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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
$$D = 1010100000_2 = 512_{10} + 128_{10} + 32_{10} = 672_{10}$$ $$D = 1010100000_2 = 512_{10} + 128_{10} + 32_{10} = 672_{10}$$
$$G = 10011_2 = 19_{10}$$ $$G = 10011_2 = 19_{10}$$
$$r = |G| - 1 = 4$$ $$r = |G| - 1 = 4$$
$$R =D \cdot 2^r \mod G = 10752 \mod 19 = 17$$ $$R =D \cdot 2^r \mod G = 10752 \mod 19 = 17_{10} = 10001_2$$
\section*{Exercise 3} \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} \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. 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} \end{document}