- Fact:
- The Stop-and-Wait protocol
was very popular
in the 1990s
- Nowadays, the Stop-and-Wait protocol is not used at all !!!
- The Stop-and-Wait protocol
was very popular
in the 1990s
- A bit of history first:
- Back in 1995,
we used modems
from home to
connect workstation on
campus
- The transmission rate of modems were about 56 kbps
- Nowadays, you have
DSL and
Cable
Internet service that provides:
- 150 Mbps ~ 600 Mbps
After computing the maximum channel utilization of Stop-and-Wait, you will understand why the Stop-and-Wait protocol is no longer used today !
- Back in 1995,
we used modems
from home to
connect workstation on
campus
- Case study 1: efficiency of
Stop-and-Wait on a
low speed
transmission link
- Find the
maximum channel utilization
of Stop-and-Wait
in the following
transmission link:
Summary of the operational parameters:
- Data rate of the transmission link = 50 kbps (kilobits/sec)
- The length of the link = 10 km
- Sender transmits data frames
of size 1 kBytes
- We will ignore the
time needed to
transmit
an ACK frame
because: tACK << tdata
- Find the
maximum channel utilization
of Stop-and-Wait
in the following
transmission link:
- First:
compute
tdata,
tprop and
tACK.
- Compute
tdata:
1 data frame = 1 kbytes ~= 1000 bytes = 8000 bits Transmission speed = 50 kps ~= 50000 bits in one sec ===> time to transmit 1 data frame = 8000/50000 = 0.16 sec tdata = 0.16 sec
- Compute
tACK:
- We assume that tACK ~= 0
- Compute
tprop:
Signal speed = 2 × 108 m/sec = 2 × 105 km/sec Distance between sender and receiver = 10 km ===> time for signal to travel 10 km = 10 /(2 × 105) = 5 × 10-5 tprop = 5 × 10-5 sec = 0.00005 sec
- Compute
tdata:
- Maximum channel utilization:
0.16 Channel utilization = ---------------------- 0.16 + 2 × 0.00005 0.16 = ---------- 0.1601 = 99.94 %
- Case study 2: efficiency of
Stop-and-Wait on a
high speed
transmission link
- Now find the
channel utilization
of Stop-and-Wait
is the
data rate = 300 Mbps
(everything else
remains unchanged)
Summary of the operational parameters:
- Data rate of the transmission link = 300 Mbps (kilobits/sec)
- The length of the link = 10 km
- Sender transmits data frames
of size 1 kBytes
- We will ignore the
time needed to
transmit
an ACK frame
because: tACK << tdata
- Now find the
channel utilization
of Stop-and-Wait
is the
data rate = 300 Mbps
(everything else
remains unchanged)
- First:
compute
tdata,
tprop and
tACK.
- Compute
tdata:
1 data frame = 1 kbytes ~= 1000 bytes = 8,000 bits Transmission speed = 300 Mbps ~= 300,000,000 bits in one sec ===> time to transmit 1 data frame = 8,000/300,000,000 = 2.67×10-5 sec tdata = 2.67×10-5 sec
- Compute
tACK:
- We assume that tACK ~= 0
- Compute
tprop:
Signal speed = 2 × 108 m/sec = 2 × 105 km/sec Distance between sender and receiver = 10 km ===> time for signal to travel 10 km = 10 /(2 × 105) = 5 × 10-5 tprop = 5 × 10-5 sec
- Compute
tdata:
- Maximum channel utilization:
2.67×10-5 Channel utilization = ------------------------- 2.67×10-5 + 2 × 5×10-5 2.67 2.67 = ----------- = ------- 2.67 + 10 12.67 = 21.07 %In other words:
- At most:
- 21% of the channel bandwidth
will be used to transmit (useful) data frames
- Or:
- at least 79% of the channel bandwidth will be wasted !!!
- At most:
- Verdict:
- Stop-and-Wait
has good channel utilization
for low speed links
- It is very inefficient on high speed links
- Stop-and-Wait
has good channel utilization
for low speed links
- Need:
- To achieve
better performance,
we must
develop more
efficient (= complicated)
protocols.
- We will next study the Sliding Window method.
- To achieve
better performance,
we must
develop more
efficient (= complicated)
protocols.