There will not be a quiz in this recitation.
In this recitation you will
do some review questions for the final exam.
Problems
- If machine performance improves by 35% per year, by what percentage will it have improved after 10 years?
- If machine A is 30% faster than machine B and machine A is 60% faster than machine C, how much faster is machine B than C?
- If machine performance has improved by a factor of 1,000 over 10 years, at what percentage did it increase each year,
assuming that each year had the same percentage increase?
- A system has 5 disk drives, each with a MTTF of 500,000 hours, and a single power supply with a MTTF of 200,000 hours.
The system will fail if one component fails. Calculate the system MTTF.
- If we reduce the CPI from 5 to 4 while increasing the clock cycle from 1 ns to 1.1 ns, what is the speedup?
- Processors A and B are identical, except that processor B executes floating point instructions 40% faster than A.
Calculate the speedup if a task spends 20% of its time executing floating point instructions on machine A.
- Processors A and B are identical, except that processor B executes floating point instructions 40% faster than A.
Calculate the speedup if a task spends 20% of its time executing floating point instructions on machine B.
- Suppose that a cache has a hit time of 1 cycle, a hit ratio of 92%, and a miss penalty of 8 cycles.
What is the average access time in cycles?
- Suppose we have a standard MIPS 5-stage pipeline which has a CPI of 1 when all memory accesses are cache hits.
A cache is used with a hit rate of 95% and a miss penalty of 8 cycles. Loads and stores are 30% of all instructions.
What is the CPI when cache misses are taken into account?
- A memory system uses three caches. The hit rates and hit times are as follows:
- L1: .95, 1 ns.
- L2: .98, 6 ns.
- L3: .99, 20 ns.
For a memory access, first the L1 cache is checked, If there is a miss, the L2 cache is checked. If there is a miss in the
L2 cache, the L3 cache is checked. Finally, a miss in the L3 cache causes an access to main memory which has an access time
of 60 ns. What is the effective memory access time?
The following are from a previous recitation which may not have been completed.
- Fill in the following table to illustrate how Tomasulo's algorithm works.
Assume the following:
- There are sufficient reservation stations.
- There is only one functional unit that does both floating point add and subtract.
- There is only one functional unit that does multiply and divide.
- Execution times are as follow: load: 1, fadd: 2, fsub: 2, fmul:5, fdiv: 10
- The L.D execution is the address calculation that uses dedicated hardware that can do the calculation in one cycle.
- Floating point units (fadd and fmul) are busy during execution and can be reused the cycle after execution completes.
This is different from the assumption we made in lecture.
- If two reservation stations are both ready and competing for the same functional unit, the one that
issued first has priority.
Instruction | Reservation Station | Execution Unit |
Issue Cycle | Ex Start Cycle | Ex End Cycle |
Memory Cycle | CDB cycle | Write dest |
L.D | F0,8(R2) | | | | | | | | |
ADD.D | F4,F2,F2 | | | | | | | | |
SUB.D | F6,F4,F0 | | | | | | | | |
MUL.D | F10,F2,F4 | | | | | | | | |
ADD.D | F12,F2,F4 | | | | | | | | |
SUB.D | F14,F10,F0 | | | | | | | | |
- Redo problem 11 assuming that there are 2 functional units that can add and subtract.
- Redo problem 11 assuming that the first instruction has a cache miss with a miss penalty of 10 cycles.
- Redo problem 11 assuming that there is only 1 add reservation station which is reserved until the cycle after the result is put the the CDB.