Solved 10 Questions of Computer Architecture II: Hardware-Software | CS 366, Exams of Computer Architecture and Organization

Download Solved 10 Questions of Computer Architecture II: Hardware-Software | CS 366 and more Exams Computer Architecture and Organization in PDF only on Docsity! What is your name?: CS 366 - Computer Architecture If Midterm Exam #2 - Prof. Reed Fal 2003 (0 points) There are two sections: I, Short Questions ......... 40 points, (20 questions, 2 points each) Il. Short Sections of Code . . .60 points; (10 questions, 6 points each) 100 points total This test is worth 20% of your final grade. This test is open book and open notes. You have 50 minutes. L True/False: (2 pts. each) a 10. Ii. ve According to our textbook, mips stands for Millions of Instructions Per Second, | shah a In addition to the 32 MIPS register file registers having mnemonic names such as $13, the registers can be referred to by their number, starting from $1 up to and including $32. In a recursive program in most cases {though not in every case) $ra needs to be stored on the stack. A carry-out at the most significant bit after an addition of two signed mumbers al ways indi- cates overflow, have an n implicit 1 corresponding to inary floating point number. TEEE 754 single-precision floating point numbers the 1 thal is stored to the left of the decimal place foi Both overflow and underflow are possible with floating point numbers. IEEE 764 single-precision floating point numbers are evenly spaced on the number line. As part of the process to add floating point numbers it is necessary to realign the smallest of the two (so it is no longer in normalized form) in order to make the exponents the same. In the IEEE 754 single precision floating-point format, just like im mathematics, any number to the Oth power is 1. In the solution for the hardwired binary multiplier we discussed in class the shift register Q has double-duty, being used for two purposes. In the solution for the hardwired binary multiplier we discussed in class a carry-out from an intermediate result gets lost, as it is not stored directly into the A shift register. . Having a CDR (Contro] Data Register) is a convenience, however it does not have any effect 2 &) y on the speed of the control unit since each instruction must run to cornpletion before the next instruction is run. . The maximum number of opcades that we can directly implement in Mythsim is 64. . There is at least one MIPS assembler instruction that can directly store a result of an ALU calculation into memory. . If necessary, the Sat register could be used by a SPIM programmer, though certain instruc- tions would have to be avoided. CS 366 - Computer Architecture IL, Midtenn Exam #2, page 1 of S fy) F 16. The value part (in this case Oxflf1) of the mstruction “ li $i, Oxf1£l can actually be a 32-bit value. Tre. Bee 17. If we find in a microprogram that we are running out of registers, one approach we can take is to temporarily store some register values on the stack. 18. Each state in an ASM diagram corresponds to one or more microprogram statements. 19. Once an IR is implemented in a control unit design, the MDR is only used to ‘yrile to mem- ory, and ali reading from memory is through the IR. ‘ak ee tj F 20. A microprogram instruction could write to all registers,in a single instruction. sf A ay ILL. Short Answer: (6 points each) Where there are multiple steps, please circle your final answer. You must show your wark for credit. 1. Given the following diagram of a latch: Is this ever capable of ever giving an output of 1? Explain why or why not. CS 366 - Computer Architecture 1, Midterm Exam #2, page 2 of 5