MTR 3013 SEM 1 2008/2009Quiz 1 – Solution ( P18 page 99)

propagation speed, s transmission rate Rlength of the link, m

a) R = 2 Mbps d prop = m/s = 10,000 km /( 2.5 x 10^8 m/s) = 10,000,000 m / (2.5 x 10^8 m/s) = 1/25 sec.

Bandwidth delay product = R * d prop = 2 Mbps * (1/25) s = 80,000 bits

b) 80,000 bits

c) The bandwidth-delay product of a link is the maximum number of bits that can be in the link in any given time.

d) m = 10,000 km Bandwidth delay product = 80,000 bits

80,000 bits 10,000 km1 bit 10,000 km / 80,000 bits

0.125 km / bits

A bit will occupy 0.125 kilometer = 125 meter length.

Football field is 100 yards = 91.44 meter, so the bit is longer than football filed.

e) width of the bit, w = m / Bandwidth delay products = m / ( R * d prop ) = m / ( R * m/s)

= s/R

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