CS455 Syllabus

Performance of the Stop-and-Wait protocol

• Efficiency: Channel utilization

• Definition: channel utilization

• Channel utilization = the fraction of the transmission capacity of a communication channel that contains data (frames) transmissions

• Another term for channel utilization:

 Channel efficiency or simply efficiency

• Maximum channel utilization of Stop-and-Wait

• 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 found when there are no transmission errors (The question implies that we assume that there are no transmission errors)

• Notations:

• tdata = time to transmit the message

• 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 message

 NOTE: the ACK message is usually very short compared to the message It is often ignored in the analysis...

• Utilization of the transmission channel:

• The following pattern is repeated indefinitely:

• 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)

• Important fact from Physics

• Facts:

 Speed of light in vacuum (or space) ~= 3×108 m/sec Speed of electrical signal in (copper) wire ~= 2×108 m/sec

• Channel utilization in Stop-and-Wait protocol for low speed and high speed links

• 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

In this case study, you will understand why the Stop-and-Wait protocol is no longer used today !

• Case study 1: efficiency of Stop-and-Wait on a low speed transmission link

• Find the channel utilization of Stop-and-Wait in the following transmission link:

Explanation:

 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 (but we do not ignore the propagation delay of ACK frames !)

• 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 ```

• 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)

• 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 ```

• 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:

 Approximately 1/5 of the time, the transmission link is used to transmit useful data Or: approximately 4/5 of the time, the cable modem channel is useless !!!

• 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.