Standard delay and throughput model from introductory networking courses (Kurose and Ross, section 1.4). Every figure in the examples, practice and solutions was recomputed in code before publishing.
Learn the four delays as a set of four, with what each one depends on. Then do the worked examples with a calculator and write the units on every line. Delay questions are almost free marks once the units are under control.
The picture: loading a truck and driving it#
Imagine a warehouse shipping boxes by truck to the next warehouse down the highway.
- First a clerk checks the paperwork on each box. That is quick.
- If other trucks are already lined up at the loading dock, your boxes wait their turn.
- Then the boxes are loaded onto the truck one at a time. A bigger shipment takes longer to load, and a faster crew loads quicker.
- Finally the truck drives to the next warehouse. That time depends only on the distance and the speed of the truck. It does not care how many boxes are on board.
Those are the four delays a packet pays at every router: check, wait, load, drive. The formal names come next.
Vocabulary, before anything uses it#
| Term | What it means |
|---|---|
| Nodal delay | The total delay a packet picks up at one router (one “node”): all four delays added together. |
| L | Packet length, in bits. Convert bytes to bits by multiplying by 8. |
| R | Link transmission rate, in bits per second (bps). |
| d | Length of the physical link, in meters. |
| s | Propagation speed of the signal in the medium, roughly 2 × 108 to 3 × 108 meters per second. |
| a | Average packet arrival rate at a router, in packets per second. |
| Traffic intensity | La/R: how many bits arrive per second, divided by how many bits the link can send per second. |
| Throughput | The rate, in bits per second, at which the receiver actually gets data. |
| Bottleneck link | The slowest link on the path. It limits the throughput of the whole path. |
The four delays at every router#
Add these four up and you get the nodal delay:
dnodal = dproc + dqueue + dtrans + dprop
| Delay | What happens | Depends on | Truck picture |
|---|---|---|---|
| Processing, dproc | Check the header for bit errors and decide which output link to use. | Router speed. Typically microseconds. | Checking the paperwork |
| Queuing, dqueue | Wait in the buffer until the output link is free. | How busy the link is. Anywhere from zero to large. | Waiting at the dock |
| Transmission, dtrans | Push every bit of the packet onto the link. | L / R: packet size and link rate. | Loading the truck |
| Propagation, dprop | Let one bit travel from one end of the link to the other. | d / s: link length and signal speed. | Driving the truck |
Transmission is L/R. Propagation is d/s. Transmission has nothing to do with distance: a 1,000 km link and a 1 m link at the same rate take the same time to load a packet onto. Propagation has nothing to do with packet size: one bit and a million bits both take d/s to cross. If a question changes the distance, only propagation changes. If it changes the packet size or the link rate, only transmission changes.
Two quick calculations
Transmission: a 1,500-byte packet on a 10 Mbps link. First convert bytes to bits: 1,500 × 8 = 12,000 bits.
dtrans = 12,000 / 10,000,000 = 0.0012 s = 1.2 ms
Propagation: a 2,500 km link where signals travel at 2.5 × 108 m/s. Convert km to meters: 2,500,000 m.
dprop = 2,500,000 / 250,000,000 = 0.01 s = 10 ms
Queuing delay, traffic intensity and loss#
Queuing delay is the odd one out: it changes from packet to packet. The first packet into an empty buffer waits zero. The tenth packet in a burst waits for nine others. So we describe it with a single ratio instead of a fixed number.
Traffic intensity = La / R
Read it as “bits arriving per second, divided by bits the link can send per second”. Take 12,000-bit packets on a 2 Mbps link:
| Arrival rate a | La/R | What happens |
|---|---|---|
| 100 packets/s | 12,000 × 100 / 2,000,000 = 0.6 | Queues form in bursts but clear. Average delay is modest. |
| 150 packets/s | 12,000 × 150 / 2,000,000 = 0.9 | Close to 1. Average queuing delay grows very fast. |
| 200 packets/s | 12,000 × 200 / 2,000,000 = 1.2 | More than 1. The queue grows without limit and packets are lost. |
The rule to remember: design so that La/R stays below 1. As it approaches 1, the delay does not rise gently; it shoots up.
Packet loss
A real buffer has a fixed size. When a packet arrives and the buffer is full, the router drops it. That packet is lost. It may be sent again by the previous node, by the sending host, or not at all, depending on the protocols in use.
A full buffer does not slow down, it throws away. Loss is what heavy queuing turns into once the waiting room is full.
Seeing delay with traceroute
Traceroute is a program that measures the delay to each router along a path. For each router, it sends three small probe packets and times how long each takes to come back. The output is one line per router, with three round-trip times on each line. Asterisks (*) mean a probe got no reply, often because the router was told not to answer. Times do not always rise hop by hop, because each probe meets different queues.
Worked example 1: end-to-end delay over three links#
A 1,500-byte packet goes from a host through two routers to a destination: three links in total. Each link is 10 Mbps and 1,000 km long, and signals travel at 2 × 108 m/s. Ignore processing and queuing. How long from the first bit leaving the host to the last bit arriving?
Step 1: Transmission per link. 1,500 × 8 = 12,000 bits.
12,000 / 10,000,000 = 1.2 ms
Step 2: Propagation per link.
1,000,000 m / 200,000,000 m/s = 5 ms
Step 3: Add per link, then multiply by the number of links. Store-and-forward means each router receives the whole packet before sending it on, so every link charges both delays again.
3 × (1.2 + 5) ms = 3 × 6.2 ms = 18.6 ms
Bytes converted to bits, both delays computed separately with units, and the count of links (three), not routers (two). Counting routers instead of links is the classic slip.
Throughput and the bottleneck link#
Delay is how long one packet takes. Throughput is how fast data actually arrives, in bits per second. A garden hose makes the idea concrete: join a wide hose to a narrow one, and the water comes out at the rate of the narrow one. The wide part does not help.
For a path made of a server link of rate Rs and a client link of rate Rc:
Throughput = min(Rs, Rc)
With more links, it is the minimum over all of them. The link that sets the minimum is the bottleneck link. To estimate how long a big file takes, divide its size by the throughput.
Two throughput words exist: instantaneous throughput is the rate at one moment, and average throughput is the file size divided by the total time taken.
A server with a 4 Mbps link sends a 40-million-bit file to a client with a 1.5 Mbps link. Throughput = min(4, 1.5) = 1.5 Mbps. Time ≈ 40,000,000 / 1,500,000 ≈ 26.7 s. The server's extra speed changes nothing.
When a link in the core is shared
Today, the bottleneck is usually at the edge, because the core is so fast. But a shared core link can be the bottleneck too. If 5 downloads share an 8 Mbps backbone link fairly, each gets 8 / 5 = 1.6 Mbps of it. With server links of 3 Mbps and client links of 2 Mbps:
Per-download throughput = min(3, 2, 8/5) = 1.6 Mbps
Here the shared backbone, not either access link, is the bottleneck.
Transmission is L over R and ignores distance, propagation is d over s and ignores packet size, and throughput is set by the slowest link on the path.
Practice#
Write the formula, then the numbers with units, then the answer. In that order, every time.
- Name the four components of nodal delay and what each one depends on.
- A 1,000-byte packet goes onto a 1 Mbps link. What is its transmission delay?
- The same link is 5,000 km long with propagation speed 2.5 × 108 m/s. What is its propagation delay?
- If the link in question 3 were twice as long, which delay changes, and which one does not?
- Packets of 8,000 bits arrive at a 1 Mbps link at an average of 100 per second. What is the traffic intensity, and what does it tell you?
- Why does a traffic intensity close to 1 cause a problem even though the link is not yet overloaded?
- A server has a 10 Mbps link and a client has a 2 Mbps link. What is the throughput, and which link is the bottleneck?
- 10 downloads share a 20 Mbps core link fairly. Each server link is 5 Mbps and each client link is 2 Mbps. What is each download's throughput?
- A traceroute line shows three times, and a later line shows smaller times than an earlier one. Why is that possible?
Solutions#
Find the first line where your working split from the solution, and redo the problem from that line. Do not just read the answer and nod.
- Processing, queuing, transmission, propagation. Processing depends on the router (checking the header and choosing an output). Queuing depends on how busy the link is. Transmission is L/R, packet size over link rate. Propagation is d/s, link length over signal speed.
- 8 ms. 1,000 × 8 = 8,000 bits. 8,000 / 1,000,000 = 0.008 s = 8 ms.
- 20 ms. 5,000 km = 5,000,000 m. 5,000,000 / 250,000,000 = 0.02 s = 20 ms.
- Propagation doubles; transmission is unchanged. Propagation becomes 40 ms because it depends on distance. Transmission stays 8 ms because it depends only on packet size and link rate.
- 0.8. La/R = 8,000 × 100 / 1,000,000 = 0.8. It is below 1, so the queue will not grow forever, but it is high enough that noticeable queuing delay is likely during bursts.
- Delay grows very fast near 1. Arrivals come in bursts. When the average load is close to the link's capacity, there is almost no spare time to clear a burst, so queues build up and the average queuing delay rises sharply.
- 2 Mbps; the client link. Throughput = min(10, 2) = 2 Mbps. The client's 2 Mbps link is the bottleneck.
- 2 Mbps. The core share is 20 / 10 = 2 Mbps. min(5, 2, 2) = 2 Mbps. The client link and the core share tie, so either can be called the bottleneck.
- Each probe meets different queues. Queuing delay changes from moment to moment, so a probe to a nearer router can happen to wait longer than a later probe to a farther one.
Test-readiness checklist#
- ☐ I can list the four delays and give the formula or cause for each.
- ☐ I can compute L/R after converting bytes to bits.
- ☐ I can compute d/s after converting kilometers to meters.
- ☐ I can say which delay changes when distance, packet size, or link rate changes.
- ☐ I can compute La/R and explain what happens near 1 and above 1.
- ☐ I can explain why packets are lost and what traceroute shows.
- ☐ I can compute end-to-end delay across several links, counting links, not routers.
- ☐ I can find throughput as the minimum link rate, including a shared core link.
Convert first. Bytes to bits, kilometers to meters, milliseconds only at the end. A correct formula with a unit slip still gets the wrong number, and delay questions are marked on the number.
Frequently asked questions#
What are the four types of delay in a computer network?
Processing delay (checking the header and choosing an output link), queuing delay (waiting in the router's buffer), transmission delay (L/R, the time to push all the packet's bits onto the link), and propagation delay (d/s, the time for a bit to travel the length of the link).
What is the difference between transmission delay and propagation delay?
Transmission delay is L/R and depends on packet size and link rate, not distance. Propagation delay is d/s and depends on link length and signal speed, not packet size.
What is traffic intensity?
La/R: the average rate at which bits arrive at a link divided by the rate the link can send them. Below 1 the queue stays bounded; near 1 the average queuing delay grows very quickly; above 1 the queue grows without limit and packets are lost.
How do you calculate end-to-end throughput?
It is the rate of the slowest link on the path, the bottleneck. For a server link Rs and client link Rc it is min(Rs, Rc). If a core link of rate R is shared fairly by n connections, include R/n in the minimum.
Why are packets lost in a network?
Router buffers have a fixed size. When packets arrive faster than the output link can send them and the buffer fills, newly arriving packets are dropped.
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