This is part 2 of part I same name just replace II with a I I WILL GIVE AS MUCH

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This is part 2 of part I same name just replace II with a I

I WILL GIVE AS MUCH TIME AS NEEDED FOR THIS QUESTION MULTIPLE DAYS IF NEEDED. I NEED JUST A LITTLE WORK SHOWN. I HAVE OTHER QUESTIONS SO IF YOUR GOOD AT THIS AND WANT TO KEEP MAKING POINTS MAKE A COMMENT BELOW WITH YOUR ANSWER AND I WILL GIVE YOU FIRST CHANCES TO ANSWER FOR NEXT ONES. Goodluck and have fun

42. You are reviewing a conceptual binary addressing scheme that uses 35 bits for network layer addressing (divided into seven 5-bit blocks that are written in dotted decimal format). The first and second subnetworks are reserved for the wire address and to support the broadcast address. The first and second host addresses on each subnet are also reserved to support the subnetwork wire address and subnet broadcast address. Network (N), subnet (s) and host (h) bits are arranged in the following manner with the left most bits of each type being the high order bits:

        h h h s s . h h s s s . N N N N N . N N N N N . N N N N N . s s s h h . s s h h h

e.       Given the following address:

20.11.30.8.0.17.4

What Major network address does this host belong to?
What subnet address does this host belong to?
What would be the broadcast address on this subnet?

f.      Determine the dotted decimal address for the 834th useable host on the 1000th useable subnet of the 7,351st major network.

Explanation / Answer

e.

this belongs to 30X8 =240 th major network

lets write the binary of each number

20 =10100

11 = 01011

30 = 11110

8   = 01000

0   = 00000

17 = 10001

4   = 00100

now subnet we will extract the subnet bits given in ques 0001110000 = 112th subnet or 110th usable subnet.

since second host address is broadcast address.

h=0 0 0 0 0 0 0 0 1 0

we will put this in host bits to obtain the address

00000 =0

00011 =3

11110 = 30

01000 =8

00000 =0

10000 =16

00010 =2

hence the broadcast address is 0.3.30.8.0.16.2

f

834th usable host is 834+2 host =836

binary 110 10 00 100

1000th usable subnet is 1000+2 =1002

binary 11 111 010 10

7351 st network in binary is

binary 001110010110111

now we arrange these bits in given format

h h h s s . h h s s s . N N N N N . N N N N N . N N N N N . s s s h h . s s h h h

11011.10111.00111.00101.10111.01000.10100

27.23.7.5.23.8.20

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