§1.4.3–1.4.4End-to-End Delay · Throughput

Adding the delays along a path

Section 1.4.1 gave the delay at one router. Now the whole path.

Suppose there are $N-1$ routers between source and destination. Assume the network is uncongested, so queuing delays are negligible. Let every node have the same processing delay $d_{\text{proc}}$ and the same outgoing rate $R$, and let every link have the same propagation delay $d_{\text{prop}}$. Then the nodal delays accumulate into the end-to-end delay end-to-end delay Simple The total time a packet takes to travel from the sending host to the receiving host. Precise The accumulation of the nodal delays along the whole path. With N−1 routers between the hosts, negligible queuing, and identical nodes: d_end-end = N(d_proc + d_trans + d_prop), where d_trans = L/R. introduced in ch. 1 — open in glossary :

where $d_{\text{trans}} = L/R$.

d_end-end = N (d_proc + d_trans + d_prop)
d_proc  = 2.00 µs   d_trans = 120 µs   d_prop = 9.00 ms
per hop = 9.12 ms
        = 5 × 9.12 ms
        = 45.6 ms
(queuing is assumed negligible — the one term that would vary)

Traceroute: measuring a real path

Traceroute traceroute Simple A program that lists every router on the path to a destination and the delay to each. Precise A program that sends N specially marked packets toward a destination; the nth router along the path returns a message to the source instead of forwarding packet n. The source thereby reconstructs the route and measures the round-trip delay to each router. Described in RFC 1393. introduced in ch. 1 — open in glossary is a simple program that runs on any host. You give it a destination name, and it tells you every router on the way and how long each one took to reach.

The trick is neat. Suppose there are $N-1$ routers between source and destination. The source sends $N$ special packets toward the destination, marked 1 through $N$. When the $n$th router receives the packet marked $n$, it does not forward it. Instead it sends a short message back to the source, carrying its own name and address. When the destination receives packet $N$, it replies too.

The source records how long each reply took, and who sent it. From that it can reconstruct the whole route and the round-trip delay to every router along it. Traceroute repeats the experiment three times, so it actually sends $3N$ packets. It is described in RFC (Request For Comments) Request For Comments The name of an IETF standards document. There are currently nearly 9000 of them. introduced in ch. 1 1393.

An asterisk means the source received fewer than three replies from that router, because a probe or a reply was lost on the way.

Other delays, outside the routers

Two more delays exist that none of the four components covers.

A host transmitting into a shared medium, such as WiFi (IEEE 802.11 wireless LAN) IEEE 802.11 wireless LAN Wireless LAN access; a user must be within a few tens of metres of the access point. WiFi is a trade name, not an abbreviation. introduced in ch. 1 or a cable modem, may deliberately delay itself. Waiting is part of the protocol for sharing that medium with others. Chapter 6 covers those protocols.

And in VoIP (Voice over IP) Voice over IP Carrying telephone calls over a packet-switched network; it suffers an extra packetization delay. introduced in ch. 1 , the sending side must first fill a packet with encoded digitised speech before it can hand the packet to the Internet. That packetization delay can be significant, and it affects how good the call sounds.

Throughput

Delay is one measure. The other is throughput throughput Simple How many bits per second actually arrive at the receiver. Precise The rate, in bits/second, at which the receiving host receives data. Instantaneous throughput is that rate at an instant; the average throughput of a transfer of F bits taking T seconds is F/T. With no competing traffic it is min{R₁, …, R_N}. introduced in ch. 1 — open in glossary .

Consider transferring a large file from host A to host B.

• Instantaneous throughput is the rate, in bits per second, at which host B is receiving the file at that instant. Many applications display it during a download.
• Average throughput is $F/T$, where the file is $F$ bits and the transfer takes $T$ seconds.

Different applications care about different things. Internet telephony wants low delay and an instantaneous throughput consistently above a threshold — over 24 kbps for some telephony applications, over 256 kbps for some real-time video. For a file transfer, delay is not critical, but the highest possible throughput is.

Two links and a router

Either way, the throughput is the rate of the slower link:

That slowest link is the bottleneck link bottleneck link Simple The slowest link on the path; it sets the speed of the whole transfer. Precise The link along a path whose transmission rate is the minimum, and which therefore determines the end-to-end throughput. A link with a high transmission rate can still be the bottleneck if many other flows share it. introduced in ch. 1 — open in glossary . For a path of $N$ links with rates $R_1, \ldots, R_N$, the same argument gives $\min\{R_1, \ldots, R_N\}$.

throughput = min{Rs, Rc}
           = min{2.00 Mbps, 1.00 Mbps}
           = 1.00 Mbps   (the client access link is the bottleneck)
transfer   = F / throughput
           = 32,000,000 bits / 1.00 Mbps = 32.0 s

Why the access network is usually the bottleneck

Now put the server and the client on a real network. The server has an access link of rate $R_s$, the client has one of rate $R_c$, and the links in the core are much faster than both.

The core of today’s Internet is over-provisioned with high-speed links and experiences little congestion. So the core is a wide pipe, and throughput is again $\min\{R_s, R_c\}$.

Unless enough flows share one link

Ten servers and ten clients, with ten downloads happening at once, and one link in the core that all ten cross. Let its rate be $R$, and let it divide its rate equally among the ten.

With $R_s = 2$ Mbps, $R_c = 1$ Mbps and $R = 5$ Mbps, each download gets 5 ÷ 10 = 500 kbps of that shared link. That is less than either access link can carry, so the bottleneck has moved into the core.

each download gets R/M of the shared link
           = 5.00 Mbps / 10 = 500 kbps
throughput = min{Rs, Rc, R/M}
           = min{2.00 Mbps, 1.00 Mbps, 500 kbps}
           = 500 kbps   (bottleneck: the shared core link)

Check yourself