Ernest Grindle
Network Fundamentals
Problem Set 1


1.
a) dprop = m/s

b) dtrans = L/R

c) end-to-end delay = dprop + dtrans = (m/s)+(L/R)

d) Just leaving host A

e) In the next host

f) On the wire heading to the next host

g) 
dprop = (m/2.5*10^8meter/second)
dtrans = (100bit/28Kb)
(m/2.5*10^8meter/second) = (100bit/28Kb)
m = (100bit/1)(1/28Kbit)(2.5*10^8meter/second)
m = (100/28000)(2.5*10^8meter/second)
m = (892857.14meter/second)
I don't know where I should have factored out the second.


2.
a)
=(2RTT/1)(100ms/1RTT)(1second/1000ms)+(1000000KB)(1second/1500000Kb)(8bit/1Byte)
=(0.2second)+(5.33second)
=(5.53second)

b)
=(0.2second)+(1000packets/1)(100ms/1RTT)(1RTT/1packet)(1second/1000ms)+
	(1000KB/1)(second/1.5Mbit)(1Mbit/1000Kbit)(8bit/1byte)
=(0.2second)+(100second)+(5.33second)
=(105.53second)

c)
=(0.2second)+(100second/20)
=(0.2second)+(5second)
=(5.2second)

d)
=0.2s+0.1s+0.1s+0.1s+0.1s+0.1s+0.1s+0.1s+0.1s+0.1s+0.1s
=(1.2second)
I know for this I should have used e or log or something, but I forget.
Fortunately it was a quick compound!


3.
a)
First switch
=(7.5*10^6bit)(second/1.5Mbit)(1Mbit/1*10^6bit)
=(7.5second/1.5)
=(5second)
Entire Transmit
=(3stops/1)(5second/stop)
=(15second)

b)
First Packet
=(1500bit/1)(second/1.5Mbit)(1Mbit/1*10^6bit)
=(0.001second)
Second Packet
=(2packt/1)(0.001second/packet)
=(0.002second)

c)
=(5000packet/1)(0.001second/packet)+(3stop/1)(0.001second/stop)
=(5second)+(0.003second)
=(5.003second)
In this case, using segmentation cut the total transmition time to 1/3.

d)Packets can get dropped.


4.
goergewbush.com
a)average = 0.579ms; standard deviation = 0.168ms
b)9 routers; same path
c)+- 3 isps (missing reverse lookup on some).  Largest time inside an ISP
d)
airshow.com.cn
average = 6.94ms
standard deviation = 0.264ms
13 routers; same path
+- 3 ISPs (missing reverse lookup on some).  Largest time was inside a section
	missing reverse lookup.