QUESTIONS CARRY 2 MARKS EACH
Consider a project with the following functional units:Number of user inputs = 50 Number of user outputs = 40 Number of user inquiries = 35 Number of user files = 06 Number of external interfaces= 04 Assuming all complexity adjustment factors and weighing factors as average, the function points for the project will be________.
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Solution
All complexity adjustment factors are average.Therefore CAF =0.65 + 0.01(∑Fi)
i =1 to 14
= (0.65 + 0.01 × (14 × 3) : Rate for average is 3 =0.65 + 0.42 = 1.07
Calculating FP
FP = UFP × CAF = 628 × 1.07 = 672
QUESTIONS CARRY 2 MARKS EACH
In a digital transmission, the receiver clock is 0.1% faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Mbps?
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Solution
At 1 Mbps, the receiver receives 1,000,000(1 + 0.001)= 1,001,000 bps instead of 1,000,000 bps. i.e. 1,000,000 bits sent and received 1,001,000 so extra bits are 1000.
QUESTIONS CARRY 2 MARKS EACH
Consider an undirected graph G with 100 nodes. The maximum number of edges to be included in G so that the graph is not connected is ________.
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Solution
If a graph with n vertices has more than ((n – 1) × (n – 2))/2 edges then it is connected.
QUESTIONS CARRY 2 MARKS EACH
A computer system has 6 tape drives,with n process competing for them. Each process may need three tape drives.The maximum value of n for which the system is guaranteed to be deadlock free is ________.
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Solution
The maximum value of n for which the system is guaranteed to be deadlock free is 2.Two processes can never lead to deadlock as the peak time demand of (3 + 3 = 6)tape drives can be satisfied.
But three processes can lead to dead lock if each process hold 2 drives and then demand one more.
Consider a system having m resources of the same type.These resources are shared by 3 processes A, B and C which have peak demands of 3, 4 and 6 respectively.For what value of m deadlock will not occur ……………?
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Solution
Consider the peak demand situation of resources (A,B,C) = (3,4,6).For a specific number of resources,to conclude there is possibility of deadlock (not deadlock free), we have to find at least one resource allocation which results in deadlock, that is an allocation that cannot completly satisy the resouce requirements of even one process .With number of shared resources m = 7, the following resource allocation (for example)among 3 process A,B,C (2,3,2).Process A holding 2 resources and waitning for 1, Process B holding 3 resource and waiting for 1 and process C holding 2 resources and waiting for 4 more resources. This is a deadlock situation.With number of shared resources m = 9, the following resource allocation (for example)among 3 process A,B,C (2,3,4).Process A holding 2 resources and waitning for 1, Process B holding 3 resource and waiting for 1 and process C holding 4 resources and waiting for 2 more resources. This is a deadlock situation.
With number of shared resources m= 10, the following resource allocation (for example)among 3 process A,B,C (2,3,5).Process A holding 2 resources and waitning for 1, Process B holding 3 resource and waiting for 1 and process C holding 5 resources and waiting for 1 more resource. This is a deadlock situation.
But for m > = 11 , the resource allocation is deadlock free.
QUESTIONS CARRY 2 MARKS EACH
If the disk head is located initially at 32, find the number of disk moves required with FCFS if the disk queue of I/O blocks requests are 98, 37, 14, 124, 65, 67 is ………. ?
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Solution
Initially head is at 32.Movement is FCFS.so disk movement are as follows. (32 to 98)+(98 to 37)+(37 to 14)+(14 to 124)+(124 to 65)+(65 to 67) then total disk movement is 321.
QUESTIONS CARRY 2 MARKS EACH
Consider the following code segment:
x =u –t ;
y = x *v ;
x =y + w ;
y = t –z ;
y = x * y ;
The minimum number of total variables required to convert the above code segment to stafic single assignment form is……….. .
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Solution
Static single assignment means assignment to register can be done one time only.so, draw the GRAPH and count number of nodes which will give the number of register required.
QUESTIONS CARRY 2 MARKS EACH
With four programs in memory and with 80% average I/O wait, the CPU utilisation is_____%?
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Solution
CPU utilisation is given by the formula
= 1 – (P)n
CPU utilisation is calculated from a probabilistic viewpoint. P stands for the fraction of time waiting for I/O to complete.
Number of processes in memory = n
The probability that all n processes are waiting for I/O is (P)n.
P = 80% = 80/100 = 0.8
n = 4
CPU Utilization = 1 – (P)n
= 1 – (0.8)4= 1 – 0.4096 = 60%
QUESTIONS CARRY 2 MARKS EACH
A relation Empdtl is defined with attributes empcode (unique),name, street, city, state and pincode. For any pincode, there is only one city and state. Also, for any given street, city and state, there is just one pincode. In normalization terms,Empdtl is a relation in
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Solution
empdt 1 {empcode, name, street, city, state, pincode}
Given functional dependency
Pincode →(city) (state)
(street) (city) (state) →Pincode
This is in 1 NF and 2 NF but not in 3NF because it contain transitive dependency. Hence correct answer is(b).
QUESTIONS CARRY 2 MARKS EACH
A relational database contains two tablets student and department in which student table has columns roll_on, name and dept_id and dept_name. The following insert statements were executed successfully to populate the empty tables:
Insert into department values (1, “Mathematics’)
Insert into department values (2, ‘Physics’)
Insert into department values (1, ‘Navin’,1)
Insert into department values (2, ‘Mukesh’, 2)
Insert into department values (3, ‘Gita’ 1)
How many rows and columns will be retrieved by the following SQL statement?
Select * from student, department
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Solution
After populating the table
Select * from student, department.This query will execute natural join first and will retrieve all the row and column. Hence 6 row and 5 column will retrieve, which is same as cartesian product of these table. Hence correct answer is(d).