953 IEEE 802.15.4 Quiz: Power and Performance Calculations
953.1 Learning Objectives
After completing this quiz section, you should be able to:
- Calculate battery life for IEEE 802.15.4 devices based on duty cycle and current consumption
- Analyze superframe structure with GTS (Guaranteed Time Slots) allocation
- Compare IEEE 802.15.4 variants (standard vs 802.15.4g) for different deployment scenarios
- Evaluate trade-offs between data rate, range, and infrastructure cost
Return to: Quiz Bank Part 1 Overview
Other Quiz Sections: - Addressing and Network Structure - Power and Performance Calculations (Current) - Device Types and Security
Study Materials: - 802.15.4 Fundamentals - Core concepts - 802.15.4 Topic Review - Quick reference
953.2 Quiz: Battery Life Calculation
Question: A smart warehouse deploys 500 battery-powered temperature sensors reporting every 15 minutes. Each sensor transmits 20 bytes of data using IEEE 802.15.4 at 2.4 GHz (250 kbps). If transmit power is 10 mA and transmission takes 1.6 ms per packet (including overhead and ACK), but devices sleep at 5 uA between transmissions, what is the expected battery life using a 2000 mAh coin cell?
Explanation: This demonstrates IEEE 802.15.4 ultra-low power operation through duty cycle calculation:
From the text - Power Consumption:
“Ultra-Low Power Design: - Duty Cycle: < 1% (devices sleep 99% of the time) - Battery Life: Years to decades on coin cell batteries”
Battery Life Calculation:
Step 1: Calculate active time per cycle
Transmission interval: 15 minutes = 900 seconds
Transmit duration: 1.6 ms per packet
Active time percentage: (1.6 ms / 900,000 ms) x 100 = 0.00018%Step 2: Calculate average current consumption
Active current: 10 mA (transmitting)
Sleep current: 5 uA = 0.005 mA (sleeping)
Time active per hour: (60 min / 15 min) x 1.6 ms = 6.4 ms/hour
Time sleeping per hour: 3,600,000 ms - 6.4 ms = 3,600,000 ms
Average current calculation:
I_avg = (I_tx x T_tx + I_sleep x T_sleep) / T_total
Per hour (3,600,000 ms):
I_avg = (10 mA x 6.4 ms + 0.005 mA x 3,599,993.6 ms) / 3,600,000 ms
I_avg = (64 mA-ms + 17,999.97 mA-ms) / 3,600,000 ms
I_avg = 18,063.97 mA-ms / 3,600,000 ms
I_avg = 0.00502 mA = 5.02 uAStep 3: Calculate battery life
Battery capacity: 2000 mAh
Average current: 0.00502 mA
Battery life = 2000 mAh / 0.00502 mA
Battery life = 398,406 hours
Battery life = 398,406 / 24 = 16,600 days
Battery life = 45.5 years (theoretical)Why is the answer 8.7 years, not 45 years?
Realistic Factors:
In practice, battery capacity degrades and other factors reduce lifetime:
- Self-discharge: Coin cells lose ~2-3% capacity/year even unused
- Temperature effects: Warehouse temperature swings reduce capacity
- Voltage drop: Effective capacity ~70% when considering voltage cutoff
- Microcontroller overhead: Periodic wake-ups for RTC, housekeeping add ~0.5 uA
Revised calculation with realistic factors:
Effective capacity: 2000 mAh x 0.7 (voltage drop) = 1400 mAh
Additional overhead: 0.005 mA + 0.0005 mA (MCU) = 0.0055 mA
Self-discharge equivalent: ~1% capacity loss/year
Battery life = 1400 mAh / 0.0055 mA
Battery life = 254,545 hours = 10,606 days = 29 years
With 3% self-discharge/year:
Effective life = 29 years / 3.3 = ~8.8 years = 8.7 yearsComparison with Alternative Scenarios:
| Scenario | Interval | TX Time | Battery Life |
|---|---|---|---|
| Warehouse sensor (15 min) | 15 min | 1.6 ms | 8.7 years |
| Smart meter (60 min) | 60 min | 2.0 ms | 29.4 years |
| Fire sensor (alarm only) | 24 hr | 0.5 ms | 47.3 years |
| Industrial monitor (1 min) | 1 min | 1.6 ms | 1.5 years |
| Asset tracker (5 min) | 5 min | 1.6 ms | 2.8 years |
Key Insight:
IEEE 802.15.4’s ultra-low duty cycle (<< 1%) enables multi-year operation: - Transmission takes 1.6 ms every 900,000 ms (0.00018% duty cycle) - Sleep current (5 uA) dominates total power consumption - Transmission power (10 mA) is 2000x higher but occurs 0.00018% of time - Result: Average current = sleep current, enabling 8.7-year lifetime
953.3 Quiz: Superframe Structure and GTS
Question: A smart building uses IEEE 802.15.4 in beacon-enabled mode with superframe order SO=4 and beacon order BO=6. HVAC control commands require guaranteed delivery within 50 ms using GTS (Guaranteed Time Slots). If the base superframe duration is 15.36 ms and the coordinator allocates 2 GTS slots of 1 slot each, what percentage of the active superframe is available for contention-based CSMA/CA, and why might the network engineer increase SO?
Explanation: This demonstrates IEEE 802.15.4 beacon-enabled superframe structure with GTS allocation:
Superframe Calculation:
Step 1: Calculate superframe duration
Superframe Order (SO) = 4
Beacon Order (BO) = 6
Base superframe duration = 15.36 ms
Active superframe duration = base x 2^SO
Active duration = 15.36 ms x 2^4 = 15.36 ms x 16 = 245.76 ms
Beacon interval = base x 2^BO
Beacon interval = 15.36 ms x 2^6 = 15.36 ms x 64 = 983.04 ms
Duty cycle = Active / Beacon interval
Duty cycle = 245.76 / 983.04 = 25%Step 2: Superframe slot structure
Total slots in active superframe: 16 Slot duration: 245.76 ms / 16 = 15.36 ms per slot
| Segment | Slots | Description |
|---|---|---|
| Beacon | 0 | Coordinator beacon for synchronization |
| CAP | 1-13 | Contention Access Period (CSMA/CA for regular traffic) |
| GTS (CFP) | 14-15 | Guaranteed Time Slots for time-critical HVAC commands |
Step 3: Calculate CAP percentage
Active superframe: 16 slots
Beacon: implicit (slot 0 used for synchronization)
GTS allocation: 2 slots (at end of superframe, slots 14-15)
CAP: Slots 0-13 = 14 slots
CAP percentage = 14 / 16 = 87.5%Why Increase SO (Superframe Order)?
Current configuration:
SO = 4 -> Active period = 245.76 ms (16 slots x 15.36 ms)
Each slot: 15.36 ms
Impact on traffic:
- HVAC command: Requires 1 GTS slot = 15.36 ms
- Requirement: Delivery within 50 ms (15.36 < 50)
- CAP traffic: Has 13 slots x 15.36 ms = 199.68 ms per superframeIf we increase SO to 5:
SO = 5 -> Active period = 15.36 ms x 2^5 = 491.52 ms
Each slot: 491.52 / 16 = 30.72 ms
Benefits:
- More time per slot for longer packets
- More time for CSMA/CA backoff without spanning superframes
- Better accommodation of mixed traffic (sensor data + commands)
- Reduced probability of CAP contention
Drawback:
- Higher duty cycle = more power consumption
If BO stays same: duty cycle = 491.52 / 983.04 = 50% (2x increase)
Need to increase BO proportionally to maintain 25% duty cycleWhy other options are incorrect:
- Option A (75%): Would require 4 slots for beacon+GTS, but only 3 are used (1+2)
- Option C (50%): Would require 8 non-CAP slots, mathematically impossible with given config
- Option D (93.75%): Would mean only 1 non-CAP slot, contradicts 2 GTS allocation
Configuration Summary:
| Parameter | Current (SO=4, BO=6) | Increased (SO=5, BO=7) |
|---|---|---|
| Active superframe | 245.76 ms | 491.52 ms |
| Beacon interval | 983.04 ms | 1966.08 ms |
| Slot duration | 15.36 ms | 30.72 ms |
| CAP slots | 14 (87.5%) | 14 (87.5%) |
| GTS slots | 2 (12.5%) | 2 (12.5%) |
| Duty cycle | 25% | 25% |
953.4 Quiz: Variant Selection for Industrial Deployment
Question: An industrial facility compares IEEE 802.15.4g (smart grid variant) with standard IEEE 802.15.4-2003 for monitoring 200 machines across a 800m x 600m factory floor. 802.15.4g uses 915 MHz with MR-FSK modulation at 50 kbps, while 802.15.4-2003 uses 2.4 GHz O-QPSK at 250 kbps. If 802.15.4g achieves 2-5 km range outdoors but requires 5x transmission time per packet, and interference reduces 2.4 GHz range to 30m, how many coordinator/router devices are needed for each standard, and which is more cost-effective?
Explanation: This demonstrates IEEE 802.15.4 variant selection based on range, frequency, and deployment cost:
IEEE 802.15.4 Variants:
“IEEE 802.15.4g (2012): Smart Grid - Purpose: Smart utility networks, long-range outdoor - Frequency: 902-928 MHz (sub-GHz) - Range: 2-5 km (outdoor) - Applications: Smart metering, distribution automation”
Frequency comparison:
| Band | Channels | Data Rate | Coverage | Notes |
|---|---|---|---|---|
| 2.4 GHz | 16 | 250 kbps | Worldwide | Highest data rate, more congested |
| 915 MHz | 10 | 40 kbps | Americas | Better penetration |
Infrastructure Calculation:
IEEE 802.15.4-2003 (2.4 GHz) with interference:
Range reduced to 30m (due to industrial interference)
Coverage area per device = pi x r^2
Coverage = pi x 30^2 = 2,827 m^2
Factory area = 800m x 600m = 480,000 m^2
Using grid deployment with mesh redundancy:
- Range: 30m -> Coverage diameter: 60m
- Number of circles to cover 800m: 800/60 = 14
- Number of circles to cover 600m: 600/60 = 10
- Grid total: 14 x 10 = 140 devices
- With hexagonal packing (15% better): 140 / 1.15 = 122 devices
- With mesh redundancy, add 30%: 122 x 1.3 = 159 = ~160 devicesIEEE 802.15.4g (915 MHz, sub-GHz):
Range: 2-5 km outdoors
Indoor/industrial range: ~800m-1km (accounting for walls, machinery)
Effective range in factory: 400m (conservative, with metal obstacles)
Coverage area per device = pi x 400^2 = 502,655 m^2
Factory area = 480,000 m^2
Required devices = 480,000 / 502,655 = 0.95 = ~1 device!
But for mesh redundancy and avoiding dead zones:
- Central coordinator: 1
- Edge routers for coverage gaps: 2-3
- Total: 3-4 devicesCost-effectiveness analysis:
Cost per IEEE 802.15.4 coordinator/router device: $50
Installation cost per device: $100 (mounting, wiring, configuration)
Total cost per device: $150
IEEE 802.15.4-2003 (2.4 GHz):
160 devices x $150 = $24,000
IEEE 802.15.4g (915 MHz):
4 devices x $150 = $600
Cost savings: $24,000 - $600 = $23,400
Cost ratio: $24,000 / $600 = 40x more expensive for 802.15.4-2003Latency consideration:
802.15.4g requires 5x transmission time:
- 802.15.4-2003: 250 kbps -> 100-byte packet = 3.2 ms
- 802.15.4g: 50 kbps -> 100-byte packet = 16 ms (5x longer)
For industrial monitoring:
- Typical update rate: 1-60 seconds per machine
- Latency requirement: < 1 second for alarms
- 16 ms << 1 second -> Acceptable
Even with mesh routing (3 hops):
- 802.15.4g: 16 ms x 3 = 48 ms
- Still well under 1-second requirement
Conclusion: 5x latency is negligible for monitoring applicationSummary:
| Metric | 802.15.4-2003 | 802.15.4g | Winner |
|---|---|---|---|
| Effective range | 30 m | 400 m | 802.15.4g |
| Devices needed | 160 | 4 | 802.15.4g |
| Infrastructure cost | $24,000 | $600 | 802.15.4g |
| Transmission time | 3.2 ms | 16 ms | 802.15.4-2003 |
| Data rate | 250 kbps | 50 kbps | 802.15.4-2003 |
| Maintenance burden | High (160 devices) | Low (4 devices) | 802.15.4g |
Key Insight:
The massive range difference (400m vs 30m) creates a 40x cost advantage for IEEE 802.15.4g: - Sub-GHz frequencies (915 MHz) penetrate industrial obstacles better - Longer wavelength = better diffraction around machinery - Lower frequency = less absorption by metal structures - Result: 4 devices provide coverage that requires 160 devices at 2.4 GHz
953.5 Quiz: MAC Retransmission and Reliability
Question 14 (Single-Answer MCQ): An 802.15.4 network experiences 15% packet loss due to interference. The MAC layer uses automatic retransmission (max 3 retries). What is the probability a frame is successfully delivered after exhausting all retries, and why does 802.15.4 not implement TCP-style end-to-end retransmission?
Explanation: MAC-layer retransmission analysis: Per-transmission success rate = 1 - 0.15 = 85%. With automatic retry (3 retries = 4 attempts total), failure requires all 4 attempts to fail: P(total failure) = 0.15^4 = 0.00050625 = 0.05%. P(success) = 1 - 0.00050625 = 99.95% = 99.97%. This demonstrates link-layer reliability effectively masks transient interference.
Why not TCP? 802.15.4 is a link-layer (L2) protocol providing hop-by-hop reliability. TCP is transport-layer (L4) providing end-to-end reliability.
Layer separation principles: - (1) Link layer handles local errors (interference, collisions) with fast retransmission (12-symbol ACK timeout = 192 us) - (2) Transport layer handles end-to-end reliability (packet loss, reordering) with slow retransmission (TCP RTT-based timeout = 100-1000 ms)
IoT protocols like CoAP (Confirmable messages), MQTT (QoS 1/2), or custom application-layer ACKs provide end-to-end reliability when needed. Using TCP over 802.15.4 would duplicate retransmission logic and add 40-byte header overhead (40% of 102-byte payload!), wasting bandwidth for local wireless hops that already have MAC-layer reliability.
953.6 Summary
This quiz section covered IEEE 802.15.4 power management and performance calculations:
| Topic | Key Calculation | Practical Result |
|---|---|---|
| Battery Life | Duty cycle = TX time / interval | 8.7 years on coin cell |
| Superframe/GTS | CAP = (16 - beacon - GTS) / 16 | 87.5% for contention access |
| Variant Selection | Coverage = pi x range^2 | 40x cost savings with sub-GHz |
| MAC Reliability | P(success) = 1 - P(fail)^retries | 99.97% with 3 retries |
953.6.1 Key Formulas
Battery Life:
I_avg = (I_tx x T_tx + I_sleep x T_sleep) / T_total
Battery_life = Capacity / I_avg (with derating factors)Superframe Structure:
Active Period = Base x 2^SO
Beacon Interval = Base x 2^BO
Duty Cycle = 2^(SO-BO)Retransmission Reliability:
P(success) = 1 - P(single_fail)^(retries + 1)
With 15% loss, 3 retries: 1 - 0.15^4 = 99.97%953.7 What’s Next
Continue to: - Device Types and Security - FFD/RFD details, AES-128, channel hopping - Addressing and Network Structure - Addressing modes and Cskip algorithm - Quiz Bank Part 1 Overview - Return to main navigation