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# System prepended metadata

title: 計網概 HW1 參考解答

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# 計網概 HW1 參考解答
## 評分標準
**每題 6 分，總分 102 分**
* 單位錯、沒寫、未標明回答哪個問題：扣 **1** 分
* 表達不清楚、用詞不精確：扣 **1～3** 分
* 沒寫、回答錯、邏輯錯：**全扣**
## Prob 25.
### a.
$d_{drop} = \dfrac{distance}{propagation\ speed} = \dfrac{2\times10^{4}\times10^{3}}{2.5\times10^{8}} = 0.08(s) \\
R\cdot D_{drop}=2\times10^{6}\cdot0.08 = 1.6\times10^{5}(bits)$
### b.
**Propagation delay** 乘上**傳輸速率**((a)小題答案)即為該 link 會出現的最高 bits 數，接著與總傳輸的 bits 相比
$\because1.6\times10^{5}(bits) < 8\times10^{5} \\
\therefore1.6\times10^{5}(bits)\ 即為答案$
### c.
bandwidth-delay product 即為在 link 中因為傳輸延遲可容納的 bits 數
### d.
$\dfrac{2\times10^{4}\times10^{3}}{1.6\times10^{5}}=125(m/bit)$
而[足球場大小](https://zh.wikipedia.org/wiki/%E8%B6%B3%E7%90%83%E5%A0%B4#%E6%A8%99%E6%BA%96%E8%B6%B3%E7%90%83%E5%A0%B4)為 90 ～ 120 公尺
故，每 bit 分到 125 公尺比足球場長
### e.
The width of a bit is $\dfrac{m}{R\times\frac{m}{S}}=\dfrac{S}{R}$

## Prob 26.
$\because \dfrac{S}{R}=m\\
\therefore R=\dfrac{S}{m} = \dfrac{2.5\times10^{8}}{2\times10^{4}\times10^{3}}=12.5(bps)$
## Prob 27.
### a.
$d_{drop}=0.08(s)\\
R\cdot d_{drop}=1\times10^{9}\times0.08=8\times10^{7}(bits)$
### b.
因為 $8\times10^{7}>8\times10^{5}$，所以在 link 中最大 bits 為檔案大小 $8\times10^{5}(bits)$ 
### c.
The width of a bit is $\dfrac{S}{R}=\dfrac{2.5\times10^{{8}}}{10^{9}}=0.25(m/bit)$
## Prob 28.
### a.
Transmission delay: $\dfrac{8\times10^5}{2\times10^6}=0.4(s)$

Propagation delay: $\dfrac{2\times10^4\times10^3}{2.5\times10^8}=0.08(s)$

Total time = $0.4 + 0.08 = 0.48(s)$
### b.
Transmission delay: $\dfrac{4\times10^4}{2\times10^6}=0.02(s)$


Propagation delay: $0.08(s)$

Total time 
$\begin{split}&= 20\times (Transmission\ delay + 2 \times Propagation\ delay)\\
&= 20 \times (0.02 + 0.16) = 3.6(s)\\
\end{split}$

### c.

3分: 單純比較 0.48(s) 跟 3.6(s) 的時間快慢
3分: 要提到 ACK 造成多出來的 20 個 Propagation delay

## Prob 31.
### a.
1. $\dfrac{8\times10^6}{2\times10^6}=4(s)$
2. $3\times4=12(s)$
### b.
1. $\dfrac{10^4}{2\times10^6}=0.005(s)$
2. $2\times0.005=0.01(s)$
### c.
1. Time 1st packet arrives destination : $3\times0.005=0.015(s)$
Time the last packet arrive destination: $0.015+799\times0.005=4.01(s)$
2. $4.01(s)<12(s)$, It costs less time when using message segmentation.
### d.
1. Without message segmentation, if bit error occurs, the whole message has to be retransmitted. 
With message segmentation, if bit error occurs, only the packets having bit error need to be retransmitted.
2. Without message segmentation, enormous packets(e.g, HD video file) are sent into the network. 
If the small packet is queueing behind enormous packet, it will suffer unfair dalays.
### e.
1. Packets have to be put in sequence at the destination.
2. Message segmentation results in many small packets, and each packet contains a header and a message segment $\Rightarrow$ the total amount of header bytes is more. 