formal answers

2 files changed, 225 insertions, 44 deletions

diff --git a/assignments/2/README.md b/assignments/2/README.md
index 09af7ca..c72f500 100644
--- a/assignments/2/README.md
+++ b/assignments/2/README.md
@@ -18,6 +18,15 @@ unreliable best‑effort service. Study related sections of the textbook
 and articles from other sources. In your own words, explain how TCP
 provides a reliable data transfer service.
 
+Answer:
+- Error detection: a 16-bit ones' complement checksum over header and payload detects bit errors.
+- Sequencing and in-order delivery: byte sequence numbers and reassembly buffers deliver a contiguous, ordered byte stream.
+- Positive acknowledgments: cumulative ACKs (and optional SACK) inform the sender of data received and gaps.
+- Retransmissions on loss: timeout-based retransmission using RTT/RTO estimation; fast retransmit after 3 duplicate ACKs repairs loss quickly.
+- Sliding window pipelining: allows multiple unacknowledged segments in flight for efficiency.
+- Flow control: receiver-advertised window (rwnd) prevents buffer overrun.
+- Connection management: three-way handshake syncs initial sequence numbers and options; graceful close ensures all data is delivered.
+
 ## 1.2 Go-Back-N Protocol (5%)
 
 > (5%) While the RDT protocols are essentially stop‑and‑wait
@@ -25,6 +34,12 @@ protocols, the GBN protocol allows the sender to send multiple packets
 without waiting for acknowledgement from the receiving parties. How does
 GBN achieve that?
 
+Answer:
+- Pipelining with a sending window: up to N unacknowledged packets may be outstanding (window size N).
+- Cumulative ACKs: the receiver ACKs the highest in-order packet; out-of-order packets are discarded (in basic GBN).
+- One timer for the oldest unacked packet: on timeout, the sender retransmits that packet and all later packets in the window (go back N).
+- As ACKs arrive, the window slides forward and new packets are sent immediately.
+
 ## 1.3 IPv6 Transition (5%)
 
 > (5%) Invention and adoption of IPv6 is a big advance in computer
@@ -32,12 +47,29 @@ networking. What problems was IPv6 intended to solve? With the large
 number of networking devices and applications using IPv4 still in use,
 how is the transition from IPv4 to IPv6 being resolved?
 
+Answer:
+- Problems IPv6 addresses:
+  - Address exhaustion: 128-bit addressing provides vastly more addresses than IPv4.
+  - Simpler, extensible header (with extension headers) and efficient forwarding.
+  - Improved autoconfiguration (SLAAC), multicast, and neighbor discovery; easier renumbering.
+  - Reduced reliance on NAT; IPsec support designed in.
+- Transition mechanisms (since IPv4 persists):
+  - Dual stack hosts/routers choose IPv4 or IPv6 per destination.
+  - Tunneling (e.g., 6rd, GRE) to carry IPv6 over IPv4 (or vice versa).
+  - Translation (NAT64/DNS64, 464XLAT) to let IPv6-only clients reach IPv4 services.
+  - Incremental, piecemeal deployment driven by ISPs, enterprises, content and CDNs.
+
 ## 1.4 SNMP Protocol (5%)
 
 > (5%) SNMP is a protocol for network management. It has seven message
 types. What are the purposes of the SNMP GetRequest and SetRequest
 messages? Why were UDP datagrams chosen to transport SNMP messages?
 
+Answer:
+- GetRequest: manager reads current values of MIB objects (OIDs) from an agent; agent replies with GetResponse.
+- SetRequest: manager writes values to writable MIB objects to configure the device.
+- UDP was chosen because it is simple and lightweight (no connection state), matches request/response with app-layer timeouts and retries, avoids TCP setup and head-of-line blocking for sporadic polls, and supports asynchronous notifications (Traps/Inform) easily.
+
 ## 1.5 SDN-Enabled Devices (5%)
 
 > (5%) In today’s market and its applications, there are many
@@ -46,12 +78,26 @@ an SDN-enabled networking device usually has?
 
 Reference: Kurose & Ross, 8th ed., Ch. 5 (SDN control plane, generalized forwarding).
 
+Answer:
+- Open southbound programming interfaces: OpenFlow, P4Runtime, or NETCONF/YANG.
+- Flexible match-action pipeline (TCAM/ACL capacity), possibly P4-programmable parsing and tables.
+- High-throughput, low-latency forwarding with QoS (queues, shaping, scheduling) and traffic engineering (ECMP, segment routing).
+- Rich telemetry: streaming telemetry (gNMI), sFlow/NetFlow/IPFIX, accurate counters and timestamps; in-band network telemetry support.
+- Secure, robust control-plane: TLS, RBAC, high availability, fast failover.
+- Virtualization and overlay support: VRFs, VXLAN/EVPN, NAT/ACL features for multi-tenancy.
+- APIs/SDKs and upgradable software for automation and programmability.
+
 ## 1.6 BGP Loop Detection (5%)
 
 > (5%) BGP is a routing protocol used for routing among ISPs. One problem
 that BGP faces is detecting loops in paths. What are the loops? Why
 should loops be avoided? How does BGP detect the loops in paths?
 
+Answer:
+- Loops: a path that revisits the same autonomous system (AS) more than once (e.g., AS1->AS2->AS3->AS1), creating a forwarding loop.
+- Why avoid: loops waste bandwidth, increase latency, may blackhole traffic, and can destabilize routing.
+- Detection: BGP is path-vector; each route carries AS_PATH. If a router sees its own AS in the received AS_PATH, it rejects the route. Within an AS, iBGP loop-prevention uses rules with route reflectors (ORIGINATOR_ID, CLUSTER_LIST).
+
 # Part 2: Long Answer Questions (70%)
 
 ## 2.1 1's Complement Checksum (10%)
@@ -68,6 +114,12 @@ errors. Suppose you have the following 8‑bit bytes: 11011001, 01010010,
 
 > d) With this checksum scheme, is it possible that any 1‑bit error will go undetected? How about a 2‑bit error? Explain your answer.
 
+Answer:
+- a) Sum the 8-bit words using ones' complement addition (end-around carry). In decimal: 217 + 82 + 202 + 164 + 89 = 754. Reduce with ones' complement modulus 255: 754 - 2*255 = 244 (binary 11110100). The checksum is the ones' complement (bitwise invert) of 244, which is 00001011 (decimal 11). Check: 244 + 11 = 255 (all 1s).
+- b) Using ones' complement of the sum makes the overall ones' complement sum of all words (data + checksum) equal to all 1s, enabling simple verification, incremental update, and better detection of many common error patterns than straight modulo-2^n addition.
+- c) The receiver computes the ones' complement sum over the received data and checksum. If the result is 11111111 (all 1s), accept; otherwise, declare a checksum error.
+- d) Any single-bit error changes the sum and is always detected. Some two-bit errors that are complementary in the same bit position across two words can cancel and go undetected; thus not all two-bit (or multi-bit) errors are detected.
+
 ## 2.2 Dijkstra's Shortest Path Algorithm (20%)
 
 > (20%) The following table is used to compute the shortest path from A
@@ -78,6 +130,17 @@ which is better known as Dijkstra’s shortest path algorithm.
 
 > b) Consider the network shown in the following diagram. With the indicated link costs, use Dijkstra’s shortest path algorithm to compute the shortest path from x to all other network nodes. Show how the algorithm works by computing a table like the one above.
 
+Answer:
+- a) Each row (Step k) shows the state after the k-th iteration of Dijkstra's algorithm. N' is the set of nodes whose shortest-path distance from the source is known (settled). For each other node n, D(n) is the current best-known distance from the source to n, and P(n) is the predecessor on that tentative path. At each iteration, the node outside N' with minimum D(n) is added to N', and distances to its neighbors are relaxed.
+- b) From source x (using the link costs in the figure), the algorithm proceeds as follows:
+  - Step 0: N' = {x}; D(u)=1 via x; D(y)=1 via x; D(w)=3 via x; D(v)=inf; D(z)=inf
+  - Step 1: add y; N' = {x,y}; relax y->w and y->z: D(w)=2 via y; D(z)=3 via y; D(u)=1 via x; D(v)=inf
+  - Step 2: add u; N' = {x,y,u}; relax u->v (cost 2): D(v)=3 via u; others unchanged
+  - Step 3: add w; N' = {x,y,u,w}; no better paths found
+  - Step 4: add v; N' = {x,y,u,w,v}; no better paths found
+  - Step 5: add z; done
+  Shortest-path costs from x: to u = 1 (x->u); to y = 1 (x->y); to w = 2 (x->y->w); to v = 3 (x->u->v); to z = 3 (x->y->z).
+
 ## 2.3 CIDR Routing (20%)
 
 > (20%) A router running classless interdomain routing (CIDR) has the following entries in its routing table:
@@ -93,6 +156,18 @@ which is better known as Dijkstra’s shortest path algorithm.
 > c) 192.53.40.6
 > d) 192.53.56.7
 
+Answer:
+- CIDR routers use longest-prefix match: among all entries that match the destination, choose the one with the longest mask; if none match, use the default route.
+- Entry coverage:
+  - 135.46.56.0/22 covers 135.46.56.0 to 135.46.59.255 -> Interface 0
+  - 135.46.60.0/22 covers 135.46.60.0 to 135.46.63.255 -> Interface 1
+  - 192.53.40.0/23 covers 192.53.40.0 to 192.53.41.255 -> Router 2
+  - Default -> Router 3
+- Per address:
+  - a) 135.46.61.10 -> matches 135.46.60.0/22 -> forward to Interface 1.
+  - b) 135.46.53.16 -> no /22 match -> forward using Default -> Router 3.
+  - c) 192.53.40.6 -> matches 192.53.40.0/23 -> forward to Router 2.
+  - d) 192.53.56.7 -> no /23 match -> forward using Default -> Router 3.
 
 ## 2.4 TCP Congestion Control (20%)
 
@@ -105,7 +180,7 @@ that each TCP segment size is 1,500 bytes; the two-way propagation delay
 of this connection is 15 msec; and this TCP connection is always in the
 congestion avoidance phase (ignore slow start).
 
-Given: Single TCP flow over a 1 Gbps link, no buffering at the link; MSS=1500 B (12,000 bits); two-way propagation delay (RTT) ≈ 15 ms; always in congestion avoidance.
+Given: Single TCP flow over a 1 Gbps link, no buffering at the link; MSS=1500 B (12,000 bits); two-way propagation delay (RTT) = 15 ms; always in congestion avoidance.
 
 > a) What is the maximum window size (in segments) that this TCP connection can achieve?
 
@@ -115,6 +190,15 @@ Given: Single TCP flow over a 1 Gbps link, no buffering at the link; MSS=1500 B
 
 > d) Assume we want the 1 Gbps link to buffer a finite number of segments and always keep the link busy sending data. How would you choose a buffer size? Justify your answer.
 
+Answer:
+- a) Maximum window without queuing equals the bandwidth-delay product (BDP) in segments.
+  - BDP(bits) = 1e9 bps * 0.015 s = 15,000,000 bits.
+  - MSS(bits) = 1,500 B * 8 = 12,000 bits.
+  - Wmax = 15,000,000 / 12,000 = 1,250 segments.
+- b) In AIMD congestion avoidance (Reno-like), cwnd sawtooths between Wmax and Wmax/2. Average window Wavg = 0.75 * Wmax = 937.5 segments. Average throughput = Wavg * MSS / RTT = 937.5 * 12,000 / 0.015 = 750,000,000 bps (0.75 Gbps).
+- c) After a loss, cwnd halves to Wmax/2 = 625 segments and then grows by ~1 segment per RTT. Time to return to Wmax is 625 RTTs = 625 * 0.015 s = 9.375 s.
+- d) To keep the link busy right after a multiplicative decrease, the queue should supply the missing in-flight data between BDP and cwnd. Provision about BDP/2 worth of buffering: ~625 segments. Then cwnd (625) + queue (625) = BDP, so the link remains fully utilized while cwnd ramps up (with some extra headroom and AQM/ECN recommended in practice).
+
 # References
 
 - Kurose, J. F., & Ross, K. W. (8th ed.). Computer Networking: A Top‑Down Approach.
diff --git a/assignments/3/README.md b/assignments/3/README.md
index 278af7f..c015629 100644
--- a/assignments/3/README.md
+++ b/assignments/3/README.md
@@ -15,84 +15,181 @@ linestretch: 1.0
 
 > (5%) What is the role of the anchor MSC in GSM networks?
 
+Answer:
+- The anchor MSC (Mobile Switching Center) is the call-control anchor for a mobile-terminated or mobile-originated call. When a GSM user moves and an inter-MSC handover occurs, the original MSC that set up the call remains the anchor MSC and keeps end-to-end call control and billing context. Subsequent serving MSCs (visited MSCs) change as the user roams, but the anchor MSC maintains the connection to external networks and coordinates handovers to preserve call continuity.
+
 ## 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?
+> (5%) What are the main characteristics of LTE radio access networks? How does LTE network differ from previous generations of cellular networks?
+
+Answer:
+- LTE RAN characteristics:
+  - OFDMA downlink and SC-FDMA uplink; flexible channel bandwidths (1.4 to 20 MHz).
+  - Flat, all-IP architecture: eNodeB integrates baseband and RNC functions; EPC core (MME, SGW, PGW).
+  - Short TTI (1 ms), HARQ, link adaptation, MIMO (spatial multiplexing/beamforming).
+  - Designed for low latency and high spectral efficiency; QoS bearers.
+- Differences vs 2G/3G (GSM/UMTS-HSPA):
+  - No circuit-switched core; voice via VoLTE/IMS instead of CS domain.
+  - WCDMA/CDMA replaced by OFDMA/SC-FDMA.
+  - RNC removed (flatter control/user plane), simpler handover signaling, pure IP transport.
 
 ## 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.
+> (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.
+
+Answer:
+- CSMA/CD: Carrier Sense Multiple Access with Collision Detection.
+- Operation: A node senses the medium; if idle for an IFG, it transmits. While transmitting, it listens (collision detect). On collision, it sends a jam signal, stops, and schedules a random backoff (binary exponential, slots of 512 bit times), then retries.
+- Importance of RTT: The Ethernet slot time (512 bit times) is set to exceed the worst-case round-trip propagation on the LAN. This guarantees a transmitter will detect a collision before it finishes sending the minimum-sized frame, enabling reliable collision detection, proper backoff, and bounding the network diameter and minimum frame size.
 
 ## 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?
+> (5%) What does CSMA/CA stand for? How does the protocol work? How can collisions be avoided in the protocol?
+
+Answer:
+- CSMA/CA: Carrier Sense Multiple Access with Collision Avoidance (802.11).
+- Operation: A station senses the channel; after DIFS and a random backoff (slots), it transmits when the counter reaches zero. Receivers ACK after SIFS. On failure, CW is doubled and the process repeats.
+- Collision avoidance: Random backoff, shorter SIFS for ACK to gain priority, and optional RTS/CTS handshake to reserve the channel and set NAVs at neighbors, mitigating hidden-node collisions.
 
 ## 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?
+> (5%) What techniques can be used for error-detection and error-correction, respectively, on the data link layer?
+
+Answer:
+- Error detection: parity bits (1D/2D), Internet checksum, CRC (common in Ethernet, 802.11).
+- Error correction: forward error correction codes (e.g., Hamming, Reed-Solomon, convolutional/LDPC in Wi-Fi), and ARQ protocols (Stop-and-Wait, Go-Back-N, Selective Repeat) for retransmission-based correction.
 
 ## 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?
+> (5%) What wireless (Wi-Fi) network standards are used in today's industries? What are the characteristics of the link specified in each standard?
+
+Answer (high level):
+- 802.11a/g: OFDM in 5 GHz (a) and 2.4 GHz (g), up to 54 Mbps PHY rate, 20 MHz channels.
+- 802.11n (Wi-Fi 4): 2.4/5 GHz, MIMO (up to 4x4), channel bonding (20/40 MHz), up to ~600 Mbps.
+- 802.11ac (Wi-Fi 5): 5 GHz, wider channels (80/160 MHz), higher-order QAM (256-QAM), MU-MIMO downlink, up to multi-Gbps PHY.
+- 802.11ax (Wi-Fi 6/6E): 2.4/5/6 GHz, OFDMA, MU-MIMO uplink/downlink, 1024-QAM, BSS coloring; better efficiency in dense environments.
+- 802.11be (Wi-Fi 7, emerging): 320 MHz channels, 4096-QAM, multi-link operation, enhanced OFDMA/MIMO; target tens of Gbps PHY.
 
 # 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.
+> (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.
+
+Answer:
+- Scheme: Direct-Sequence CDMA (DS-CDMA).
+- How it works:
+  - Each user is assigned a unique spreading code (chip sequence) with low cross-correlation to others (e.g., orthogonal Walsh codes or pseudo-noise sequences).
+  - A user multiplies its data bits by the spreading code at a higher chip rate, spreading each bit over many chips (processing gain). Multiple users transmit simultaneously over the same band.
+  - The receiver correlates the composite received signal with the intended user’s code to despread and recover that user’s data while treating other users as near-white interference.
+- Advantages vs TDM/FDM:
+  - Soft capacity: more users can be admitted with graceful degradation (increased interference) rather than hard limits per time/frequency slot.
+  - Robustness to narrowband interference and multipath (RAKE receivers exploit multipath diversity).
+  - Asynchronous access without precise global slot/symbol alignment; frequency reuse factor ~1 (same band in adjacent cells with code planning and power control).
+  - Security through spreading (low probability of intercept) and resilience to jamming.
+- Sources consulted:
+  - Kurose & Ross, Computer Networking: A Top-Down Approach (Ch. 7 overview of CDMA).
+  - 3GPP specifications (conceptually: IS-95/3G CDMA systems use DS-CDMA with power control and RAKE reception).
 
 ## 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.
+> (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.
+
+Answer:
+- Arrange the 16 bits as a 4x4 matrix (closest to a square to minimize overhead):
+  Row1: 1011
+  Row2: 0110
+  Row3: 1010
+  Row4: 1011
+- Even-parity row bits (add one parity bit per row):
+  - Row1: 1011 has 3 ones -> parity 1
+  - Row2: 0110 has 2 ones -> parity 0
+  - Row3: 1010 has 2 ones -> parity 0
+  - Row4: 1011 has 3 ones -> parity 1
+  Row parity vector = 1 0 0 1
+- Even-parity column bits (one per column over data bits only):
+  - Col1: 1,0,1,1 -> 3 ones -> parity 1
+  - Col2: 0,1,0,0 -> 1 one  -> parity 1
+  - Col3: 1,1,1,1 -> 4 ones -> parity 0
+  - Col4: 1,0,0,1 -> 2 ones -> parity 0
+  Column parity vector = 1 1 0 0
+- Checksum field (row parity followed by column parity): 1001 1100 (8 bits).
+- Minimality: With R rows and C columns, parity overhead is R + C bits and must satisfy R*C >= 16. The sum R+C is minimized when R and C are as close as possible (by AM-GM inequality), i.e., R=C=4, giving 8 bits. Any other integer factorization (e.g., 2x8 or 1x16) yields R+C >= 10 or 17.
+- 1-bit error detect/correct proof: A single flipped bit toggles the parity of exactly one row and exactly one column. The unique pair (row_i, col_j) that fails parity identifies the error location (i,j). Correct by flipping that bit back. Hence, any 1-bit error is both detectable and correctable.
 
 ## 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.
+> (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.
-
+>
 > 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?
 
+Answer:
+- Given:
+  - Link rate R = 1 Gbps -> 1 bit time = 1 ns.
+  - Cable length = 180 m; propagation speed = 2e8 m/s.
+  - 3 repeaters; each adds 20 bit times = 20 ns.
+  - Frame length L = 1,024 bits -> transmit time T_tx = 1,024 ns = 1.024 us.
+  - Jam length = 48 bits -> 48 ns; IFG = 96 bit times -> 96 ns.
+
+- a) One-way propagation including repeater delays:
+  - Cable propagation: 180 m / (2e8 m/s) = 9.0e-7 s = 0.90 us.
+  - Repeaters: 3 * 20 ns = 60 ns = 0.06 us.
+  - One-way total t_prop_one = 0.90 us + 0.06 us = 0.96 us (9.6e-7 s).
+  Collision at t = 0: both start; collision detected after t_prop_one = 0.96 us; each sends 48-bit jam (48 ns). The medium becomes idle after the last jam bit propagates: 0.96 us + 48 ns + 0.96 us = 1.968 us. Add IFG 96 ns -> A can retransmit at t = 2.064 us. With K=0, A starts immediately; B has K=1 and defers.
+  Last bit arrival time at B:
+  - A finishes sending at 2.064 us + 1.024 us = 3.088 us.
+  - Add one-way propagation 0.96 us -> B receives last bit at 4.048 us.
+  Answer: one-way propagation incl. repeaters = 9.6e-7 s; A's frame completely delivered at B at t ≈ 4.048e-6 s.
+
+- b) Replace 3 repeaters with 3 switches (store-and-forward) and only A transmits. Each switch adds: full-frame store-and-forward (1.024 us) plus 8-bit processing (8 ns). Total time to last bit at B (no collisions):
+  - 4 serial transmissions of the 1,024-bit frame (A->S1, S1->S2, S2->S3, S3->B): 4 * 1.024 us = 4.096 us.
+  - 3 switch processing delays: 3 * 8 ns = 24 ns = 0.024 us.
+  - Cable propagation across 180 m (split over links): ~0.90 us total.
+  - Sum: 4.096 us + 0.024 us + 0.90 us = 5.020 us.
+  Answer: A's packet delivered (last bit) at B at t ≈ 5.020e-6 s.
+
 ## 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?
+> (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?
+
+Answer (802.11ac assumptions):
+- PHY/MAC parameters assumed:
+  - SIFS = 16 us; slot time = 9 us; DIFS = SIFS + 2*slot = 34 us. Backoff = 0 (idle channel, single attempt).
+  - Control (RTS/CTS/ACK) sent at legacy OFDM 24 Mbps with legacy preamble 20 us. Frame times: RTS 20 B, CTS 14 B, ACK 14 B.
+    - OFDM symbol at 24 Mbps carries N_DBPS = 96 bits; each legacy PPDU adds 16 service + 6 tail bits.
+    - T_RTS = 20 us + ceil((16+8*20+6)/96)*4 us = 20 us + 2*4 us = 28 us.
+    - T_CTS = 20 us + ceil((16+8*14+6)/96)*4 us = 20 us + 2*4 us = 28 us.
+    - T_ACK = same as CTS = 28 us.
+  - Data at 802.11ac VHT, 80 MHz, 1 spatial stream, MCS 9, short GI (0.4 us): N_DBPS = 234 subcarriers * 8 bits * 5/6 = 1,560 bits/symbol; symbol time = 3.6 us. VHT preamble ≈ 40 us. Payload = 1024 B (8192 bits); add 16 service + 6 tail = 8214 bits.
+    - N_sym = ceil(8214 / 1560) = 6 symbols; payload time = 6 * 3.6 us = 21.6 us.
+    - T_DATA = 40 us (VHT preamble) + 21.6 us = 61.6 us.
+- Timeline and results:
+  - Data transmission completes at: DIFS + T_RTS + SIFS + T_CTS + SIFS + T_DATA
+    = 34 + 28 + 16 + 28 + 16 + 61.6 ≈ 183.6 us.
+  - ACK reception finishes at: previous time + SIFS + T_ACK
+    = 183.6 + 16 + 28 ≈ 227.6 us.
+  Thus, the station completes transmitting the data at ≈ 183.6 microseconds and finishes receiving the ACK at ≈ 227.6 microseconds after t = 0.
 
 ## 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.
+> (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.
+
+Answer:
+- Classic Bluetooth (BR/EDR) baseband packet:
+  - Access Code (72 bits): used for synchronization and identification.
+  - Header (54 bits): AM_ADDR (3 bits), Type (4 bits), Flow, ARQN (ACK/NACK), SEQN (seq), and HEC.
+  - Payload (0 to 2,745 bits depending on packet type), with optional CRC.
+- Bluetooth Low Energy (BLE) link-layer PDU (advertising/data):
+  - Preamble (1 byte), Access Address (32 bits), Header (16 bits), Payload (0–255 bytes), optional MIC (4 bytes), CRC (24 bits). BLE uses different channelization/hopping and frame structure than BR/EDR.
+- Features/limitations:
+  - Short headers with built-in error detection (HEC/CRC), ARQ with SEQN/ARQN.
+  - Frequency hopping spread spectrum for interference robustness.
+  - BR/EDR piconet limit: The AM_ADDR field is 3 bits, allowing 7 non-zero addresses for active slaves plus the master -> at most 7 active slaves per piconet (8 active devices including the master). Parked or sniff/hold states can increase associated but not simultaneously active devices. BLE does not use AM_ADDR and can support many connections by scheduling, but practical limits arise from controller resources and timing.
 
 # 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 (conceptual background for CDMA in IS-95/3G systems).