Packet Switching and Network Performance

A web request starts in one application and ends in another. The internet supplies the links and switching equipment between them, while shared rules let separately built devices exchange useful messages.

Hosts, links and switches

A host, or end system, runs applications. Phones, laptops, sensors and servers are all hosts; many servers are grouped in data centers. A client requests a service, such as a browser requesting a page. A server provides it. These names describe roles in a communication, not two different physical kinds of computer.

Communication links carry signals over copper, coaxial cable, optical fiber or radio. Packet switches pass data from one link to another. Routers join networks and form much of the network core; link-layer switches connect devices mainly within local and access networks. The sequence of links and switches crossed by a packet is its path, or route.

The network edge consists of the end systems and the access networks that attach them. The network core is the mesh of switches and links interconnecting those access networks. For a home download, the laptop belongs to the edge even when the server is thousands of kilometers away: geographic distance does not turn an end system into a core router.

The service and the rules

Viewed as a service, the internet carries data between distributed applications: web browsers and web servers, callers, games and email programs. A program hands data to the operating system through a socket interface. The network then carries that data toward the receiving program.

A protocol defines message formats, message order and the actions taken when messages are sent or received. HTTP governs web exchanges, while TCP and IP perform other jobs in the TCP/IP stack. Agreeing on a cable alone is insufficient: the communicating systems must also interpret the messages and react to them consistently.

Convert sizes before calculating time

A sender divides application data into chunks and adds headers, producing packets. A link's rate tells you how many bits it can place onto the medium each second. Decimal prefixes apply: 1 kbps is 1,000 bit/s, 1 Mbps is one million bit/s, and 1 Gbps is one billion bit/s.

File sizes often use bytes. One byte contains 8 bits; 1 kB is 1,000 bytes and 1 MB is one million bytes. A MiB is a different quantity: 1,048,576 bytes, about 5% larger than an MB.

To push a 1 MB file onto a 20 Mbps link:

1. Convert the size: $1\text{ MB}=8\times {10}^6\text{ bits}$.
2. Express the rate: $20\text{ Mbps}=20\times {10}^6\text{ bit/s}$.
3. Divide size by rate: $8\times {10}^6\mathrm{/}(20\times {10}^6)=0.4\text{ s}$.

This is the time to put the bits onto that link. Travel time and waiting elsewhere are separate delays. At 100 Mbps the corresponding maximum byte rate is $100\mathrm{/}8=12.5$ MB/s. Reading Mbps as MB/s makes a transfer-time estimate eight times too small.

A home connection has two stretches: a local network from your devices to the home router, then an access connection into the ISP. A fast local Wi-Fi connection cannot remove a limit on that second stretch.

Copper telephone lines and cable networks

A home router often combines a router, a Wi-Fi access point and a small Ethernet switch. A separate modem may connect it to the access medium. That connection leads into the ISP's network and its first, or edge, router.

DSL reuses a copper telephone pair. The home DSL modem communicates with a DSLAM at the provider's central office. In ADSL, frequency bands separate voice, upstream data and downstream data. The voice band occupies 0–4 kHz; the downstream band is wider than the upstream band. Each home has its own copper pair to the office.

Home DSL services usually offer more downstream capacity than upstream capacity. The DSL family also has symmetric forms used mainly by businesses. Longer copper lines weaken the signal and collect more noise, lowering the available rate. Fiber to a street cabinet shortens the copper stretch; VDSL often uses less than a kilometer of copper.

Cable access uses hybrid fiber-coax, or HFC. Fiber reaches neighborhood nodes; coaxial cable then reaches typically hundreds of homes. A home's cable modem exchanges data with a CMTS at the head end. DOCSIS 4.0, published in 2020, specifies up to 10 Gbps downstream and 6 Gbps upstream capacity for the shared channel. Those figures describe channel capacity, not a rate reserved for each home.

Which capacity is shared?

Cable downstream signals reach every home on the node. Upstream users must take turns. FTTH brings fiber all the way home; in a passive optical network (PON), an unpowered splitter commonly serves 32 or 64 homes. The provider has an OLT, and each home an ONT. Each ONT selects its own downstream data. FTTH supports very high access rates, often 1 Gbps or more, with symmetric services common.

Access without a new cable

5G fixed wireless access uses a stationary home modem and a nearby cellular base station. Satellite access can reach remote homes. A GEO satellite orbits at 35,786 km, so the shortest ground–satellite–ground trip is 71,572 km. At $3\times {10}^8$ m/s, this adds about 0.24 s one way and 0.48 s for a request and reply, before other delays.

A LEO satellite at about 550 km gives a minimum up-and-down distance of about 1,100 km, or 3.7 ms. Actual paths also depend on geometry and ground connections. The advertised access rate alone therefore cannot predict a download: shared use, a long DSL line or a slower link farther along the path can reduce the delivered rate.

Inside an office, a packet may start on copper or radio and leave the building on fiber. A mobile packet also enters the same internet core after its wireless access stretch. Each link has its own medium and its own limits.

Local and wide-area access

A local area network (LAN) joins nearby devices to an edge router. With Ethernet, a computer commonly connects over twisted-pair copper to a switch at 1 Gbps. Servers often connect at 10 Gbps or more. Links between switches and the router join these local connections into a usable path.

Wi-Fi, based on IEEE 802.11, exchanges packets by radio with an access point connected to the LAN. It covers distances of tens of meters. Devices sharing its radio channel compete for transmission time, so adding active users reduces the capacity available to each.

Cellular access spans a wider area. A phone exchanges packets with an operator's base station, sometimes several kilometers away. 4G LTE and 5G provide this access; 5G supports higher rates and lower delay, with higher-frequency bands offering more capacity over shorter ranges. The rate a phone achieves still depends on signal conditions and cell load.

The radio connection does not lead into a separate wireless internet. Beyond the access point or base station, packets cross ordinary interconnected networks. Cellular backhaul, the connection from a base station into the operator's network, can itself use microwave radio as well as wired media.

Match the medium to the job

A guided medium confines a signal to a physical path. An unguided medium lets it spread through air or space.

Unshielded twisted pair is common in LANs. Category 6A supports 10 Gbps Ethernet over up to 100 m. Fiber carries tens to hundreds of Gbps on a link, with more capacity possible through multiple wavelengths. Its low signal loss suits long-distance backbone and undersea links. Optical equipment can cost more than copper equipment on short links, and tapping fiber is difficult but possible.

Radio is affected by path loss as signals weaken over distance and through obstacles. Multipath occurs when reflections arrive by different routes at different times. Other transmitters add interference. These effects make wireless rates and errors less predictable than those of wired links.

Choose for the installation

A mobile inventory scanner needs freedom to move, making radio valuable even where Ethernet is faster. A fixed link between distant buildings may favor fiber for reach and resistance to electrical interference. Installation labor often costs more than the cable itself, so new buildings commonly receive fiber and spare cable together. The useful comparison weighs rate, distance, installation cost, interference and mobility, not just a technology's peak rate.

Packet switching divides a message into independently forwarded packets. Each outgoing link sends one packet at its full link rate, while other packets may be crossing other links at the same time.

Wait for the whole packet

Under store-and-forward, a switch receives the complete packet before sending its first bit on the next link. For packet length $L$ bits and rate $R$ bit/s, one transmission takes $L\mathrm{/}R$ seconds.

A 1,500-byte packet contains 12,000 bits. On a 10 Mbps link:

Across three equal-rate links, the packet is transmitted three times. Ignoring propagation, processing and queuing, it arrives after $3\times 1.2=3.6$ ms. Three links means two intervening routers: count the connections, including the source's and destination's connections.

Build a packet pipeline

Now send five equal packets back to back along those three links. The source sends packet 2 while the first router forwards packet 1. After the first packet fills the path, later packets arrive one transmission time apart.

For $P$ equal packets over $N$ equal-rate links, with no other delays, the last arrival is:

The first packet pays for all $N$ transmissions; each remaining packet adds one more transmission interval. Multiplying $P\times N\times L\mathrm{/}R$ would force every link to wait until the previous packet reached the destination, eliminating the parallel work that the network can perform.

The five packets contain 60,000 bits altogether. Sending those bits as one unbroken message would take 6 ms per link and 18 ms over three store-and-forward links. The 8.4 ms packetized transfer is faster because forwarding begins before the entire message has arrived. Extra packet headers use capacity too; they are separate from this comparison of equal data totals.

Decide where the packet goes

A router examines the destination IP address in a packet's header. Its forwarding table maps destination address ranges to outgoing links. Routing protocols populate those tables; forwarding is the local act of consulting a table and sending this packet through the chosen interface.

Each outgoing link has an output buffer, or queue. If that link is busy, newly arrived packets wait there. Their waiting time is queuing delay, which changes with the traffic reaching that output. A quiet input does not prevent congestion if several other inputs feed the same output.

Buffers have finite space. When a buffer fills, the router drops an arriving or already queued packet, depending on its policy. This is packet loss. Store-and-forward explains when forwarding can begin; it does not guarantee that the outgoing link or buffer will be available.

Circuit switching reserves resources along an entire path for a communication session. The sender first establishes a circuit, then uses the reserved rate until the session ends. Traditional telephone networks illustrate this design.

A reservation across shared links

A circuit is a reserved share of link and switch capacity. It need not be a separate physical wire. Many circuits can share the same cable while keeping their reserved resources separate. Once a circuit is established, other users' data does not create output queues for its reserved share, so the sender gets a constant rate and predictable delay.

Multiplexing lets several circuits occupy one physical link. Two common methods separate them by frequency or by time.

In FDM, several bands exist at once. FM radio stations provide a familiar example. Analog telephone trunks assigned calls bands about 4 kHz wide. A band's bandwidth is a frequency width measured in hertz. A bit rate is measured in bit/s: the two units describe different quantities.

In TDM, time repeats in frames, each containing fixed slots. A circuit gets one slot every frame. For equal slots with no frame overhead, divide the link rate by the number of slots to obtain that circuit's rate. A circuit transmits during its own slot and waits for the next occurrence; its average rate is its reserved fraction of the link.

Calculate the circuit rate first

A 2.048 Mbps link has 32 equal slots per frame. Circuit setup takes 400 ms, and the sender has a 960,000-bit file.

1. Convert the full link rate to 2,048,000 bit/s.
2. Divide among slots: $\mathrm{2,048,000}\mathrm{/}32=\mathrm{64,000}\text{ bit/s}$.
3. Divide file size by circuit rate: $\mathrm{960,000}\mathrm{/}\mathrm{64,000}=15\text{ s}$.
4. Add setup: $15+0.4=15.4\text{ s}$, ignoring propagation.

The sender does not obtain 2.048 Mbps merely because other users happen to be quiet. Its circuit still owns one slot in each frame. Also, the file transmission time is not multiplied by the number of links on the path. The reserved circuit carries a flowing stream of bits; the entire file is not stored and then forwarded as one packet at every switch. Add propagation separately when it is part of a task.

Account for unused reservations

The guarantee has a cost. When a caller is silent or a user stops to read a loaded page, the circuit's share sits idle. Another user cannot take that share on demand. If all circuits are allocated, a new connection may be blocked even while some existing connections send no data. This is a limit on admitting sessions, distinct from dropping packets from a full buffer.

Web traffic often alternates between short downloads and long pauses. A reservation sized for each user's busy moment leaves capacity unused during the pauses. Packet switching lets other users use that capacity as their packets arrive.

Share capacity when demand occurs

Statistical multiplexing is on-demand sharing of a link among packets from different users. A waiting packet can use the next available transmission opportunity, so one user's silence does not reserve unused capacity. This works especially well for bursty traffic, whose active periods are separated by quiet periods.

Consider ten users sharing a 10 Mbps link. One sends 2,000 packets of 1,000 bits while the other nine are silent. Its total is $\mathrm{2,000}\times \mathrm{1,000}=2\times {10}^6$ bits. With ten equal TDM reservations, its 1 Mbps circuit takes 2 s. With packet switching, the full 10 Mbps link sends the burst in 0.2 s, ignoring other delays.

Compare peak demand with average demand

Suppose each active user needs 1 Mbps and is active independently 20% of the time. A 10 Mbps circuit-switched link can guarantee that rate to at most ten users. Packet switching can serve 20 users with a small chance that simultaneous demand exceeds capacity.

Let $X$ count active users. It follows a binomial distribution: 20 independent users, each active with probability 0.2. Exactly $k$ active users have probability:

The binomial coefficient counts which users are active. Overload begins at 11 active users, because ten fit exactly. Add the probabilities for 11 through 20:

That is about 0.06% of the time. The average demand is only 4 Mbps, but that average alone would not tell you the overload probability. Independence matters: users who become active together produce larger coincident bursts than this calculation predicts.

Match the design to the traffic

Packet switching supports efficient use of capacity and comparatively simple switching equipment. Variable waiting and loss are its central service costs. Real-time applications can use playout buffers to smooth arrival timing, while retransmission can recover missing data at the cost of added delay.

For steady traffic that needs a guaranteed rate, a circuit can provide better service after setup. Packet switching is not always faster: a heavily loaded output can add a long wait. The internet uses packet switching, and telephone services have increasingly moved from circuits to voice over IP.

Connecting to an access ISP gives a host a first network to use. Reaching hosts elsewhere requires agreements between networks. These independently operated networks together form the internet.

Why interconnect through providers?

Directly connecting every access ISP to every other would need $n(n-1)\mathrm{/}2$ links for $n$ ISPs. With 10,000 ISPs, that is 49,995,000 separate links. Maintaining that many relationships and physical connections would not scale well.

Instead, access ISPs buy transit from larger providers. A customer pays a provider to carry traffic between the customer's network and the rest of the internet, with charges commonly related to exchanged traffic. Regional providers can buy transit from still larger providers. Some access ISPs connect directly to the top level.

A tier-1 ISP reaches all internet destinations without buying transit. Roughly a dozen large international networks are commonly described this way, and they interconnect with one another. Tier-1 describes a reachability and business position; it is not a claim that every packet must cross one.

Distinguish transit from peering

A peer does not, merely by peering, provide transit to every other network it can reach. That distinction matters when tracing possible paths. Networks often peer at an internet exchange point (IXP), a shared switching facility where many ISPs and content providers meet. Large exchanges in cities such as Frankfurt, Amsterdam and London connect hundreds of networks.

A point of presence (PoP) is a provider site containing routers where customers can connect. An IXP brings many networks together to exchange traffic; a PoP is a site within a provider's footprint. A network can also be multi-homed, connecting to two or more providers for resilience. If one provider fails, another connection can keep it reachable, provided the remaining path and the customer's own equipment work.

Follow a path and the payments

Ana uses cable ISP A. A pays regional ISP R, which pays tier-1 T1. Ben uses mobile operator B, which pays tier-1 T2. T1 and T2 peer. A packet can follow:

A → R → T1 → T2 → B.

The payments follow customer–provider links, not every direction in which a packet travels. B pays T2 for connectivity even when this packet is arriving at B. T1 and T2 commonly exchange their traffic without a settlement payment.

If A and B join an IXP and agree to peer, their customer traffic can instead cross directly between A and B. They avoid upstream transit for that traffic and often shorten the path. The transit connections remain useful for destinations outside the peers' networks and customers.

An autonomous system (AS) is an independently operated network, or part of one, with its own routing policy and AS number. An ISP may operate several ASes. This gives the network-of-networks picture an administrative unit: separately managed ASes interconnect through the business relationships just described.

A video service can deliver data more efficiently when it controls a network close to its viewers. Large content companies combine private backbone networks, direct ISP connections and nearby copies of popular files.

Content companies build networks too

Companies such as Google, Microsoft, Amazon and Meta operate content provider networks linking their data centers. These private global networks let them control more of the route their services use and reduce reliance on purchased transit.

A content network can peer directly with an access or regional ISP at an IXP or through a private connection. Traffic then enters that ISP without first climbing the full provider hierarchy. For destinations not reached by these arrangements, the content network can buy transit from an upstream provider. That provider need not be tier-1; some large content networks rely mostly on direct connections.

The economic reason and the performance reason reinforce each other. Direct exchange can avoid transit charges for that traffic, and a shorter path usually crosses fewer links. However, owning a backbone and peering are not the same as keeping a copy of every file near every user. Server placement is a separate part of delivery.

Put content near the request

A content delivery network (CDN) places servers with copies of popular content inside or near access networks and at exchange points. Videos, web files and software updates can then come from a nearby server instead of a distant origin data center. This reduces long-distance traffic and the propagation component of delay.

A city viewer requesting a cached video may cross only a few hops inside the access ISP. Compare a data center 8,000 km away. Even an ideal fiber route of that length contributes:

in one direction. A request and reply add at least $2\times 0.04=0.08$ s, or 80 ms, from propagation alone. Queuing, transmission, equipment and a longer real cable route can add more. Raising the link rate does not remove that distance; using a closer content copy reduces it.

Read the flatter network picture

These arrangements coexist. Access, regional and tier-1 providers still form a loose hierarchy, but direct content connections carry a large share of traffic. Multi-homing, private peering and IXPs add further links across the levels.

No single organization owns or operates this entire system. A route reflects the available interconnections between independently managed networks. When tracing a download, first ask where its serving content copy sits and which networks have direct agreements. A CDN inside the user's ISP can explain a short path without any tier-1 hop at all.

A router's contribution to delay is not one indivisible pause. Distinguishing its four parts shows whether a proposed upgrade can address the delay you observe.

Four parts of nodal delay

The nodal delay is:

Processing includes examining the header, checking for bit errors and selecting an outgoing link. It is typically microseconds or less in a high-speed router. Queuing is waiting for packets ahead at the output. It can be zero for an idle link and empty queue, or many milliseconds under load. Unlike the fixed components for a particular packet and link, it varies with other traffic.

The final two parts are often confused:

Transmission is pushing the packet onto the medium. Propagation is a bit traveling through it. Fiber and copper use an approximate signal speed of $2\times {10}^8$ m/s; radio through air or space is close to $3\times {10}^8$ m/s. In fiber, each kilometer adds about 5 µs of propagation.

A long link

Send a 1,500-byte packet on a 100 Mbps fiber link 2,000 km long.

1. Convert packet size: $\mathrm{1,500}\times 8=\mathrm{12,000}$ bits.
2. Divide by rate: $\mathrm{12,000}\mathrm{/}{10}^8=0.00012$ s, or 0.12 ms of transmission.
3. Convert distance: 2,000 km is 2,000,000 m.
4. Divide by propagation speed: $2\times {10}^6\mathrm{/}(2\times {10}^8)=0.01$ s, or 10 ms.

Propagation dominates. When the sender finishes placing the packet onto the link after 0.12 ms, its first bit has traveled only $0.00012\times 2\times {10}^8=\mathrm{24,000}$ m, or 24 km. The packet can therefore be fully on the link before its first bit arrives.

A short link at the same rate

Across 10 m of cable, propagation takes $10\mathrm{/}(2\times {10}^8)=50$ ns. The packet still needs 0.12 ms to enter the link. By the time its first bit arrives, only $50\times {10}^{-9}\times {10}^8=5$ bits have left the sender. Here transmission dominates.

Increasing the long link to 1 Gbps cuts its transmission time to 0.012 ms. It leaves propagation at 10 ms. A higher bit rate packs more bits into each second; it does not make a bit travel faster through the fiber.

Keep the units consistent before substituting: multiply bytes by 8, kilometers by 1,000 and Mbps by one million. Dividing bytes directly by bit/s loses the factor of eight, while using kilometers with m/s loses a factor of 1,000.

For one packet, store-and-forward makes the path a sequence of transmissions. Add each link's contribution and each router's work to find when the last bit reaches the destination.

Sum links and routers separately

For $N$ identical links and $N-1$ intervening routers, with empty queues:

Add any endpoint processing separately. For unequal links, calculate each transmission and propagation term using that link's rate and distance. Add queuing wherever it occurs; it need not be the same at every router.

Consider host → R1 → R2 → server. A 12,000-bit packet crosses 100 Mbps over 10 km, 1 Gbps over 1,000 km, then 50 Mbps over 5 km. Signal speed is $2\times {10}^8$ m/s. Each router processes the packet for 20 µs, and the queues are empty.

Two routers add $2\times 0.02=0.04$ ms. The last bit arrives after $0.372+5.075+0.04=5.487$ ms. The middle link's propagation alone accounts for over 90% of this total. Counting transmission only on the slowest link would give an incomplete one-packet delay.

Distinguish arrival time and round-trip time

On a single link, an $F$-bit message's last bit arrives after $F\mathrm{/}R+d\mathrm{/}s$. The last bit must first leave the sender, then travel to the receiver. Earlier bits propagate while later bits are still being transmitted.

Round-trip time (RTT) runs from sending a probe until its reply returns. It includes forward and return delays plus the destination's response time. The ping command measures it. Doubling a one-way estimate works only when the reverse path and response assumptions support that calculation.

Count bits in flight

The bandwidth-delay product for a link is $R\times d_{\text{prop}}$. A continuously transmitting 1 Gbps link with 10 ms propagation can hold ${10}^9\times 0.01={10}^7$ bits, or 1.25 MB, in flight. One bit occupies $s\mathrm{/}R=0.2$ m of that link. The product is a quantity of data, not a delay.

Include delays before transmission

A voice source producing 64 kbps takes time to fill a 160-byte payload. It must collect $160\times 8=\mathrm{1,280}$ bits, requiring $\mathrm{1,280}\mathrm{/}\mathrm{64,000}=20$ ms. This is packetization delay, incurred before the packet can be sent. Shared access such as Wi-Fi can add a wait for the channel too.

Short LAN paths can be dominated by transmission and processing, long paths by propagation, and congested paths by queues. Moving a serving CDN copy closer reduces the distance term; increasing a link rate mainly addresses transmission and, when demand is unchanged, queuing.

An output queue grows whenever arrivals temporarily demand more work than the link can finish. Both the average load and the timing of arrivals matter, especially when the buffer has limited space.

Compare arrival load with service capacity

Traffic intensity is:

$I=La\mathrm{/}R$,

where $a$ is average packet arrivals per second, $L$ is packet length in bits, or mean length for variable packets, and $R$ is link rate. The numerator is incoming bit/s, so intensity is a ratio without units.

When $I>1$ persists, arrivals outpace service. An unlimited queue would grow without bound; a real finite buffer instead fills and drops packets. Keeping average intensity below 1 is a basic design rule, but it does not eliminate bursts.

The same average can create different waits

Starting with an empty queue and idle link, one arrival every $L\mathrm{/}R$ seconds lets each packet find the link ready. For batches of $N$ packets every $NL\mathrm{/}R$ seconds, the batch fits the same average capacity but its packets wait for one another. Packet $n$ waits $(n-1)L\mathrm{/}R$, giving an average wait of $(N-1)L\mathrm{/}(2R)$.

For five packets and a service time of 1.2 ms, waits are 0, 1.2, 2.4, 3.6 and 4.8 ms. Their total is 12 ms, giving a 2.4 ms average. Waiting excludes each packet's own transmission time.

Random arrivals near full load

Random arrivals produce uneven queues. One model, M/M/1, assumes independent Poisson arrivals, independent exponentially distributed packet lengths with mean $L$, one server and an unlimited buffer. For a steady load below 1, its mean queue wait is:

With a mean service time of 1.2 ms, increasing intensity from 0.5 to 0.8 raises the mean wait from 1.2 ms to 4.8 ms. At 0.9 it is 10.8 ms; at 0.96 it is 28.8 ms. The shrinking spare capacity makes delays grow sharply near full load.

For 800 packets/s, mean length 12,000 bits and a 10 Mbps link, intensity is $\mathrm{12,000}\times 800\mathrm{/}{10}^7=0.96$. The model gives $0.96\mathrm{/}0.04\times 1.2=28.8$ ms. Upgrading to 20 Mbps makes intensity 0.48 and mean service time 0.6 ms. The new mean wait is $0.48\mathrm{/}0.52\times 0.6\approx 0.55$ ms. Both factors change, so the waiting time falls by much more than half.

Finite buffers trade loss against delay

A full buffer cannot hold every arrival. A router drops a packet, producing loss; wireless bit errors can cause loss too. Bursts can fill a buffer even when average intensity is below 1. A larger buffer absorbs more of a burst, but persistent long queues create bufferbloat: high delay despite fewer drops.

The router does not resend a packet it dropped for lack of space. TCP can detect missing data and have the source retransmit it end to end. UDP supplies no transport-layer retransmission, so an application must handle the missing data, perhaps as a brief video glitch.

A receiver's delivered bit rate is its throughput. It describes how much arrives per second, while delay describes how long an individual packet or exchange takes.

Measure the delivered rate

Instantaneous throughput is the rate at a particular moment. Average throughput is the amount received divided by the elapsed transfer time:

A voice call needs low delay with a modest bit rate; a large download benefits most from sustained throughput. A high-capacity distant link can support a fast download while still producing a noticeable conversational delay.

Find the bottleneck

For a long transfer across an otherwise idle path, the slowest link limits the sustained rate:

Link rates are capacities in series, not numbers to add or average. An upstream link can send faster briefly, but its excess packets must wait at the slower output. That queue cannot make the receiver obtain data faster than the slow link transmits it.

A server access link runs at 100 Mbps and a client's at 25 Mbps. Their path supports 25 Mbps when no other link imposes a lower limit. A 500 MB file contains $500\times 8=\mathrm{4,000}$ Mbit, so the capacity-based estimate is $\mathrm{4,000}\mathrm{/}25=160$ s, ignoring startup, propagation and protocol overhead.

If the client upgrades to 1 Gbps, the server's 100 Mbps link becomes the bottleneck. The estimate becomes $\mathrm{4,000}\mathrm{/}100=40$ s. Improving the client's connection again would not increase this path's capacity.

Include shared capacity

A core link's total rate need not be available to one transfer. If $M$ transfers share a link equally, each receives a share $R\mathrm{/}M$. Combine that share with the endpoint links:

Fifty transfers sharing 1 Gbps equally receive 20 Mbps each. For the original 100 Mbps server and 25 Mbps client, the shared core becomes the bottleneck at 20 Mbps. The 500 MB file now takes about $\mathrm{4,000}\mathrm{/}20=200$ s.

Access links commonly limit throughput, but bottlenecks can occur anywhere and move as demand changes. Packet headers consume capacity, losses require recovery and transport protocols can reduce sending rates during congestion. Application throughput can therefore be below the raw link-rate estimate.

Choose the correct transfer-time calculation

When average throughput has already been measured over a complete transfer, $T=F\mathrm{/}\text{average throughput}$ directly gives elapsed time. Adding RTT again would count delay twice.

When estimating from a bottleneck capacity, $F\mathrm{/}{R}_{\text{bottleneck}}$ omits startup and path delays. Those additions matter most for small objects. A 10 kB object needs only 3.2 ms of transmission at 25 Mbps, so a round trip of tens of milliseconds can dominate its fetch time. For a large file, the amount of data divided by the sustained rate usually dominates.

traceroute discovers responding routers toward a destination and records a round-trip time to each. Windows calls the command tracert. Both exploit a field that normally prevents packets from circulating forever.

Make probes expire at successive routers

IPv4's time to live (TTL) is a hop count; IPv6 calls the corresponding field hop limit. A router decreases the value before forwarding. If the value reaches zero, the router discards the packet and returns an ICMP Time Exceeded message when its configuration and filtering allow it.

Traceroute sends probes starting with TTL 1. The first router expires these probes and can identify itself through its reply. Probes with TTL 2 can pass that router and expire at the next. Increasing the value exposes successive responding hops. By default, three probes are sent per TTL, producing three RTT measurements at each hop.

For host → R1 → R2 → R3 → server:

With all replies received, this four-hop trace sends twelve probes. Each RTT starts when its particular probe leaves the source and ends when that probe's reply returns. The value is not the time to cross just the link from the previous printed hop.

Recognize the destination's response

The default Linux and macOS method sends UDP probes to an unlikely destination port, beginning at 33434 and increasing the port per probe. A destination with no listener returns ICMP Port Unreachable. Despite its error-like name, that response tells traceroute that the probe reached the destination.

Windows tracert sends ICMP Echo Requests; reaching the destination elicits an Echo Reply. Intermediate expired probes still produce Time Exceeded messages. Linux traceroute and Windows tracert default to a maximum of 30 hops. A trace ends when the destination responds or the limit is reached.

Run and identify the output

Use traceroute example.org on Linux or macOS, or tracert example.org on Windows. The source address of a reply identifies a router interface. DNS lookups may add readable names. Use traceroute -n example.org or tracert -d example.org to skip those lookups and see numeric addresses without waiting for name resolution.

Router names chosen by operators sometimes contain a city or airport code, but a name is a hint, not a geographic measurement. A router also has multiple interfaces, so different traces can show different addresses for the same device.

The resulting listing records replies to these probes at this time. Routes can change between runs, and load balancing can send probes along different paths. The probes travel forward toward the destination, while each measured RTT also includes a return journey that the listing does not reveal.

Read a trace as a set of timed replies. A row reports the hop number, responding interface names or addresses and the RTTs of its probes. Silence and variation need interpretation before they become a diagnosis.

Read a complete trace

This trace uses private and documentation addresses:

1 gw.home.example (192.168.1.1)              1.9 ms  1.6 ms  1.7 ms
2 10.20.0.1                               8.4 ms  8.9 ms  8.1 ms
3 core1.city-a.isp.example (198.51.100.17)   9.6 ms 10.2 ms  9.8 ms
4 * * *
5 ix1.city-b.transit.example (203.0.113.40) 21.5 ms 21.1 ms 22.0 ms
6 sub1.ocean.transit.example (203.0.113.73) 84.3 ms 83.9 ms 84.6 ms
7 www.example.org (192.0.2.80)             85.2 ms 85.0 ms 86.1 ms

The home router replies at about 1.7 ms, the next hop at about 8.4 ms, and the destination at about 85 ms. These are source-to-hop round trips. The three numbers on a row belong to separate probes, not three pieces of one packet's delay.

Handle missing replies

An asterisk means a reply did not arrive before the timeout: normally 4 s on Windows and up to 5 s on Linux with default settings. A router may suppress or rate-limit ICMP replies; a firewall may filter them; a probe or its reply may be lost.

Hop 5 replies after silent hop 4. The trace therefore shows that probes can travel beyond that point. It does not establish that ordinary traffic is being dropped at hop 4. Asterisks at the end can mean that the destination or a nearby firewall does not answer probes, but an actual reachability problem is also possible. Check other evidence before choosing the cause.

Look for a lasting change

The minimum RTT rises from 21.1 ms at hop 5 to 83.9 ms at hop 6: a 62.8 ms increase that persists at hop 7. This suggests added path delay near that part of the route.

If the entire increase were propagation on one extra fiber link used in both directions, the extra one-way time would be $62.8\mathrm{/}2=31.4$ ms. At $2\times {10}^8$ m/s, that corresponds to $0.0314\times 2\times {10}^8=\mathrm{6,280,000}$ m, or 6,280 km. This estimate depends on the stated path model. Different reply routes, queues and router response times can alter the increase; the trace does not identify cable type or measure link length.

Separate forwarding from replying

One router can reply slowly while forwarding packets quickly. Its ICMP processing may have low priority, and its reply can take a different return path. A high RTT that disappears at later hops usually points to a slow reply, while an increase that persists deserves investigation.

Repeated minima help estimate fixed delay, while a spread of RTTs can reveal variable queuing. Neither is a perfect separation: response processing and path changes can also vary. Compare runs at different times, and use repeated ping measurements to check the destination over time.

Load balancing can show several addresses at one hop. Paris traceroute keeps flow-identifying header fields constant to reduce per-flow path changes, but per-packet balancing can still split probes. MPLS tunnels can hide internal routers, so printed hops need not reveal every physical router.