计算机网络-哈尔滨工业大学mooc-第4周作业（网络应用 下）解答

第4周作业 网络应用（下）

@[.ComputerNetwork]

文章目录

• 第4周作业 网络应用（下）
• 客户-服务器模型
• p2p分发模型
• 画图脚本

最小分发时间图标如下：

客户-服务器模型

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$d_i = 2M_{bps}$di​=2Mbps​

$F = 15Gb$F=15Gb

$u_s = 30M_{bps}$us​=30Mbps​

$t_{min} =max( \frac{N*F}{u_s},\frac{F}{d_i})$tmin​=max(us​N∗F​,di​F​)

$N = 10 时， t_{min} = max(\frac{10*15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}} )= max(5000,7500)=7500s$N=10时，tmin​=max(30Mbps​10∗15Gb​,2Mbps​15Gb​)=max(5000,7500)=7500s

$N = 100 时， t_{min} = max(\frac{100*15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}} )= max(50000,7500)=50000s$N=100时，tmin​=max(30Mbps​100∗15Gb​,2Mbps​15Gb​)=max(50000,7500)=50000s

$N = 1000 时， t_{min} = max(\frac{1000*15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}} )= max(500000,7500)=500000s$N=1000时，tmin​=max(30Mbps​1000∗15Gb​,2Mbps​15Gb​)=max(500000,7500)=500000s

p2p分发模型

$d_i = 2M_{bps}$di​=2Mbps​

$F = 15Gb$F=15Gb

$u_s = 30M_{bps}$us​=30Mbps​

$d_{P2P}=max(\frac{F}{u_s},\frac{F}{min(d_i)},\frac{NF}{u_s+\sum{u_i}})$dP2P​=max(us​F​,min(di​)F​,us​+∑ui​NF​)

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当$u=500kbps$u=500kbps时

$N = 10 时，d_{P2P}=max(\frac{15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}},\frac{10*15Gb}{500K_{bps}*10+30M_{bps}})=max(500,7500,4286)=7500s$N=10时，dP2P​=max(30Mbps​15Gb​,2Mbps​15Gb​,500Kbps​∗10+30Mbps​10∗15Gb​)=max(500,7500,4286)=7500s

$N = 100 时，d_{P2P}=max(\frac{15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}},\frac{100*15Gb} {500K_{bps}*100+30M_{bps}})=max(500,7500,18750) = 18750s$N=100时，dP2P​=max(30Mbps​15Gb​,2Mbps​15Gb​,500Kbps​∗100+30Mbps​100∗15Gb​)=max(500,7500,18750)=18750s

$N = 1000 时，d_{P2P}=max(\frac{15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}},\frac{1000*15Gb} {500K_{bps}*1000+30M_{bps}})=max(500,7500,28301) = 28301s$N=1000时，dP2P​=max(30Mbps​15Gb​,2Mbps​15Gb​,500Kbps​∗1000+30Mbps​1000∗15Gb​)=max(500,7500,28301)=28301s

当$u=1Mbps$u=1Mbps时

$N = 10 时，d_{P2P}=max(\frac{15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}},\frac{10*15Gb}{1M_{bps}*10+30M_{bps}})=max(500,7500,3750s)=7500s$N=10时，dP2P​=max(30Mbps​15Gb​,2Mbps​15Gb​,1Mbps​∗10+30Mbps​10∗15Gb​)=max(500,7500,3750s)=7500s

$N = 100 时，d_{P2P}=max(\frac{15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}},\frac{100*15Gb} {1M_{bps}*100+30M_{bps}})=max(500,7500,11538) = 11538s$N=100时，dP2P​=max(30Mbps​15Gb​,2Mbps​15Gb​,1Mbps​∗100+30Mbps​100∗15Gb​)=max(500,7500,11538)=11538s

$N = 1000 时，d_{P2P}=max(\frac{15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}},\frac{1000*15Gb} {1M_{bps}*1000+30M_{bps}})=max(500,7500,14563) = 14563s$N=1000时，dP2P​=max(30Mbps​15Gb​,2Mbps​15Gb​,1Mbps​∗1000+30Mbps​1000∗15Gb​)=max(500,7500,14563)=14563s

当$u=2Mbps$u=2Mbps时

$N = 10 时，d_{P2P}=max(\frac{15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}},\frac{10*15Gb}{2M_{bps}*10+30M_{bps}})=max(500,7500,3000s)=7500s$N=10时，dP2P​=max(30Mbps​15Gb​,2Mbps​15Gb​,2Mbps​∗10+30Mbps​10∗15Gb​)=max(500,7500,3000s)=7500s

$N = 100 时，d_{P2P}=max(\frac{15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}},\frac{100*15Gb} {2M_{bps}*100+30M_{bps}})=max(500,7500,6522) = 7500s$N=100时，dP2P​=max(30Mbps​15Gb​,2Mbps​15Gb​,2Mbps​∗100+30Mbps​100∗15Gb​)=max(500,7500,6522)=7500s

$N = 1000 时，d_{P2P}=max(\frac{15Gb}{30M_{bps}},\frac{15Gb}{2M_{bps}},\frac{1000*15Gb} {2M_{bps}*1000+30M_{bps}})=max(500,7500,7389) = 7500s$N=1000时，dP2P​=max(30Mbps​15Gb​,2Mbps​15Gb​,2Mbps​∗1000+30Mbps​1000∗15Gb​)=max(500,7500,7389)=7500s

画图脚本

import matplotlib.pyplot as plt
import numpy as np

N = np.linspace(10, 1000, 990)

y = (N*15*1000/30)
y1=(N*15000)/(0.5*N+30)
y2 =(N*15000)/(1*N+30)
y3 =(N*15000)/(2*N+30)


np.putmask(y1, y1 <= 7500, 7500)
np.putmask(y2, y2 <= 7500, 7500)
np.putmask(y3, y3 <= 7500, 7500)
np.putmask(y, y <= 7500, 7500)




l, = plt.plot(N, y, color='black', linewidth=1.0, label='CS')

l1, = plt.plot(N, y1, color='red',label='u=500kps', linewidth=1.0)
l2, = plt.plot(N, y2, color='green', linewidth=1.0, label='u=1Mps')


l3, = plt.plot(N, y3, color='blue', linewidth=1.0,label='u=2Mps')

plt.xlabel('N')
plt.ylabel('T')


plt.legend(loc='best')

plt.show()