A Microprocessor Instruction Cycle Analysis

Microprocessor Instruction Cycle Analysis

A microprocessor provides an instruction capable of moving a string of bytes from one area of memory to another. The fetching and intel decoding of the instruction takes 15 clock cycles. Thereafter, it takes 20 clock cycles to transfer each byte. The microprocessor is clocked at a rate of 10 GHZ.

a. Determine the length of the instruction cycle for the case of a string of 56 bytes:

b. What is the worst-case delay for acknowledging an interrupt if the instruction is non-interruptible?

c. Repeat part (b) assuming the instruction can be interrupted at the beginning of each byte transfer.

Final answer: The instruction cycle time for a string of 56 bytes is 113.5 nanoseconds. If the instruction is non-interruptible, the worst case delay is also 113.5 nanoseconds. However, if the instruction can be interrupted at the beginning of each byte transfer, the worst-case delay would be 2 nanoseconds.

Explanation: The cycle length of the instruction for a string of 56 bytes can be calculated as the sum of the initial 15 clock cycle, 20 cycles for each byte, multiplied by the total number of bytes. Total cycles = 15 + (20 * 56) = 1135 cycles. Given that the processor clock speed is 10GHZ, The instruction cycle time = 1135 cycles / 10^9 (frequency in Hertz) = 113.5 nanoseconds. The worst-case delay for acknowledging an interrupt, if the instruction is non-interruptible, would be the total time of execution which is 113.5 nanoseconds. If the instruction can be interrupted at the beginning of each byte transfer, the worst-case delay would be the time to transfer one byte, which is 20 clock cycles or 2 nanoseconds.

a. Determine the length of the instruction cycle for the case of a string of 56 bytes. b. What is the worst-case delay for acknowledging an interrupt if the instruction is non-interruptible? c. Repeat part (b) assuming the instruction can be interrupted at the beginning of each byte transfer. The length of the instruction cycle for a string of 56 bytes is 113.5 nanoseconds. If the instruction is non-interruptible, the worst-case delay is also 113.5 nanoseconds. However, if the instruction can be interrupted at the beginning of each byte transfer, the worst-case delay would be 2 nanoseconds.
← The friendship adjacency list analysis Optimistic outlook on educational services officer eso data →

More Posts: