- The Boolean expression
is given by
ans is A
3.The circuit shown in the given figure is
4.When two numbers are added in excess-3 code and the sum is les than 9,then in order to get the correct answer it is necessary to
(a)subtract 0011 from the sum
(b)add 0011 to the sum
(c)subtract 0110 from the sum
(d)add 0110 to the sum
5 . The characteristic equation of an SR flip-flop is given by
6.A graph with “n” vertices and n-1 edges that is not a tree is
7.If a graph requires k different colors for its proper coloring,then the chromatic number of graph is
Explanation: The chromatic number of a graph is the smallest number of colors needed to color the vertices so that no two adjacent vertices share the same color.
8.A read bit can be read
(a)and written by CPU
(b) and written by peripheral
(c)by peripheral and written by CPU
(d)by CPU and written by peripheral
10.The term aging refers to
(a)booting up the priority of the process in multi-level of queue without feedback
(b)gradually increasing the priority of jobs that wait in the system for a long time to remedy infinite blocking
(c)keeping track of the following the page has been in memory for the purpose of LRU replacement
(d)letting job reside in memory for a certain amount of time so that the number of pages required can be estimated accurately
11.Consider a set of n tasks with known runtimes r1 ,r2…………..rn to be run on a uniprocessor macine .Which of the following processor scheduling algorithms will result in maximum throughput ?
b)Shortest job first
c)Highest response ratio next
d)first cum first served
Throughput means total number of tasks executed per unit time. Shortest Job First has maximum throughput because in this scheduling technique shortest jobs are executed first hence maximum number of tasks are completed.
Note: Highest-Response-Ratio-Next policy favors shorter jobs, but it also limits the waiting time of longer jobs.
12. Consider a job scheduling problem with 4 jobsand with corresponding deadlines: . Which of the following is not a feasible schedule without violating any job schedule?
TO OPTIMIZE AND TO GET A FEASIBLE SOLUTION WE HAVE TO FINISH ALL THE JOBS WITH IN THEIR DEAD LINE.
SINCE THE DEADLINE OF J2 AND J4 ARE 2.
SO THEY MUST BE COMPLETED BY 2.SO WE CAN SCHEDULE ANY ONE
THAT IS J2 OR J4..
IN OPTION C,D J4 IS SCHEDULED FIRST AND THEN J2 AND
IN OPTION A, FIRST J2 THEN J4 SO WE CAN ABLE TO SCHEDULE OTHER 2 JOBS
NOW CONSIDER OPTION B
J4 IS SCHEDULED AND NEED TO BE COMPLETED ….BY 2 —NO PROBLEM
J1 IS SCHEDULED AND NEED TO BE COMPLETED ….BY 4 –NO PROBLEM
BUT NOW AS J2 SHOULD COME BEFORE TIME 2..BUT WE SCHEDULED J1….SO J2 CANT BE SCHEDULED
13. By using an eight bit optical encoder the degree of resolution that can be obtained is (approximately)
A digital optical encoder is a device that converts motion into a sequence of digital pulses.Resolution is defined as the smallest discernible quantity.
Hence its is 360/2^8
Ans is 1.4 degree.
14 The principle of locality of reference justifies the use of
a) virtual memory b) interrupts c) main memory d)cache memory
locality of reference, also known as the principle of locality, is a term for the phenomenon in which the same values, or related storage locations, are frequently accessed, depending on the memory access pattern.
There are two basic types of reference locality –
temporal and spatial locality.
15 . Consider the following pseudo-code
x:=1; i:=1; while (x <= 1000) begin x:=2^x; i:=i+1; end;
What is the value of i at the end of the pseudo-code?
st time … x = 2 and i = 2
2nd time …x=4 and i=3
3rd time…..x=16 and i = 4
4th time ……before entering the loop value of x < 1000 ….it is 16 so it will enter the loop
and make i = 5
5th time x>1000 so while loop doesn’t execute…..
SO the answer is i = 5