On a non-pipelined sequential processor, a program segment, which is a part of the interrupt service routine, is given to transfer 500 bytes from an I/O device to memory.Initialize the address register Initialize the count to 500 LOOP: Load a byte from device Store in memory at address given by address register Increment the address register Decrement the count If count != 0 go to LOOPAssume that each statement in this program is equivalent to a machine instruction which takes one clock cycle to execute if it is a non-load/store instruction. The load-store instructions take two clock cycles to execute. The designer of the system also has an alternate approach of using the DMA controller to implement the same transfer. The DMA controller requires 20 clock cycles for initialization and other overheads. Each DMA transfer cycle takes two clock cycles to transfer one byte of data from the device to the memory. What is the approximate speedup when the DMA controller based design is used in place of the interrupt driven program based input-output?

To calculate the speedup, we compare the execution time (in clock cycles) of both approaches for transferring 500 bytes:

1. Interrupt Service Routine (ISR) Approach:

   • Initialization:

     • Initialize the address register: 1 cycle

     • Initialize the count to 500: 1 cycle

     • Initialization overhead = $1 + 1 = 2$ cycles

   • Loop body (executed 500 times):

     • Load a byte from device (load instruction): 2 cycles

     • Store in memory at address... (store instruction): 2 cycles

     • Increment the address register (non-load/store): 1 cycle

     • Decrement the count (non-load/store): 1 cycle

     • If count != 0 go to LOOP (non-load/store): 1 cycle

     • Cycles per loop iteration = $2 + 2 + 1 + 1 + 1 = 7$ cycles

   • Total cycles for ISR $= 2 + (500 \times 7) = 2 + 3500 = 3502$ clock cycles.

2. DMA Controller Approach:

   • Initialization and overheads = $20$ cycles

   • Transfer time for 500 bytes $= 500 \times 2 = 1000$ cycles

   • Total cycles for DMA $= 20 + 1000 = 1020$ clock cycles.

3. Speedup Calculation:

   $$\text{Speedup} = \frac{\text{Time}_{\text{ISR}}}{\text{Time}_{\text{DMA}}} = \frac{3502}{1020} \approx 3.433$$

Thus, the approximate speedup is 3.4, which corresponds to option (A) [p1].