Availability or downtime of the servers can be found out

numericals

1. Availability
•Availability can be expressed as a percent
uptime per year, month, week, day, or hour,
compared to the total time in that period
–For example:
•24/7 operation
•Network is up for 165 hours in the 168-hour week
•Availability is 98.21%
•Different applications may require different
levels
•Some enterprises may want 99.999% or
“Five Nines” availability

2. Availability
Downtime in Minutes
4.32
1.44
.72
.01
30
10
5
.10
157799.70%
52699.90%
26399.95%
599.999%
Per HourPer DayPer WeekPer Year
.18
.06
.03
.0006
.29 210599.98% .012

3. 99.999% Availability May Require
Triple Redundancy
Enterprise
ISP 1 ISP 2 ISP 3
•Can the customer afford this?

4. Availability
•Availability can also be expressed as a
mean time between failure (MTBF) and
mean time to repair (MTTR)
•Availability = MTBF/(MTBF + MTTR)
–For example:
•The network should not fail more than once every 4,000
hours (166 days) and it should be fixed within one hour
•4,000/4,001 = 99.98% availability

5. Scalability
•Scalability refers to the ability to grow
•Some technologies are more scalable
–Flat network designs, for example, don’t scale well
•Try to learn
–Number of sites to be added
–What will be needed at each of these sites
–How many users will be added
–How many more servers will be added

6. Network Performance
•Common performance factors include
–Bandwidth
–Throughput
–Bandwidth utilization
–Offered load
–Accuracy
–Efficiency
–Delay (latency) and delay variation
–Response time

7. Bandwidth Vs. Throughput
•Bandwidth and throughput are not the same
thing
•Bandwidth is the data carrying capacity of a
circuit
•Usually specified in bits per second
•Throughput is the quantity of error free data
transmitted per unit of time
•Measured in bps, Bps, or packets per second (pps)

8. Bandwidth, Throughput, Load
Offered Load
T
h
r
o
u
g
h
p
u
t
Actual
I d
e
a
l
100 % of Capacity
100 % of Capacity

9. Other Factors that Affect Throughput
•The size of packets
•Inter-frame gaps between packets
•Packets-per-second ratings of devices that forward packets
•Client speed (CPU, memory, and HD access speeds)
•Server speed (CPU, memory, and HD access speeds)
•Network design
•Protocols
•Distance
•Errors
•Time of day, etc., etc., etc.

10. Delay from the User’s Point of View
•Response Time
–A function of the application and the equipment the
application is running on, not just the network
–Most users expect to see something on the screen in
100 to 200 milliseconds

11. Delay from the Engineer’s Point of
View
•Transmission delay (also known as
serialization delay)
•Propagation delay
•Processing delay
•Queuing delay

12. Transmission delay
time taken to transmit a packet from the host to the
transmission medium
T
d
= L/B
This delay depends upon the following factors:
•If there are multiple active sessions, the delay will become
significant.
•Increasing bandwidth decreases transmission delay.

13. Propagation delay
The time taken by the last bit of the packet to reach the destination.
Tp = Distance / Velocity
Note: 
 
Velocity =3 X 10^8 m/s (for air)
Velocity= 2.1 X 10^8 m/s (for optical fibre)

14. Processing delay
the time required by intermediate routers to decide where to
forward the packet, update TTL, perform header checksum
calculations.
depends upon the speed of the processor.

15. Queueing delay
When the packet is received by the destination, the packet will not
be processed by the destination immediately.
It has to wait in a queue in something called a buffer.
So the amount of time it waits in queue before being processed is
called queueing delay.
Note:
If the size of the queue is large, the queuing delay will be huge. If
the queue is empty there will be less or no delay.
If more packets are arriving in a short or no time interval, queuing
delay will be large.
The less the number of servers/links, the greater is the queuing
delay.

16. Example
•A packet switch has 5 users, each offering
packets at a rate of 10 packets per second
•The average length of the packets is 1,024 bits
•The packet switch needs to transmit this data
over a 56-Kbps WAN circuit
–Load = 5 x 10 x 1,024 = 51,200 bps
–Utilization = 51,200/56,000 = 91.4%
–Average number of packets in queue =
(0.914)/(1-0.914) = 10.63 packets

17. Circuit switching Network:
a dedicated path is established between the sending and receiving device.
In this physical links connect via a set of switches.
computer connect via 4 switches with a point to point connections.

18. Packet switching Network:
the message is divide into packets. Each packet contains a header which
includes the source address, destination address, and control
information.
In the above figure, it shows how a data gram approach is used to
deliver four packets from station A to station D.

20. In the good old days of cables that consisted of a large number of copper wires, and
switching stations with relays that effectively implemented a big crossbar, circuit
switching meant that the provider put all the switches in the position that made a
real (electrical) circuit from one client to another. The two literally had a cable for
their own private use (for as long as then connection was established).
Virtual circuit switching gives the clients the impression that they have a dedicated
cable, while in fact they are allocated time-slots (or frequency bands, or some
other shared slice) in all the cables that make up the connection.
This better matches the current economic reality that a cable is very expensive, but
has a bandwidth that far exceeds the needs of most clients, hence sharing is
economical.
The difference between a virtual switched circuit and packet switching is that for a
virtual circuit all the steps that make up the connection are determined when the
connection is established, and the buffers and bandwidth at each step is claimed
and thus guaranteed.
For (plain) packet switching there is no such circuit creation: each packet finds its
own way through the network, and has to hope that sufficient buffers and
bandwidth is available at each step.

21. Timing in Circuit Switching
Assume:
Number of hops = M
Per-hop processing delay = P
Link propagation delay = L
Transmission speed = W bit/s
Message size = B bits
Total Delay = total propagation
+ total transmission
+ total processing
= 4ML + B/W + (M-1)P
P
L
B/W
Total Delay

22. Timing in Datagram Packet
Switching
Assume:
Number of hops = M
Per-hop processing delay = P
Link propagation delay = L
Packet transmission delay = T
Message size = N packets
Total Delay = total propagation
+ total transmission
+ total store&forward
+ total processing
= ML + NT + (M-1)T + (M-1)P
P
T
L
Total
Delay
P
T

23. Timing in Virtual Circuit Packet
Switching
Assume:
Number of hops = M
Per-hop processing delay = P
Link propagation delay = L
Packet transmission delay = T
Message size = N packets
Total Delay = total propagation
+ total transmission
+ total store&forward
+ total processing
= 4ML + NT + (M-1)T + 4(M-1)P
P
T
L
Total
Delay
P
T
P

24. Remark
•We are often interested only in the delay elapsed from the time the
first bit was sent to the time the last bit was received (i.e., we exclude
the time involved in acknowledging connection termination).
•If this is the case, the delay will be given as follows:
–Circuit Switching:
Delay = 3ML + B/W + (M-1)P
–Datagram packet switching:
Delay = ML + NT + (M-1)T + (M-1)P (same as before)
–Virtual circuit packet switching:
Delay = 3ML + NT + (M-1)T + 3(M-1)P

25. Solved Exercises
Q1: It’s 1989. Alice and Bob are 4 hops apart on a datagram packet-switched network where
each link is 100 mile long. Per-hop processing delay is s. Packets are 1500 bytes long.
All links have a transmission speed of 56kbit/s (original speed of Internet backbone links in the
80s). The speed of light in the wire is approximately 125,000 miles/s.
If Bob sends a 10-packet message to Alice, how long will it take Alice to receive the message
up to the last bit (measured from the time Bob starts sending)?
•Answer: We know the following:
–Number of hops M=4,
–Number of packets N=10,
–Per-hop processing delay P=5s=0.000005s,
–Link propagation delay L = distance/speed of light = 100/125,000 = 0.0008s,
–Packet size = 1500 bytes = 1500*8=12,000 bits,
–Packet transmission delay T = packet size/transmission speed = 12,000/56000

=0.214s.
Delay = ML + NT + (M-1)T + (M-1)P =0.0032 + 2.14 + 0.642 + 0.000015 = 2.785s.
•Note that the total delay is dominated by the transmission delay which depends on link
speed. A link with a higher transmission speed can reduce the delay dramatically.

26. Solved Exercises
Q2: Alice and Bob 22 years later. All is the same, except that link transmission speed now
is 1Gbit/s. How long will it take Alice to receive the message up to the last bit
(measured from the time Bob starts sending)?
•Answer: As before, we know the following:
–Number of hops M=4,
–Number of packets N=10,
–Per-hop processing delay P=5s,
–Link propagation delay L = distance/speed of light = 100/125,000 = 800s,
–Packet size = 1500 bytes = 1500*8=12,000 bits,
–Packet transmission delay T = packet size/transmission speed = 12,000/10
9
=12s.
Delay = ML + NT + (M-1)T + (M-1)P =3200 + 120 + 36 + 15 = 3371s = 3.371ms.
•Note that the total delay is now dominated by the propagation delay which cannot be
improved because it is constrained by the speed of light. Hence, it is unlikely that future
technologies will significantly reduce the delay of Bob’s message at this point (unless
we break the speed of light)!

27. Q3: Repeat Q1 and Q2, assuming that the network uses circuit switching
instead of datagram packet switching. Bob’s message is the same length
as before.
•Answer: Year 1989: Let link transmission speed be W = 56kbit/s,
Number of hops M=4,
Message size B = 10 * 1500 * 8 =120,000 bits (it is not packetized)
Per-hop processing delay P=0.000005s,
Link propagation delay L = distance/speed of light = 100/125,000 =
0.0008s,
Delay = 3ML + B/W + (M-1)P =0.0096 + 2.14 + 0.000015 = 2.1496s
•Note that the delay improved over the case of datagram packet switching
for the same link speed. Why?
•Year 2011: Let link transmission speed be W = 1Gbit/s
Delay = 3ML + B/W + (M-1)P =9600 + 120 + 15 = 9735 s = 9.735ms

•Note that the delay is worse than in the case of datagram packet
switching. Why?

28. Observations
•With the advances in transmission speed total delays are dominated by
propagation delays which are bound by the speed of light.
•Circuit switching adds an extra roundtrip over datagram packet
switching, but eliminates store and forward delays. We have two cases:
–When links are slow, the bottleneck is transmission speed on the
link.
–Eliminating the need to store-and-forward helps a lot. The extra
roundtrip adds only negligible delay. Hence, using circuit switching
results in a net reduction in delay.
–When links are fast, the bottleneck is propagation delay. Adding a
roundtrip hurts a lot.
–Eliminating the need for store-and-forward saves a negligible
amount of time. Hence, using circuit switching results in a net
increase in delay.