Ch1 delays, loss, and throughput l5

Delay, Loss, and Throughput in Packet-Switched Networks
Ideally, we would like Internet services to be able to move as much data as we want between any
two end systems, instantaneously, without any loss of data. Alas, this is a lofty goal, one that is
unachievable in reality. Instead, computer networks necessarily constrain throughput (the amount
of data per second that can be transferred) between end systems, introduce delays between end
systems, and can actually lose packets.
Overview of Delay in Packet-Switched Networks
As a packet travels from one node (host or router) to the subsequent node (host or router) along
this path, the packet suffers from several types of delays at each node along the path. The most
important of these delays are the nodal processing delay, queuing delay, transmission delay,
and propagation delay; together, these delays accumulate to give a total nodal delay.
Figure 1: The nodal delay at router A
Processing Delay
The time required to examine the packet’s header and determine where to direct the packet is
part of the processing delay. The processing delay can also include other factors, such as the
time needed to check for bit-level errors in the packet that occurred in transmitting the packet’s
bits from the upstream node to router A.
Queuing Delay
At the queue, the packet experiences a queuing delay as it waits to be transmitted onto the link.
The length of the queuing delay of a specific packet will depend on the number of earlier-
arriving packets that are queued and waiting for transmission onto the link. If the queue is empty
and no other packet is currently being transmitted, then our packet’s queuing delay will be zero.
On the other hand, if the traffic is heavy and many other packets are also waiting to be
transmitted, the queuing delay will be long.
Transmission Delay
Assuming that packets are transmitted in a first-come-first-served manner, as is common in
packet-switched networks, our packet can be transmitted only after all the packets that have
arrived before it have been transmitted. Denote the length of the packet by L bits, and denote the
transmission rate of the link from router A to router B by R bits/sec. For example, for a 10 Mbps
Ethernet link, the rate is R = 10 Mbps; for a 100 Mbps Ethernet link, the rate is R = 100 Mbps.
The transmission delay is L/R. This is the amount of time required to push (that is, transmit) all
of the packet’s bits into the link.

Propagation Delay
Once a bit is pushed into the link, it needs to propagate to router B. The time required to
propagate from the beginning of the link to router B is the propagation delay. The bit
propagates at the propagation speed of the link. The propagation speed depends on the physical
medium of the link (that is, fiber optics, twisted-pair copper wire, and so on) and is in the range
of
2 *108
meters/sec to 3 *108
meters/sec
which is equal to, or a little less than, the speed of light. The propagation delay is the distance
between two routers divided by the propagation speed. That is, the propagation delay is d/s,
where d is the distance between router A and router B and s is the propagation speed of the link.
Comparing Transmission and Propagation Delay
The transmission delay is the amount of time required for the router to push out the packet; it is a
function of the packet’s length and the transmission rate of the link, but has nothing to do with
the distance between the two routers. The propagation delay, on the other hand, is the time it
takes a bit to propagate from one router to the next; it is a function of the distance between the
two routers, but has nothing to do with the packet’s length or the transmission rate of the link.
For example, the time from when the caravan is stored in front of a tollbooth until the caravan is
stored in front of the next tollbooth is the sum of transmission delay and propagation delay
Figure 2: Caravan analogy
If we let dproc, dqueue, dtrans, and dprop denote the processing, queuing, transmission, and
propagation delays, then the total nodal delay is given by
dnodal = dproc + dqueue + dtrans + dprop
Queuing Delay and Packet Loss
If 10 packets arrive at an empty queue at the same time, the first packet transmitted will suffer no
queuing delay, while the last packet transmitted will suffer a relatively large queuing delay
(while it waits for the other nine packets to be transmitted). Therefore, when characterizing
queuing delay, one typically uses statistical measures, such as average queuing delay, variance of
queuing delay, and the probability that the queuing delay exceeds some specified value.
Packet Loss
With no place to store such a packet, a router will drop that packet; that is, the packet will be
lost.
From an end-system viewpoint, a packet loss will look like a packet having been transmitted into
the network core but never emerging from the network at the destination. The fraction of lost

packets increases as the traffic intensity increases. Therefore, performance at a node is often
measured not only in terms of delay, but also in terms of the probability of packet loss.
End-to-End Delay
Our discussion up to this point has focused on the nodal delay, that is, the delay at a single
router. Let’s now consider the total delay from source to destination. To get a handle on this
concept, suppose there are N -1 routers between the source host and the destination host. Let’s
also suppose for the moment that the network is uncongested (so that queuing delays are
negligible), the processing delay at each router and at the source host is dproc, the transmission
rate out of each router and out of the source host is R bits/sec, and the propagation on each link is
dprop. The nodal delays accumulate and give an end-to-end delay,
dend-end = N (dproc + dtrans + dprop)
where, once again, dtrans = L/R, where L is the packet size.
Traceroute
Traceroute is a simple program that can run in any Internet host. When the user specifies a
destination hostname, the program in the source host sends multiple, special packets toward that
destination. As these packets work their way toward the destination, they pass through a series of
routers. When a router receives one of these special packets, it sends back to the source a short
message that contains the name and address of the router.
When the destination host receives the Nth packet, it too returns a message back to the source.
The source records the time that elapses between when it sends a packet and when it receives the
corresponding return message; it also records the name and address of the router (or the
destination host) that returns the message.
Throughput in Computer Networks
To define throughput, consider transferring a large file from Host A to Host B across a computer
network. This transfer might be, for example, a large video clip from one peer to another in a
P2P file sharing system. The instantaneous throughput at any instant of time is the rate (in
bits/sec) at which Host B is receiving the file.
If the file consists of F bits and the transfer takes T seconds for Host B to receive all F bits, then
the average throughput of the file transfer is F/T bits/sec.
Consider the throughput for a file transfer from the server to the client. Let Rs denote the rate of
the link between the server and the router; and Rc denote the rate of the link between the router
and the client.
Clearly, the server cannot pump bits through its link at a rate faster than Rs bps; and the router
cannot forward bits at a rate faster than Rc bps. If Rs < Rc, then the bits pumped by the server will
“flow” right through the router and arrive at the client at a rate of Rs bps, giving a throughput of
Rs bps. If, on the other hand, Rc < Rs, then the router will not be able to forward bits as quickly as
it receives them. In this case, bits will only leave the router at rate Rc, giving an end-to-end
throughput of Rc.

Figure 3: Throughput for a file transfer from server to client
Thus, for this simple two-link network, the throughput is min{Rc, Rs}, that is, it is the
transmission rate of the bottleneck link.
For a specific example, suppose you are downloading an MP3 file of F = 32 million bits, the
server has a transmission rate of Rs = 2 Mbps, and you have an access link of Rc = 1 Mbps. The
time needed to transfer the file is then 32 seconds.
Figure 4: End-to-end throughput: (a) Client downloads a file from server;
(b) 10 clients downloading with 10 servers

Ch1 delays, loss, and throughput l5

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