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Problem 7.) Program size and Execution Time INSTRUCTION Addressing Mode # of Bytes # of Clock Cycles MOVW R1, #0×4BFC DIR IMM, IND,RE, 4 1 MOVT R1,#0x2000 DIR IMM, IND,RE, 4 1 LDRB RO, [R1,#5] DIR IMM, IND,RE, 4 1 ADD R1.R1,#1 DIR IMM, IND,RE, 4 1 LDRB R2, [R1] DIR IMM, IND,RE, 2 1 ADD R3, RO,R2 DIR IMM, IND,RE, 2 1 STRB R3, [R1} DIR IMM, IND,RE, 2 KEY: IMM=> Immediate, IND=> Indexed, REL=> Relative a) For each instruction on the table above please CIRCLE the correct addressing mode used. b) How much memory does this code segment occupy? c) Assume that the program is run on a microcontroller with a CPU clock of 80 MHz, how long in milliseconds would the six instructions take to execute?

Problem 7.) Program size and Execution Time INSTRUCTION Addressing Mode # of Bytes # of Clock Cycles MOVW R1, #0×4BFC DIR IMM, IND,RE, 4 1 MOVT R1,#0x2000 DIR IMM, IND,RE, 4 1 LDRB RO, [R1,#5] DIR IMM, IND,RE, 4 1 ADD R1.R1,#1 DIR IMM, IND,RE, 4 1 LDRB R2, [R1] DIR IMM, IND,RE, 2 1 ADD R3, RO,R2 DIR IMM, IND,RE, 2 1 STRB R3, [R1} DIR IMM, IND,RE, 2 KEY: IMM=> Immediate, IND=> Indexed, REL=> Relative a) For each instruction on the table above please CIRCLE the correct addressing mode used. b) How much memory does this code segment occupy? c) Assume that the program is run on a microcontroller with a CPU clock of 80 MHz, how long in milliseconds would the six instructions take to execute?

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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**Problem 7: Program Size and Execution Time** ### Instructions Table | **INSTRUCTION** | **Addressing Mode** | **# of Bytes** | **# of Clock Cycles** | |-----------------|---------------------|----------------|-----------------------| | MOVW R1, #0x4BFC | DIR IMM, IND, RE, | 4 | 1 | | MOVT R1, #0x2000 | DIR IMM, IND, RE, | 4 | 1 | | LDRB R0, [R1, #5] | DIR IMM, IND, RE, | 4 | 1 | | ADD R1, R1, #1 | DIR IMM, IND, RE, | 4 | 1 | | LDRB R2, [R1] | DIR IMM, IND, RE, | 2 | 1 | | ADD R3, R0, R2 | DIR IMM, IND, RE, | 2 | 1 | | STRB R3, [R1] | DIR IMM, IND, RE, | 2 | 1 | **KEY**: - **IMM** => Immediate - **IND** => Indexed - **REL** => Relative ### Questions a) For each instruction on the table above, please CIRCLE the correct addressing mode used. b) How much memory does this code segment occupy? **Answer**: To calculate the total memory occupied by the code segment, sum up the number of bytes for each instruction: - MOVW : 4 bytes - MOVT : 4 bytes - LDRB : 4 bytes - ADD : 4 bytes - LDRB : 2 bytes - ADD : 2 bytes - STRB : 2 bytes Total memory = 4 + 4 + 4 + 4 + 2 + 2 + 2 = 22 bytes c) Assume that the program is run on a microcontroller with a CPU clock of 80 MHz, how long in milliseconds would the six instructions take to execute? **Answer**: Each instruction takes 1 clock cycle to execute. With a 80 MHz clock, each clock cycle is: \[ \text{Time per cycle} = \frac{1}{80 \text{ MHz}} = \frac{1
Transcribed Image Text:**Problem 7: Program Size and Execution Time** ### Instructions Table | **INSTRUCTION** | **Addressing Mode** | **# of Bytes** | **# of Clock Cycles** | |-----------------|---------------------|----------------|-----------------------| | MOVW R1, #0x4BFC | DIR IMM, IND, RE, | 4 | 1 | | MOVT R1, #0x2000 | DIR IMM, IND, RE, | 4 | 1 | | LDRB R0, [R1, #5] | DIR IMM, IND, RE, | 4 | 1 | | ADD R1, R1, #1 | DIR IMM, IND, RE, | 4 | 1 | | LDRB R2, [R1] | DIR IMM, IND, RE, | 2 | 1 | | ADD R3, R0, R2 | DIR IMM, IND, RE, | 2 | 1 | | STRB R3, [R1] | DIR IMM, IND, RE, | 2 | 1 | **KEY**: - **IMM** => Immediate - **IND** => Indexed - **REL** => Relative ### Questions a) For each instruction on the table above, please CIRCLE the correct addressing mode used. b) How much memory does this code segment occupy? **Answer**: To calculate the total memory occupied by the code segment, sum up the number of bytes for each instruction: - MOVW : 4 bytes - MOVT : 4 bytes - LDRB : 4 bytes - ADD : 4 bytes - LDRB : 2 bytes - ADD : 2 bytes - STRB : 2 bytes Total memory = 4 + 4 + 4 + 4 + 2 + 2 + 2 = 22 bytes c) Assume that the program is run on a microcontroller with a CPU clock of 80 MHz, how long in milliseconds would the six instructions take to execute? **Answer**: Each instruction takes 1 clock cycle to execute. With a 80 MHz clock, each clock cycle is: \[ \text{Time per cycle} = \frac{1}{80 \text{ MHz}} = \frac{1
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