Self-checkedTier 1 6 min readcn.performance-metrics

Delay, throughput and bandwidth-delay product

The four sources of delay (processing, queueing, transmission, propagation) and their formulas, bandwidth against throughput and the bottleneck link, the bandwidth-delay product as the amount of data in flight, round-trip time, and the utilisation of a stop-and-wait protocol and how many frames must be in flight to fill a pipe; numbers are exact fractions cross-checked by a simulation.

Why it exists

"How long does this transfer take?" and "how fast is this link really?" are the first questions about any network, and GATE asks them as arithmetic problems. The answers depend on telling four different delays apart, on the difference between the rate of a link and the throughput you get, and on the bandwidth-delay product, which says how much data can be on its way at once.

Prerequisites

You need packets and store and forward (switching).

Precise definitions

The four delays at a node (per packet):

delay	cause	formula
processing $d_{proc}$	examining the header, checking errors, choosing the output link	microseconds, usually given
queueing $d_{queue}$	waiting in the output queue for packets ahead	depends on traffic; zero if the link is idle
transmission $d_{trans}$	pushing all $L$ bits of the packet onto a link of rate $R$	$L / R$
propagation $d_{prop}$	a bit travelling over a link of length $D$ at signal speed $s$	$D / s$

The delay of one packet over one link is the sum $d_{proc} + d_{queue} + d_{trans} + d_{prop}$ . Transmission delay depends on the packet size and the rate, propagation delay on the distance and the medium (not on the rate). Typical speed: about $2 \times 10^8$ m/s in copper and fibre.

Bandwidth (here the rate of a link) is measured in bits per second. Throughput is the rate at which data actually get from the sender to the receiver; for a path of several links it is limited by the bottleneck, the slowest link: throughput $\le \min_i R_i$ (and lower if other traffic shares the link or the protocol waits).

Round-trip time (RTT): time from sending a packet to receiving its acknowledgement; at least $2 d_{prop}$ (plus transmission and processing time). Latency is used for the delay of a packet or a message.

Bandwidth-delay product (BDP): $R \times d$ where $d$ is the one-way propagation delay (or $R \times \text{RTT}$ for the data in flight in a round trip). It is the number of bits that fit in the "pipe" between the sender and the receiver. Examples differ in whether the one-way delay or the RTT is used; the problem says which.

How it works

Transmission against propagation

Transmission against propagation delay: start

Computed by a script, not typed

link	transmission delay L/R	propagation delay d/s	larger
1500-byte frame, 100 Mbps, 100 m of cable
1500-byte frame, 100 Mbps, 2000 km of fibre
1500-byte frame, 10 Mbps, 2.5 km (classic Ethernet)
1000-byte packet, 1 Mbps, 36 000 km (satellite hop)
40-byte ack, 1 Gbps, 2000 km

Step 0

The signal speed 2 × 10^8 m/s is the usual approximation for copper and fibre; exam questions state the value to use.

0 / 5

computed by script: exact arithmetic (fractions), CRC long division and exhaustive enumeration of error patterns. python3 scripts/gen-cn1-traces.py

On a short fast link transmission dominates; over long distances, or for tiny packets, propagation dominates and making the link faster does not help: a $40$ -byte acknowledgement on a $1$ Gbps link over $2000$ km is sent in $0.00032$ ms but needs $10$ ms to travel.

The pipe

A sender that transmits continuously at rate $R$ has $R \cdot d$ bits "on the wire" at any moment. If a protocol sends one frame and then waits for its acknowledgement (stop-and-wait), the sender is busy for $L/R$ and idle for $2d$ in every cycle (acknowledgement assumed instantaneous and tiny):

U = \frac{L/R}{L/R + 2d}.

The utilisation falls as the bandwidth-delay product $Rd$ grows relative to the frame size. To keep the pipe full the sender needs $1 + \lceil 2dR / L \rceil$ frames in flight at once (the window size of the sliding window protocols, sliding window).

Bandwidth-delay product and stop-and-wait utilisation: start

Computed by a script, not typed

link	round-trip time	bandwidth-delay product	in bytes	stop-and-wait utilisation	frames needed to fill the pipe
1 Mbps, one-way delay 10 ms
100 Mbps, one-way delay 10 ms
1 Gbps, one-way delay 25 ms
1 Gbps, one-way delay 0.1 ms

Step 0

Acknowledgements are taken to be negligibly short; frames needed = 1 + ⌈2dR/L⌉ (the window size for 100 % utilisation).

0 / 4

computed by script: exact arithmetic (fractions), CRC long division and exhaustive enumeration of error patterns. python3 scripts/gen-cn1-traces.py

On a $1$ Mbps link with $10$ ms one-way delay and frames of $8000$ bits stop-and-wait uses $28.6$ percent of the capacity; on a $1$ Gbps link with $25$ ms delay and $1500$ -byte frames utilisation is $0.024$ percent, and $4168$ frames would have to be in flight. This is why high-speed long-distance links need large windows.

Throughput of a path

For a message of $M$ bits sent over links with rates $R_1, \dots, R_N$ the long-run throughput is $\min R_i$ . With several users sharing the bottleneck, the fair share is $R/n$ .

Worked examples

EX 1Transmission and propagation

Easy

A $1500$ -byte ( $12000$ -bit) frame on a $100$ Mbps link: $d_{trans} = 12000/10^8 = 0.12$ ms. Over $2000$ km at $2 \times 10^8$ m/s: $d_{prop} = 10$ ms. Total for one link: $10.12$ ms.

EX 2Stop-and-wait utilisation

Medium

Frame $1000$ bytes ( $8000$ bits), rate $1$ Mbps, one-way delay $10$ ms. $L/R = 8$ ms; cycle $8 + 20 = 28$ ms; $U = 8/28 = 28.57$ percent. Throughput $= U \times R \approx 285.7$ kbps.

EX 3Frames to fill the pipe

Medium

Rate $1$ Gbps, one-way delay $0.1$ ms, frames of $1500$ bytes: $L/R = 12$ µs, $2d = 200$ µs, so $1 + \lceil 200/12 \rceil = 1 + 17 = 18$ frames; the BDP (with RTT) is $200000$ bits $= 25000$ bytes.

EX 4Typical questions

GATE level

(a) "Does doubling the bandwidth halve the total delay?" Only the transmission part. (b) "Which delay depends on the distance?" Propagation. (c) "Throughput of a path with links of $10$ , $2$ and $5$ Mbps?" $2$ Mbps. (d) "Window size to fill a link with BDP $50000$ bits and $1000$ -bit frames?" $50 + 1 = 51$ frames if the BDP is the RTT product (one frame for the one being sent).

Cheat box

$d_{trans} = L/R$ , $d_{prop} = D/s$ ; total per link = processing + queueing + transmission + propagation.
Throughput $\le$ bottleneck rate. BDP $= R \times$ delay (or RTT).
Stop-and-wait: $U = (L/R)/(L/R + 2d)$ .
Frames to fill the pipe: $1 + \lceil 2dR/L \rceil$ .
Units: $k = 10^3$ , $M = 10^6$ , $G = 10^9$ for rates.

Traps and GATE patterns

Connections

Builds on

Circuit, packet and virtual-circuit switching

Unlocks

Circuit, packet and virtual-circuit switchinguses Medium access: ALOHA, CSMA and CSMA/CDPrerequisite of Medium access: ALOHA, CSMA and CSMA/CDuses Sliding window: stop-and-wait, Go-Back-N, Selective RepeatPrerequisite of Sliding window: stop-and-wait, Go-Back-N, Selective Repeatuses TCP flow controluses TCP congestion controluses HTTPuses

Across subjects

No cross-subject links yet.

Within this subject

Circuit, packet and virtual-circuit switchinguses Sliding window: stop-and-wait, Go-Back-N, Selective Repeatimplemented-by Medium access: ALOHA, CSMA and CSMA/CDuses TCP flow controluses

Medium access protocols compare frame time with propagation delay (medium access); sliding window protocols exist because of this table (sliding window); TCP's window must be at least the BDP to use a fast path (TCP flow control).

PYQs

9 previous year questions from your book. I re-solved each one myself (9 by running or simulating, 0 by reasoning only); the book's key is shown next to my answer. All PYQs

GATE 2026 · MCQ · 1 markQ1 · page 533 of the book

Consider a file of size 4 million bytes being
transferred between two hosts connected via a path
consisting of three consecutive links of bandwidth
2 Mbps, 500 kbps, and 1 Mbps, respectively.

All processing delays and propagation delays are
negligible. Assume there is no other background
traffic over the path and no additional overhead to
transfer the file.

Which one of the following is the total time

(in seconds) to transfer the file?

Note: 1M = 106, 1k = 103

(a)731
(b)64
(c)8
(d)16

GATE 2022 · NAT · 2 marksQ6 · page 533 of the book

Consider a 100 Mbps link between an earth station
(sender) and a satellite (receiver) at an altitude of

COMPUTER NETWORK
2100 km. The signal propagates at a speed of 3×108
m/s. The time taken (in milliseconds, rounded off
to two decimal places) for the receiver to
completely receive a packet of 1000 bytes
transmitted by the sender is: ________

GATE 2016 · MCQQ15 · page 535 of the book

Consider two hosts X and Y ,connected by a single
direct link of rate 106 bits /sec .The distance
between the two hosts is 10,000 km and the
propagation speed along the link is 2 × 108 m/sec
.Host X sends a file of 50,000 bytes as one large
message to host Y continuously .Let the
transmission
and
propagation
delays
be
p
milliseconds and q milliseconds respectively.Then
the value of p and q are:

(a)p = 50 and q = 100
(b)p = 50 and q = 400
(c)p = 100 and q = 50
(d)p = 400 and q = 50

GATE 2015 · NATQ23 · page 536 of the book

Two hosts are connected via a packet switch with 107
bits per second links. Each link has a propagation
delay of 20 microseconds. The switch begins
forwarding a packet 35 microseconds after it receives
the same. If 10000 bits of data are to be transmitted
between the two hosts using a packet size of 5000
bits, the time elapsed between the transmission of the
first bit of data and the reception of the last bit of the
data in microseconds is:

COMPUTER NETWORK

GATE 2012 · MCQ · 2 marksQ30 · page 537 of the book

Consider a source computer (S) transmitting a file of
size 106 bits to a destination computer (D) over a
network of two routers (R1 and R2) and three links
(L1, L2, and L3).

COMPUTER NETWORK

L1 connects S to R1; L2 connects R1 to R2; and L3
connects R2 to D. Let each link be of length 100
km.

Assume signals travel over each link at a speed of
108meters per second.

Assume that the link bandwidth on each link is 1
Mbps. Let the file be broken down into 1000
packets, each of size 1000 bits.

Find the total sum of transmission and propagation
delays in transmitting the file from S to D?

(a)1005 ms
(b)1010 ms
(c)3000 ms
(d)3003 ms

GATE 2007 · MCQ · 2 marksQ36 · page 539 of the book

In a token ring network, the transmission speed is
107 bps and the propagation speed is 200 metres/μs.
The 1-bit delay in this network is equivalent to:

(a)500 metres of cable
(b)200 metres of cable
(c)20 metres of cable
(d)50 metres of cable

GATE 2024 · MCQQ14 · page 549 of the book

Consider a network path P → Q → R between
nodes P and R via router Q. Node P sends a file of
size 106 bytes to R via this path, splitting the file
into chunks one after the other, each of size 103
bytes. Node P sends these chunks without any wait
time between the successive chunk transmissions.
Assume that the size of the extra headers added to
these chunks is negligible, and that the chunk size
is less than the MTU. Each of the links P → Q and
Q → R has a bandwidth of 106 bits/sec, and
negligible
propagation
latency.
Router
Q
immediately transmits every packet it receives
from P to R, with negligible processing and
queuing delays.

Assume P starts transmitting the chunks at time t =
0. Which one of the following options gives the
time (in seconds, rounded off to 3 decimal places)
at which R receives all the chunks of the file?

(a)16.000
(b)8.000
(c)8.008
(d)15.992

GATE 2016 · NATQ50 · page 555 of the book

For a host machine that uses the token bucket
algorithm for congestion control, the token bucket has
a capacity of 1 megabyte and the maximum output
rate is 20 megabytes per second. Tokens arrive at a
rate to sustain output at a rate of 10 megabytes per
second. The token bucket is currently full and the
machine needs to send 12 megabytes of data. The

COMPUTER NETWORK
minimum time required to transmit the data is
______ seconds.

GATE 2008 · MCQ · 2 marksQ84 · page 561 of the book

A computer on a 10 Mbps network is regulated by a
token bucket. The token bucket is filled at a rate of 2
Mbps. It is initially filled to capacity with 16
Megabits. What is the maximum duration for which
the computer can transmit at the full 10 Mbps?

(a)1.6 seconds
(b)2 seconds
(c)5 seconds
(d)8 seconds COMPUTER NETWORK

Self-check

Question 1

Frame $4000$ bits, rate $2$ Mbps, one-way delay $5$ ms. Stop-and-wait utilisation?

Show answer

$L/R = 2$ ms; cycle $2 + 10 = 12$ ms; $U = 2/12 = 16.67$ percent.

Question 2

How many bits are in flight on a $100$ Mbps link with $10$ ms one-way delay?

Show answer

$10^8 \times 0.01 = 10^6$ bits (one way); $2 \times 10^6$ for the round trip.

Question 3

Which delay grows if the packet gets longer: transmission or propagation?

Show answer

Transmission ( $L/R$ ).

Revision card

One-screen revision

Four delays: processing, queueing, transmission $L/R$ , propagation $D/s$ .
Throughput $\le$ slowest link. BDP $= R \cdot$ delay.
Stop-and-wait $U = (L/R)/(L/R + 2d)$ ; window to fill the pipe $1 + \lceil 2dR/L \rceil$ .

References

·Kurose, Ross, Computer Networking: A Top-Down Approach (8th edition), Chapter 1, Section 1.4 (Delay, loss and throughput in packet-switched networks)
·Tanenbaum, Wetherall, Computer Networks (5th edition), Chapter 1, Section 1.5 and Chapter 3 (Data link layer: bandwidth-delay product)

Verification

scripts/enum-checks/cn_basic_check.py: delays against an event simulation (see the switching page), stop-and-wait utilisation = (L/R)/(L/R + 2d) recomputed as busy time over cycle time on 300 random links; all values on the page are exact fractions printed with at most six digits. MODEL (as in the usual exam problems): the acknowledgement is of negligible length and is not delayed, the signal speed is the value given in the problem (2 x 10^8 m/s is used in the tables), there are no errors, no queueing and no processing delay unless stated. Units: bandwidth in bits per second with k = 10^3, M = 10^6, G = 10^9 (the usual convention in networking), whereas memory sizes use powers of two; exam problems state the convention to use. Textbook chapter references are from memory. SECOND PASS (2026-11-02, done by the assistant, not a human check): every Self-check question was solved before the printed answer was read and all agreed; the definitions, worked examples, cheat box and traps were read claim by claim and the numbers re-derived by hand or by a small script (CRC remainders, binomial tail, checksums); long C programs were not re-read line by line (they are compiled and compared by npm run verify:c, and their printed outputs were checked against the prose). No defect found. Status raised to self-checked; verified needs a textbook or human check.

last verified: never

My notes (saved in this browser)

Circuit, packet and virtual-circuit switching

Error detection: parity, checksum and CRC