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Message

   Consider a packet switching network with the packet

size as 512 bits,

message size as 1Kbits,propagation delay as 2sec/hop 

and the number of hops are 5.

Then find the time required for the last it of the 

last packet to reach the 

receiver.(data transmission rate as 1 Kbit/sec)

a.4058 sec  b.2058sec    c.3058 sec d.None

my approach is this.........

total message---1000 bits

1 pkt--------512 bits

pkt2---------488 bits

1000 bit transmitted in 1 sec

512 ....................1/1000X512=.512 sec with delay 5X2=10 sec

488 ....................1/1000X488=.488 sec with delay  5X2=10 sec

total time=(.512+10+.488+10)sec....

correct me plz.

A system that uses a 2 level page table has 2^12 byte 

as page size and 32 bit virtual address.The 1st 8 bits

 of the address serve as the 

index into the 1st level page table.

How many pages are there in virtual address???

a.2^10    b.2^12      c.2^20                 d.None

Author	Message
parvati123 Newbie Joined: 04Aug2007 Online Status: Offline Posts: 1	Topic: unsolve question Posted: 04Aug2007 at 3:24am
	Consider a packet switching network with the packet size as 512 bits, message size as 1Kbits,propagation delay as 2sec/hop and the number of hops are 5. Then find the time required for the last it of the last packet to reach the receiver.(data transmission rate as 1 Kbit/sec) a.4058 sec b.2058sec c.3058 sec d.None my approach is this......... total message---1000 bits 1 pkt--------512 bits pkt2---------488 bits 1000 bit transmitted in 1 sec 512 ....................1/1000X512=.512 sec with delay 5X2=10 sec 488 ....................1/1000X488=.488 sec with delay 5X2=10 sec total time=(.512+10+.488+10)sec.... correct me plz. A system that uses a 2 level page table has 2^12 byte as page size and 32 bit virtual address.The 1st 8 bits of the address serve as the index into the 1st level page table. How many pages are there in virtual address??? a.2^10 b.2^12 c.2^20 d.None Post Resume: Click here to Upload your Resume & Apply for Jobs

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