Respuesta :
Answer:
a. (210^6)((210^7)/(2.510^8)) = 1.610^5 bits or 160,000 bits
b. 1.6*10^5 bits or 160,000 bits
c. Bandwidth delay product of link is maximum number of bits that can be in the link
d. Width of bit = Length of link / bandwidth-delay product so 1 bit is 125 meters long. Yes, this is longer than a football field.
e. Width of bit = s/R
Hope this helps :)
For point a:
The distance from of the two hosts A to B = 20,000 km
[tex]=2\times 10^{7} meters \left ( since\ 1km=10^{3}m \right )[/tex]
Direct connection rate (R) from A to B =2Mbps
[tex]=2\times 10^{6}bps \left ( 1Mbps =10^{6}bps \right )[/tex]
Speed(S) of propagation of a link from A to B [tex]=2.5\times 10^{8} meters/sec[/tex]
Calculate the time limit for propagation:
[tex]d_{prog}=\frac{Distance}{Speed} =\frac{2\times 10^{7}}{2.5\times10^{8} } =0.08sec[/tex]
bandwidth-delay product calculation:
[tex]R \times d_{prog} =2\times 10^{6}\times 0.08 =16\times 10^{4}bits[/tex]
The delay product is therefore [tex]160000\ bit[/tex]
For point b:
File size[tex]=800000\ bits =8\times 10^{5} \ bits[/tex]
Direct connection rate (R) from A to B =2Mbps
[tex]=2\times 10^{6}\ bps \left ( 1Mbps =10^{6}bps \right )[/tex]
product bandwidth-delay:
[tex]R \times d_{prog} =2\times 10^{6}\times 0.08 =16\times 10^{4}\ bits[/tex]
The maximum number of bits will thus be [tex]160000\ bit[/tex] at a given time.
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