computer

1. Supporting Document
For the Question to be challenge
By Priyanka Chatterjee

2. In this given table q2 is making a self loop. So the machine will stuck here. The table is incorrect.
It can not produce a^n b^m c^n d^m . Check diagram for more clarification.
QBID:
187096
187097
187098
187099
187100

4. QBID:187099
Correct ans 1 (q5,Y1,L)

5. 130.56.0.0/16 means 16 bit allocated for NET ID and 16 bit for host.
To create 1024 subnet we need to borrow more 10 bit (2^10 = 1024) form Host part.
Remaining 6 bit is there for host so we can create 2^6 = 64 host in each subnet.
But 1st ip is always dedicated for Subnet id and last one for Broad cast so total usable address
Should be 62. It can be challenged.
130.56.0.0/16 means 16 bit allocated for NET ID and 16 bit for host.
First 16 bit is fixed for Net id so no change in 1st byte so prefix 130.56.
This is correct ans.

6. 130.56.0.0/16 means 16 bit allocated for NET ID and 16 bit for host.
To create 1024 subnet we need to borrow more 10 bit (2^10 = 1024) form Host part.
Remaining 6 bit is there for host so we can create 2^6 = 64 host in each subnet.
In 1st sub net there will be 1st 64 address so 1st subnet range is
130.56.00000000.00000000 to 130.56.00000000.00111111
130.56.0.0 to 130.56.0.63
But 1st ip is always dedicated for Subnet id and last one for Broad cast so
Ans should be 130.56.0.1 to 130.56.0.62
It can be challenged

7. 130.56.0.0/16 means 16 bit allocated for NET ID and 16 bit for host.
To create 1024 subnet we need to borrow more 10 bit (2^10 = 1024) form
Host part. Remaining 6 bit is there for host so we can create 2^6 = 64 host in
each subnet.
In last sub net there will be last 64 address so last subnet range is
130.56.11111111.11000000 to 130.56.11111111.11111111
130.56.255.192 to 130.56.255.255
But 1st ip is always dedicated for Subnet id and last one for Broad cast so
Ans should be 130.56.255.193 to 130.56.255.254
It can be challenged
130.56.0.0/16 means 16 bit allocated for NET ID and 16 bit for host.
To create 1024 subnet we need to borrow more 10 bit (2^10 = 1024) form
Host part. Remaining 6 bit is there for host so we can create 2^6 = 64 host in
each subnet.
In Subnet mask all net and subnet bit should be 1 and host bit 0 so.
11111111.11111111.11111111.11000000
255.255.255.192 so answer is definitely wrong. It should be drop.
No option is correct here.

8. QBID:187050
We know that so option a and c both are correct

9. QBID:187081
A module connected to other
module is in coupling not
cohesion. So reason in false

10. Minimum string in abcd
And any number a, b,c,d possible because of *.
We can generate string like
aaabbccccdddddd
No comparison in number like of a=b and c=d Or a=c and c=d or a=b=c=d
So option a and c is wrong.
Option b is also not completely correct but partially
Either it should be drop or only B correct. Means option 2.