SME
Computer Networks
Network performance
Use this for “A 1500-byte packet waits 2 ms in a queue, then crosses a 2000 km, 100 Mb/s link. How long does it take?”
Diagram
The packet leaves bit by bit for dₜ, and each bit takes dₚ to cross the link. total = 2 ms + 120 μs + 10 ms = 12.12 ms: the last bit arrives dₜ + dₚ after the first leaves.
The shorter time is drawn wider than to scale so it shows.
This is an example. Type your own values to start a new problem.
dₜₒₜₐₗ (ms)
=
dq (ms)
+
dₜ (μs)
+
dₚ (ms)
L
Packet size
example · B
R
Link rate
example
d
Distance
example
s
Signal speed
example · m/s
Formulas
dₜ
=
8
×
L
R
0.00012
=
8
×
1,500
100,000,000
dₚ
=
d
s
0.01
=
2,000,000
2 ×
10
8
dₜₒₜₐₗ = dq + dₜ + dₚ
0.01212 = 0.002 + 0.00012 + 0.01
Formulas are worked in b/s, s, m. The step-by-step shows the conversions.
Assumptions
•
The signal travels at about s = 2 × 10⁸ m/s in fiber or copper.
•
The packet waits dq in the queue, then its bits go out one after another at the rate R.
•
The receiver has the packet when its last bit arrives; 1 Mb/s is 10⁶ bits a second.
Step-by-step
WE KNOW
L = 1,500 B, R = 100 Mb/s, d = 2,000 km, s = 2 × 10⁸ m/s, dq = 2 ms
FIND
transmission delay (dₜ), propagation delay (dₚ), total delay (dₜₒₜₐₗ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to b/s, s, m).
R = 100 Mb/s = 100,000,000 b/s (1 Mb/s = 10⁶ b/s)
d = 2,000 km = 2,000,000 m (1 km = 1000 m)
dq = 2 ms = 0.002 s (1 s = 1000 ms)
Step 2 · Find transmission delay (dₜ)
Use dₜ = 8 × L ÷ R
The
packet’
s
bits,
eight
a
byte,
go
out
at
R
bits
a
second.
dₜ
=
8
×
1,500
100,000,000
dₜ
=
12,000
100,000,000
dₜ = 0.00012 s
Step 3 · Find propagation delay (dₚ)
Use dₚ = d ÷ s
Distance over speed: the time one bit takes to cross the link.
dₚ
=
2,000,000
2 ×
10
8
dₚ = 0.01 s
Step 4 · Find total delay (dₜₒₜₐₗ)
Use dₜₒₜₐₗ = dq + dₜ + dₚ
The delays follow one another, so they add.
dₜₒₜₐₗ = 0.002 + 0.00012 + 0.01
dₜₒₜₐₗ = 0.01212 s
Step 5 · Convert the answers
Change each answer back to the unit you chose.
dₜ = 0.00012 s = 120 μs (1 s = 10⁶ μs)
dₚ = 0.01 s = 10 ms (1 s = 1000 ms)
dₜₒₜₐₗ = 0.01212 s = 12.12 ms (1 s = 1000 ms)
Check
Put the numbers back in. Both sides should match.
0.00012
=
8
×
1,500
100,000,000
✓
0.01
=
2,000,000
2 ×
10
8
✓
0.01212 = 0.002 + 0.00012 + 0.01 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Bandwidth-delay product
Use this for “How many bytes are in flight on a 100 Mb/s path with a 20 ms round trip?”
Stop-and-wait utilization
Use this for “A frame takes 0.12 ms to send and the round trip is 20 ms. How busy is a stop-and-wait link?”
Queueing delay
Use this for “Packets reach a router at 800 a second and it serves 1000 a second. Find the average delay and number queued.”
Bottleneck link
Use this for “A 4 MB file crosses a 10 Mb/s link, then a 2 Mb/s link. How long does the transfer take?”