For the following diagram, what seems to be trueSpeed decreases down the pyramidStorage increases down the pyramidSpeed increases down the pyramidStorage decreases down the pyramidConsider these processors:Processor 1: Memory Mapped I/O has 90% hit rate and cache access time of 90 ns, in case of miss the extra time is 180 ns.Processor 2: 40% miss rate and double main memory access time if cache time is 40 ns.Which of the following statements are true?Processor 1 is faster than Processor 2Processor 2 is faster than Processor 1Effective access time of processor 1 is 1.5 times of processor 2Effective access time of processor 2 is 1.5 times of processor 1

The question seems to be incomplete. However, I can help you calculate the effective access time for both processors.

For Processor 1: The hit rate is 90%, which means the miss rate is 10%. The cache access time is 90 ns and the extra time in case of a miss is 180 ns. So, the effective access time can be calculated as follows:

Effective Access Time = Hit Time + Miss Rate * Miss Penalty = 90 ns + 0.10 * 180 ns = 90 ns + 18 ns = 108 ns

For Processor 2: The miss rate is 40%, which means the hit rate is 60%. The cache access time is 40 ns and the main memory access time is double if cache time is 40 ns, which means it's 80 ns. So, the effective access time can be calculated as follows:

Effective Access Time = Hit Time + Miss Rate * Miss Penalty = 40 ns + 0.40 * 80 ns = 40 ns + 32 ns = 72 ns

From these calculations, it seems that Processor 2 is faster than Processor 1 because it has a lower effective access time. However, without more context or information, it's hard to definitively say which statements are true.