1 EE:450 – Computer Networks Discussion Session #32 Some Terminology n Bit (b) n Basic unit of information in computers n Binary : 0 or 1 n Byte (B) n 8 bits in one byte n Bit Rate n Number of bits transmitted in a time unit n Typical unit is bits-per-second (bps) n Used to measure transmission speed in digital transmissions3 Terminology continued… n 1K Bytes = 1000 Bytes = 8000 bits n Similarly, 1M Bytes = 1,000,000 Bytes However, n 1 Kbps ≠ 210 bps 1 Kbps = 1000 bps n Similarly, 1 Mbps = 106 bps In this course, the approximation 1KB ~ 1000 Bytes is always allowed4 Terminology n Delay/Latency: Time it takes a message to travel from one end of a link to another n It is a very important performance parameter n End to End delay consists of several components n Transmission time n Propagation delay n Nodal Processing time n Queuing delay5 Transmission time n How long does it take to transmit a message (usually in KB) over a link with bit rate (usually in Mbps)? n Steps: n 1. Convert message size to bits n 1KB = 1000 bytes n 1MB = 1,000,000 bytes n 1 Byte = 8 bits n Key is the difference between “B” and “b”6 Transmission time ctd. n 2. To obtain the transmission time, divide the message size (in bits) by the bit rate a.k.a. bandwidth (in bps) Transmission time = Message size/Bit rate7 Transmission time example Ex: How long does it take to transmit a 4KB file over a link with 1Mbps bandwidth? Solution: n Step 1: Convert the file size to bits 4 KB = 4 x 1000 Bytes = 4000 Bytes = 32000 bits 1Mbps = 106 bps n Step 2: Transmission Time = file size / bandwidth ttrans = 32000 bits / 106 bps = 32ms8 Propagation delay n Propagation delay: The time it takes for a bit to traverse from one end of the link to the other end tprop = Link length (m) / Vprop (m/s) Where Vprop is the speed with which the bit travels in the medium - same as the speed of light in the given medium9 Propagation delay example Ex: What is the propagation time of a message in a link of 2.5 Km long? The speed of light in the cable is 2.3 x 108 m/s. Solution: tprop = Link length/ Vprop = 2500 m / 2.3 x 108 m/s = 10.9 µs Attention: tprop is independent of message size and bit rate of the link.10 Message Transfer Time n Message transfer time (txfr) : Time taken from the point when the sender starts transmitting the message till the receiver receives the entire message. Also known as end – to – end delay txfr = ths + ttrans + tprop + tqueuing/processing Where: n ths is the handshake time (time it takes for the initial connection establishment phase) n tqueuing/processing is the queuing and processing delay in the network. n We will assume the latter as zero most of the time.11 Round Trip Time (RTT) n Round Trip Time: The time to send a message from a sender to the receiver and receive a response back n Depends on the message size, length of link, direction of propagation, propagation velocity (speed), node processing delay, network traffic load etc. n We will assume RTT = 2 x tprop n May not be true if the message and the response choose different links to traverse n The other delay components are ignored here.12 Bit Duration n Bit Duration: duration (in time) of a pulse representing a bit – depends on bit rate (bandwidth) of the link. n Bit Duration= 1 / Bandwidth n A bit is 1 µs wide in a 1 Mbps channel 1/(106 bps) = (1 x 10-6) seconds per bit n A bit is 0.5 µs wide in a 2 Mbps channel13 Bit Length(bit Width) n Bit length: The length occupied by a bit on a transmission link n Bit length = Bit durationx Prop. Speed = (sec) x (meters/sec) = (meters)14 Bandwidth Delay Product n Product of Bandwidth and link latency (propagation delay) n Represents the maximum number of bits present in the link at given time n Analogy n A Pipe: delay is the length : bandwidth is the width n Bandwidth Delay product gives the volume15 Example #1 Ex: A terminal sends a 1 MB file to another computer through a link of 10 Mbps. The distance between the two terminals is 2000 Km and the propagation speed in the cable is 2x108 m/s. a) What is the RTT? b) What is the Bandwidth Delay Product? (Use RTT as the delay) c) What is the bit duration? d) Assume a handshake period of 2 RTT’s and no processing/queuing delay, what is the total transfer time of the file?16 Example contd… a) RTT = 2 tprop tprop = 2 x 106 m / 2 x 108 m/s = 10 msec. Therefore, RTT = 20 msec b) Bandwidth X Delay = 10 Mbps x 20 ms = 200000 bits ~ 25000 Bytes = 25 KB17 Example contd… c) Bit duration= 1 / Bandwidth = 1 / (10 Mbps) = 10-7 sec./bit = 0.1µs/bit d) txfr = ths + ttrans + tprop ttrans = 1 MB / (10 Mbps) = 8 Mb / (10 Mbps) = 800 msec ths = 2RTT = 40 msec tprop = 10 msec txfr = 40 + 800 + 10 = 850 msec18 Example #2: Bandwith or Delay Sensitive? n For each of the Following operations on a remote file server, discuss whether they are more likely to be delay-sensitive or bandwidth-sensitive: n Open a file n Read the contents of a file n List the contents of a directory n Display the attributes of a file19 Solution n Delay-sensitive; the messages exchanged are short. n Bandwidth-sensitive, particularly for large files. (Technically this does presume that the underlying protocol uses a large message size or window size; stop-and-wait transmission (as in Section 2.5 of the text) with a small message size would be delay-sensitive.) n Delay-sensitive; directories are typically of modest size. n Delay-sensitive; a file’s attributes are typically much smaller than the file itself (even on NT file systems).20 Example #3 n Hosts A and B are each connected to a switch via 10 Mbps links as shown in the figure. The propagation delay on each link is 20µs. S is a store and forward device; it begins transmitting a received packet 35µs after it has finished receiving it. Calculate the total time required to transmit 10,000 bits from A to B n As a single packet n As 2 5000-bit packets sent one right after another21 Solution A