also @ TechSpot: HP Envy/Pavilion revamp, more touchscreens, 3200x1800 LCD, 20" tablet

IP fragmentation and total length

Discussion in 'Storage and Networking' started by Jskid, Oct 26, 2012.

Post New Reply

  1. Jskid TechSpot Enthusiast Posts: 322

    I am so confused by the answer to this question. If the MTU is 601 bytes how come the total length for the first two are 596, shouldn't they be 601?

    Shouldn't the fragment offset for the second fragment be 601/8=75.125?

    How do you know what the MF flag should be?

    I looked up all these terms on wikipedia and understand them but I still don't get this question.

    [IMG]
    upload picture
    Jskid, Oct 26, 2012
    #1

  2. jobeard TS Ambassador Posts: 12,228   +122

    Hmm; goofy question in the first place as MTU must be multiples of 8 :)

    second, the no headers option MUST apply to optional stuff, as the packet destination is in the header portion of the packet while total length = (payload + header) <= MTU

    That's all I can offer, ie never took TCP 101 :sigh:
    jobeard, Oct 26, 2012
    #2

Add New Comment

Social Login & Guest Posting

TechSpot Members
Login here or sign up for free,
it takes about a minute.
Get complete access to the TechSpot community. Join thousands of technology enthusiasts that contribute and share knowledge in our forum. Get a private inbox, upload your own photo gallery and more.