GATE CS 2016 Set 1 Q31 — I/O Interface: Interrupt and DMA Mode

To find the minimum number of times the DMA controller needs to acquire control of the system bus, we calculate the maximum amount of data that can be transferred in a single DMA operation.

1. Maximum Transfer Size per DMA Operation:
Since the memory is byte-addressable and the data count register of the DMA controller is $16\text{ bits}$, the maximum number of bytes that can be transferred in one go is: $$\text{Max Bytes} = 2^{16} \text{ bytes} = 65,536 \text{ bytes}$$ Converting this to kilobytes (where $1\text{ KB} = 1024\text{ bytes}$): $$\text{Max Transfer Size} = \frac{65,536}{1024} \text{ KB} = 64 \text{ KB}$$

2. Total File Size to Transfer:
$$\text{Total File Size} = 29,154 \text{ KB}$$

3. Number of DMA Transfers Required:
Each DMA transfer requires the DMA controller to request and obtain control of the system bus once. The total number of transfers is: $$\text{Number of transfers} = \left\lceil \frac{\text{Total File Size}}{\text{Max Transfer Size}} \right\rceil = \left\lceil \frac{29,154 \text{ KB}}{64 \text{ KB}} \right\rceil$$ $$\text{Number of transfers} = \lceil 455.53125 \rceil = 456$$

Thus, the DMA controller needs to get control of the system bus at least 456 times.