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Lecture 2 - Weeks 2 3.pptx
Lecture_2_-_Weeks_2___3.pptx
Showing 41-46 out of 70
Lecture 2 - Weeks 2 3.pptx-Data Communication an...
Lecture_2_-_Weeks_2___3.pptx-Data Communication and Computer Networks 2.
Lecture 2 - Weeks 2 3.pptx-Data C...
Lecture_2_-_Weeks_2___3.pptx-Data Communication and Computer Networks 2.
Page 41
Application Layer A 2-41
More about Web caching
❖
cache acts as
both client and
server
▪
server for original
requesting client
▪
client to origin server
❖
typically cache is
installed by ISP
(university,
company,
residential ISP)
why Web caching?
❖
reduce response
time for client
request
❖
reduce traffic on an
institution’s access
link
❖
reduce traffic in the
Internet as a whole
❖
Internet dense with
caches: enables
“poor” content
providers to
effectively deliver
content (so too
Page 42
Application Layer A 2-42
Caching example:
origin
servers
public
Internet
institutiona
l
network
100 Mbps LAN
15 Mbps
access link
assumptions:
❖
100 Mbps
LAN
❖
15 Mbps
access
between institution
router and Internet
router
❖
avg object size:
1 Mbits
❖
avg request rate from
browsers to origin
servers:
15 request/sec
❖
HTTP requests are too
small; hence represent
negligible
load on both
institutional and public
links
❖
Internet delay (RTT from
Internet-side router to
Page 43
Application Layer A 2-43
Caching example:
origin
servers
public
Internet
institutiona
l
network
100 Mbps LAN
15 Mbps
access link
Q: what is the total
response time of this
network?
❖
Total response time (from
time browser requests an
object and it receives it)
is:
❖
LAN delay
+
❖
access delay between
the two routers
+
❖
Internet delay
Page 44
Application Layer A 2-44
Caching example:
traffic intensity (La/R) on the LAN is:
origin
servers
public
Internet
institutiona
l
network
100 Mbps LAN
15 Mbps
access link
(15 requests/sec) . (1
Mbits/request) / (100 Mbps) =
0.15
traffic intensity on access link
(between the
two routers) is:
(15 requests/sec) . (1 Mbits/request) /
(15 Mbps) = 1
❖
traffic intensity of 15% on LAN
➔ small delay (µsecs to10s of
msec)
❖
traffic intensity of 1 on access link
➔ delay grows without bound
order of minutes, if not more!
❖
total delay
= Internet delay +
access delay + LAN delay
=
2 sec + minutes + µsecs ➔
TERRIBLE!
Page 45
Application Layer A 2-45
Caching example:
Solution 1 (expensive solution):
❖
Increase access rate from 15
Mbps to, say, 100 Mbps
traffic intensity on LAN and
access link is:
(15 requests/sec) . (1 Mbits/request) / (100
Mbps) = 0.15
❖
traffic intensity of 15%
➔ small delay (µsecs to10s of
msec)
❖
total delay
= Internet delay +
access delay + LAN delay
=
2 sec + msecs + µsecs
≈
2 seconds
origin
servers
public
Internet
institutiona
l
network
100 Mbps LAN
100
Mbps
access link