Essay On Sp Mail

Decent Essays

Problem 1: a).sol SMTP is a protocol used to transfer emails, so this is command to specify the MAIL SENDER mail from: ( sender ) rcpt to: ( recipient ) so by using such commands email servers communicate with each other. In mail message MIME is shown which shows u the email contents, there is no relation amoung MAIL FROM and FROM b). SMTP uses a line containing only a period to mark the end of a message body. HTTP uses “Content-Length header field” to indicate the length of a message body. No, HTTP cannot use the method used by SMTP, because HTTP message could be binary data, whereas in SMTP, the message body must be in 7 bit ASCII format. c). The name server (NS) resource record …show more content…

The complete time for the message from source host to destination host = 4 sec× 3 hops = 12 sec. b) Time for first packet from source host to first packet switch = 1*10^4/2*10^6 = 5 m sec. Time for the second packet is received at the first switch = time at which first packet is received at the second switch = 2 × 5m sec = 10 m sec c) Time of the packet received at the destination host = 5 m sec× 3 hops = 15 m sec. After this, every 5msec one packet will be received; thus time at which last (800 th) packet is received = 15 m sec + 799 * 5m sec = 4.01 sec. d) i. Without message division, if bit mistakes are not endured, if there is a solitary piece blunder, the entire message must be re transmitted (instead of a solitary bundle). ii. Without message division, enormous bundles (containing HD recordings, for instance) are sent into the system. Switches need to suit these gigantic packets. Littler packets need to line behind huge bundles and endure out of line delays. e) i. Packets must be placed in grouping at the goal. ii. Message division brings about numerous littler bundles. Since header measure is typically the same for all packets paying little mind to their size, with message division the aggregate sum of header bytes is more. Problem 4 R: ∆ = (850,000 bits)/(15,000,000 bits/sec) = .0567 sec β∆=(16 requests/sec)(.0567 sec/request) = 0.907. Average access delay is (.0567 sec)/(1 - .907) ≈ .6 seconds. The total average response

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