Practice Question
Duration: 3 min
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AI Summary
An AI-generated summary of this video lecture.
The video features an instructor solving a computer architecture problem regarding instruction set architecture (ISA) and opcode expansion. The problem involves a 16-bit instruction format with a 128-word memory size, supporting both 1-address and 2-address instructions. Given two 2-address instructions, the goal is finding supported 1-address instructions. The instructor derives x * 2^7 + y * 2^14 = 2^16, where x and y are instruction counts. He substitutes y=2 and solves for x, finding the answer to be 256. He breaks down bit allocation for addresses and opcodes.
Chapters
0:00 – 2:00 00:00-02:00
The instructor begins by reading the problem statement displayed on the screen, highlighting critical constraints: 16 bit instruction size, 128 words memory, and the existence of two, 2-address instruction. He establishes the total instruction space as 2^16. He explains that since memory has 128 words, each address field requires 7 bits (2^7 = 128). He formulates the equation x * 2^7 + y * 2^14 = 2^16, where x represents the number of 1-address instructions and y represents the number of 2-address instructions. He substitutes y = 2 into the equation, leading to x * 2^7 + 2 * 2^14 = 2^16. He simplifies 2 * 2^14 to 2^15. The equation becomes x * 2^7 = 2^16 - 2^15. He factors out 2^15 to get x * 2^7 = 2^15(2 - 1), which simplifies to x * 2^7 = 2^15. Solving for x, he finds x = 2^8, which equals 256. He circles the result x = 256 on the board.
2:00 – 2:39 02:00-02:39
In the final segment, the instructor double-checks his work. He writes 256 / 2^7 at the top left, dividing the number of 1-address instructions by the address space per instruction. This yields 2, representing the number of available opcode prefixes. This matches the unused 2-address opcodes (4 total - 2 used). He confirms the steps, and the video ends with x = 256 visible.
The lecture demonstrates opcode expansion in instruction set design. By allocating bits for addresses based on memory size (7 bits for 128 words), remaining bits are used for opcodes. The instructor calculates capacity for different instruction types by subtracting used space from total available space. The verification step confirms the derived instructions align with available opcode prefixes.