Core CS · Computer Networks

Ten milliseconds that no amount of bandwidth will buy back

Q: Name the four components of end-to-end delay and say what each one depends on.

Transmission delay is L / R , packet size over the rate of the link it is being pushed onto, so it depends on packet size and link rate and not at all on distance. Propagation delay is d / v , distance over the signal speed in the medium, roughly 2 × 10⁸ m/s in copper and fibre, and it depends on nothing but distance. Queuing delay is time waiting in a router’s outbound buffer and depends on how busy that link already is. Processing delay is the header check and next-hop lookup at each node, a small near-constant per router.

Bandwidth, throughput and latency are three separate measurements that students collapse into the word “speed” and interviewers pull apart in one question. Push a single 1500-byte packet from Delhi to Chennai with the units written down at every step, and the four delays stop being a list you memorised.

Set the packet size, link rate and distance, then step the sum →

Bandwidth is how many bits per second the link can carry. Latency is how long one bit’s journey takes. Widening a pipe does not shorten it, which is exactly why a satellite link is high bandwidth and high latency at the same time.

01 The idea

Four numbers, and only one of them is a rate you can buy

A student says “my internet is slow” and means one of at least three unrelated things. The pipe is narrow. The pipe is wide but nothing is filling it. Or everything arrives perfectly and each thing takes a quarter of a second to start. Networking gives those three complaints three different names, three different units and three different fixes. Using one word for all of them is the fastest way to lose a networking round.

Bandwidth is a property of the link. It is the number of bits the link can carry per second, fixed by the hardware and the standard: Fast Ethernet is 100 Mbps whether you are sending anything or not, Gigabit Ethernet is 1000 Mbps. Bandwidth is a capacity, not a speed, and the pipe picture is the right one. A wider pipe carries more water per second. It does not make the water arrive sooner.

Throughput is what you actually measured on a real transfer. It is also in bits per second, so it is directly comparable with bandwidth, and it is always at or below it. Below, because of the sender’s window, a slower link further along the path, competing traffic, and anything that had to be sent a second time. Strip out the protocol headers and the retransmissions and count only the application bytes that arrived once and stayed, and that number is called goodput. It is the only one of the three the user ever feels.

Latency is a duration, measured in milliseconds. It is not a rate and it is not comparable with the other two at all. It is how long one bit’s journey takes, and it is made of four separate delays: the time to push the packet onto the wire, the time for the signal to physically cross the distance, the time spent waiting in a router’s buffer, and the time each node spends reading the header. Section 03 gives each one its formula and, more usefully, tells you what each one ignores.

Keeping them apart matters because they respond to different money. Bandwidth is bought. Latency is mostly geography, and the one part of it that is not geography is the part that changes every time the network gets busy.

Bandwidth is how wide the pipe is. Throughput is how much actually came out of it. Latency is how long the trip took. You can buy a wider pipe. You cannot buy a shorter trip.

BandwidthThe capacity of a link in bits per second, fixed by the hardware and the standard. Link rates use powers of ten, so 1 Mbps is 10⁶ bits per second and not 2²⁰. It exists whether traffic is flowing or not.

Throughput and goodputThroughput is the rate you actually achieved, in bits per second, always at or below bandwidth. Goodput is throughput with the headers and the retransmissions taken out, so it counts only application bytes that arrived once and stayed.

LatencyHow long one bit’s journey takes, in milliseconds. Transmission plus propagation plus queuing plus processing. Round-trip time is roughly two one-way latencies plus whatever the responder takes to answer.

02 Worked example

One 1500-byte packet, Delhi to Chennai, with every unit written down

This one path carries the whole lesson: the flow below, the console in section 04, and every number in the cheat sheet. It is a plain fibre route between two data centres, and the settings are the second preset in the console, so you can reproduce every line of arithmetic on screen.

Quantity	Symbol	Value on this path
Packet size	L	`1500 bytes`, one full Ethernet MTU
Rate of every link	R	`1 Gbps = 1 × 10⁹ bit/s`
End-to-end distance	D	`2000 km` of fibre
Signal speed in fibre	v	`2 × 10⁸ m/s`, about two-thirds of the speed of light in vacuum
Links in the path	N	`4`, so 3 routers sit between the two hosts
Processing at each router	d_proc	`0.020 ms`, given
Queuing at each router	d_queue	`0.100 ms`, given, a lightly loaded path

Those seven givens produce four delays. Read the flow left to right in the order the packet actually meets them at each node, and watch which one is enormous.

ProcessingThe router reads the header and looks up the next hop. 0.020 ms each, 3 routers. 0.060 ms

→

QueuingWaiting behind whatever is already in the outbound buffer. 0.100 ms each, 3 routers. 0.300 ms

→

TransmissionPushing all 12,000 bits onto the wire at 1 Gbps. 0.012 ms per link, 4 links. 0.048 ms

→

PropagationThe bits crossing 2,000,000 m at 2 × 10⁸ m/s. Paid once, and it is 96.1% of the trip. 10.000 ms

→

One way0.048 + 10.000 + 0.060 + 0.300, all in ms. 10.408 ms

Now the arithmetic, one substitution at a time. Every line shows the expression with the numbers put in before it shows a result, because a result you cannot re-derive is a result you will not reproduce in an interview.

Put everything into base unitsBytes to bits, Mbps to bits per second, kilometres to metres. Link rates are always quoted in bits and file sizes in bytes, and that factor of eight is where most of these sums die.L = 1500 B × 8 bit/B R = 1000 Mbps × 10⁶ D = 2000 km × 1000 m/kmL = 12,000 bits R = 1,000,000,000 bit/s D = 2,000,000 m

Transmission delay, paid once on every linkHow long the sender takes to push the whole packet onto the wire. It depends on the packet size and on the rate of the link it is being pushed onto. Distance does not appear in it at all. Each router pushes the same packet onto its own outgoing link, so a 4-link path pays it four times.d_trans = L / R = 12,000 bits ÷ 1,000,000,000 bit/s= 0.000012 s = 0.012 ms per link → 4 links × 0.012 ms = 0.048 ms

Propagation delay, paid once over the whole distanceHow long the signal takes to physically cross the fibre. It depends only on distance and the medium. Packet size and link rate appear nowhere in it, which is the single most useful fact in this lesson.d_prop = D / v = 2,000,000 m ÷ 200,000,000 m/s= 0.01 s = 10.000 ms, paid once for the whole path

Processing delay, once at every routerHeader check and next-hop lookup. A four-link path has three routers between the two hosts, and the two hosts themselves are not counted here.d_proc total = (N − 1) × 0.020 ms = 3 × 0.020 ms= 0.060 ms

Queuing delay, once at every router, and the only one that movesTime spent in the outbound buffer behind packets that arrived first. This is the only one of the four that changes when the network gets busy, and it is the reason latency is unstable under load.d_queue total = (N − 1) × 0.100 ms = 3 × 0.100 ms= 0.300 ms on this lightly loaded path (the same three routers congested at 5.000 ms each would add 15.000 ms)

Add the fourSame units on every term, so the addition is safe.0.048 ms + 10.000 ms + 0.060 ms + 0.300 msone-way latency = 10.408 ms, of which propagation alone is 10.000 ms = 96.1%

Round tripRTT is what ping reports. It is roughly two one-way latencies. Roughly, because the return path may take a different route, the queues in the two directions differ at any instant, and the responder needs time of its own to answer.RTT ≈ 2 × 10.408 ms= 20.816 ms, before the responder has done any work

Bits in flight: the bandwidth-delay productMultiply the link rate by the round-trip time and you get how many bits the sender can have outstanding before the first acknowledgement could possibly return. That is the volume of the pipe, and it is the size the sending window must reach for the link to be busy rather than idle.BDP = R × RTT = 1,000,000,000 bit/s × 0.020816 s= 20,816,000 bits = 2,602,000 bytes ≈ 2.60 MB ≈ 1,735 packets of 1500 bytes

Look at where the time went. 10.000 ms of the 10.408 ms is propagation, and propagation is the one term in the list that a faster link cannot touch. Swap every 1 Gbps link for 10 Gbps and transmission falls from 0.048 ms to 0.0048 ms, so the one-way delay moves from 10.408 ms to 10.365 ms. That is a 0.4% improvement for a ten times faster network. The console in section 04 lets you make exactly that change and watch the total refuse to move.

The mirror image matters as much. Run the same 1500-byte packet across an office LAN at 100 Mbps over 200 m, two links through one idle switch, and the numbers invert. Transmission becomes 2 × 0.120 ms = 0.240 ms, propagation becomes 0.001 ms, and with the switch’s 0.020 ms of processing and no queuing the one-way delay is 0.261 ms, of which transmission is 92.0%. Here ten times the bandwidth genuinely helps: the one-way delay falls to 0.045 ms, an improvement of 82.8%. Those are the first preset in the console, so you can step them line by line too. Neither situation is the general rule. The ratio between packet size, link rate and distance decides which term wins, and that is the thing the next section makes systematic.

03 Mechanics

What each number measures, what it depends on, and what it ignores

Three of the four words in this lesson are rates in bits per second and one is a duration in milliseconds. Getting that straight is half the topic, so start there before touching a formula.

Term	What it measures	Unit	Its ceiling	The lever that moves it
Bandwidth	The capacity of one link. A property of the hardware, present with no traffic on it at all.	`bit/s`	The standard: Fast Ethernet 100 Mbps, Gigabit Ethernet 1000 Mbps, 802.11 and cellular vary by mode.	Buy a faster link. Nothing you do in software raises it.
Throughput	The rate you actually achieved end to end on a real transfer.	`bit/s`	`the narrowest link on the path`	Remove whichever thing is actually limiting: the sending window, congestion, or that one slow hop.
Goodput	Application bytes delivered per second, once headers and resent data are excluded.	`bit/s`	`throughput, always strictly below it`	Bigger payloads behind the same fixed headers, and fewer losses to retransmit.
Latency	The time one bit takes to get there. Not a rate, and not comparable with the three above.	`ms`	`none; it is a duration, so nothing caps it`	Move the endpoints closer. Of its four parts, only queuing responds to anything else.

Now split latency into its four parts. The column that wins interviews is not the formula column, it is the one that says what each delay ignores, because almost every wrong answer in this topic comes from putting a variable into a formula it does not belong in.

Component	Formula	Depends on	Ignores completely	On our path	Moves with load?
Transmission	`L / R`	Packet size L and the rate R of the link it is being pushed onto, which is the sender’s outgoing link, then each router’s.	Distance. Entirely.	`0.012 ms per link, 0.048 ms over 4`	`no`
Propagation	`d / v`	Distance d and the medium. v is about 2 × 10⁸ m/s in copper and fibre and 3 × 10⁸ m/s for radio through near-vacuum.	Packet size and link rate. Entirely.	`10.000 ms`	`no`
Queuing	No closed form. Grows with traffic intensity `La/R`.	How much is already in the outbound buffer, so on everybody else’s traffic and not on yours.	Nothing useful. This is the term that knows about the rest of the world.	`0.300 ms light, 15.000 ms congested`	`yes, the only one`
Processing	A per-router constant.	Header check and next-hop lookup, so on router hardware and table size.	Packet size, for all practical purposes.	`0.020 ms each, 0.060 ms over 3`	`barely`

Why transmission is paid once per link and propagation is not. Routers are store-and-forward devices: a router does not begin sending a packet onto its outgoing link until it has received the last bit of that packet and checked it. So the packet is pushed onto a wire once at the source and once again at every router, which is why an N-link path costs N × (L / R) and not one L / R. Propagation behaves the opposite way: it is charged over the total distance, and splitting 2000 km into four 500 km links or one 2000 km link gives the same 10 ms either way. Put the two together with the per-router terms and the whole path is d_total = N(L/R) + D/v + (N−1)(d_proc + d_queue) when every link runs at the same rate. When the links differ, drop the shortcut and add L/R separately for each one.

Why queuing is the only component that moves, and why it moves badly. A router’s outbound queue fills when bits arrive faster than the link can drain them. The measure is traffic intensity, La/R, where L is the packet size in bits, a is the arrival rate in packets per second and R is the link rate in bits per second. While that ratio stays well under 1 the queue stays short and queuing delay is negligible. As it approaches 1 the average queue length grows without bound, so delay does not degrade gently, it falls off a cliff. Push it past 1 and the buffer overflows and packets are dropped instead of delayed. That is why the same ping to the same server reads 12 ms at 3 a.m. and 300 ms at 9 p.m. with nothing about the cables having changed, and why oversized buffers make it worse rather than better, a failure with its own name: bufferbloat.

Round-trip time is not a clean doubling. RTT ≈ 2 × one-way + the responder’s own service time. Three things spoil the exact doubling: the return route may differ in distance and hop count, the queues in the two directions are not the same at the same instant, and the far end has to actually produce an answer. ping reports the RTT and never the one-way delay, so you cannot halve a ping and call it a one-way latency unless you know the path is symmetric. This is also the reason one-way delay measurement needs synchronised clocks at both ends and is genuinely hard.

The bandwidth-delay product is the volume of the pipe. BDP = R × RTT, and its unit is bits, not bits per second, because a rate multiplied by a time is a quantity. It is how many bits can be on the wire and unacknowledged at one moment. Its use is window sizing: a sender allowed at most W bits outstanding gets at most W / RTT of throughput regardless of what the link can carry. On our path, the largest window the original 16-bit TCP window field can express is 65,535 bytes, which is 524,280 bits, and 524,280 ÷ 0.020816 s = 25.2 Mbps. That is 2.5% of a 1 Gbps link, and nothing is broken. This is exactly what the window scaling option of RFC 7323 exists to fix, and it is why a link that is both fast and long is the case where window size matters most.

The two conclusions the arithmetic actually delivers. First: adding bandwidth never reduces propagation delay. Only the L / R term has an R in it, so a ten times faster network changes one of four terms and leaves the other three untouched. A geostationary satellite link makes the point brutally: about 239 ms one way from the distance alone, with a transponder happily carrying tens of megabits per second, so it is high bandwidth and high latency in the same breath. Second: for a small packet over a long distance, propagation dominates and the link rate is nearly irrelevant, while for a large transfer over a short distance the reverse holds and bandwidth is almost the whole answer. The interview-safe version is that the ratio of packet size to link rate against the ratio of distance to signal speed decides which one you are in, and you can compute it in ten seconds.

05 Cheat sheet

The numbers they ask you to produce out loud

Every row is something an interviewer can ask you to state or compute in under ten seconds, with no paper. The right-hand column is the specific wrong answer that gets given, not a general warning.

What they ask	The answer	The trap
Transmission delay	`d_trans = L / R`	Putting distance in it, or using the receiver’s download rate instead of the rate of the link the packet is being pushed onto.
Propagation delay	`d_prop = d / v, with v ≈ 2 × 10⁸ m/s in copper and fibre`	`using 3 × 10⁸ m/s, which is only right for radio through vacuum`
Which component depends on packet size	`transmission only`	Claiming a bigger packet takes longer to propagate. It does not; only the pushing takes longer.
Which component varies with load	`queuing only`	`blaming bandwidth for a ping that changes during the day`
1500 bytes onto a 1 Gbps link	`1500 × 8 = 12,000 bits ÷ 10⁹ = 0.012 ms`	Forgetting the × 8 and being wrong by a factor of eight.
2000 km of fibre	`2 × 10⁶ m ÷ 2 × 10⁸ m/s = 10 ms`	Dividing kilometres by metres per second and landing 1000 times off.
Our four-link path, one way	`0.048 + 10.000 + 0.060 + 0.300 = 10.408 ms`	`paying transmission once for the path instead of once per link`
Round-trip time on it	`RTT ≈ 2 × 10.408 = 20.816 ms`	Treating it as exactly double, with no responder time and an assumed symmetric route.
Bandwidth-delay product on it	`10⁹ × 0.020816 = 20,816,000 bits ≈ 2.60 MB`	`multiplying by the one-way delay instead of the round trip`
Throughput ceiling of a fixed window W	`W / RTT`	Quoting the link rate instead, when a sender that waits for an acknowledgement before sending more never gets close to it.
Mbps against MB	`1 Mbps = 10⁶ bit/s and 1 MB = 8 × 10⁶ bits`	Dividing a link rate by 1024. Link rates are powers of ten; only memory sizes are powers of two.
Goodput	`throughput minus headers and retransmissions`	Using it as a synonym for throughput, then being asked which of the two a packet capture shows.

A rate and a duration are not comparableBandwidth is bit/s, latency is ms. The only legitimate way to combine them is to multiply, and that product is the bandwidth-delay product measured in bits. Any sentence that adds or compares them directly is wrong.

Only queuing movesPacket size, link rate, distance and router hardware do not change between 3 a.m. and 9 p.m. If a measured latency changes during the day, it is queuing until you have proven otherwise, and the fix is capacity or scheduling, never a faster cable.

Bits, bytes and powers of tenDivide by 8 to go from bits to bytes. Link rates use powers of ten, so 1 Gbps is exactly 10⁹ bit/s. Memory sizes use powers of two, so 1 GiB is 2³⁰ bytes. Mixing the two conventions in one sum is the classic error here.

06 Where & why

Where these four numbers are decisions somebody already made

None of this is arithmetic invented for exams. Each of the four systems below exists because somebody worked out which term dominated and then attacked that term specifically.

Starlink

Lower orbit is the only way to cut propagation

A geostationary satellite sits 35,786 km up, so the up-and-down path is 71,572 km and radio at 3 × 10⁸ m/s takes about 239 ms one way whatever the modem cost. Starlink’s first shell orbits at roughly 550 km, so the same up-and-down path is about 1100 km and about 3.7 ms. The radios did not get faster. The distance got smaller, and that is the only lever propagation has.

Cloudflare

A CDN is a latency fix wearing a caching costume

Propagation is distance divided by a speed nobody can raise, so the only remaining lever is distance. Cloudflare answers your request from a data centre in your own city and uses anycast so the same address routes to the nearest one, collapsing d from thousands of kilometres to tens. The bytes returned are identical; only the geography changed. This is also why “buy more bandwidth” never fixes a site that feels sluggish.

RFC 7323 · TCP window scaling

The bandwidth-delay product is a real configuration number

The window field in a TCP header is 16 bits, so unaided a sender may have at most 65,535 bytes outstanding, which on a 20.816 ms round trip caps one connection near 25 Mbps no matter what the link carries. RFC 7323 negotiates a scale factor in the handshake, up to 14, lifting the ceiling to 65,535 × 2¹⁴ = 1,073,725,440 bytes. Linux enables it by default via net.ipv4.tcp_window_scaling and sizes the buffers from net.ipv4.tcp_rmem.

ping and iperf3

Two tools, because they measure two different things

ping sends one small packet and reports the round-trip time, so it measures latency and is almost unaffected by bandwidth. iperf3 opens a stream and reports bits per second, so it measures throughput and says nothing about the trip time. Run both: a high ping with healthy iperf3 numbers is a distance problem, while a low ping with poor iperf3 numbers is a window, congestion or bottleneck problem, and the two have no fix in common.

Two sentences to be able to defend on the spot. Adding bandwidth shortens the transmission term and nothing else, so on a long path it barely moves the total. And queuing is the only component that responds to load, so any latency that changes during the day is queuing until proven otherwise.

07 Interview questions

What they actually ask

This is the topic interviewers use to find out whether you understand networks or have memorised them, because it can be checked with mental arithmetic in ten seconds. Expect to be handed a packet size, a link rate and a distance and asked for a delay out loud.

What is the difference between bandwidth and throughput?

Bandwidth is the capacity of the link in bits per second and it is a property of the hardware, present whether you send anything or not. Throughput is the rate you actually achieved on a real transfer, measured in the same unit, and it is always at or below bandwidth. The gap between them comes from the sender’s window, the narrowest link on the path, competing traffic, and anything that had to be sent twice. Take the headers and retransmissions out of throughput and what remains is goodput, the application bytes that arrived once and stayed.

A link is sold as 100 Mbps and my download manager reports 11 MB per second. Is somebody lying?

No, those are the same number in two different units. 100 Mbps is 100 × 10⁶ bits per second; divide by 8 and the ceiling is 12.5 × 10⁶ bytes per second, so 11 MB per second is 88% of line rate and entirely normal once headers are counted. The trap is that link rates are always quoted in bits and file sizes always in bytes, so a factor of eight sits between them. Note as well that link rates use powers of ten while file sizes often use powers of two, which is a second, smaller discrepancy on top of the first.

Name the four components of end-to-end delay and say what each one depends on.

Transmission delay is L / R, packet size over the rate of the link it is being pushed onto, so it depends on packet size and link rate and not at all on distance. Propagation delay is d / v, distance over the signal speed in the medium, roughly 2 × 10⁸ m/s in copper and fibre, and it depends on nothing but distance. Queuing delay is time waiting in a router’s outbound buffer and depends on how busy that link already is. Processing delay is the header check and next-hop lookup at each node, a small near-constant per router.

Transmission delay uses a link rate. Whose link rate?

The rate of the link the packet is currently being pushed onto. For the first hop that is the sender’s outgoing link, and after that it is each router’s outgoing link in turn. It is never the receiver’s download rate and never the end-to-end throughput. This matters on a path made of links with different rates, because you must add a separate L / R for each one, and the slowest link both dominates that sum and becomes the bottleneck that caps throughput.

Which of the four varies with load, and why does that matter?

Queuing, and only queuing. The other three are pinned by things that do not change during the day: packet size, link rate, distance and router hardware. Queuing tracks traffic intensity, roughly La/R, the arrival rate of bits divided by the service rate. While that ratio stays well under 1 the queue is short, but as it approaches 1 the average queue grows without bound, so delay does not degrade gently, it collapses. Past 1 the buffer overflows and packets are dropped instead of delayed.

Two links run over the same 100 km of fibre, one at 1 Gbps and one at 10 Mbps. Which has the lower propagation delay?

Neither. They are identical, because the propagation formula has no rate term in it at all: 100,000 m ÷ 2 × 10⁸ m/s = 0.5 ms on both. What the faster link buys is a smaller transmission delay, since it finishes pushing the packet onto the wire sooner, and more bits per second once the pipe is full. This is the cleanest one-line demonstration that bandwidth and latency are independent quantities.

Why is a satellite link called high bandwidth and high latency at the same time? Is that not a contradiction?

It is not, because the two words measure different things. A geostationary satellite sits 35,786 km up, so the signal travels 71,572 km up and back down and radio at 3 × 10⁸ m/s takes about 239 ms one way, so roughly 477 ms of any round trip is propagation before a single other delay is counted, and no amount of bandwidth shortens a distance. Meanwhile the transponder can carry tens or hundreds of megabits per second, so bulk transfers are fine while anything interactive is painful. Say the rule out loud: bandwidth is a rate, latency is a duration, and a wider pipe is not a shorter one.

What is RTT and why is it not exactly twice the one-way delay?

Round-trip time is the interval from sending something to receiving the answer, and it is what ping reports. It is roughly two one-way latencies, but three things break the exact doubling: the return path may take a different route with a different distance and hop count, the queues in the two directions are not equal at the same instant, and the responder needs its own time to produce a reply. The honest form is RTT ≈ 2 × one-way + responder time, and you cannot derive a one-way delay from a ping unless you already know the path is symmetric.

What is the bandwidth-delay product and what is it used for?

It is the link rate multiplied by the round-trip time, and its unit is bits, not bits per second, because a rate times a time is a quantity. It is the number of bits that can be on the wire and unacknowledged at one instant. Its use is window sizing: a sender allowed at most W bits outstanding can achieve at most W / RTT however fast the link is, so a window smaller than the bandwidth-delay product leaves the link idle waiting for acknowledgements. Links that are both fast and long are exactly where this bites, and it is why TCP window scaling exists.

What is goodput, and how would you see the difference between it and throughput?

Hard

Two data centres are 2000 km apart on fibre at 2 × 10⁸ m/s. Three measurements: a small ping has a steady 21.0 ms round trip when the path is idle and never reads lower; the same ping reads 41.0 ms when the path is busy; and one TCP connection using a fixed 65,535-byte window measures about 25 Mbps, while the operator insists the link is 1 Gbps. Answer three things. (a) How much of the idle 21.0 ms round trip is propagation, and what is left over for everything else. (b) The busy path adds 20.0 ms to the round trip and the path crosses 4 routers in each direction, so give the average extra delay per router pass, and give the throughput that same window now achieves at 41.0 ms. (c) Show where the 25 Mbps figure comes from, and give the window size in bytes that would be needed to fill 1 Gbps on the idle path.

Follow-up

Nothing in the three measurements proves the operator wrong. Two of the three numbers are properties of the sender and of the traffic rather than of the link, and the one number that really is a property of the link never appears directly in any measurement you were given.

Show the hint

Work out what a single 65,535-byte window can deliver in one round trip before you accuse anybody of underselling the link, and remember that a round trip crosses every router twice.