Exercises
ex-ch25-01
EasyFor 802.11ax with 20 MHz bandwidth:
(a) What is the FFT size? (b) How many data subcarriers per OFDM symbol? (c) What is the subcarrier spacing? (d) What is the OFDM symbol duration (useful part only)?
802.11ax uses a larger FFT than 802.11a/n/ac for the same bandwidth.
FFT size
(a) FFT size: 256 (compared to 64 in 802.11a/n/ac for 20 MHz).
Data subcarriers
(b) 234 data subcarriers (out of 256 total: 8 pilot + 3 DC null
- 11 guard null = 22 non-data).
Subcarrier spacing
(c) kHz.
Symbol duration
(d) s.
Compare to 802.11a: s. The longer symbol improves multipath tolerance.
ex-ch25-02
EasyCompute the peak PHY data rate for each Wi-Fi standard with a single spatial stream, highest MCS, and shortest GI:
(a) 802.11a: 20 MHz, 64-QAM , GI = 0.8 s. (b) 802.11n: 40 MHz, 64-QAM , GI = 0.4 s. (c) 802.11ac: 80 MHz, 256-QAM , GI = 0.4 s. (d) 802.11ax: 160 MHz, 1024-QAM , GI = 0.8 s.
Use the data rate formula .
802.11a
(a) , s, s. Mbps.
802.11n
(b) (40 MHz), s. Mbps.
802.11ac
(c) (80 MHz), s. Mbps.
802.11ax
(d) (160 MHz), s. Mbps.
Peak rate growth: Mbps (per spatial stream).
ex-ch25-03
EasyA 20 MHz 802.11ax channel uses 256-point FFT.
(a) What is the maximum multipath delay (in s) that can be tolerated with the 3.2 s guard interval? (b) What indoor environment delay spread does this correspond to? (c) Compare with the 0.8 s GI of 802.11a — how much more delay spread tolerance does 802.11ax provide?
The GI must exceed the maximum channel delay spread.
Maximum delay
(a) The 3.2 s GI tolerates multipath delays up to 3.2 s, corresponding to a path length difference of m.
Indoor environment
(b) Typical indoor RMS delay spread: 20--100 ns. Rule of thumb: GI . Maximum ns. This covers even large indoor venues and some outdoor scenarios.
Comparison
(c) 802.11a: GI = 0.8 s max ns. 802.11ax: GI = 3.2 s max ns.
802.11ax provides more delay spread tolerance, enabling outdoor and large-venue operation.
ex-ch25-04
EasyIdentify which Wi-Fi generation introduced each feature:
(a) OFDM modulation. (b) 40 MHz channel bonding. (c) Multi-user MIMO (downlink). (d) OFDMA resource units. (e) 320 MHz channels. (f) Multi-link operation.
Each feature was introduced in a different 802.11 amendment.
Feature mapping
(a) OFDM: 802.11a (1999). (b) 40 MHz bonding: 802.11n (2009). (c) DL MU-MIMO: 802.11ac (2013). (d) OFDMA: 802.11ax (2020). (e) 320 MHz: 802.11be (2024). (f) MLO: 802.11be (2024).
ex-ch25-05
MediumUsing the Bianchi model, compute the saturation throughput for and 802.11ax stations with:
- ,
- Payload = 1500 bytes, PHY rate = 86 Mbps (MCS 7)
- s, SIFS = 16 s, DIFS = 34 s
- Preamble = 40 s
Compare the MAC efficiency in each case.
Solve the fixed-point equations and iteratively.
.
Frame durations
Data duration: s. s. s.
Case $n = 5$
Iteration converges to , .
. .
Mbps.
MAC efficiency: %.
Case $n = 20$
Iteration converges to , .
. .
Mbps.
MAC efficiency: %.
Throughput drops by 13% as stations increase .
ex-ch25-06
MediumAnalyse the RTS/CTS overhead trade-off:
(a) Compute the additional overhead of RTS/CTS in terms of time (RTS = 20 bytes, CTS = 14 bytes at 6 Mbps, plus 2 SIFS intervals). (b) For a 1500-byte data frame at MCS 7 (86 Mbps PHY rate), compute the throughput with and without RTS/CTS. (c) What is the minimum frame size for which RTS/CTS provides a net benefit, assuming a hidden node collision probability of 10%?
RTS/CTS overhead = RTS + SIFS + CTS + SIFS.
The benefit is avoiding long data frame collisions.
RTS/CTS overhead
(a) RTS: s (preamble + data). CTS: s. Total overhead: s.
Throughput comparison
(b) Without RTS/CTS: s. Throughput: Mbps.
With RTS/CTS: s. Throughput: Mbps.
RTS/CTS costs 24% throughput when there are no hidden nodes.
Break-even frame size
(c) With 10% collision probability on data frames: Without RTS/CTS: effective time per success = s. With RTS/CTS: collisions only on short RTS: s.
Break-even: approximately 1500 bytes at 10% collision rate. For larger frames or higher collision rates, RTS/CTS is beneficial.
ex-ch25-07
MediumConsider an 802.11ax network with EDCA and two traffic classes: AC_VO (voice: 64-byte packets every 20 ms per user) and AC_BE (best effort: saturated traffic).
(a) With 5 voice users and 5 BE users, estimate the collision probability between voice and BE traffic. (b) Can all voice users achieve their required throughput of 25.6 kbps each? (c) What is the remaining throughput available for BE traffic?
AC_VO has AIFSN=2, ; AC_BE has AIFSN=3, .
Voice traffic load
(a) Each voice user: kbps. Total voice: kbps.
Voice transmission probability per slot is very low () because voice packets are small and infrequent. AC_VO wins contention almost always against AC_BE due to shorter AIFS (AIFS_VO = 16 + 2 9 = 34 s vs. AIFS_BE = 16 + 3 9 = 43 s) and smaller (4 vs. 16).
Collision probability for voice % (voice packets rarely coincide, and voice wins over BE).
Voice feasibility
(b) Yes, easily. Total voice load is 128 kbps out of 35 Mbps available capacity. Voice uses % of the channel, and EDCA priority ensures voice packets experience minimal delay.
Remaining BE throughput
(c) BE throughput Mbps. Voice has negligible impact on BE throughput. The EDCA priority differentiation is meaningful only when voice traffic is bursty or when latency (not throughput) matters.
ex-ch25-08
MediumA hidden node scenario: Station A and C both communicate with AP B. A and C are outside each other's carrier sense range.
(a) If A transmits a frame of duration s, and C starts transmitting uniformly at random during a window of duration , what is the collision probability? (b) With RTS/CTS, what is the effective collision probability, assuming C can hear B's CTS? (c) Quantify the throughput loss from the hidden node compared to the case where both nodes can hear each other.
Without RTS/CTS, a collision occurs if C starts transmitting within the vulnerable period of A.
Collision probability without RTS/CTS
(a) The vulnerable period is (C must not start during A's frame or start a frame that overlaps with A's frame). If C starts uniformly in : Collision occurs if C starts within of A's start. .
With RTS/CTS
(b) If C hears B's CTS, it defers for the entire duration indicated in the NAV. The only vulnerable period is the RTS duration (47 s): .
RTS/CTS reduces collision probability by .
Throughput loss
(c) Without RTS/CTS: effective throughput . With RTS/CTS: throughput (accounting for RTS/CTS overhead of 25%). No hidden node: throughput .
Hidden node causes 25--50% throughput loss depending on mitigation.
ex-ch25-09
MediumCompute the channel bonding efficiency:
(a) For 802.11ac, compute the ratio of data subcarriers to total subcarriers (FFT size) for 20, 40, 80, and 160 MHz channels. (b) For 802.11ax, compute the same ratios. (c) Which standard has better spectral efficiency from channel bonding? Explain why.
802.11ac uses 312.5 kHz SCS; 802.11ax uses 78.125 kHz SCS.
802.11ac ratios
(a) 802.11ac (312.5 kHz SCS): 20 MHz: 52/64 = 81.3% (48 data + 4 pilot). 40 MHz: 114/128 = 89.1%. 80 MHz: 242/256 = 94.5%. 160 MHz: 484/512 = 94.5%.
802.11ax ratios
(b) 802.11ax (78.125 kHz SCS): 20 MHz: 242/256 = 94.5%. 40 MHz: 484/512 = 94.5%. 80 MHz: 996/1024 = 97.3%. 160 MHz: 1992/2048 = 97.3%.
Comparison
(c) 802.11ax has better spectral efficiency at all bandwidths because the narrower SCS means the null subcarriers (DC, guard) consume a smaller fraction of the total.
At 20 MHz: 802.11ax gains % more data subcarriers than 802.11ac.
ex-ch25-10
HardCompare the efficiency of OFDMA vs. CSMA/CA for a dense lecture hall with 100 students, each generating 50-byte HTTP ACK packets every 100 ms. 802.11ax, 80 MHz, MCS 4 (16-QAM, ).
(a) With CSMA/CA: compute the average access delay and total throughput consumption. (b) With OFDMA using 26-tone RUs: compute the channel time per multi-user transmission. (c) What fraction of channel capacity does this IoT-like traffic consume under each scheme?
CSMA/CA has per-frame overhead of DIFS + backoff + preamble + SIFS + ACK.
OFDMA can serve up to 37 users per OFDM symbol group on 80 MHz.
CSMA/CA analysis
(a) Per frame overhead: DIFS (34) + avg backoff (72) + preamble (40)
- data (508/(86M) s) + SIFS (16) + ACK (44) = 211 s per frame.
100 users 10 frames/s = 1000 frames/s. Channel time: s/s = 21.1%.
With collisions (, high collision rate 40%): effective channel time %.
OFDMA analysis
(b) With 37 users per group on 80 MHz using 26-tone RUs: Need multi-user transmissions per 100 ms.
Per MU transmission: trigger (40) + SIFS (16)
- data ( symbols 6 13.6 = 81.6 s) + SIFS (16) + ACK (60) = 213.6 s.
Channel time: s/s = 0.64%.
Comparison
(c) CSMA/CA: 35% of channel capacity for trivial traffic. OFDMA: 0.64% of channel capacity.
OFDMA is more efficient for this dense small-packet scenario. This demonstrates why 802.11ax was designed for dense environments.
ex-ch25-11
HardAnalyse BSS coloring and spatial reuse in a dense apartment building with 20 APs on the same 80 MHz channel, arranged in a 4 5 grid with 5 m spacing.
(a) Without BSS coloring: if the carrier sense threshold is dBm and the path loss exponent is 3.5, how many APs does each AP sense (assuming 20 dBm TX power)? (b) With BSS coloring and OBSS/PD = dBm: how many APs does each AP defer to? (c) Estimate the throughput improvement from BSS coloring.
Path loss: dB for distance in meters.
Without BSS coloring
(a) Received power at distance : dBm.
Carrier sense at dBm: m.
In the 45 grid with 5 m spacing, all 20 APs are within 59 m of each other. Each AP senses all 19 others.
With BSS coloring
(b) OBSS/PD = dBm (10 dB higher than default): m.
With 5 m spacing, APs within 30.5 m: approximately 12 APs (those within 6 grid positions). Each AP ignores 7 distant APs. With TX power reduced by 10 dB (to 10 dBm) per OBSS/PD rules, the sensing range further decreases.
Throughput improvement
(c) Without BSS coloring: each AP shares the channel with 19 others. Per-AP throughput .
With BSS coloring: effectively sharing with 12 APs. Per-AP throughput .
Improvement: (% gain). In practice, gains of -- are reported because BSS coloring also reduces collision probability.
ex-ch25-12
HardDesign a TWT schedule for an IoT sensor network with 200 sensors, each sending a 50-byte report every 5 seconds, on an 802.11ax AP with 20 MHz bandwidth.
(a) Compute the total traffic load in kbps. (b) Design a TWT schedule using broadcast TWT with groups of 9 sensors per OFDMA slot (26-tone RUs). (c) What is the duty cycle for each sensor? (d) If the sensor draws 100 mW when awake and 0.1 mW when sleeping, what is the average power consumption?
Group sensors into TWT service periods spaced evenly over the 5-second interval.
Traffic load
(a) Per sensor: bps. Total: kbps. This is trivial for 802.11ax (% of capacity).
TWT schedule design
(b) groups. TWT interval: 5 s. Space groups evenly: ms apart.
Each group's TWT service period: trigger (40 s)
- SIFS (16) + UL data (2 OFDM symbols = 27.2 s)
- SIFS (16) + ACK (60) = 159.2 s. Add margin: 500 s TWT service period.
Duty cycle
(c) Each sensor is awake for 500 s every 5 seconds. Duty cycle: %.
Average power
(d) mW.
Compared to always-on (100 mW): power reduction. With a 1000 mAh battery at 3.3 V: battery life hours years.
ex-ch25-13
HardCompute the theoretical peak PHY rate for 802.11be with:
- 320 MHz bandwidth (4 996-tone RU 3920 data tones)
- 16 spatial streams
- MCS 13 (4096-QAM, )
- GI = 0.8 s
(a) Compute bits per OFDM symbol. (b) Compute the peak PHY data rate. (c) What SNR is required for 4096-QAM at BER ? (d) At what distance from the AP is this rate achievable (assuming free-space path loss at 6 GHz, TX power = 20 dBm, noise figure = 6 dB)?
Required SNR for -QAM at BER : .
Bits per OFDM symbol
(a) Per stream: data bits. Total (16 streams): bits.
Peak rate
(b) Symbol duration: s. Gbps.
Required SNR
(c) For 4096-QAM: required SNR dB per stream. With MIMO and 16 streams, the total required SNR is even higher (beamforming gain partially compensates).
Range estimate
(d) Free-space loss at 6 GHz: .
Required dBm.
dB. m.
4096-QAM is only usable within 5 m of the AP with line of sight.
ex-ch25-14
HardAnalyse Multi-Link Operation (MLO) latency for a video conferencing application with the following setup:
- Link 1: 5 GHz, 80 MHz, average contention delay = 3 ms (congested)
- Link 2: 6 GHz, 160 MHz, average contention delay = 0.5 ms
- Video packet: 1200 bytes every 10 ms
(a) Compute the average and 99th-percentile latency on each link individually (assume exponential access delay distribution). (b) With MLO (send on fastest available link), compute the average and 99th-percentile latency. (c) If the application requires ms latency for 99% of packets, can each individual link meet this requirement? Can MLO?
For exponential distribution: .
of independent exponentials has rate .
Individual link latency
(a) Link 1: mean = 3 ms. 99th percentile: ms.
Link 2: mean = 0.5 ms. 99th percentile: ms.
MLO latency
(b) Rate of minimum: ms. Mean: ms. 99th percentile: ms.
Requirement check
(c) ms at 99th percentile: Link 1 alone: ms — fails. Link 2 alone: ms — passes. MLO: ms — passes with large margin.
MLO reduces the 99th-percentile latency by compared to the congested link and provides robustness against single-link congestion.
ex-ch25-15
MediumAn 802.11be AP operates on a 320 MHz channel with 16 spatial streams, MCS 13 (4096-QAM, ), and guard interval s.
(a) Compute the peak PHY data rate.
(b) If the MAC efficiency (ratio of application throughput to PHY rate) is 65%, what is the achievable application throughput?
(c) The AP serves 20 users via MU-MIMO/OFDMA. Assuming equal resource sharing, what is the per-user throughput?
Use approximately 3920 data subcarriers for 320 MHz.
Data rate formula: .
Peak PHY rate
Numerator: bits.
Gbps.
Application throughput
Gbps.
Per-user throughput
Gbps per user.
This exceeds Gigabit Ethernet, demonstrating that Wi-Fi 7 can serve as a viable wired-replacement technology in enterprise environments.