f 44,209,694 bits each second. Now the fun begins. First, assume we will be using 48/53-byte cell structure. Next, understand and analyze this picture. 53bytes of 8bits equals 424 bits in each cell. There are 44,209,694 bits to put ATM cells into. A little simple arithmetic reveals that the available bandwidth can accommodate 104,268 of these cells or carriers. This quantity requires 44,209,632 bits-no more, no less. The difference of 62 is not 53. It is possible the available bandwidth could carry as many as 104,268 53-byte cells. Yes, but not in every successive second. So what has to happen is organized transfer with some organized bit stuffing. One simple approach could be to transfer 104,269 cells and follow that with the 62 bits of stuffing. In practice, that's just fine because the utilization is almost 100%. After the ATM channel is fitted to the PDH/SONET/SDH channel, next comes how to fit the program content into the ATM structure. Much like the TDM structure, the ATM structure wants to see 48 bytes of payload that it attaches its overhead to and sends it on its way. At the other end, the reverse happens, whereby the payload is subjected to a reversal process and extracted from its transport struc- ture. Simple arithmetic tells us those 104,268 cells per second, with APPENDIX 424 bits in each, amounts to a maximum content payload potential of 33,783,156 bits per second. Estimating and configuring an ATM network to transport program content becomes a matter of defining picture quality and sound fidelity in terms of aggregate program stream bandwidth and selecting the size of the ATM channel to accommodate that stream. One or more program streams or payloads can be accommodated on a single E3/T3/J3 access facility as long as the total aggregate of all streams does not exceed the 33.7Mbs maximum. IP Transmission IP transport is an entirely different matter. First, early IP networks- the Internet-were implemented on ATM backbone networks. This meant that the basic path a set of information traversed include