Ans .
(B) A4
Starting from the one has the highest priority i.e. 000. Thus the second highest priority is 001.
Thus the number is 101 001 00 = A4
Ans .
(B) 82H
In 8085 programming, the result of an operation is stored in the accumulator. So output is 82H.
Ans .
(D) INTR
TRAP
RST7.5
RST6.5
RST5.5
INTR
INTA is not an interrupt. INTA is used by the Microprocessor for sending the acknowledgement. TRAP has highest priority and RST 7.5 has second highest priority and so on.
Ans .
(D) 0.32
in 1ms : refresh = 32 times
Memory cycle = 1ms/250ns = 106ns/250ns = 4000 times
Therefore, % of refresh time = (32 x 100ns)/(4000 x 250ns)
= 3200ns/1000000 x 100% = 0.32%
Ans .
(B) 0.12%
The DMA combines one word from four consecutive characters (bytes) so we get 4800 chars/s = 4800 bytes/s = 1200 words/s (one word = 32 bits = 4 bytes)
If we assume that one CPU instruction is one word wide
then 1 million instructions/s = 1 million words/s = 106 word/s
So we have 1200 words received during one second and (106-1200) words processed by the CPU (while DMA is transferring a word, the CPU cannot fetch the instruction so we have to subtract the number of words transferred by DMA).
While DMA transfer CPU executes only 106 - 1200 = 998800 instructions
[998800 / 106] * 100 = 99.88 %
Slowdown = 100 - 99.88 = 0.12%
The CPU will be slowed down by 0.12%
Ans .
(A) by checking interrupt register after execution of each instruction
Ans .
(A) 1
Ans .
(C) Data driven, Goal driven
Backward Chaining is a form of reverse engineering, which is very applicable in situations where there are so many rules that could be applied to a single problem, the system could be there all day sifting through rules before it gets anywhere. Forward Chaining is a problem solving technique used by Expert Systems when they are faced with a scenario and have to give a solution or conclusion to this scenario. The system will work its way through the rules, finding which ones fit and which leads to which using Deductive Reasoning
Ans .
(D) (ii) (iv) (i) (iii)
Ans .
(B) socket address
Socket address is a combination of IP address and Port address.
Ans .
(A) 2 Mbps
In data transmission, throughput is the amount of data moved successfully from one place to another in a given period of time, and typically measured in bits per second (bps), or in megabits per second (Mbps) or gigabits per second (Gbps). Here, Throughput = 15000 x 8000/60 = 2 Mbps
Ans .
(A) 25
Each router needs 10 entries for local routers, 8 entries for routing to other regions within its own cluster, and 7 entries for distant clusters, for a total of 25 entries.
Ans .
(A) refers to the current network
Ans .
(D) MIME
MIME stands for Multi-purpose Internet Mail Extensions or Multimedia Internet Mail Extensions . It is a encoding protocol like BinHex in Mac and UUEncode in UNIX. At first it was used as a way of sending more than just text via email. Later the protocol was extended to manage file typing by Web servers.
Email attachment
Let’s talk about email first. By using MIME, you can enclose the following types of binary file to your text-based e-mail message:
character sets other than ASCII
enriched text
images
sounds
other messages (reliably encapsulated)
tar files
PostScript
FTPable file pointers
The easiest way to send and receive MIME attachment is to use a MIME-aware Email client such as Netscape Mail or Microsoft Outlook. The procedure is selfexplanatory. When a MIME ready e-mail system receives a MIME encoded file, the binary file should show up as an attachment and the proper software in your computer should be capable of reading the file. For example, if the attached file is a PostScript file, your LaserWriter Utilities should be able to download it to the printer. If the file is a graphic, DeBabelizer or PhotoShop should be able to open the file. However, you must pay attention to the file extension. For instance, if someone sends you a JPEG image from a Mac with an extension as “JPEG,” you would not be able to open it in a PC unless you change the extension to “JPG,” and vice versa.
Ans .
(C) 400
1,00,000 characters = 1,00,000 x 8 bits = 8,00,000 bits
8,00,000 bits/2000 bps = 400 seconds
Ans .
(C) si≥fj or sj≥fi
Ans .
(D)
A subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. A sequence G is said to be a common subsequence of X and Y , if Z is a subsequence of both X and Y .
Here X = , the sequences ,
Given a second sequence of symbols Y = , then ,
However, the longest common subsequence of X and Y is .
Ans .
(D) O(d(n+k))
Ans .
(D) None of the above
Ans .
(C) Dynamic programming, θ(V3)
Floyd-warshall algorithm uses Dynamic programming approach to solve the allpairs shortest paths problem on a directed graph θ(V3) in time.
Ans .
(B)
Ans .
(C) closed, closed
Ans .
No option matched
Ans .
(C) Four
Ans .
(C) Usability
Ans .
(C) (a+b)*(bbb)(a+b)*
Ans .
(D) 5832
N * N^(N*M)* 2^N (N is number of states , M is number of alphabet ) the first N is for no. of way to select initial state . Here N will be 1 because they told us X is only initial state . N^(N*M) is for no. of transition functions from a set of N*M elements to a set of N elements. . 2^N No. of final state we can choose . so , 1 * 3^(3*2)* 2^3 =5832
Ans .
(A) L1 is context free language and L2 is not context free language
Ans .
(3) Identical with average revenue
Ans .
(B) (iii) (iv) (i) (ii)
Ans .
(C) (a), (c) and (d)
Ans .
(A) For x=0, y=0; x=1, y=3; x=2, y=6; x=3, y=9
two end points A(0,0) and B(6,18) are given
The equation of straight line is y =mx + c where c is y intercept and m = slope of line we know m= (y2-y1)/(x2-x1)
m=(18-0)/(6-0) m =3, y intercept c = y-mx
for y intercept we take x= x1 and y = y1 here x1 =0 and y1 =0, c = y1 – mx1
c=0
first point is x=0,y=0
second point is x step incremented by 1 so x=1
y = mx+c =3*1+0 so y=3
x=1,y=3
Third point is x step incremented by 1 so x=2
y = mx+c =3*2+0 so y=6 x=2,y=6
Fourth point is x step incremented by 1 so x=3
y = mx+c =3*3+0 so y=9
x=3,y=9
so the option is (1) i.e For x=0, y=0; x=1,y=3; x=2,y=6; x=3, y=9 is the correct answer
Ans .
(B) (a) and (b)
Ans .
(A) (a) only
Ans .
(D) (a) and (c)
Ans .
(C) Both (a) and (b) are true
Ans .
(D) (ii) (iii) (iv) (i)
Ans .
(A) Address translation
Ans .
(B) unsafe state
At the time T1
Maximum need P1 P2 P3 is 10 4 9 respectively
Allocated P1 P2 P3 is 5 2 3 respectively
Total tape drives = 12
Available tape drives = 12-(Allocated tape drives) = 12 – (10)=2
Need P1 P2 P3
5 2 6
So process P2 needs 2 tap drives and we have available 2.
So P2 satisfies is need at t1 After this Available becomes 4
So we can see neither P1 nor P3 need satisfies so system no longer becomes safe and System goes to unsafe state at time t1
Ans .
(B) provide a mechanism to map physical device to file names
Ans .
(A) (iv) (i) (ii) (iii)
Ans .
(C) Processor can not be pre-empted
Ans .
(D) A clause that has at most one positive literal
Ans .
(B) Q
Ans .
(A) Forward chaining, backward chaining and problem reduction
Ans .
(A) Functional programming
Ans .
(A) Arithmatic equations and inequalities that bind the values of variables
Ans .
(C) It is a business investment
Ans .
(B) -plog(p)-(1-p)log(1p)
Ans .
(D) Arithmetic coding
the type of compression we’ve been discussing here is called lossless compression, because it lets you recreate the original file exactly. All lossless compression is based on the idea of breaking a file into a “smaller” form for transmission or storage and then putting it back together on the other end so it can be used again. Lossy compression works very differently. These programs simply eliminate “unnecessary” bits of information, tailoring the file so that it is smaller. This type of compression is used a lot for reducing the file size of bitmap pictures, which tend to be fairly bulky. To see how this works, let’s consider how your computer might compress a scanned photograph.
Ans .
(A) Combination of blur identification and image restoration
Ans .
(A) degenerate
In Linear Programming (LP) a basic feasible solution is one that also belong to the feasible region or problem area can be represented by a feasible solution in implementing the Simplex Method satisfying nonnegative conditions.
Ans .
(D) (a), (b) and (c)
Ans .
(B) 1465
Ans .
(C) 16 bit UNICODE
Ans .
(D) All of the above
Ans .
(A) Remove file chap0[1 - 3]
Ans .
(A) (a), (b) and (c)
Ans .
(A) 5
Ans .
(D) 4 and 2 respectively
Ans .
(C) 10
Ans .
(C) Tightly coupled software on loosely coupled hardware and message passing, respectively.
Ans .
(A) (a)
Ans .
(D) x1+x2-0.5=0
Ans .
(C)
Ans .
(B)
Ans .
(D) All the above
Ans .
(D) None of the above
Ans .
(C)
Ans .
(A) (-1/2, 0), (-1/2, 1), (-3/2, 1), (-3/2, 0)
Ans .
(C) (a), (b), (c) and (d)
Ans .
(D) In white-box testing, the test cases are decided from the specifications or the requirements.
Ans .
(C) 0011
if n=2
[n/2] = [1] = 1 i.e. Language is : 011
if n=3
[n/2] = [1.5] = 1 i.e. Language is : 0111
if n=4
[n/2] = [2] = 2 i.e. Language is : 001111
Hence option A is possible, B is possible, D is possible but C is not possible.
Ans .
(C) “Modelling-oriented approach” is one of the methods for specifying the functional specifications.
Ans .
(A)