- Question:
- A sender transmits data frames
to a receiver using the
Stop-and-Wait protocol:
- What is the maximum channel utilization ?
- In other words: what fraction of time on the time line in the above figure is yellow (= data frames)
Note:
- The maximum channel utilization
is experienced
when there
are no transmission errors
- When there are transmission errors, there will be extra overhead to recover from the error(s)
(So the question implies that we must assume that there are no transmission errors)
- A sender transmits data frames
to a receiver using the
Stop-and-Wait protocol:
- Notations:
- tdata =
time to transmit the
data frame
- tprop =
time for the signal to
propagate from
the sender to the receiver
- This delay is called the propagation delay
- tACK = time to transmit the ACK frame
- tdata =
time to transmit the
data frame
- NOTE:
-
the ACK frame is
usually very short
compared to a
data frame
- Therefore:
tACK ------ << 1 tdata
- It is often assumed
that:
tACK ------ = 0 tdata
-
the ACK frame is
usually very short
compared to a
data frame
- Maximum channel utilization
of the
transmission channel:
- The following pattern
is repeated
indefinitely
when there are
no transmission errors:
- The duration
of each cycle is:
tdata + tprop + tACK + tprop = tdata + tACK + 2×tprop
- The amount of time in the
cycle used to transmit
data frames is:
tdata
- The fraction
of time used to
transmit
data frames is:
tdata Channel efficiency = --------------------- tdata + tACK + 2 tprop tdata ~= --------------- tdata + 2 tprop(tACK is very small and is usually discarded in the approximation)
- The following pattern
is repeated
indefinitely
when there are
no transmission errors: