update

1 files changed, 16 insertions, 15 deletions

diff --git a/assignments/3/README.md b/assignments/3/README.md
index cc2ab43..fbc9d34 100644
--- a/assignments/3/README.md
+++ b/assignments/3/README.md
@@ -26,8 +26,8 @@ The anchor MSC (Mobile Switching Center) keeps a call "anchored" to a single MSC
 
 - E-UTRAN with a flat architecture: eNodeBs handle radio and many control functions; no RNC.
 - Air interface: OFDMA (DL), SC-FDMA (UL); flexible channel bandwidths (1.4 to 20 MHz).
-- MIMO and advanced scheduling; adaptive modulation and coding.
-- All-IP core (EPC); low latency (short TTI), high throughput, QoS bearers.
+- MIMO and advanced scheduling; adaptive modulation and coding; HARQ; 1 ms TTI.
+- All-IP core (EPC); low latency, high throughput, QoS bearers and dedicated/default 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.
 
 ## 1.3 CSMA/CD Protocol (5%)
@@ -77,14 +77,14 @@ Chosen scheme: UMTS W-CDMA (FDD).
 - 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 (examples):
-- 3GPP TS 25.213: Spreading and Modulation (FDD).
-- Holma & Toskala, WCDMA for UMTS.
-- Proakis & Salehi, Digital Communications (CDMA chapters).
+Sources consulted:
+- 3GPP TS 25.213, Spreading and Modulation (FDD), v. specify latest.
+- K. Holma and A. Toskala, WCDMA for UMTS: HSPA Evolution and LTE, Wiley.
+- J. G. Proakis and M. Salehi, Digital Communications, McGraw-Hill.
 
 ## 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... even parity ... minimize length ... Show all your work and prove why 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.
 
 Arrange the 16 data bits in a 4x4 matrix (row-major):
 
@@ -109,9 +109,10 @@ Why shortest: For an m x n arrangement, checksum length is m + n + 1. For 16 bit
 
 ## 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... propagation speed 2*10^8 m/s; backoff multiples of 512 bits; each repeater inserts 20-bit delay. At t=0 both transmit. After first collision, A draws K=0 and B draws K=1 after sending 48-bit jam.
-> 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 only A has a packet and repeaters are replaced with switches. Each switch has an 8-bit processing delay in addition to a store-and-forward delay. At what time is A’s packet delivered at B? Include all delays and show work.
+> (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?
+> 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.
@@ -123,12 +124,12 @@ 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 (ignoring IFG, as typical in homework timing unless specified). B would attempt after 1 slot (0.512 microseconds) but will sense busy and defer.
+- 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.
+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.
@@ -141,9 +142,9 @@ Answer (b): about 5.020 microseconds.
 
 > (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):
+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 (as commonly used in examples).
+- 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.