Io Interface Practice Questions

The data transfer rate for DMA is calculated by the amount of data transferred per unit of time.
Given: CPU clock frequency = 4 MHz = $4 \times 10^6$ cycles/second.
One CPU cycle time = $1 / (4 \times 10^6)$ seconds = $0.25 \times 10^{-6}$ seconds = 0.25 µsec.
DMA uses 1% of processor cycles. So, in 100 cycles, 1 cycle is used by DMA.
Time for 1 DMA cycle = 0.25 µsec.
In this 1 DMA cycle, 8 bytes are transferred.
8 bytes = $8 \times 8 = 64$ bits.
So, 64 bits are transferred in 0.25 µsec.
Data transfer rate = $\frac{64 \text{ bits}}{0.25 \text{ \mu sec}} = \frac{64 \text{ bits}}{0.25 \times 10^{-6} \text{ seconds}}$.
Data transfer rate = $\frac{64}{0.25} \times 10^6 \text{ bits/second} = 256 \times 10^6 \text{ bits/second} = 256,000,000 \text{ bits/second}$.

Wait, let's recheck the calculation of $256 \times 10^6$.
Data transfer rate = $\frac{64}{0.25} \text{ bits/\mu sec} = 256 \text{ bits/\mu sec}$.
To convert to bits/second: $256 \text{ bits/\mu sec} = 256 \times 10^6 \text{ bits/second}$.
The options are given in numbers like 2,56,000. Let's check the units.
$256 \times 10^6$ bits/second = $256,000,000$ bits/second. This does not match option C (2,56,000).

Let's look at the solution provided. It says "64 bits / 25 µsec" for the rate.
If 1% of processor cycles are used for DMA, it means 1 DMA cycle occurs for every 100 CPU cycles.
So, 8 bytes (64 bits) are transferred every 100 CPU cycles.
Time for 100 CPU cycles = $100 \times 0.25 \text{ \mu sec} = 25 \text{ \mu sec}$.
So, 64 bits are transferred in 25 µsec.
Data transfer rate = $\frac{64 \text{ bits}}{25 \text{ \mu sec}} = \frac{64}{25} \times 10^6 \text{ bits/second}$.
$\frac{64}{25} = 2.56$.
Data transfer rate = $2.56 \times 10^6 \text{ bits/second} = 2,560,000 \text{ bits/second}$.
This matches option (C) 2,56,000 (if the comma notation represents million, i.e., 2.56 million).
The given solution in the PDF uses 25 µsec, which is indeed 100 CPU cycles (1% usage).
So, 64 bits in 25 µs means $64 / (25 \times 10^{-6})$ bits/sec = $2.56 \times 10^6$ bits/sec = 2,560,000 bits/sec.
This matches option (C) if it is interpreted as 2,560,000. Option C in the question is 25,60,000, which is $2.56 \times 10^6$.

To achieve the highest throughput for bulk data transfer from a hard disk to main memory, we need a mechanism that minimizes CPU involvement and overhead.
(A) DMA (Direct Memory Access) based I/O transfer: DMA controllers allow peripherals to transfer data directly to and from main memory without involving the CPU. This frees the CPU for other tasks, significantly increasing throughput for large data transfers.
(B) Interrupt-driven I/O transfer: The CPU is interrupted upon I/O completion, which is more efficient than polling but still requires CPU intervention for each I/O event. This is less efficient for bulk data.
(C) Polling based I/O transfer: The CPU continuously checks the status of an I/O device. This is very CPU-intensive and inefficient for any data transfer, especially bulk data.
(D) Programmed I/O transfer: The CPU is directly involved in transferring each byte or word of data between the device and memory. This is also very CPU-intensive and has low throughput for bulk data.
Therefore, DMA is the most efficient method for bulk data transfer with the highest throughput.

Given:
DMA transfers one 8-bit character per CPU cycle.
CPU clock frequency = 2 MHz = $2 \times 10^6$ cycles/second.
0.5% of processor cycles are used for DMA.

Time for one CPU cycle = $\frac{1}{2 \times 10^6}$ seconds = 0.5 $\mu$s.
DMA uses 0.5% of CPU cycles, so in 100 CPU cycles, 0.5 cycles are used by DMA.
This means for every 100 CPU cycles, DMA transfers 0.5 characters.
Number of CPU cycles per second = $2 \times 10^6$.
Number of DMA cycles per second = $0.005 \times 2 \times 10^6 = 10^4$ cycles/second.
Since 1 DMA cycle transfers 1 character (8 bits),
Data transfer rate = $10^4 \text{ characters/second} \times 8 \text{ bits/character} = 80000 \text{ bits/second}$.

The DMA controller's data count register size is 16 bits. This means the maximum amount of data that can be transferred in a single DMA operation is $2^{16}$ bytes.
$2^{16}$ bytes = 65,536 bytes = 64 kilobytes (KB).
The total size of the file to be transferred is 29,154 KB.
To find the minimum number of times the DMA controller must take control of the bus, we divide the total file size by the maximum size per transfer:
Number of transfers = $\lceil \frac{29,154 \text{ KB}}{64 \text{ KB}} \rceil = \lceil 455.53125 \rceil = 456$.
Each transfer, including the final partial one, requires the DMA controller to acquire the bus.

To calculate the speedup, we first determine the total clock cycles for both the interrupt-driven I/O and DMA approaches.

1. Clock Cycles for Interrupt-Driven I/O:

• Initialization (address register, count): 2 non-load/store instructions = 2 cycles.
• Loop (500 iterations):
  • Load byte: 2 cycles.
  • Store byte: 2 cycles.
  • Increment address: 1 cycle.
  • Decrement count: 1 cycle.
  • If count != 0 (branch): 1 cycle.
  • Total per iteration: $2 + 2 + 1 + 1 + 1 = 7$ cycles.
• Total cycles for interrupt-driven I/O: $2 + (500 \times 7) = 2 + 3500 = 3502$ cycles.

2. Clock Cycles for DMA:

• Initialization/Overheads: 20 cycles.
• Data transfer (500 bytes): Each byte transfer takes 2 cycles.
• Total cycles for DMA: $20 + (500 \times 2) = 20 + 1000 = 1020$ cycles.

3. Calculate Speedup:
Speedup is the ratio of execution time of the slower method (interrupt-driven I/O) to the faster method (DMA).
$\text{Speedup} = \frac{\text{Cycles (Interrupt-Driven I/O)}}{\text{Cycles (DMA)}} = \frac{3502}{1020} \approx 3.433$
The approximate speedup is 3.4.