, 17, 8, 4, 15
D
40, 35, 20, 10, 15, 16, 17, 8, 4, 30
Question 34 Explanation: 
Given max. heap is

Heapification:

Array representation of above max-heap is (BFS)
40,35,20,10,30,16,17,8,4,15
Question 35
A
a=6,b=4
B
a=4,b=6
C
a=3,b=5
D
a=5,b=3
Question 35 Explanation: 
Given λ1=-1 and λ2=7 are eigen values of A
By properties,

⇒ 6=1+a and -7=a-4b
⇒ a=5 ⇒ -7=5-4b
⇒ b=3
Question 36
Let G = (V, E) be a simple undirected graph, and s be a particular vertex in it called the source.  For x ∈ V , let d(x)denote the shortest distance in G from s to x. A breadth first search (BFS) is performed starting at s. Let T be the resultant BFS tree. If (u, v) is an edge of G that is not in T, then which one of the following CANNOT be the value of d(u- d(v) ?   
A
-1
B
0
C
1
D
2
Question 36 Explanation: 
In an undirected graph if (u, v) be the edge then (u, v) also the edge. So shortest path that can be obtained by the (u, v) of (u, v).
Then the difference between the d(u) and d(v) is not more than '1'.
In the option 'D' the difference is given as '2' it is not possible in the undirected graph.
Question 37
A
Both commutative and associative
B
Commutative but not associative
C
Not commutative but associative
D
Neither commutative nor associative
Question 37 Explanation: 
It is clear that from the truth table, the binary operation # is equivalent to XOR i.e., ⊕, which satisfies both commutative and associative i.e., p#q q#p and p#(q#r) (p#q)#r
Question 38
 
A
4
B
6
C
7
D
8
Question 38 Explanation: 


Question 39
A
69
B
70
C
71
D
72
Question 39 Explanation: 

⇒ Total sum = 10 + 9 + 2 + 15 + 7 + 16 + 4 + 6 = 69
Question 40

Consider a disk pack with a seek time of 4 milliseconds and rotational speed of 10000 rotations per minute (RPM). It has 600 sectors per track and each sector can store 512 bytes of data. Consider a file stored in the disk. The file contains 2000 sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is _________.

A
14020
B
14021
C
14022
D
14023
Question 40 Explanation: 
Given
Seek time = 4ms
60s→ 10000 rotations

∴ Rotational latency =1/2×6ms=3ms
1track→600sectors
⇒6ms←600sectors (1 rotation means 600 sectors (or)1 track)
1 track→6ms/600=0.01ms
2000sector→2000(0.01)=20ms
∴total time needed to read the entire file is
=2000(4+3)+20
=8000+6000+20
=14020ms
Question 41
 
A
-1
B
-2
C
-3
D
-4
Question 41 Explanation: 
Question 42
A
5
B
6
C
7
D
8
Question 43
 
A
n3
B
n(log n) 2
C
nlog n
D
nlog (log n)
Question 43 Explanation: 
int fun1 (int n) {
int i, j, k, p, q = 0;
for (i=1; i
return q;
}
∴ nlog(logn)
Question 44
Let an represent the number of bit strings of length n containing two consecutive 1s. What is the recurrence relation for an?
A
an-2+an-1+2n-2
B
an-2+2an-1+2n-2
C
2an-2+an-1+2n-2
D
2an-2+2an-1+2n-2
Question 44 Explanation: 
For string of length 1, there is '0' consecutive 1's.
So, a1=0
For string of length 2,
Similarly, a3 = 3
a4= 8
Only (A) will satisfy the above values.
Question 45
A
p, s, u
B
r, s, u
C
r, u
D
q, v
Question 45 Explanation: 
In compilers, live variable analysis is a classic data flow analysis to calculate the variables that are live at each point in the program.
A variable is live at some point if it holds a value that may be needed in the future, of equivalently if its value may be read before the next time the variable is written to.
→ '1' can be assigned by the p and s and there is no intermediate use of them before that.
→ And p and s are not to be live in the both 2 & 3.
→ And q also assigned to u not live in 2 & 3.
→ And v is live at 3 not at 2.
→ u is live at 3 and also at 2, if we consider a path of length 0 from 2-8.
Finally r, u is the correct one.
Question 46
Consider a uniprocessor system executing three tasks T1, T2 and T3, each of which is composed of an infinite sequence of jobs (or instances) which arrive periodically at intervals of 3, 7 and 20 milliseconds, respectively. The priority of each task is the inverse of its period and the available tasks are scheduled in order of priority, with the highest priority task scheduled first. Each instance of T1, T2 and T3 requires an execution time of 1, 2 and 4 milliseconds, respectively. Given that all tasks initially arrive at the beginning of the 1st millisecond and task preemptions are allowed, the first instance of T3 completes its execution at the end of ______________ milliseconds.
A
13
B
12
C
15
D
16
Question 46 Explanation: 
All the processes or tasks are available at the begining of 1st millisecond, means at t=0. So, Gantt chart will be as follows:

∴ At the end of 12 milliseconds, 1st instance of T3 will complete its execution.
Question 47
A
2
B
3
C
4
D
5
Question 47 Explanation: 
Output table is
Question 48

A positive edge-triggered D flip-flop is connected to a positive edge-triggered JK flipflop as follows. The Q output of the D flip-flop is connected to both the J and K inputs of the JK flip-flop, while the Q output of the JK flip-flop is connected to the input of the D flip-flop. Initially, the output of the D flip-flop is set to logic one and the output of the JK flip-flop is cleared. Which one of the following is the bit sequence (including the initial state) generated at the Q output of the JK flip-flop when the flip-flops are connected to a free-running common clock? Assume that J = K = 1 is the toggle mode and J = K = 0 is the state-holding mode of the JK flip-flop. Both the flip-flops have non-zero propagation delays.

A
0110110...
B
0100100...
C
011101110...
D
011001100...
Question 48 Explanation: 
The circuit for the given data is

The characteristic equations are
QDN=D=QJK

The state table and state transition diagram are as follows:

Consider QDQJK=10 as initial state because in the options QJK=0 is the initial state of JK flip-flop.
The state sequence is

0 → 1 → 1 → 0 → 1 → 1
∴ Option (a) is the answer.
Question 49
Let G be a connected planar graph with 10 vertices. If the number of edges on each face is three, then the number of edges in G is _______________.
A
24
B
25
C
26
D
27
Question 49 Explanation: 
By Euler’s formula,
|V|+|R|=|E|+2 ------(1) where |V|, |E|, |R| are respectively number of vertices, edges and faces (regions)
Given |V|=10 ------(2) and number of edges on each face is three
∴3|R|=2|E|⇒|R|=2/3|E| ------(3)
Substituting (2), (3) in (1), we get
10+2/3|E|=|E|+2⇒|E|/3=8⇒|E|=24
Question 50

Consider S2, S3 and S4. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmit in the same slot. The probabilities of generation of a frame in a time slot by S1, S2, S3 and S4 are 0.1, 0.2, 0.3 and 0.4, respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is _________.

A
0.4404
B
0.463
C
0.464
D
0.465
Question 50 Explanation: 
S1→0.1
S2→0.2
S3→0.3
S4→0.4
The probability of sending a frame without any collision by any of these stations is
Question 51

Suppose the following disk request sequence (track numbers) for a disk with 100 tracks is given: 45, 20, 90, 10, 50, 60, 80, 25, 70. Assume that the initial position of the R/W head is on track 50. The additional distance that will be traversed by the R/W head when the Shortest Seek Time First (SSTF) algorithm is used compared to the SCAN (Elevator) algorithm (assuming that SCAN algorithm moves towards 100 when it starts execution) is _________ tracks.

A
10
B
11
C
12
D
13
Question 51 Explanation: 

∴ Total Head moments = 10+10+10+10+10+55+20+5+10 = 140
SSTF:

∴ Total Head moments = 5+15+10+10+10+65+5+10 = 130
∴ Additional distance that will be traversed by R/w head is = 140-130 = 10
Question 52

Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is _________.

A
320
B
321
C
322
D
323
Question 52 Explanation: 
Given B = 64 kbps
Tp = 20 ms
η ≥ 50%
For η≥50%⇒L≥BR
⇒ L=64×103×2×20×10-3
=2560bits
=320bytes
Question 53

Consider a main memory with five page frames and the following sequence of page references: 3, 8, 2, 3, 9, 1, 6, 3, 8, 9, 3, 6, 2, 1, 3. Which one of the following is true with respect to page replacement policies First-In-First Out (FIFO) and Least Recently Used (LRU)?

A
Both incur the same number of page faults
B
FIFO incurs 2 more page faults than LRU
C
LRU incurs 2 more page faults than FIFO
D
FIFO incurs 1 more page faults than LRU
Question 53 Explanation: 

∴ Number of page faults = 9

∴ Number of page faults = 9
Question 54
A
Both {f} and {g} are functionally complete
B
Only {f} is functionally complete
C
Only {g} is functionally complete
D
Neither {f} nor {g} is functionally complete
Question 54 Explanation: 
A function is functionally complete if (OR, NOT) or (AND, NOT) are implemented by it.
f(X,X,X)=X'XX'+XX'+X'X'
=0+0+X'
=X'
Similarly, f(Y,Y,Y)=Y' and f(X,Z,Z)=Z'
f(Y',Y',Z')=(Y')'Y'Z'+Y'(Y')'+(Y')'(Z')'
=YY'Z'+Y'Y+YZ
=0+0+YZ
=YZ
We have derived NOT and AND. So f(X,Y,Z) is functionally complete.
g(X,Y,Z)=X'YZ+X'YZ'+XY
g(X,X,X)=X'XX+X'XZ'+XX
=0+0+X
=X
Similarly, g(Y,Y,Y)=Y and g(Z,Z,Z)=Z
NOT is not derived. Hence, g is not functionally complete.
Question 55
 
A
0.99
B
1.00
C
2.00
D
3.00
Question 55 Explanation: 

=2-1/1(2)+3-2/2(3)+4-3/3(4)+…+100-99/99(100)
=1/1-1/2+1/2-1/3+1/3…+1/98-1/99+1/99-1/100
=1-1/100
=99/100
=0.99
There are 55 questions to complete.