67 lines
4.0 KiB
TeX
67 lines
4.0 KiB
TeX
\documentclass[12pt,a4paper]{article}
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\usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry}
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\usepackage{amstext} \usepackage{amsmath} \usepackage{array}
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\newcommand{\lra}{\Leftrightarrow}
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\title{Howework 3 -- Computer Networking}
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\author{Claudio Maggioni}
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\begin{document} \maketitle
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\section*{Exercise 1}
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The EUI-48 MAC address space can have $2^{48} = 281474976710656$ possible addresses. The IPv4 address space has $2^32 = 4294967296$
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addresses while IPv6 has = $2^{128} \approx 3,402823669209385 \cdot 10^{38}$ addresses. In practice, some of the addresses in
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these spaces may be reserved for special purposes (e.g.
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\texttt{FF:FF:FF:FF:FF:FF} as the broadcast MAC address or the
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\texttt{127.0.0.0/8} subnet reserved for loopback networks
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in IPv4) so the number of usable addresses is smaller than
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these figures.
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\section*{Exercise 2}
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$$D = 1010100000_2 = 512_{10} + 128_{10} + 32_{10} = 672_{10}$$
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$$G = 10011_2 = 19_{10}$$
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$$r = |G| - 1 = 4$$
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$$R =D \cdot 2^r \mod G = 10752 \mod 19 = 17_{10} = 10001_2$$
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\section*{Exercise 3}
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For 10 Mbps:
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$$t = \frac{100 \cdot 512 b}{10^7\frac{b}{s}} = 5.12 ms$$
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For 100 Mbps:
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$$t = \frac{100 \cdot 512 b}{10^8\frac{b}{s}} = 0.512 ms$$
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\section*{Exercise 4}
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A mechanism that an AP can use to maintain constant reliability (e.g. constant BER) while communicating with a station that is getting farther away from it is to use rate adaptation: once transmission errors are detected through ARQ checks, the AP can switch to a lower transmission rate (akin to TCP's congestion control mechanisms) to counteract interference and lowering signal strength.
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The AP may also decide to boost its transmission power if possible, thus increasing signal strength. However, this technique does not provide a linear increase in signal strength since obstacles or powerful interferences may be present.
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In alternative, if the station is in a ESS and another AP is present in the same subnet, the station may decide to hop from the first AP to the new AP if the latter has a better RSSI. If the DS uses switches, those should be informed of the change of BSS so as to keep their switching tables updated.
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\section*{Exercise 5}
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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.
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\section*{Exercise 6}
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Let $d$ be the DIFS time and $s$ be the SIFS time.
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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:
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$$ d + \frac{20b}{11 \cdot 10^6 \frac{b}{s}} \approx d + 1.82 us$$
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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:
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$$ d + 1.82us + s + \frac{14b}{11 \cdot 10^6 \frac{b}{s}} \approx d + s + 1.82us + 1.27us = d + s + 3.09us$$
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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:
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$$d + s + 3.09us + s + \frac{1060b}{11 \cdot 10^6 \frac{b}{s}} \approx d + 2s + 3.09us + 96.36us =
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d + 2s + 99.45us$$
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Finally, the ACK frame will be sent by the AP after $s$ units transmitting $2+2+6+4=14$ bytes, making the final formula:
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$$d+2s+99.45us + s + \frac{14b}{11 \cdot 10^6 \frac{b}{s}} = d + 3s + 100.72us$$
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\end{document}
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