From ldunn@cisco.com Tue Aug 10 12:51:47 1999 Date: Mon, 09 Aug 1999 10:47:31 -0500 Subject: Re: normal-burst and excess-burst ============================================================================= Clarence have provided a good explanation. But the algorithm is tricky to visualize, I think. The following example is from Bob Olsen. It helps sort out the "actual" and "compound" terminology. [snip] ************************************************************************** Okay, here is a simple example. Assume token rate is 1 data_unit/time_unit, normal burst size is 2 data_units (DUs) and extended burst is 4 DUs. Further assume 2 DUs arrive per time unit. So, after 2 time units, the stream has used up its normal burst and must begin borrowing one DU per time unit, beginning at time unit 3. Time DU arrivals Actual Debt Compounded Debt ------------------------------------------------------- 1 2 0 0 2 2 0 0 3 2 1 1 4 2 2 3 5 2 3 (temp) 6 (temp) At this time a packet is dropped because the new compounded debt (6) would exceed the extended burst limit (4). This causes compounded debt to effectively become 0, and lowers actual debt back down to 2. The values 3 and 6 were only temporary and do not remain valid in the case where a packet is dropped. The final values for time unit 5 are given below. 5 2 2 (*) 0 6 2 3 3 7 2 4 (temp) 7 (temp) Again a packet is dropped and the debt values are adjusted accordingly. 7 2 3 (*) 0 ************************************************************************** (*) NOTE (Tiziana Ferrari): after a packet drop the actual debt value is set back to the corresponding value before the arrival of the dropped packet (i.e. to the value at instant 4 and 6 respectively in this example). ************************************************************************** [end]