The CPU of a system having 1 MIPS execution rate needs 4 machine cycles on an…
2015
The CPU of a system having 1 MIPS execution rate needs 4 machine cycles on an average for executing an instruction. The fifty percent of the cycles use memory bus. A memory read/write employs one machine cycle. For execution of the programs, the system utilizes 90% of the CPU time. For block data transfer, an IO device is attached to the system while CPU executes the background programs continuously. What is the maximum IO data transfer rate if programmed IO data transfer technique is used ?
- A.
500 Kbytes/sec
- B.
2.2 Mbytes/sec
- C.
125 Kbytes/sec
- D.
250 Kbytes/sec
Attempted by 180 students.
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Correct answer: D
Solution overview:
Step 1 — machine cycles per second:
CPU instruction rate = 1 MIPS = 1,000,000 instructions/sec.
Average cycles per instruction = 4, so machine cycles per second = 1,000,000 × 4 = 4,000,000 cycles/sec.
Step 2 — memory-bus cycles available in total:
Fifty percent of machine cycles use the memory bus, so total memory-bus cycles = 0.5 × 4,000,000 = 2,000,000 memory cycles/sec.
Step 3 — memory-bus cycles consumed by regular program execution:
CPU is utilized 90% for running programs, so machine cycles used by programs = 0.9 × 4,000,000 = 3,600,000 cycles/sec.
Of those, 50% use the memory bus, so memory-bus cycles used by programs = 0.5 × 3,600,000 = 1,800,000 memory cycles/sec.
Step 4 — remaining memory-bus cycles available for programmed I/O:
Available memory cycles for I/O = total memory-bus cycles − memory-bus cycles used by programs = 2,000,000 − 1,800,000 = 200,000 cycles/sec.
A memory read/write uses one machine cycle, so each available memory cycle can transfer one data unit (assumed one byte here).
Step 5 — compute programmed I/O transfer rate:
Maximum transfer rate = 200,000 bytes/sec = 200 Kbytes/sec.
Note about the provided choices:
The detailed calculation yields 200 Kbytes/sec. None of the supplied options exactly equals 200 Kbytes/sec. The listed choice of 250 Kbytes/sec is the closest available option; therefore it may have been marked as the intended answer in the original question set, but the correct computed maximum is 200 Kbytes/sec. If this question is used in assessment, adjust the provided choices to include 200 Kbytes/sec or correct the stated result.