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Delay. Advanced Computer Networks تذکر: جوابهای هر سوال را در اسلاید بعدی بنویسید. Q1.
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Delay Advanced Computer Networks تذکر: جوابهای هر سوال را در اسلاید بعدی بنویسید
Q1 • How long does it take a packet of length 1,000 bytes to propagate over a link of distance 5,000 km, propagation speed 2.5 * 108 m/s. and transmission rate 1 Mbps? Generally, how long does it take a packet of length L to propagate over a link of distance d, propagation speed s, and transmission rate R bps? Does this delay depend on packet length? Does this delay depend on transmission rate?
Q1 Answer Total delay=transmitiondelay+propagte delay Total delay= (L/R)+(d/s) Total delay=(8000/10^6)+(5*10^6/2.5*10^8) Total delay=0.008+0.02 second this delay depend on packet length and transmission rate.
Q2 • How long does it take to transmit a packet of length 1,000 bytes a link of distance 5,000 km, propagation speed and transmission rate 1 Mbps? Generally, how long does it take to transmit a packet of length L over a link of distance d, propagation speed s, and trans-mission rate R bps? Does this delay depend on the length of the link? Does this delay depend on the propagation speed of the link?
Q2 Answer • Total delay= (L/R)+(d/s) • Total delay=(8000/10^6)+(5*10^6/10^6)= • 0.008+5 second • this delay depend on the length of the link and propagation speed of the link.
Q3 • Consider two packet switches directly connected by a link of 5000 km, propagation speed 2.5 * 108 m/s and transmission rate 1 Mbps. How long does it take to move a packet of length 1,000 bytes from one packet switch to the other packet switch? Gen-erally, how long does it take to move a packet of length L over a link of dis-tance d, propagation speed s, and transmission rate R bps?
Q3 Answer • Total delay= (L/R)+(d/s) • Total delay= (8000/10^6)+(5*10^6/2.5*10^6) • Total delay= 0.008+0.02 second
Q4 Answer • Total delay= [(L/R1)+(d1/s1)]+[(L/R2)+(d2/s2)] • Total delay=[(8000/10^6)+(4*10^6/2.5*10^8)]+[(8000/10^6)+(10^6/2.5*10^8)]=0.008+0.016+0.008+0.004