# Homework 1 (FCN)

## Ivan Christian (1003056)

# 1
a) $10 M / 200 kb = 50s$
50 users can be supported at 1 time using circuit switching
b) 150 users (packet switching)

$$
P(user) = (^{150}_{i})* 0.05^i * 0.95 ^ {150-i}
$$
c) sum of all p from 17 to 150

$$
P(X ≥ 17) = \sum_{i=17}^{150} (^{150}_{i})0.5^{i} *0.95^{150-i}
$$
$$
P(X ≥ 17) = 0.00143912
$$
# 2
x kB (NB. B = bytes, cf. b = bits) over a y Mbps link

$$
t = \frac {8x}{10^3 y} s
$$
# 3

> a) Delay Sensitive
> b) Bandwidth Sensitive
> c) Bandwidth Sensitive
> d) Delay Sensitive

# 4

a) minimum RTT is the time take for a return trip of a packet.

$$
t= 2*\frac {3.8*10^5*10^3}{2.8*10^8} s
$$

$t=2.7142 s$

b) Delay = $2.7142 s$
Delay Bandwidth product = $Delay * 500 Mb/s$
Delay Bandwidth product = $1.357 Gb$
c) Delay Bandwidth product is a measurement of how many bits can fill up a network link. It gives the maximum amount of data that can be transmitted by the sender at a given time before waiting for acknowledgment. Thus it is the maximum amount of unacknowledged data. Any more data would be dropped by the network.
d) $t_{delay} = \frac{30*10^6*8}{500*10^6}s$
$t_{delay} = 0.48 s$

Total Delay = $2.7142 s + 0.48 s$ = $3.1942s$

# 5
a)
$$
t = \frac {15000}{10^8}s
$$

Total delay = $2*(t+12) μs$ = $324μs$

b) 4 switches along the way means 5 delay since the the switches come with those.

$$
Total = 5*(t + 12) μs = 810 μs
$$

c) For every 300 bits there is a delay of $\frac{300}{100*10^6}s$

**Total Delay** = $3 μs + 2*12μs + 150 μs$ = $177 μs$

d)


# 6

Packet loss probability $p$

$P(received) = (1-p)^N$ is the probability that the packet is received

$P($Not received$) =1 - (1-p)^N=q$
$P(retransmission) = q^N (1-q)$ Because loss is need to do a restransmission and there is only a need to have 1 received packet to stop
$E(retransmission) = \sum_{i} i*[q^i(1-q)]$

$E(retransmission)=(1-q)\sum_{i}^N i*q^i]$

$E(retransmission)=(1-q)[q + 2q^2+ 3q^3+...+n*q^n]$
$E(retransmission)=q-q^2 +2q^2 -2q^3 + 3q^3-3q^4 + ... + nq^n -(n+1)q^n$
$E(retransmission) = q+q^2+q^3+q^4+...+q^n$
$E(retransmission) = q \sum_{i=0}q^i$
$E(restransmission)= \frac{q}{1-q}$ for $|q|<1$(which is always true because $q$ is a probability) , where $q=1 - (1-p)^N$

# 7

a)
Day 1
traceroute to www.example.com

18  93.184.216.34 (93.184.216.34)  192.431 ms  192.619 ms  192.721 ms
18  93.184.216.34 (93.184.216.34)  192.559 ms  192.737 ms  192.647 ms
18  93.184.216.34 (93.184.216.34)  192.538 ms  192.531 ms  192.525 ms
18  93.184.216.34 (93.184.216.34)  192.699 ms  192.585 ms  192.579 ms
18  93.184.216.34 (93.184.216.34)  192.234 ms  192.223 ms  192.581 ms
18  93.184.216.34 (93.184.216.34)  192.142 ms  192.203 ms  192.121 ms
18  93.184.216.34 (93.184.216.34)  192.955 ms  192.538 ms  192.531 ms
18  93.184.216.34 (93.184.216.34)  192.631 ms  192.356 ms  192.349 ms
18  93.184.216.34 (93.184.216.34)  192.631 ms  192.356 ms  192.349 ms
18  93.184.216.34 (93.184.216.34)  192.094 ms  192.863 ms  192.184 ms

Day 2

18  93.184.216.34 (93.184.216.34)  193.031 ms  192.646 ms  192.692 ms
18  93.184.216.34 (93.184.216.34)  192.462 ms  192.449 ms  192.570 ms
18  93.184.216.34 (93.184.216.34)  192.183 ms  192.465 ms  192.458 ms
18  93.184.216.34 (93.184.216.34)  192.471 ms  192.553 ms  192.762 ms
18  93.184.216.34 (93.184.216.34)  193.176 ms  193.168 ms  192.161 ms
18  93.184.216.34 (93.184.216.34)  192.455 ms  192.879 ms  192.692 ms
18  93.184.216.34 (93.184.216.34)  192.184 ms  192.111 ms  192.804 ms
18  93.184.216.34 (93.184.216.34)  192.623 ms  192.616 ms  192.363 ms
18  93.184.216.34 (93.184.216.34)  192.409 ms  192.401 ms  192.468 ms
18  93.184.216.34 (93.184.216.34)  192.431 ms  192.424 ms  192.581 ms

Day 3

18  93.184.216.34 (93.184.216.34)  193.884 ms  192.682 ms  192.673 ms
18  93.184.216.34 (93.184.216.34)  192.406 ms  193.295 ms  192.391 ms
18  93.184.216.34 (93.184.216.34)  192.365 ms  192.511 ms  192.493 ms
18  93.184.216.34 (93.184.216.34)  192.535 ms  192.517 ms  192.474 ms
18  93.184.216.34 (93.184.216.34)  192.523 ms  193.194 ms  193.178 ms
18  93.184.216.34 (93.184.216.34)  193.251 ms  193.047 ms  193.315 ms
18  93.184.216.34 (93.184.216.34)  192.781 ms  192.271 ms  192.510 ms
18  93.184.216.34 (93.184.216.34)  192.575 ms  192.521 ms  192.798 ms
18  93.184.216.34 (93.184.216.34)  203.890 ms  203.883 ms  203.965 ms
18  93.184.216.34 (93.184.216.34)  192.500 ms  192.493 ms  192.487 ms

Average 1 = $192.48373333333$
Standard deviation 1= $0.21897700538844$

Average 2 = $192.55626666667$
Standard Deviation 2 = $0.26189577231325$

Average 3 = $193.84693333333$
Standard Deviation 3 = $3.374768346947$

Total Average = $192.8608$ ms

Standard deviation = $1.7940394672359$ ms

b) 18 hops and no changes to the hops
c) 

There are 3 different ISPs between source computer and the destination site.

myrepublic.com.sg
sprintlink.net
edgecastcdn.net

Hop 12 to 13 is extremely large, meaning that the distance is also large.