DMA Cycle Stealing Numerical Explained (MIPS vs Data Rate Problem)

Numerical questions on DMA (Direct Memory Access) are very common in exams like GATE and UGC NET. These questions test your understanding of cycle stealing and processor slowdown.

Let’s solve a real exam-level problem step by step.

🔵 Given Data

Processor speed = 1 MIPS
Device transfer rate = 9600 bits/sec
DMA mode = Cycle Stealing

🔴 Step 1: Convert Data Rate

We need data in bytes per second:

9600 bits/sec ÷ 8 = 1200 bytes/sec

🔴 Step 2: Understand Processor Speed

1 MIPS =

1 million instructions/sec

So, time per instruction:

$T = \frac{1}{10^6} \text{ sec } = 1 \, \mu s$ T=1061 sec =1μs

🔴 Step 3: Concept of Cycle Stealing

DMA steals 1 cycle per byte transfer
Total bytes transferred per second = 1200

👉 So CPU loses 1200 cycles per second

🔴 Step 4: Total Time Lost

Time lost:

$\text{Time lost} = 1200 \times 1\,\mu s = 1200\,\mu s$ Time lost=1200×1μs=1200μs

Convert to milliseconds:

1200 μs = 1.2 ms

🎯 Final Answer

👉 Processor slowdown = 1.2 ms

⚡ Exam Shortcut Trick

Convert bps → bytes/sec
Find time per instruction (1 / MIPS)
Multiply:

Time lost = bytes/sec × time per cycle

🧠 Quick Formula

$\text{Slowdown} = \text{Data Rate (bytes/sec)} \times \text{Cycle Time}$ Slowdown=Data Rate (bytes/sec)×Cycle Time

🔥 Common Mistakes

❌ Forgetting to convert bits → bytes
❌ Ignoring DMA mode (cycle stealing assumed here)
❌ Using wrong unit (ms vs μs)

🎯 Conclusion

DMA numerical problems become easy if you:

Understand cycle stealing concept
Convert units carefully
Apply simple multiplication

🧠 DMA Numerical Problem (Processor Slowdown Explained Step-by-Step)