Gate 2006-IT
Question 1 |
In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25°C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25°C, or at/below 25°C. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25°C?
0.4 | |
0.6 | |
0.8 | |
0.9 |
0.6 = (0.5×0.4) + (0.5×P(rain at temp>25)
0.6 = (2) + (0.5×P(rain at temp>25)
P(rain at temp>25) = 0.8
Question 2 |
- f (x, y) = x + y - 3
- f (x, y) = max(x, y)
- f (x, y) = xy
1 and 2 only | |
2 and 3 only | |
1 and 3 only | |
None of these |
y = 3
Here, identity element is 3.
(2) f(x, y) = max(x, y) = x = max(y, x)
⇒ y = 1
Here, identity element = 1
(3) f(x, y) = xny is not equal to f(y, x) = ynx
So, no identity element.
Question 3 |

The automaton accepts u and v but not w | |
The automaton accepts each of u, v, and w | |
The automaton rejects each of u, v, and w | |
The automaton accepts u but rejects v and w |
where t is final state
(ii) v = bab
s is not final state
(iii) w = aabb
s is not final state
Question 4 |
aaaa | |
baba | |
abba | |
babaaabab |
Given string accepts all palindromes.
Option B → baba is not palindrome.
So, this is not accpeted by S.
Question 5 |

(a + b)* a(a + b)b | |
(abb)* | |
(a + b)* a(a + b)* b(a + b)* | |
(a + b)* |

Question 6 |
f (x1, x2, …, xn) = x1’f(x1, x2, …, xn) + x1f(x1, x2, …, xn) | |
f (x1, x2, …, xn) = x2f(x1, x2, …, xn) + x2’f(x1, x2, …, xn) | |
f (x1, x2, …, xn) = xn’f(x1, x2, …, 0) + xnf(x1, x2, …,1) | |
f (x1, x2, …, xn) = f(0, x2, …, xn) + f(1, x2, .., xn) |
LHS: f(x1) = 0 where x1 = 0
LHS: f(x1) = 1 when x1 = 1
RHS: f(0) + f(1) = 0 + 1 = always 1
Question 7 |
0001 and an overflow | |
1001 and no overflow | |
0001 and no overflow | |
1001 and an overflow |
2's complement of 1100 = 1100
Add = 1111
Now convert 1111 to normal form.
⇒ 0000 (1's complement)
⇒ 0001 (2's complement) No carry bit.
Question 8 |
Transparent DMA and Polling interrupts | |
Cycle-stealing and Vectored interrupts | |
Block transfer and Vectored interrupts | |
Block transfer and Polling interrupts
|
→ In case of cycle stealing in each cycle time derive send data then wait again few CPU cycle it sends to memory. So, option B is wrong.
→ In case of polling CPU takes the initiative. So I/O bandwidth cannot be high. So, option D is wrong.
→ Consider block transfer in each single block device. Send data so bandwidth must be high. This makes option (C) correct,
Question 9 |
10 | |
11 | |
12 | |
15 |
No. of 1 degree nodes = 5
No. of 2 degree nodes = 10
Total no. of edges = (1*5) + (2*10) = 5 + 20 = 25
So, Total no. of edges = 25 + 1 = 26 (No. of nodes in a tree is 1 more than no. of edges)
Total no. of leaf nodes (node with 0 degree) = 26 - 5 - 10 = 11
Question 10 |
It can be reduced to the 3-SAT problem in polynomial time | |
The 3-SAT problem can be reduced to it in polynomial time | |
It can be reduced to any other problem in NP in polynomial time | |
Some problem in NP can be reduced to it in polynomial time
|
Question 11 |
Hamiltonian cycle | |
grid | |
hypercube | |
tree |
If all edge weights of an undirected graph are positive, then any subset of edges that connects all the vertices and has minimum total weight is minimum spanning tree.
Question 12 |
- It initiates another process if there are enough extra frames.
- It selects a process to suspend if the sum of the sizes of the working-sets exceeds the total number of available frames.
1 only | |
2 only | |
Neither 1 nor 2 | |
Both 1 and 2 |
Question 13 |
It is a multiprogrammed operating system | |
It uses preemptive scheduling | |
It uses non-preemptive scheduling | |
It is a multi-user operating system
|
Question 14 |
Consider the relations r1(P, Q, R) and r2(R, S, T) with primary keys P and R respectively. The relation r1 contains 2000 tuples and r2 contains 2500 tuples. The maximum size of the join r1⋈ r2 is :
2000 | |
2500 | |
4500 | |
5000 |
Question 15 |
- Relational algebra
- Tuple relational calculus restricted to safe expressions
- Domain relational calculus restricted to safe expressions
2 and 3 only | |
1 and 2 only | |
1 and 3 only | |
1, 2 and 3 |
Question 16 |
an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at most once | |
a lower bound for the number of tests that must be conducted to ensure that all statements have been executed at most once | |
an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at least once | |
a lower bound for the number of tests that must be conducted to ensure that all statements have been executed at least once |
Question 17 |
- E - N + P
- E - N + 2
- P + 2
- P + 1
1 or 3 | |
2 or 3 | |
2 or 4 | |
1 or 4 |
Question 18 |
FTP and HTTP | |
TELNET and POP3 | |
HTTP and TELNET | |
SMTP and FTP |
Question 19 |
Both Ethernet frame and IP packet include checksum fields | |
Ethernet frame includes a checksum field and IP packet includes a CRC field | |
Ethernet frame includes a CRC field and IP packet includes a checksum field | |
Both Ethernet frame and IP packet include CRC fields |

IP packet:

IP Datagram:

Question 20 |
- A hash function takes a message of arbitrary length and generates a fixed length code.
- A hash function takes a message of fixed length and generates a code of variable length.
- A hash function may give the same hash value for distinct messages.
1 only | |
2 and 3 only | |
1 and 3 only | |
2 only |
(2) Statement-2 is wrong, refer statement-1.
(3) Statement-3 is correct, for example hash function N%10, this will generate same values for 1 as well as 2!
Question 21 |
satisfiable and valid | |
satisfiable and so is its negation | |
unsatisfiable but its negation is valid | |
satisfiable but its negation is unsatisfiable |
Question 22 |
When a coin is tossed, the probability of getting a Head is p, 0<p<1. Let N be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is
1/p | |
1/(1-p) | |
1/p2 | |
1/(1-p2) |
Multiply both sides with (1 - p) and subtract,
E - (1 - p) E = 1 × p + (1 - p) p + (1 - p) (1 - p) p + ......
E - (1 - p) E = p/(1 - (1 - p))
(1 - 1 + p) E = 1
pE = 1
E = 1/p
Question 23 |
1 only | |
2 only | |
Neither 1 nor 2 | |
Both 1 and 2 |

P = {1, 2, 4, 5}
Q = {2, 3, 5, 6}
R = {4, 5, 6, 7}
(1) P Δ (Q ∩ R) = (P Δ Q) ∩ (P Δ R)
P Δ ({2,3,5,6} ∩ {4,5,6,7}) = ({1,2,4,5} Δ {2,3,5,6} ∩ {1,2,4,5} Δ {4,5,6,7})
P Δ {5,6} = ({1,2,3,4,5,6} - {2,5}) ∩ ({1,2,4,5,6,7} - {4,5})
({1,2,4,5} Δ {5,6}) = {1,3,4,6} ∩ {1,2,6,7}
{1,2,4,5,6} - {5} = {1,6}
{1,2,4,6} ≠ {1,6}
Statement-1 is False.
(2) P ∩ (Q ∩ R) = (P ∩ Q) Δ (P Δ R)
LHS:
{1,2,4,5} ∩ {5,6} = {5}
RHS:
({1,2,4,5} ∩ {2,3,5,6}) Δ ({1,2,4,5} Δ {4,5,6,7})
{2,5} Δ ({1,2,4,5,6,7} - {4,5})
{2,5} Δ {1,2,6,7}
{1,2,5,6,7} - {2}
{1,5,6,7}
LHS ≠ RHS
Statement - 2 is also wrong.
Question 24 |
28 | |
33 | |
37 | |
44 |
A = set of numbers divisible by 2
B = set of numbers divisible by 3
C = set of numbers divisible by 5
n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)
= ⌊123/2⌋ + ⌊123/3⌋ + ⌊123/5⌋ - ⌊123/6⌋ - ⌊123/15⌋ - ⌊123/10⌋ + ⌊123/30⌋
= 61 + 41 + 24 - 20 -12 - 8 + 4
= 90
Total no. that are not divisible
= n - n(A∪B∪C)
= 123 - 90
= 33
Question 25 |
Consider the undirected graph G defined as follows. The vertices of G are bit strings of length n. We have an edge between vertex u and vertex v if and only if u and v differ in exactly one bit position (in other words, v can be obtained from u by flipping a single bit). The ratio of the chromatic number of G to the diameter of G is
1/(2n-1) | |
1/n | |
2/n | |
3/n |
That will give us a bipartite graph, with chromatic number = 2
Also from the same we can conclude that we need for a 'n' bit string, to traverse no more than (n-1) edges or 'n' vertices to get a path between two arbitary points. So the ratio is (2/n).
Question 27 |

1-R, 2-S, 3-P, 4-Q | |
1-S, 2-R, 3-Q, 4-P | |
1-S, 2-Q, 3-R, 4-P | |
1-S, 2-P, 3-Q, 4-R |
Question 29 |
{w ∈ (a + b)* | #a(w) is even) and {w ∈ (a + b)* | #a(w) is odd} | |
{w ∈ (a + b)* | #a(w) is even) and {w ∈ (a + b)* | #b(w) is odd} | |
{w ∈ (a + b)* | #a(w) = #b(w) and {w ∈ (a + b)* | #a(w) ≠ #b(w)} | |
{ϵ}, {wa | w ∈ (a + b)* and {wb | w ∈ (a + b)*}
|

⇒ This results even number, no. of a's.
Question 30 |
Every language has a regular superset | |
Every language has a regular subset | |
Every subset of a regular language is regular | |
Every subset of a finite language is regular
|
Question 31 |
{anbncn ∣ n≥0} | |
{albmcn ∣ l≠m or m≠n} | |
{anbn ∣ n≥0} | |
{ambn∣ m,n≥0} |
To compare both conditions at the same time, we need a NPDA.
Question 32 |
always regular | |
never regular | |
always a deterministic context-free language | |
always a context-free language
|
Question 33 |
{albmcn | l = m = n} | |
{albmcn | l = m} | |
{albmcn | 2l = m+n} | |
{albmcn | m=n} |

For every 'a' we put two X in stacks [at state S].
After that for every 'b' we pop out one X [reach to state t].
After that for every 'c' we pop out one X [reach to state u].
If all X are popped out then reached to final state f, means for every 'b' and 'c' there is 'a'. 'a' is followed by 'b' and 'b' is followed by 'c'.
Means,
Sum of no. of b's and no. of c's = twice of no. of a's
i.e., {albmcn | 2l = m+n}
Question 34 |
((a + b)* b)* | |
{ambn | m ≤ n} | |
{ambn | m = n} | |
a* b* |
Option C&D:
→ abb accepted by given grammar but option C & D are not accepting.
Question 35 |

QRS | |
PQS | |
PQ'S' | |
Q'S' |
Question 36 |

XOR, AND | |
XOR, XOR | |
OR, OR | |
OR, AND |
Majority means at least two inputs should be 1.
Whenever the majority of the variables have value 1, the output is 1.
The following combinations produce output 1.
110, 101, 111, and 011
The circuit has 3 inputs namely x, y and z with x as the selection line to MUX.
Consider Majority cases.
Case 1: x=1
x y z
1 1 0 When x is 1 then at least one of y and z should be 1. Then P is OR gate.
1 0 1
1 1 1
Case 2: x=0
0 1 1
When x is 0 both y and z should be 1. So Q is AND gate
Question 37 |

S+ = S’ . y’ + S . x | |
S+ =S. x . y’ + S’ . y . x’ | |
S+ =x . y’ | |
S+ =S’ . y + S . x’col |

From the table:
S' = S'y' + Sx
Question 38 |

2Y and Y | |
-2Y and 2Y | |
-2Y and 0 | |
0 and Y |

⇒ -2Y and 0
Question 39 |
It enables reduced instruction size | |
It allows indexing of array elements with same instruction | |
It enables easy relocation of data | |
It enables faster address calculations than absolute addressing
|
Question 40 |
Memory location 1000 has value 20 | |
Memory location 1020 has value 20 | |
Memory location 1021 has value 20 | |
Memory location 1001 has value 20
|
Rd ← 1
Rd ← 1001
Store in address 1001 ← 20.
Question 41 |


32, 5, 010 | |
5, 32, 010 | |
5, 31, 011 | |
5, 31, 010 |
→ So, there may be only 31, is for an unsigned even integer.
→ And 31 left shifts are needed to determine the number of 1's.
Question 42 |
A cache line is 64 bytes. The main memory has latency 32 ns and bandwidth 1 GBytes/s. The time required to fetch the entire cache line from the main memory is
32 ns | |
64 ns | |
96 ns | |
128 ns |
→ So, for 64 bytes it will take 64*1/109 = 64ns.
Main memory latency = 32
Total time required to place cache line is
64+32 = 96 ns
Question 43 |

1K × 18-bit, 1K × 19-bit, 4K × 16-bit | |
1K × 16-bit, 1K × 19-bit, 4K × 18-bit | |
1K × 16-bit, 512 × 18-bit, 1K × 16-bit | |
1K × 18-bit, 512 × 18-bit, 1K × 18-bit |
Bits to represent blocks = m
No. of words in a block = 2n
Bits to represent a word = n
Tag bits = (length of physical address of a word) - (bits to represent blocks) - (bits to represent a word)
⇒ Each block will have its own tag bits.
So, total tag bits = no. of blocks × tag bits.
Question 44 |
{23, 17, 14, 6, 13, 10, 1, 12, 7, 5} | |
{23, 17, 14, 6, 13, 10, 1, 5, 7, 12} | |
{23, 17, 14, 7, 13, 10, 1, 5, 6, 12} | |
{23, 17, 14, 7, 13, 10, 1, 12, 5, 7} |

In this every children and parent satisfies Max heap properties.
Question 45 |
Suppose that we have numbers between 1 and 100 in a binary search tree and want to search for the number 55. Which of the following sequences CANNOT be the sequence of nodes examined?
{10, 75, 64, 43, 60, 57, 55} | |
{90, 12, 68, 34, 62, 45, 55} | |
{9, 85, 47, 68, 43, 57, 55} | |
{79, 14, 72, 56, 16, 53, 55} |
Question 46 |

{P, Q, R, S}, {T}, {U}, {V} | |
{P, Q, R, S, T, V}, {U} | |
{P, Q, S, T, V}, {R}, {U} | |
{P, Q, R, S, T, U, V} |
From given graph {P, Q, R, S, T, V} and {U} are strongly connected components.
Question 47 |
There is only one connected component | |
There are two connected components, and P and R are connected | |
There are two connected components, and Q and R are connected | |
There are two connected components, and P and Q are connected |
Question 48 |
fdheg | |
ecgdf | |
dchfg | |
fehdg |

∴ 110111100111010 = fdheg
Question 49 |
#include <stdio.h> struct test { int i; char *c; }st[] = {5, "become", 4, "better", 6, "jungle", 8, "ancestor", 7, "brother"}; main () { struct test *p = st; p += 1; ++p -> c; printf("%s,", p++ -> c); printf("%c,", *++p -> c); printf("%d,", p[0].i); printf("%s n", p -> c); }
jungle, n, 8, ncestor | |
etter, u, 6, ungle | |
cetter, k, 6, jungle | |
etter, u, 8, ncestor |
Line 1 - main ( )
Line 2 - {
Line 3 - struct test *p = st;
Line 4 - p += 1;
Line 5 - ++p → c;
Line 6 - printf("%s", p++→ c);
Line 7 - printf("%c", +++p → c);
Line 8 - printf("%d", p[0].i);
Line 9 - printf("%s\n", p → c);
Line 10 - }
Now,
Line 3: Initially p is pointing to st, i.e., first element of st which is {5, "become"}
Line 4: Now p is pointing to {4, "better"}
Line 5: ++(p → c), since → has higher precedence, so p → c points to 'e' of "better".
Line 6: prints 'enter' and p now points to {6, "jungle"}
Line 7: ***(p → c), since → has higher precedence. So, prints 'u'.
Line 8: p → i, which is 6 so prints '6'.
Line 9: prints 'ungle' since p is pointing to 'u'.
So, output is "enter, u, 6, ungle".
Question 50 |
#include void swap (int *x, int *y) { static int *temp; temp = x; x = y; y = temp; } void printab () { static int i, a = -3, b = -6; i = 0; while (i <= 4) { if ((i++)%2 == 1) continue; a = a + i; b = b + i; } swap (&a, &b); printf("a = %d, b = %dn", a, b); } main() { printab(); printab(); }
a = 0, b = 3 a = 0, b = 3 | |
a = 3, b = 0 a = 12, b = 9 | |
a = 3, b = 6 a = 3, b = 6 | |
a = 6, b = 3 a = 15, b = 12 |
Inside print 'a' and 'b' are added to odd integers from 1 to 5, i.e., 1+3+5=9. So, in first call to print ab,
a = -3+9 = 6
b = -6+9 = 3
Static variable have one memory throughout the program run (initialized during program start) and they keep their values across function calls. So during second call to print ab,
a = 6+9 = 15
b = 3+9 = 12
Question 51 |
#include int a1[] = {6, 7, 8, 18, 34, 67}; int a2[] = {23, 56, 28, 29}; int a3[] = {-12, 27, -31}; int *x[] = {a1, a2, a3}; void print(int *a[]) { printf("%d,", a[0][2]); printf("%d,", *a[2]); printf("%d,", *++a[0]); printf("%d,", *(++a)[0]); printf("%dn", a[-1][+1]); } main() { print(x); }
8, -12, 7, 23, 8 | |
8, 8, 7, 23, 7 | |
-12, -12, 27, -31, 23 | |
-12, -12, 27, -31, 56 |
It returns the value of 3rd element in a1.
First printf print 8.
2) *a[2] = *(*(a+2))
It returns the value of 1st element in a3.
Second printf print -12.
3) *++a[0] = *(++(*(a+0)))
a[0] is pointing to 1st element in a1.
++a[0] - after preincrement performed, now a[0] is pointing to 2nd element in a1.
*++a[0] return the value of 2nd element in a1.
Third printf print 7.
4) *(++a)[0]
++a - after preincrement is performed 'a' is pointing to a2.
(++a)[0] is pointing to 1st element in a2.
*(++a)[0] returns the value of 1st element in a2.
Fourth printf print 23.
5) a[-1][+1] = *(*(a-1)+1)
(a-1) is pointing to a1.
*(a-1) is pointing to the 2nd element in a1, because in 3rd printf already a1 was incremented by 1.
*(a-1)+1 is pointing 3rd element in a1.
*(*(a-1)+1) returns the value of 3rd element in a1, i.e., 8.
Question 52 |
int func(int m, int n) { if (E) return 1; else return(func(m -1, n) + func(m - 1, n - 1)); }In the above function, which of the following is the correct expression for E?
(n == 0) || (m == 1) | |
(n == 0) && (m == 1) | |
(n == 0) || (m == n) | |
(n == 0) && (m == n) |
mC0 = 1
nCn = 1
Question 53 |
(i) - B, (ii) - D, (iii) - E, (iv) - F, (v) - G, (vi) - A | |
(i) - C, (ii) - A, (iii) - E, (iv) - D, (v) - H, (vi) - F | |
(i) - C, (ii) - F, (iii) - H, (iv) - A, (v) - G, (vi) - D | |
(i) - B, (ii) - E, (iii) - C, (iv) - F, (v) - G, (vi) - H |
Question 54 |
The arrival time, priority, and duration of the CPU and I/O bursts for each of three processes P1, P2 and P3 are given in the table below. Each process has a CPU burst followed by an I/O burst followed by another CPU burst. Assume that each process has its own I/O resource.
The multi-programmed operating system uses preemptive priority scheduling. What are the finish times of the processes P1, P2 and P3 ?
11, 15, 9 | |
10, 15, 9 | |
11, 16, 10 | |
12, 17, 11 |

Hence, finish times of process.
P1 - 10
P2 - 11
P3 - 9
Question 55 |
- Interchanging Wait (F) and Wait (S) in the Producer process
- Interchanging Signal (S) and Signal (F) in the Consumer process
1 only | |
2 only | |
Neither 1 nor 2 | |
Both 1 and 2 |
Now if Wait(S) in producer succeeds, then producer will wait for Wait(F) which is never going to succeed as consumer would be waiting for Wait(S). So deadlock, can happen.
If Signal(S) and Signal(F) are interchanged in consumer, deadlock won't happen. It will just give priority to a producer compared to the next consumer waiting.
Question 56 |

P < S < T | |
S < P < T | |
S < T < P | |
T < S < P |
For P1,
Page size is 1KB. So, no. of pages required for P1=195. An entry in page table is of size 4 bytes and assuming an inner level page table takes the size of a page, we can have upto 256 entries in second level page table and we require only 195 for P1. Thus only 1 second level page table is enough. So, memory overhead = 1KB (for first level) + 1KB for second level = 2KB.
For P2 and P3 also, we get 2KB each and for P4 we get 1+2=3KB as it requires 1 first level page table and 2 second level page tables (364 > 256). So, total overhead for their concurrent execution = 2×3+3 = 9KB. Thus P = 9KB.
Case-2 (For segmentation method):
P1 uses 4 segments → 4 entries in segment table = 4×8 = 32Bytes
Similarly, for P2, P3 and P4 we get 5×8, 3×8 and 8×8 Bytes respectively and the total overhead will be
32+40+24+64 = 160 Bytes
So, S = 160 Bytes
Case-3 (For segmentation with paging):
Here, we segment first and then page. So, we need the page table size. We are given maximum size of a segment is 256KB and page size is 1KB and thus we require 256 entries in the page table. So, total size of page table = 256 × 4 = 1024 Bytes (exactly one page size).
So, now for P1 we require 1 segment table of size 32 Bytes plus 1 page table of size 1KB.
Similarly,
P2 - 40 Bytes and 1 KB
P3 - 24 Bytes and 1 KB
P4 - 64 Bytes and 1KB
Thus, total overhead = 160 Bytes + 4KB = 4256 Bytes
So, T = 4256 Bytes
So, answer should be S < T < P.
Question 57 |
-
- x is a condition variable,
- mutex is a semaphore initialized to 1,
- x_sem is a semaphore initialized to 0,
- x_count is the number of processes waiting on semaphore x_sem, initially 0, next is a semaphore initialized to 0,
- next_count is the number of processes waiting on semaphore next, initially 0. The body of each procedure that is visible outside the monitor is replaced with the following:
P(mutex); body of procedure if (next_count > 0) V(next); else V(mutex);
-
- Each occurrence of x.wait is replaced with the following:
x_count = x_count + 1; if (next_count > 0) V(next) else V(mutex); ------------------------------------------------------------ E1; x_count = x_count - 1;
-
- Each occurrence of x.signal is replaced with the following:
if (x_count > 0) { next_count = next_count + 1; ------------------- E2; P(next), next_count = next_count - 1; }
- For correct implementation of the monitor, statements E1 and E2 are, respectively,
P(x_sem), V(next) | |
V(next), P(x_sem) | |
P(next), V(x_sem) | |
P(x_sem), V(x_sem) |
x_count is incremented and decremented in x_wait, which shows that in between them wait(x_sem) must happen which is P(x_sem). Correspondingly V(x_sem) must happen in x_signal. So, D choice.
Question 58 |
A software program consists of two modules M1 and M2 that can fail independently, but never simultaneously. The program is considered to have failed if any of these modules fails. Both the modules are ‘repairable’ and so the program starts working again as soon as the repair is done. Assume that the mean time to failure (MTTF) of M1is T1 with a mean time to repair (MTTR) of R1. The MTTF of M2 is T2 with an MTTR of R2. What is the availability of the overall program given that the failure and repair times are all exponentially distributed random variables?
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Question 59 |

III and IV | |
I and IV | |
II and IV | |
I, II and IV |
Question 60 |
VXZ | |
VXY | |
VWXY | |
VWXYZ |
Candidate keys are
VXY, WXY, ZXY
Question 61 |
In a database file structure, the search key field is 9 bytes long, the block size is 512 bytes, a record pointer is 7 bytes and a block pointer is 6 bytes. The largest possible order of a non-leaf node in a B+ tree implementing this file structure is
23 | |
24 | |
34 | |
44 |
n × p + (n - 1) × k ≤ B (for non-leaf node)
Here, n = order, p = tree/block/index pointer, B = size of block
So,
n × p + (n - 1) × k ≤ B
n × 6 + (n - 1) × 9 ≤ 512
n ≤ 34.77
∴ n = 34
Question 62 |
<!ELEMENT Univ (Course+, Prof+)> <!ELEMENT Course (Title, Eval*)> <!ATTLIST Course Number ID #REQUIRED Instructor IDREF #IMPLIED> <!ELEMENT Title (#PCDATA)> <!ELEMENT Eval (#PCDATA)> <!ATTLIST Eval Score CDATA #REQUIRED> <!ELEMENT Prof EMPTY> <!ATTLIST Prof Name ID #REQUIRED Teaches IDREF #IMPLIED>What is returned by the following XQuery? let $as := / /@Score for $c in /Univ/Course[Eval] let $cs := $c/Eval?@Score where min($cs) > avg($as) return $c
The professor with the lowest course evaluation | |
Professors who have all their course evaluations above the university average | |
The course with the lowest evaluation | |
Courses with all evaluations above the university average |
Question 63 |


eth0 | |
eth1 | |
eth2 | |
eth3 |
144.16.68.117 = 144.16.68.01110101 AND 255.255.255.224 = 255.255.255.11100000
= 144.16.68.96(Not matching with destination)
Now, take 255.255.255.0
144.16.68.117 AND 255.255.255.0
= 144.16.68.0 (matched)
Hence, option (C) is correct.
Question 64 |
Suppose that it takes 1 unit of time to transmit a packet (of fixed size) on a communication link. The link layer uses a window flow control protocol with a window size of N packets. Each packet causes an ack or a nak to be generated by the receiver, and ack/nak transmission times are negligible. Further, the round trip time on the link is equal to N units. Consider time i > N. If only acks have been received till time i(no naks), then the goodput evaluated at the transmitter at time i(in packets per unit time) is
1 – N/i | |
i/(N + i) | |
1 | |
1 – e(i/N) |
So, successful delivery of packet can be assured if ack has been received for it.
So till time 'i' we would have transmitted 'i' packets but only (i - N) can be acknowledged as minimum time for a packet to get acknowledged is N (since RTT is N which is equal to the window size, there is no waiting for the sender).
So, successfully delivered packets = (i - N)
Time for transmission = i
Goodput = Successfully delivered data/Time
= (i - N)/i
= 1 - N/i
Question 65 |
In the 4B/5B encoding scheme, every 4 bits of data are encoded in a 5-bit codeword. It is required that the codewords have at most 1 leading and at most 1 trailing zero. How many such codewords are possible?
14 | |
16 | |
18 | |
20 |
Codeword with first two bits '0'
= 0 0 x x x
= 23
= 8
Codeword with last two bits '0'
= x x x 0 0
= 23 = 8
Codeword with first two and last two bits '0'
= 0 0 x 0 0
= 2
Codeword with first or last two bits '0'
= 8 + 8 - 2
= 14
Therefore possible codewords
= 32 - 14
= 18
Question 66 |
A router has two full-duplex Ethernet interfaces each operating at 100 Mb/s. Ethernet frames are at least 84 bytes long (including the Preamble and the Inter-Packet-Gap). The maximum packet processing time at the router for wirespeed forwarding to be possible is (in microseconds)
0.01 | |
3.36 | |
6.72 | |
8 |
Tt = 84×8/10×106 = 6.72μs
But since a router has two full-duplex ethernet interfaces, so the maximum processing time should be,
6.72/2 μs = 3.36μs
Question 67 |
A link of capacity 100 Mbps is carrying traffic from a number of sources. Each source generates an on-off traffic stream; when the source is on, the rate of traffic is 10 Mbps, and when the source is off, the rate of traffic is zero. The duty cycle, which is the ratio of on-time to off-time, is 1:2. When there is no buffer at the link, the minimum number of sources that can be multiplexed on the link so that link capacity is not wasted and no data loss occurs is S1. Assuming that all sources are synchronized and that the link is provided with a large buffer, the maximum number of sources that can be multiplexed so that no data loss occurs is S2. The values of S1 and S2 are, respectively,
10 and 30 | |
12 and 25 | |
5 and 33 | |
15 and 22 |
Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.
Since data should not be lost, we calculate for the extreme case when all sources are ontime (that is transmitting).
10Mbps × n-station ≤ 100Mbps
n = 10 = S1
In the next part of question, it is given that the link is provided with large buffer and we are asked to find out large no. of stations.
For that we calculate expected value of bandwidth usage,
E = 1/3 × 10 + 1/3 × 10 + .......+ ....... n-station times ≤ 100Mbps
⇒ 1/3 × 10 × n-station ≤ 100Mbps ⇒ n-station = 30 = S2
So option (A) is answer.
Question 68 |
On a wireless link, the probability of packet error is 0.2. A stop-and-wait protocol is used to transfer data across the link. The channel condition is assumed to be independent from transmission to transmission. What is the average number of transmission attempts required to transfer 100 packets?
100 | |
125 | |
150 | |
200 |
So here it would be for one frame = 1/(1-0.2) = 1/0.8
So for 100 frames = 100/0.8 = 125
Question 69 |
A program on machine X attempts to open a UDP connection to port 5376 on a machine Y, and a TCP connection to port 8632 on machine Z. However, there are no applications listening at the corresponding ports on Y and Z. An ICMP Port Unreachable error will be generated by
Y but not Z | |
Z but not Y | |
Neither Y nor Z | |
Both Y and Z |
Question 70 |
is necessarily 255.255.224.0 | |
is necessarily 255.255.240.0 | |
is necessarily 255.255.248.0 | |
could be any one of 255.255.224.0, 255.255.240.0, 255.255.248.0 |
Broadcast address for subnet is
.95.255 or .01011111.11111111
(as in class B, 16 bits each are used for network and host)
So, we can take minimum 3 bits (from left) as subnet and make rest as host bits (as they are 1)
.224.0 → 11100000.00000000 (leftmost 3 bits for subnet)
.240.0 → 11110000.00000000 (leftmost 4 bits for subnet)
.248.0 → 11111000.00000000 (leftmost 5 bits for subnet)
Question 71 |
⌊i/2⌋ | |
⌈(i-1)/2⌉ | |
⌈i/2⌉ | |
⌈i/2⌉ - 1 |
Question 72 |
O(n) | |
O(log n) | |
O(n log n) | |
O(n log log n) |
Question 73 |
⌊log2 i⌋ | |
⌈log2 (i + 1)⌉ | |
⌊log2 (i + 1)⌋ | |
⌈log2 i⌉ |
Question 74 |
void swap(float* A1, float* A2) { float temp; if (*A1 = = *A2) return; temp = *A1; *A1 = *A2; *A2 = temp; return; }The program volume for the above module using Halstead's method is
60 | |
63 | |
66 | |
69 |
Question 75 |
void swap(float* A1, float* A2) { float temp; if (*A1 = = *A2) return; temp = *A1; *A1 = *A2; *A2 = temp; return; }The program effort for the above module using Halstead's method is
315 | |
330 | |
393 | |
403 |
Question 76 |
0.5 | |
0.75 | |
1.5 | |
2.0 |

Question 77 |
it will converge | |
it will diverse | |
it will neither converge nor diverse | |
It is not applicable |
|1| + |1/2| <= |9|
and |3| + |1| <= |10|
Question 78 |
A pipelined processor uses a 4-stage instruction pipeline with the following stages: Instruction fetch (IF), Instruction decode (ID), Execute (EX) and Writeback (WB). The arithmetic operations as well as the load and store operations are carried out in the EX stage. The sequence of instructions corresponding to the statement X = (S - R * (P + Q))/T is given below. The values of variables P, Q, R, S and T are available in the registers R0, R1, R2, R3 and R4 respectively, before the execution of the instruction sequence.
The number of Read-After-Write (RAW) dependencies, Write-After-Read( WAR) dependencies, and Write-After-Write (WAW) dependencies in the sequence of instructions are, respectively,
2, 2, 4 | |
3, 2, 3 | |
4, 2, 2 | |
3, 3, 2 |
I1 - I2 (R5)
I2 - I3 (R6)
I3 - I4 (R5)
I4 - I5 (R6)
WAR:
I2 - I3 (R5)
I3 - I4 (R6)
WAW:
I1 - I3 (R5)
I3 - I4 (R6)
Question 79 |

10 | |
12 | |
14 | |
16 |

Question 80 |
Both I and IV | |
Only I | |
Only IV | |
Both II and III |
Half (L), Suffix (L) and Prefix (L) are regular languages.
Question 81 |
(a + b)* | |
{ϵ, a, ab, bab} | |
(ab)* | |
{anbn | n ≥ 0} |
1) L should be regular due to demand of question.
2) L should be an infinite set of strings.
3) L should have more than one alphabet in its grammar, otherwise repeat(L) would be regular.
∴ (a + b)* is the perfect example to support the conclusions of last questions.
Question 82 |
P1-P2-P4, 1 day | |
P1-P3-P4, 1 day | |
P1-P3-P4, 2 days | |
P1-P2-P4, 2 days |
Question 83 |
100 and 1000 | |
100 and 1200 | |
150 and 1200 | |
200 and 2000 |
Question 84 |


Karthikeyan, Boris | |
Sachin, Salman | |
Karthikeyan, Boris, Sachin | |
Schumacher, Senna
|
did = {22, 22, 31, 31, 64}
For colour = "Green"
did = {22, 31, 74}
Intersection of Red and Green will be = {22, 31}, which is Karthikeyan and Boris.
Question 85 |

36 - 40 | |
44 - 48 | |
60 - 64 | |
100 - 104 |
red did : 22, 31, 64
green did : 22, 31, 74
(6) for intersection
(1) for searching 22 in driver relation, and (3) for searching 31.
Total: 38 + 6 + 4 = 48