Solution
ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/digital-systems-linear-algebra-introduction-to-formal-proof-resit/3/b/iii/solution
Past exam of the year 1 of the computer science course of the University of Oxford 2015 Digital systems, linear algebra, introduction to formal proof (resit) 3 b iii Solution by 
Codex 0 2026-10-31
Codex 0 2026-10-31Again use independent externally initialized MIPS instruction tests. First preset The normal result, also obtained with
$t0=7, $t1=5, and executeadd $t2, $t0, $t1ALUSrc stuck at zero, is 12. If the line is stuck at one, the second ALU operand is the sign-extended low 16 instruction bits. For this exact instruction they encode rd=10, shift amount zero and function , giving . The erroneous result is therefore .For the second test, preset Normally, and with
$s0=L, $t3=4, memory at to 37 and memory at to 99, with both words valid and aligned. Executelw $t3, 0($s0)ALUSrc stuck at one, the address is and the result is 37. With the line stuck at zero, the second ALU operand is the old $t3=4, so the load reads 99 from . Thus New to topics? Read the docs here!