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hw3: done

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Claudio Maggioni (maggicl) 2020-06-01 14:27:38 +02:00
parent 921065923b
commit c632e27e59
2 changed files with 19 additions and 1 deletions
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@ -22,7 +22,7 @@ addresses while IPv6 has = $2^{128} \approx 3,402823669209385 \cdot 10^{38}$ add
$$D = 1010100000_2 = 512_{10} + 128_{10} + 32_{10} = 672_{10}$$
$$G = 10011_2 = 19_{10}$$
$$r = |G| - 1 = 4$$
$$R =D \cdot 2^r \mod G = 10752 \mod 19 = 17$$
$$R =D \cdot 2^r \mod G = 10752 \mod 19 = 17_{10} = 10001_2$$
\section*{Exercise 3}
@ -44,5 +44,23 @@ In alternative, if the station is in a ESS and another AP is present in the same
\section*{Exercise 5}
Collision detection is \textit{avoided} in IEEE 802.11 since collisions, contrary to IEEE 802.3 Ethernet, are expensive to detect since the power of a received message is significantly lower than the power of transmission. In addition, if collision detection was used scenarios like the hidden terminal problem would not have been avoided.
\section*{Exercise 6}
Let $d$ be the DIFS time and $s$ be the SIFS time.
The station that wants to transmit will transmit an RTS frame after $d$ units and such frame will be $2 + 2 + 6 + 6 + 4 = 20$ bytes long. Therefore we start waiting:
$$ d + \frac{20b}{11 \cdot 10^6 \frac{b}{s}} \approx d + 1.82 us$$
Then the AP will wait for $s$ units and then send a CTS frame $2 + 2 + 6 + 4 = 14$ bytes long. The station will receive the CTS at the instant:
$$ d + 1.82us + s + \frac{14b}{11 \cdot 10^6 \frac{b}{s}} \approx d + s + 1.82us + 1.27us = d + s + 3.09us$$
Then, after $s$ units the station will send the DATA frame. This frame will contain 3 MACs: 1 for the station itself (TX address), 1 for the BSSID of the network (the MAC of the AP, the RX address), and 1 for the MAC address of the destination, which might be in the distribution system (e.g. on a 802.3 network to which the AP is connected to. I will assume for simplicity that there is no fourth address (which might mean that the ultimate destination of this frame's data is on a 802.11 network as well). Therefore, the frame is $2 + 2 + 6 + 6 + 6 + 2 + 1032 + 4 = 1060$ bytes long. So, the AP will receive the entire DATA frame at:
$$d + s + 3.09us + s + \frac{1060b}{11 \cdot 10^6 \frac{b}{s}} \approx d + 2s + 3.09us + 96.36us =
d + 2s + 99.45us$$
Finally, the ACK frame will be sent by the AP after $s$ units transmitting $2+2+6+4=14$ bytes, making the final formula:
$$d+2s+99.45us + s + \frac{14b}{11 \cdot 10^6 \frac{b}{s}} = d + 3s + 100.72us$$
\end{document}