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Sagot :
Answer:
0.095 sec
Explanation:
From the information given:
Packet size = 2000 bytes
The propagation speed on both the links = [tex]3.0*10^8 \ m/s[/tex]
The transmission rates of all three links = 2 Mbps
Packet switch processing delay = 1 msec
The length of the 1st link = 8000 km
The length of the 2nd link = 4000 km
The length of the 3rd link = 2000 km
The length of the 4th link = 1000 km
The length of the 5th link = 1000 km
The 1st end system that needs to transmit the packet onto the1st link = [tex]L/R_1[/tex]
[tex]= \dfrac{2000*8}{2*10^6}[/tex]
= 0.008 sec
The packet propagates over the 1st link in [tex]d1/s1[/tex] is;
[tex]= \dfrac{8000*10^3}{3.0 \times 10^8}[/tex]
= 0.027 sec
The packet switch generates a delay of [tex]d_{proc}= 1msec[/tex], after receiving the whole packet
The 1st end system that needs to transmit the packet onto the 2nd link = [tex]L/R_2[/tex]
[tex]= \dfrac{2000*8}{2*10^6}[/tex]
= 0.008 sec
The packet propagates over the 2nd link in [tex]d2/s2[/tex] is;
[tex]= \dfrac{4000*10^3}{3.0 \times 10^8}[/tex]
= 0.013 sec
Again, the packet switch generates a delay of [tex]d_{proc}= 1msec[/tex], after receiving the whole packet
The 1st end system that needs to transmit the packet onto the 3rd link = [tex]L/R_3[/tex]
[tex]= \dfrac{2000*8}{2*10^6}[/tex]
= 0.008 sec
The packet propagates over the 3rd link in [tex]d3/s3[/tex] is;
[tex]= \dfrac{2000*10^3}{3.0 \times 10^8}[/tex]
= 0.007 sec
Again, the packet switch generates a delay of [tex]d_{proc}= 1msec[/tex], after receiving the whole packet
The 1st end system that needs to transmit the packet onto the 4th link [tex]= L/R_4[/tex]
[tex]= \dfrac{2000*8}{2*10^6}[/tex]
= 0.008 sec
The packet propagates over the 4th link in [tex]d4/s4[/tex] is;
[tex]= \dfrac{1000*10^3}{3.0 \times 10^8}[/tex]
= 0.003 sec
Again, the packet switch generates a delay of [tex]d_{proc}= 1msec[/tex], after receiving the whole packet
The 1st end system that needs to transmit the packet onto the 5th link = [tex]L/R_5[/tex]
[tex]= \dfrac{2000*8}{2*10^6}[/tex]
= 0.008 sec
The packet propagates over the 5th link in [tex]d5/s5[/tex] is;
[tex]= \dfrac{1000*10^3}{3.0 \times 10^8}[/tex]
= 0.003 sec
The end-to-end delay = [tex]L/R_1+L/R_2+L/R_3+L/R_4+L/R_5 + d_1/s_1+d_2/s_2+d_3/s_3+d_4/s_4+d_5/s_5+d_{proc}+d_{proc}[/tex]
=0.008+0.008+0.008+0.008+0.008+0.027+0.013+0.007+0.003+0.003+0.001+0.001
= 0.095 sec
Hence, the end to end delay = 0.095 sec
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