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diff --git a/assignments/3/README.md b/assignments/3/README.md index 27724f4..278af7f 100644 --- a/assignments/3/README.md +++ b/assignments/3/README.md @@ -15,194 +15,84 @@ linestretch: 1.0 > (5%) What is the role of the anchor MSC in GSM networks? -The anchor MSC (Mobile Switching Center) keeps a call anchored to a single MSC for its entire duration while the mobile moves. It: -- Coordinates with HLR/VLR for authentication, location, and subscriber data. -- Manages intra- and inter-MSC handovers by setting up and tearing down trunks to visited MSCs while keeping the original call leg. -- Maintains continuity so numbering, charging, and signaling remain stable as the user roams. - -Reference: 3GPP TS 23.002; TS 23.009 (high-level architecture and handover procedures in GSM/UMTS networks). - ## 1.2 LTE Network Characteristics (5%) -> (5%) What are the main characteristics of LTE radio access networks? How does LTE network differ from previous generations of cellular networks? - -- E-UTRAN with a flat architecture: eNodeBs handle radio and many control functions; no RNC. All-IP core (EPC). -- Air interface: OFDMA (DL), SC-FDMA (UL); flexible channel bandwidths (1.4 to 20 MHz). HARQ; 1 ms TTI. -- MIMO and advanced scheduling; adaptive modulation and coding; QoS bearers (default/dedicated EPS bearers). -- Compared to 3G (UMTS/WCDMA): LTE moves from wideband CDMA to OFDM-based access, eliminates circuit-switched core (CS services via IMS/VoLTE), reduces latency, increases spectral efficiency, and simplifies the RAN. - -Reference: 3GPP TS 36.300 (E-UTRAN overall description); Kurose & Ross, 8th ed., Ch. 7 (cellular overview). +> (5%) What are the main characteristics of LTE radio access networks? How +does LTE network differ from previous generations of cellular networks? ## 1.3 CSMA/CD Protocol (5%) -> (5%) What does CSMA/CD stand for? How does the protocol work? Explain why RTT on an Ethernet LAN is an important parameter for the CSMA/CD protocol to work properly. - -- CSMA/CD: Carrier Sense Multiple Access with Collision Detection. -- Operation: Sense channel idle, transmit; while transmitting, detect collision (signal energy). On collision, send jam signal and abort; back off using binary exponential backoff (choose K in {0..2^n-1} slots; slot = 512 bit times); retry. -- RTT/slot-time importance: To guarantee any collision is detected while transmitting, frames must be at least 512 bits so transmission lasts at least one round-trip propagation time (including repeater delays). If frames were shorter than the slot time, a station could finish sending without detecting a collision. - -Reference: Kurose & Ross, 8th ed., Sec. 6.3.2 (CSMA/CD), Sec. 6.4.2 (Ethernet frame and slot time). +> (5%) What does CSMA/CD stand for? How does the protocol work? Explain +why RTT on an Ethernet LAN is an important parameter for the CSMA/CD +protocol to work properly. ## 1.4 CSMA/CA Protocol (5%) -> (5%) What does CSMA/CA stand for? How does the protocol work? How can collisions be avoided in the protocol? - -- CSMA/CA: Carrier Sense Multiple Access with Collision Avoidance (802.11 DCF). -- Operation: After sensing idle for DIFS, choose random backoff; decrement timer while idle; transmit when backoff reaches zero; receiver sends ACK after SIFS. Collisions are inferred by missing ACK. -- Avoidance: Random backoff spaces transmissions; RTS/CTS reserves the channel and mitigates hidden terminals; interframe spacing (SIFS/DIFS) prioritizes control/ACK. - -Reference: Kurose & Ross, 8th ed., Ch. 7 (802.11 MAC/DCF, RTS/CTS). +> (5%) What does CSMA/CA stand for? How does the protocol work? How can +collisions be avoided in the protocol? ## 1.5 Data Link Layer Error Detection/Correction (5%) -> (5%) What techniques can be used for error-detection and error-correction, respectively, on the data link layer? - -- Detection: Parity (single, two-dimensional), checksum (1's complement), CRC. -- Correction: Hamming codes, Reed-Solomon, convolutional/turbo/LDPC codes. ARQ/HARQ combine FEC with retransmissions. - -Reference: Kurose & Ross, 8th ed., Sec. 6.2 (parity, checksum, CRC). +> (5%) What techniques can be used for error-detection and +error-correction, respectively, on the data link layer? ## 1.6 Wi-Fi Network Standards (5%) -> (5%) What wireless (Wi-Fi) network standards are used in today's industries? What are the characteristics of the link specified in each standard? - -- 802.11n (Wi-Fi 4): 2.4/5 GHz, 20/40 MHz channels, MIMO, up to 600 Mbps PHY. -- 802.11ac (Wi-Fi 5): 5 GHz, 20/40/80/160 MHz, MU-MIMO (DL), higher-order QAM, multi-Gbps PHY. -- 802.11ax (Wi-Fi 6/6E): 2.4/5/6 GHz, OFDMA, MU-MIMO (UL/DL), BSS coloring, TWT; higher efficiency and capacity. -- 802.11be (Wi-Fi 7, emerging): 2.4/5/6 GHz, 320 MHz channels, multi-link operation, higher-order MIMO and QAM; very high throughput. - -Reference: IEEE 802.11 standard family (e.g., 802.11-2016 and amendments); Kurose & Ross, 8th ed., Ch. 7. - ---- +> (5%) What wireless (Wi-Fi) network standards are used in today's +industries? What are the characteristics of the link specified in each +standard? # Part 2: Long Answer Questions (70%) ## 2.1 Code Division Multiple Access (CDMA) (15%) -> (15%) Begin with reading about the simple CDMA protocol... choose one CDMA scheme and explain how it works. Describe the advantages that CDMA has over other coding schemes, such as TDM and FDM. Include in your answer the titles and sources of the articles/documents you consulted. - -Chosen scheme: UMTS W-CDMA (FDD). -- Spreading: User data is spread by orthogonal variable spreading factor (OVSF) channelization codes to a common chip rate (3.84 Mcps). Scrambling codes separate cells/users at the cell and UE level. -- Power control: Fast closed-loop power control combats near-far and fading (e.g., 1500 Hz downlink commands). -- Receiver: RAKE combining aligns and coherently combines energy from multipath components (fingers), exploiting multipath diversity. -- Handover: Soft/softer handover maintains simultaneous links to multiple cells/sectors; diversity improves robustness. -- Advantages over TDM/FDM: Interference averaging (soft capacity), graceful degradation as load increases, strong multipath resilience via RAKE, flexible rate adaptation by varying spreading factor and coding, inherent security through spreading. - -Sources consulted: -- 3GPP TS 25.213, Spreading and Modulation (FDD). -- Holma & Toskala, WCDMA for UMTS: HSPA Evolution and LTE. -- Proakis & Salehi, Digital Communications. -- Kurose & Ross, 8th ed., Ch. 7 (high-level cellular/CDMA context). +> (15%) Begin with reading about the simple CDMA protocol... choose one +CDMA scheme and explain how it works. Describe the advantages that CDMA +has over other coding schemes, such as TDM and FDM. Include in your +answer the titles and sources of the articles/documents you consulted. ## 2.2 Two-Dimensional Checksum (15%) -> (15%) Suppose host A has payload 1011 0110 1010 1011 to send to host B, and A wants to use a two-dimensional checksum for host B to detect and correct any 1-bit error that may occur during the transmission. Furthermore, host A wants to minimize the length of the checksum to conserve bandwidth of the communication channel. What would the value of the checksum field be if an even parity scheme is used? Show all your work and prove why the checksum you have worked out is the shortest. Prove that any 1-bit error can be detected and corrected. - -Arrange the 16 data bits in a 4x4 matrix (row-major): - -Row1: 1 0 1 1 (sum 3) -> row parity pR1 = 1 (to make even) -Row2: 0 1 1 0 (sum 2) -> pR2 = 0 -Row3: 1 0 1 0 (sum 2) -> pR3 = 0 -Row4: 1 0 1 1 (sum 3) -> pR4 = 1 - -Column parities: -- Col1: 1,0,1,1 (sum 3) -> pC1 = 1 -- Col2: 0,1,0,0 (sum 1) -> pC2 = 1 -- Col3: 1,1,1,1 (sum 4) -> pC3 = 0 -- Col4: 1,0,0,1 (sum 2) -> pC4 = 0 - -Overall parity p0: even parity over all 16 data bits plus 8 parity bits. Data ones = 10 (even). Row parity ones = 2. Column parity ones = 2. Total so far = 14 (even) -> p0 = 0. - -Checksum bits (one possible ordering): [pR1 pR2 pR3 pR4 pC1 pC2 pC3 pC4 p0] = 1 0 0 1 1 1 0 0 0. - -Why shortest: For an m x n arrangement, checksum length is m + n + 1. For 16 bits, possibilities include 1x16 (18), 2x8 (11), 4x4 (9), 8x2 (11), 16x1 (18). Thus 4x4 is minimal. - -1-bit detect/correct: A single flipped data bit causes exactly one row parity and one column parity to fail, and toggles overall parity (odd). The intersection of the failing row and column identifies the bit to correct. If a parity bit flips, overall parity detects it while rows/columns localize which parity bit erred. - -Reference: Kurose & Ross, 8th ed., Sec. 6.2.1 (two-dimensional parity). +> (15%) Suppose host A has payload 1011 0110 1010 1011 to send to +host B, and A wants to use a two-dimensional checksum for host B +to detect and correct any 1-bit error that may occur during the +transmission. Furthermore, host A wants to minimize the length of +the checksum to conserve bandwidth of the communication channel. What +would the value of the checksum field be if an even parity scheme is +used? Show all your work and prove why the checksum you have worked out +is the shortest. Prove that any 1-bit error can be detected and corrected. ## 2.3 CSMA/CD Ethernet Analysis (20%) -> (20%) Assume a 1 Gbps Ethernet has two nodes, A and B, connected by a 180 m cable with three repeaters in between, and they each have one frame of 1,024 bits to send to each other. Further assume that the signal propagation speed across the cable is 2*10^8 m/sec; CSMA/CD uses back-off intervals of multiples of 512 bits; and each repeater will insert a store-and-forward delay equivalent to 20-bit transmission time. At time t = 0, both A and B attempt to transmit. After the first collision, A draws K = 0 and B draws K = 1 in the exponential backoff protocol after sending the 48 bits jam signal. -> a) What is the one-way propagation delay (including all repeater delays) between A and B in seconds? At what time is A's packet completely delivered at B? -> b) Now suppose that only A has a packet to send and that the repeaters are replaced with switches. Suppose that each switch has an 8-bit processing delay in addition to a store-and-forward delay. At what time, in seconds, is A's packet delivered at B? +> (20%) Assume a 1 Gbps Ethernet has two nodes, A and B, connected by +a 180 m cable with three repeaters in between, and they each have one +frame of 1,024 bits to send to each other. Further assume that the +signal propagation speed across the cable is 2*10^8 m/sec; CSMA/CD +uses back-off intervals of multiples of 512 bits; and each repeater +will insert a store-and-forward delay equivalent to 20-bit transmission +time. At time t = 0, both A and B attempt to transmit. After the first +collision, A draws K = 0 and B draws K = 1 in the exponential backoff +protocol after sending the 48 bits jam signal. > In your calculations for a and b, you must include all the delays that occur according to CSMA/CD, and you must show the details of your work. -Given: -- Rate = 1e9 bps. Frame length = 1024 bits. Jam = 48 bits. Slot = 512 bits. -- Cable propagation = 180 m / (2e8 m/s) = 0.9 microseconds. -- 3 repeaters x 20-bit times = 60 ns = 0.06 microseconds (treated as added delay per the problem statement). -- One-way propagation including repeaters: 0.9 + 0.06 = 0.96 microseconds. - -a) Timeline with collision at t = 0: -- Both start transmitting at t = 0. -- Each detects collision when the other's signal arrives: t = 0.96 microseconds. -- Each sends 48-bit jam: duration 48 ns = 0.048 microseconds. Local jam ends at 1.008 microseconds. -- The last bit of B's jam arrives at A at 1.008 + 0.96 = 1.968 microseconds (channel becomes idle then). -- Backoff: A draws K = 0, B draws K = 1. A begins retransmission immediately when idle; B would attempt after 1 slot (0.512 microseconds) but will sense busy and defer. -- A's frame transmission time: 1024 bits / 1e9 = 1.024 microseconds. Last bit leaves A at 1.968 + 1.024 = 2.992 microseconds. -- Last bit arrives at B after one-way propagation: 2.992 + 0.96 = 3.952 microseconds. - -Answer (a): one-way delay = 0.96 microseconds; A's packet completely delivered at B at about 3.952 microseconds. - -b) Replace 3 repeaters with 3 store-and-forward switches (8-bit processing per switch): -- Per switch delay = frame time + processing = 1.024 microseconds + 8 ns = 1.032 microseconds. -- End-to-end: A transmits 1.024 microseconds, then 3 switches each add 1.032 microseconds, plus cable propagation (0.9 microseconds). Delivery time (from t = 0): - 1.024 + 3*(1.032) + 0.9 = 1.024 + 3.096 + 0.9 = 5.020 microseconds. - -Answer (b): about 5.020 microseconds. +> a) What is the one-way propagation delay (including all repeater delays) between A and B in seconds? At what time is A's packet completely delivered at B? -Reference: Kurose & Ross, 8th ed., Sec. 6.3.2 (CSMA/CD timing, jam, backoff) and Sec. 6.4.3 (store-and-forward switching). +> b) Now suppose that only A has a packet to send and that the repeaters are replaced with switches. Suppose that each switch has an 8-bit processing delay in addition to a store-and-forward delay. At what time, in seconds, is A's packet delivered at B? ## 2.4 802.11 RTS/CTS Transmission (10%) -> (10%) Suppose an 802.11 station is configured to always reserve the channel with RTS/CTS. At t = 0 it wants to transmit 1024 bytes. All other stations are idle. At what time will the station complete the transmission? At what time can the station receive the acknowledgement? - -Assumptions (matching common textbook values and ignoring PHY preamble/PLCP overheads unless specified): -- Data rate = 11 Mbps. DIFS = 50 microseconds. SIFS = 10 microseconds. -- Frame sizes: RTS = 160 bits; CTS = 112 bits; ACK = 112 bits; DATA payload = 1024 bytes = 8192 bits; assume MAC header + FCS bring the data bits to 8416 bits. - -Timing (all at 11 Mbps except interframe spaces): -- DIFS: 50.000 microseconds. -- RTS: 160/11e6 = 14.545 microseconds. -- SIFS: 10.000 microseconds. -- CTS: 112/11e6 = 10.182 microseconds. -- SIFS: 10.000 microseconds. -- DATA: 8416/11e6 = 765.091 microseconds. -- SIFS: 10.000 microseconds. -- ACK: 112/11e6 = 10.182 microseconds. - -Totals: -- Data transmission complete at: 50 + 14.545 + 10 + 10.182 + 10 + 765.091 = 859.818 microseconds. -- ACK received at: 859.818 + 10 + 10.182 = 880.000 microseconds. - -Reference: Kurose & Ross, 8th ed., Ch. 7 (802.11 timing and RTS/CTS). +> (10%) Suppose an 802.11 station is configured to always reserve the +channel with RTS/CTS. At t = 0 it wants to transmit 1024 bytes. All +other stations are idle. At what time will the station complete the +transmission? At what time can the station receive the acknowledgement? ## 2.5 Bluetooth Frame Format Analysis (10%) -> (10%) Conduct research about Bluetooth technology and describe and comment on the format of the Bluetooth frame. Focus on its features and limitations. Is there anything in the frame format that inherently limits the number of active nodes in a network to eight active nodes? Explain. - -Classic Bluetooth baseband frame: -- Access Code (72 bits): preamble, sync word (based on lower address), trailer; used for timing and identification. -- Header (54 bits transmitted with 1/3 FEC over an 18-bit header): AM_ADDR (3 bits), Type (4), Flow (1), ARQN (1), SEQN (1), HEC (8); robust via FEC. -- Payload (0 to 2745 bits) with 16-bit CRC; optional FEC (2/3, 3/5) and whitening; various packet types define payload length and structure. - -Features and limitations: -- Robust short-range links with FEC, ARQ, fast frequency hopping; low power design. -- Payload and slot structure constrain throughput; tight timing and hopping sequence complexity. -- Piconet concurrency: AM_ADDR is 3 bits, yielding up to 7 active slaves addressed by the master (plus the master itself). Additional devices can be parked (no AM_ADDR) and polled in/out, and multiple piconets can form scatternets, but a single piconet supports at most 7 active slaves concurrently due to the 3-bit address field in the frame header. - -References: Bluetooth Core Specification (e.g., v5.x); Kurose & Ross, 8th ed., Ch. 6/7 (MAC/link framing context). - ---- +> (10%) Conduct research about Bluetooth technology and describe and +comment on the format of the Bluetooth frame. Focus on its features and +limitations. Is there anything in the frame format that inherently limits +the number of active nodes in a network to eight active nodes? Explain. # References - Kurose, J. F., & Ross, K. W. (8th ed.). Computer Networking: A Top-Down Approach. (See Ch. 6: Link Layer and LANs; Ch. 7: Wireless and Mobile Networks.) -- 3GPP TS 23.002: Network Architecture; 3GPP TS 23.009: Handover Procedures. -- 3GPP TS 36.300: E-UTRAN Overall Description. -- 3GPP TS 25.213: Spreading and Modulation (FDD) for W-CDMA. -- IEEE 802.11 Standard (e.g., 802.11-2016 and amendments for 11n/ac/ax/be context). -- Bluetooth SIG, Bluetooth Core Specification (e.g., v5.x). -- Proakis, J. G., & Salehi, M. Digital Communications. (CDMA fundamentals.) -- Holma, H., & Toskala, A. WCDMA for UMTS: HSPA Evolution and LTE. |