Help! Anyone good at comp. sci.?

Discussion in 'Tech Talk' started by Stan, Oct 8, 2003.

  1. Stan

    Stan
    [OP]
    fashionably late

    Joined:
    Jun 11, 2003
    Messages:
    5,905
    Likes Received:
    0
    Location:
    Calgary
    Data communication and networks:

    A transport layer message consisting of 1500 bits of data and 160 bits of header is sent to an internet layer, which appends another 160 bits of header. This is the transmitted through two networks, each of which uses a 24-bit packet header. The destination network has a max packet size 800 bits. How many bits, including headers, are delivered to the network-layer protocol at the destination?

    :confused:

    Stan
     
  2. slouse

    Well-Known Member

    Joined:
    Jun 8, 2003
    Messages:
    22,776
    Likes Received:
    252
    i don't know anything about computers, but my guess is 1868

    the only thing i know about data communication and networks is that my DSL line is better than 56k
     
  3. Mitch

    Well-Known Member

    Joined:
    Aug 28, 2003
    Messages:
    24,167
    Likes Received:
    560
    Location:
    Bedford
    my guess is 1031.5
    btw, I know nothing about computer either, sorry Stan
     
  4. Kat

    Kat
    Active Member

    Joined:
    Oct 1, 2003
    Messages:
    4,498
    Likes Received:
    0
    Location:
    NB
    Ask ken i think he took that in University. He'd know anyways he's a brain when it comes to computers!
     
  5. Dural

    Insert something witty

    Joined:
    Jul 2, 2003
    Messages:
    4,904
    Likes Received:
    0
    Location:
    Newmarket
    Been two years since I did Data Comm :(
     
  6. Eisenflower

    .

    Joined:
    Jun 3, 2003
    Messages:
    24,486
    Likes Received:
    1
    Thank me later:

    To the Network Layer, packets from the IP layer are considered as data, we call it IP data.

    IP data = User Data + Transport Header + IP Header
    = 1500 + 160 + 160 bits
    = 1820 bits


    The maximum packet size for the network is 800 bits and network header is 24 bits. So the maximum data size the network can handle is (800 – 24) = 776.

    Since we have 1820 bits to be delivered, it is too large to fix in one packet. So fragmentation is needed.

    Define
    n = number of fragments
    d = total number of data to be transmitted at the network layer = IP data = 1820
    m = maximum data size for the network = 776

    n= [ d/m]

    = [1820/776]

    = [2.3]

    = 3

    Define
    t = total number of delivered to the destination
    h = total overhead = network header x n = 24 x 3 = 72

    t = h + d

    = 72 + 1820

    = 1892
     
  7. Boots

    www.reality-check.ca

    Joined:
    Jun 3, 2003
    Messages:
    73,558
    Likes Received:
    3,801
    Location:
    Halifax
    Stan, your Asian... your pre-programmed for that shit. What's the problem ?! :p
     
  8. Eisenflower

    .

    Joined:
    Jun 3, 2003
    Messages:
    24,486
    Likes Received:
    1
  9. Shafft

    RC-OG

    Joined:
    Jun 4, 2003
    Messages:
    10,660
    Likes Received:
    51
    Location:
    HRM
    nice work :eek:



    I might have been able to do that in school... but i can't remember if we did it .. :lol:
     

Share This Page