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# System prepended metadata

title: Intro. to Computer Networks Assignment 1
tags: [homework]

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tags: homework
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# Intro. to Computer Networks Assignment 1

![](https://i.imgur.com/35f5Mtd.jpg)

## P25

### a.

$R\cdot d_{prop}=2Mb/s\cdot \frac{2\times 10^7m}{2.5\times10^8m/s}=0.16Mb$

### b.

$160000\text{ bits}$

### c.

$\text{when time }t_{init}\text{, Host A start to send the file}$
$\text{when }t_{final}\text{, the initial bit arrived}$
$\text{therefore, in the time sequence }t_{final}-t_{init}=d_{prop},\\\text{if the file is big enough,}\\ R\cdot t_{prop}\text{ bit will be sending in the link at the same time}$

$\text{Hence, the bandwidth-delay product can be interpreted to}\\\text{"the maximum bits can be sending in the link at the same time"}$

### d.

$\frac{2.5\times 10^8m}{2\times 10^6bit}=125m/bit$

$\text{a football field is usually 105m long}$

$\Rightarrow \text{yes, it's longer than a football field}$

### e.

$\text{the width of a bit} = \frac{s}{R}$

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## P26

$\text{width of the link}=\frac{s}{R}\\\Rightarrow R=\frac{2.5\times10^8m/s}{2\times10^7m/bit}=12.5bit/s$

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## P27

### a.

$R\cdot d_{prop}=1Gbps\cdot\frac{2\times 10^7m}{2.5\times10^8m/s}=80Mb$

### b.

$\min(8\times 10^7,8\times 10^5)=8\times10^5$

### c.

$\frac{2.5\times 10^8m}{10^9bit}=0.25m/bit$

## P28

### a.

$\frac{800000bit}{2Mbps}+\frac{2\times 10^7m}{2.5\times10^8m/s}=0.4s+0.08s=0.48s$

### b.

$20(\frac{40000bit}{2Mbps}+\frac{2\times 10^7m}{2.5\times 10^8m/s})=20(0.02s+0.16s)=3.6s$

### c.

broken up the file into packages will extended the transmission time because the delay time will multiply by the package amount.

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![](https://i.imgur.com/GBOSUuO.jpg)

## P31

### a.

$\text{from source to packet switch 1}$
$\Rightarrow\frac{8\times 10^6bit}{2Mbps}=4s$

$\text{from packet switch 1 to 2 and to destination cost the same time}$

$\text{total time: }3\times 4s=12s$

### b.

$\text{first packet from source to the first switch}$
$\Rightarrow \frac{10000bit}{2Mbps}=0.005s$

$\text{second packet from source to the second}$
$\Rightarrow 0.005+\frac{10000bit}{2Mbps}=0.005s+0.005s=0.01s$


### c.


$0.005\times(800+2)=4.01s$

The require time using message segmentation compare to the one not using is much shorter.

Because of the packet switches don't need to wait for the entire file sending completely anymore, message segmentation can decrease the require time.

### d.

The buffer in the packet switches will not being used as much if using message segmentation instead of sending the full message directly.

### e.

If the other delays(processing delay, queuing delay, propagation delay, etc.) is significant, using message segmentation will increase those delays.