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What is the Average Access Time for a machine with the Cache rate of 80% and cache access time of 5ns and Main memory at time of 100s, a. Hierarchical Memory Organization b. Simultaneous Memory Organization

What is the Average Access Time for a machine with the Cache rate of 80% and cache access time of 5ns and Main memory at time of 100s, a. Hierarchical Memory Organization b. Simultaneous Memory Organization

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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### Calculating Average Access Time To find the Average Access Time (AAT) for a machine with the given parameters, we must consider both hierarchical and simultaneous memory organizations. #### Given: - Cache hit rate \(h\) = 80% - Cache access time \(t_c\) = 5ns - Main memory access time \(t_m\) = 100ns #### Formulas: 1. **Hierarchical Memory Organization:** \( \text{Average Access Time (AAT)} = h \times t_c + (1 - h) \times (t_c + t_m) \) 2. **Simultaneous Memory Organization:** \( \text{Average Access Time (AAT)} = h \times t_c + (1 - h) \times t_m \) #### Calculations: (a) **Hierarchical Memory Organization:** \[ \text{AAT} = 0.8 \times 5ns + (1 - 0.8) \times (5ns + 100ns) \] \[ \text{AAT} = 4ns + 0.2 \times 105ns \] \[ \text{AAT} = 4ns + 21ns \] \[ \text{AAT} = 25ns \] (b) **Simultaneous Memory Organization:** \[ \text{AAT} = 0.8 \times 5ns + 0.2 \times 100ns \] \[ \text{AAT} = 4ns + 20ns \] \[ \text{AAT} = 24ns \] Thus, the Average Access Time for hierarchical memory organization is 25ns, while for simultaneous memory organization, it is 24ns.
Transcribed Image Text:### Calculating Average Access Time To find the Average Access Time (AAT) for a machine with the given parameters, we must consider both hierarchical and simultaneous memory organizations. #### Given: - Cache hit rate \(h\) = 80% - Cache access time \(t_c\) = 5ns - Main memory access time \(t_m\) = 100ns #### Formulas: 1. **Hierarchical Memory Organization:** \( \text{Average Access Time (AAT)} = h \times t_c + (1 - h) \times (t_c + t_m) \) 2. **Simultaneous Memory Organization:** \( \text{Average Access Time (AAT)} = h \times t_c + (1 - h) \times t_m \) #### Calculations: (a) **Hierarchical Memory Organization:** \[ \text{AAT} = 0.8 \times 5ns + (1 - 0.8) \times (5ns + 100ns) \] \[ \text{AAT} = 4ns + 0.2 \times 105ns \] \[ \text{AAT} = 4ns + 21ns \] \[ \text{AAT} = 25ns \] (b) **Simultaneous Memory Organization:** \[ \text{AAT} = 0.8 \times 5ns + 0.2 \times 100ns \] \[ \text{AAT} = 4ns + 20ns \] \[ \text{AAT} = 24ns \] Thus, the Average Access Time for hierarchical memory organization is 25ns, while for simultaneous memory organization, it is 24ns.
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