Q # 5 Recall that ATM uses 53-byte packets consisting of 5 header bytes and 48 p

Question

Q # 5 Recall that ATM uses 53-byte packets consisting of 5 header bytes and 48 payload bytes. Fifty-three bytes is unusually small for fixed- length packets; most networking protocols (IP, Ethernet, frame relay, and so forth) use packets that are, on average, significantly larger. One of the drawbacks of a small packet size is that a large fraction of link bandwidth is consumed by overhead bytes; in the case of ATM, almost 10 percent of the bandwidth is "wasted" by the ATM header. In this problem we investigate why such a small packet size was chosen. To this end, suppose that the ATM cell consists of P bytes (possibly different from 48) and 5 bytes of header a. Consider sending a digitally encoded voice source directly over ATM. Suppose the source is encoded at a constant rate of 64 kbps. Assume each cell is entirely filled before the source sends the cell into the network. The time required to fill a cell is the packetization delay. In terms of L, determine the packetization delay in milliseconds b. Calculate the store-and-forward delay at a single ATM switch for a link rate of R = 155 Mbps (a popular link speed for ATM) for L = 1,500 bytes, and for L-48 bytes c. Packetization delays greater than 20 msec can cause a noticeable and unpleasant echo. Determine the packetization delay for L-1,500 bytes (roughly corresponding to a maximum-sized Ethernet packet) and for L-48 (corresponding to an ATM cell). d. Comment on the advantages of using a smal cell size

Explanation / Answer

Dear User,
a). The source is encoded atconstant rate or transmission rate =64kbps
                                                                                                 =64,000bps

     The time requiredto fill a cell is the packetization delay

      Hence,Total cell size = 8.L bits

     Packetization delay= 8.L/64000=0.125L msec

     Suppose L=1,500

      Packetization delay=0.125x1, 500

                                    =187.5msec    

b).

      For L=1,500bytes

    Packetizationdelay=0.125x1,500

                                   p1 =187.5msec

      For L=48bytes

      Packetization delay=0.125x48

                                    p2 =6msec

     Since,packetization delay greater then 20 msec causes noticeable andunpleasant echo.

     Hence, p1will causes noticeable and unpleasant echo

                 p2 will not causes noticeable and unpleasant echo

c).

    Store and forward delay=(L.8 + 40)/R

    ATM switch for a linkrate R=155Mbps

     ForL=1,500
     The Store andForward Delay=(1,500 x 8+40)/155Mbps
                                                    =(12000+40)/155Mbps

                                                     =7.767xsec                                         

     ForL=48

      The Storeand Forward Delay=(48 x 8+40)/155Mbps

                                                    =(384+40)/155Mbps

                                                     = 2.735 sec

If it is for ATM, the delay is relative to thepacket’s length in bits. The store-and-forward delay ortransmission delay for a link is equal to the number of bits in thepacket divided by the transmission rate of the link.

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