Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

a. P15: read 120 b. P15: read 118 P15: write 110 P15: write 130 c. P:1 read 110 P1: read 108 P1: read 130 d.

a. P15: read 120

b. P15: read 118

P15: write 110

P15: write 130

c. P:1 read 110

P1: read 108

P1: read 130

d. P0: read 100

P0: write 108

P0: write 128

image text in transcribed

image text in transcribed

image text in transcribed

image text in transcribed

CPU write hits that generate an invalidate incur Ninvalidate stall cycles. - a writeback of a block, either due to a conflict or another processor's re quest to an exclusive block, incurs an additional Nwriteback stall cycles. Consider two implementations with different performance characteristics sum marized in Figure 4.38. Consider the following sequence of operations assuming the initial cache state in Figure 4.37. For simplicity, assume that the second operation begins after the first completes (even though they are on different processors): P1: read 110 P15: read 110 For Implementation 1, the first read generates 80 stall cycles because the read is satisfied by PO's cache. Pl stalls for 70 cycles while it waits for the block, and PO stalls for 10 cycles while it writes the block back to memory in response to PI's request. Thus the second read by P15 generates 100 stall cycles because its miss is satisfied by memory. Thus this sequence generates a total of 180 stall cycles. For the following sequences of operations, how many stall cycles are generated by each implementation? CPU write hits that generate an invalidate incur Ninvalidate stall cycles. - a writeback of a block, either due to a conflict or another processor's re quest to an exclusive block, incurs an additional Nwriteback stall cycles. Consider two implementations with different performance characteristics sum marized in Figure 4.38. Consider the following sequence of operations assuming the initial cache state in Figure 4.37. For simplicity, assume that the second operation begins after the first completes (even though they are on different processors): P1: read 110 P15: read 110 For Implementation 1, the first read generates 80 stall cycles because the read is satisfied by PO's cache. Pl stalls for 70 cycles while it waits for the block, and PO stalls for 10 cycles while it writes the block back to memory in response to PI's request. Thus the second read by P15 generates 100 stall cycles because its miss is satisfied by memory. Thus this sequence generates a total of 180 stall cycles. For the following sequences of operations, how many stall cycles are generated by each implementation

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Students also viewed these Databases questions

Question

Discuss about training and development in India?

Answered: 1 week ago

Question

Explain the various techniques of training and development.

Answered: 1 week ago

Question

Explain the various techniques of Management Development.

Answered: 1 week ago