CS455 Syllabus

### Performance of the Stop-and-Wait protocol in low and high data rate links

• History of the Stop-and-Wait protocol

• Fact:

 The Stop-and-Wait protocol was very popular in the 1990s Nowadays, the Stop-and-Wait protocol is not used at all !!!

• 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 !

• Performance of Stop-and-Wait on low data rate (bandwidth) links

• 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 ```

• 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 ```

• Maximum channel utilization:

 ``` 0.16 Channel utilization = ---------------------- 0.16 + 2 × 0.00005 0.16 = ---------- 0.1601 = 99.94 % ```

• Performance of Stop-and-Wait on high data rate (bandwidth) links

• 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 ```

• 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 ```

• 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 !!!

• Verdict:

 Stop-and-Wait has good channel utilization for low speed links It is very inefficient on high speed links

• More efficient reliability protocols

• Need:

 To achieve better performance, we must develop more efficient (= complicated) protocols. We will next study the Sliding Window method.