Download University at Buffalo CSE 490| CSE 490/590 – Midterm Exam –2025 V1 Complete Solution. and more Exams Computer Science in PDF only on Docsity!
Tag Cache Content 10b Mem(0x8) 01b Mem(0x5) 10b Mem(0xA ) 11b Mem(0xF) Tag Cache Content 01b M(0x4) 01b M(0x5) 10b M(0xA) 11b M(0xF) Tag Cache Content 01b M(0x4) 01b M(0x5) 01b M(0x6) 11b M(0xF) Tag Cache Content 01b M(0x4) 01b M(0x5) 01b M(0x6) 10b M(0xB) Tag Cache Content 11b M(0xC) 01b M(0x5) 01b M(0x6) 10b M(0xB) .CSE 490/590 – Midterm Exam – 2025 V1 Solution [Question 1] (10 Points) Assume the presence of the following memories in a MIPS system: i. L1 cache ii. Main Memory iii. L3 cache iv. Registers v. SSDs vi. L2 cache vii. Hard drive Show the memory hierarchy and order them in terms of (State Increasing or decreasing order): a. Speed Registers - > L1 - > L2 - > L3 - > Main Memory - > SSDs - > Hard Drive (decreasing) b. Cost per byte Registers - > L1 - > L2 - > L3 - > Main Memory - > SSD - > Hard Drive (decreasing) a. Memory capacity (Size) Registers - > L1 - > L2 - > L3 - > Main Memory - > SSD - > Hard Drive (increasing) [Question 2] (10 Points) a) Consider a direct-mapped cache of size 4 bytes. Each index in the cache can hold only 1 word (here 1 word = 1 byte). Fill in the missing cache blocks at each step according to the address reference and specify whether it is a hit or a miss. The initial state of the cache is provided. Address references are (in order): Reference: 0x4 0x6 0xB 0x hit / miss hit / miss hit / miss hit / miss b) Calculate the hit rate for part a.
Hit rate = No. of hits / total requests = 1/4 = 0. [Question 3] (12 Points) Consider three different processors P1, P2, and P3 executing the same instruction set. Processor Clock Speed in GHz CPI P1 1.5 2. P2 3.0 1. P3 4.0 2. Which processor has the highest performance expressed in instructions per second? (1 GHz = 10^9 Hz). Show your work. IPS = Cycles per second/CPI P1: 1.5/2.5 = 0. P2: 3/1.5 = 2 P3: 4/2.5 = 1. P2 has the highest performance [Question 4] (12 Points) In the following instruction sequence, find all hazards. Rename the registers to eliminate the anti-dependencies and output dependences. Assume you have FP registers up to $F16 available. Show your work. div.s F0, F2, F mult.s F4, F0, F add.s F0, F2, F sub.s F8, F14, F Potential hazards RAW : F0 - > div.s and mult.s; WAR : F4 - > div.s and mult.s, F0 - > mult.s and add.s WAW : F0 - > div.s and add.s div.s F0, F2, F mult.s F9, F0, F add.s F10, F2, F sub.s F8, F14, F
In a pipelined processor, multiple instructions overlap in execution. The execution time is determined by the number of cycles required to complete all instructions, assuming each stage takes 300 ps. Total 7 cycles are required to execute given instructions. Execution time of a pipelined datapath = 300 * 7 = 2100 ps Execution time(pipelined) < Execution time(non-pipelined) It shows that pipelining improved performance. [Question 7] (16 Points) Assume a 5 - stage, static dual-issue MIPS processor that can perform data forwarding. The 2 - issue MIPS helps us issue arithmetic and load/store instructions in parallel. The ideal 2 - issue MIPS, with no data dependencies will have a pipelining diagram as follows: For the instruction sequence given below, draw the pipelining diagram while taking all data dependencies into account. Show the data forwarding if necessary. Find the total number of clock cycles taken for the execution of the instruction sequence. An empty table for pipelining is given below for convenience add $s1, $s2, $s add $s3, $s2, $s sw $s1, 4($t2) add $s5, $s3, $s Address Instruction Pipeline Stages add s1,s2,s2 IF ID EX MEM WB add s3,s2,s3 IF ID EX MEM WB sw s1, 4(t2) IF ID EX MEM WB add s5, s3,s1 IF ID EX MEM WB
7 clock cycles [Question 8] (16 Points) Consider the following floating-point instruction sequence on a processor (shown below) which uses Tomasulo’s Algorithm to dynamically schedule instructions (dual-issue per cycle – no speculation). The processor has the following non-pipelined execution units: ● A 2 - cycle FP add unit ● A 3 - cycle FP multiply unit Assume instructions can begin to execute in the same cycle as soon as it is dispatched and resides in Reservation Stations. Trace the execution by showing the contents of Reservation Stations and FP Registers at the start of each cycle, after instructions have been issued for that cycle. Initial register values are given in the cycle 1 tables. As completed, result is placed in Common Data Bus, and is ready for use in the next cycle. w: add r0, r6, r x: mul r4, r0, r y: add r0, r2, r
3 Adder Reservation Stations Tag 1 S^ Tag 2 S 1 2 y 3.0 8. 3 Instruction Executed: w x y Multiply Reservation Stations Tag 1 S^ Tag 2 S 4 x 8.0 3. 5 Instruction Executed: w x y FP Registers Busy Tag Data 0 yes 2 4. 2 3. 4 yes 4 6. 6 2. 4 Adder Reservation Stations Tag 1 S Tag 2 S 1 2 y 3.0 8. 3 Instruction Executed: w x y Multiply Reservation Stations Tag 1 S Tag 2 S 4 x 8.0 3. 5 Instruction Executed: w x y FP Registers Busy Tag Data 0 yes 2 4. 2 3. 4 yes 4 6. 6 2. 5 Adder Reservation Stations Tag 1 S Tag 2 S 1 2 3 Instruction Executed: w x y Multiply Reservation Stations Tag 1 S Tag 2 S 4 x 8.0 3. 5 Instruction Executed: w x y FP Registers Busy Tag Data 0 11. 2 3. 4 yes 4 6. 6 2. 6 Adder Reservation Stations Tag 1 S^ Tag 2 S 1 2 3 Instruction Executed: w x y Multiply Reservation Stations Tag 1 S^ Tag 2 S 4 5 Instruction Executed: w x y FP Registers Busy Tag Data 0 11. 2 3. 4 24. 6 2. Tag 1 S Tag 2 S 1 w 2.0 6. 2 y 3.0 1 3 Instruction Executed: w x y Tag 1 S Tag 2 S 4 x 1 3. 5 Instruction Executed: w x y Busy Tag Data 0 yes 2 4. 2 3. 4 yes 4 6. 6 2.
