Network /Propagation delay Suppose two hosts, A and B, are separated by 20,000 k
ID: 3887131 • Letter: N
Question
Network/Propagation delay
Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 x 108 meter/sec. Note that 1M bits = 1,000,000 bits. Show your work.
a. Calculate the bandwidth-delay product, R x dprop. (Explain your answer)
b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximum number of bits that will be in the link at any given time? (Explain)
c. Provide an interpretation of the bandwidth-delay product.
d. What is the width (in meters) of a bit in the link? Is it longer than a football field?
e. Derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate R, and the length of the link m. (explain)
Explanation / Answer
Hi,
given distance m=20000km,
speed s= 2.5 x 10^8 meter/sec.
rate r=2Mbps
now, propogation delay is given by
dprop=m/s
bandwidth dealy product- as the name indicates, its the product of bandwidth- i,e the amount of bits the link can carry
and the delay of propogation. as r is given, we can calculate dprop as below,
therefore bandwidth delay= r* dprop = 2* (20,000 *1000 m/2.5*10^8 m/s) = 160,000 bits
Now, since we know the bandwidth of link is 160kbits, any data we send will be in parts of this, hence ,
b. given 800kbits which is more than the calcualated 160kbit, therefore max is 160,000 bits
c.It shows total numbers of bits in link at full go i,e full capacity
d.width is total length/number of bits= 20,000km/1600000 bits = 125 m/bit, the lenght of a football field is approx 120 yards or 110 meters therefore more than it.
e. we got d above as m/ (m/s)*R the denominator is from a part
so width= s/r
Thumbs up if this was helpful, otherwise let me know in comments. Good Day.