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CS 6290 High-Performance Computer Architecture - Project 0: Simulator

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CS 6290: High-Performance Computer Architecture

Project 0

This project is intended to help you set up and use the simulator for the other three projects, and also to help you understand the importance of branch prediction a little bit better. To complete this project, you will need to set up VirtualBox and the CS6290 Project virtual machine (VM). CourseNana.COM

You will put your answers in the red-ish boxes in this Word document, and then submit it in Canvas (the submitted file name should still be PRJ0.docx). You should not modify any other text in this document, and your answers should be in the red-ish boxes, in red letters and with a red underline. The font settings for the red-ish boxes are already set that way, so you should be OK if you just put your answers in and don’t change the font colors or underline settings. Georgia Tech students have full access to Office 365 so opening and editing of this document should not be a problem for anyone, and we will not accept answers in any other format. If you use another tool to edit and save the file, please make sure that it preserves at least the font colors and underline settings, otherwise graders might not find your answers easily, and points lost for that reason will stay lost. In each answer box, you must first provide your answer to the actual question (e.g. a number). You can then use square brackets to provide any explanations that the question is not asking for but that you feel might help us grade your answer. E.g. answer 9.7102 may be entered as 9.7102 [Because 9.71+0.0002 is 9.7102]. For questions that are asking “why” and/or “explain”, the correct answer is one that concisely states the cause for what the question is describing, and also states what evidence you have for that. Guesswork, even when entirely correct, will only yield 50% of the points on such questions. CourseNana.COM

Additional files to upload are specified in each part of this document. Do not archive (zip, rar, or anything else) the files when you submit them – each file should be uploaded separately, with the file name specified in this assignment. You will lose up to 10 points for not following the file submission and naming guidelines, and if any files are missing you will lose all the points for answers that are in any way related to that file (yes, this means an automatic zero score if the PRJ0.docx file is missing). Furthermore, if it is not VERY clear which submitted file matches which requested file, we will treat the submission as missing that file. The same is true if you submit multiple files that appear to match the same requested file (e.g. several files with the same name). In short, if there is any ambiguity about which submitted file(s) should be used for grading, the grading will be done as if those ambiguous files were not submitted at all. CourseNana.COM

As explained in the course rules, this is an individual project : no collaboration with other students or anyone else is allowed . CourseNana.COM

Part 1 [60 points]: Simulator set-up and simulation

Install the Oracle VirtualBox VMM (Virtual Machine Manager), then download the CS6290 Project virtual machine (DO NOT get the VM somewhere else, e.g. at the Udacity website for CS 6290 OMS, because those are likely older obsolete versions). CourseNana.COM

Then import this VM into VirtualBox. Once you start (boot) the CS6290 Project VM from within VirtualBox, you should see a Linux desktop. Click on the the topmost icon on the left-side ribbon and in the “Type Your Command” field type “Terminal”. Now you should get a terminal window – click on it and use it to issue commands we need. CourseNana.COM

The simulator in the “sesc” directory within our home directory, and we first need to build it. We will go into that directory and build the simulator: CourseNana.COM

cd ~/sesc

In the ~/sesc directory you should see (use ‘ls’) sesc.opt and sesc.debug links to executable files. We only compiled sesc.opt now, so only that link works, but that is enough for now because we will not be modifying the simulator in this project. CourseNana.COM

Now we need some application to run inside the simulator. The simulator simulates processors that use the MIPS ISA, so we need an application compiled for a MIPS-based Linux machine. Since these are hard to find, we have set up the Splash2 benchmark suite in the ‘apps’ subdirectory. For now, let’s focus on the lu application: CourseNana.COM

cd apps/Splash2/lu

There is a Makefile for building the application from its source code, which is in the Source subdirectory: CourseNana.COM


Now you should see (‘ls’) a lu.mipseb executable in the application’s directory. This is an executable for a big-endian MIPS processor – just the right thing for our simulator. So let’s simulate the execution of this program! CourseNana.COM

Well, the simulator can simulate all sorts of processors– you can change its branch predictor, the number of instructions to execute per cycle, etc. So for each simulation we need to tell it exactly what kind of system to simulate. This is specified in the configuration file. SESC already comes with a variety of configuration files (in the ~/sesc/confs/ directory), and we will use the one described in the cmp4-noc.conf file. So we use the following command line (this is all one line): CourseNana.COM

~/sesc/sesc.opt  –fn32.rpt  -c  ~/sesc/confs/cmp4-noc.conf
-olu.out     -elu.err lu.mipseb –n32 -p1

Note that the dashes in the above command line are “minus” signs – pasting them into this document may have resulted in changing these to longer dashes, in which case you should change them back to “minus” signs. Also, note this is one command line that just does not fit in a single row in this document. Overall, be very careful if you copy-paste this from the document into the VM. CourseNana.COM

In this command line, the –f option tells the simulator to create a report file that ends with a dot and the specified string. The name of the report file is “sesc_” followed by the name of the executable for the simulated application (lu.mipseb in this case), followed by a dot and the string we specify here. So, in this case sesc_lu.mipseb.n32.rpt will be the file where the simulator stores simulation results. It is very important to delete the simulation file before we run a simulation that writes to the same file – if there is already a report file with the same name, the simulator appends its report to the file, and that causes the reporting scripts to combine the results of the simulations, and that means you will not get the results you need for the simulation you just ran! CourseNana.COM

The –c option tells the simulator which configuration file to use. The configuration file tells the simulator all sorts of parameters for the simulated processor, including what its branch predictor looks like, how big are its caches, etc. The –o and –e options tell it to save the standard and error output of the application into lu.out and lu.err files, respectively. This is needed because the simulator itself prints stuff, and we want to be able to look at the output and possible errors from the application to tell it executed OK. Note that there should be a space between “-c” and the configuration file name, but there should be no space between the “-o” and its file name, and there should be no space between “-e” and its file name. CourseNana.COM

Then the command line tells the simulator which (MIPS) executable to use, and what parameters to pass to that executable. The –n32 parameter tells the LU application (which does LU decomposition of a matrix) to work on a 32x32 matrix. The –p1 parameter tells the application to use only one core – we will want to use only one core until Project 3! Now you should look at the output of the application – the lu.err file should be empty and the lu.out should have the “Total time without initialization: 0” line at the end. If you do not check, you may get incorrect simulation results without knowing it , e.g. because you had a typo in the command line that causes the application to abort without actually doing much. Applications not really finishing is a very common pitfall in benchmarking – we are running the applications just to see how the hardware performs, so we tend to not look at the output. However, if a run got aborted half-way through because of a problem (e.g. mistyped parameter, running out of memory, etc.), the hardware might look really fast for that run because execution time will be much shorter! So we need to check for each run whether it actually completed! CourseNana.COM

After the run completes, we also get a simulation report file (sesc_lu.mipseb.n32 for the simulation we just ran). This report file contains the command line we used, a copy of the hardware configuration we used, and then lots of information about the simulated execution we have just completed. You can read the report file yourself, but since the processor we are simulating is very sophisticated, there are many statistics about its various components in the report file. To get the most relevant statistics out of the report file, you can use the report script that comes with SESC: ~/sesc/scripts/report.pl sesc_lu.mipseb.n32.rpt CourseNana.COM

You can specify multiple files (or use wildcards), in which case the script prints the summary for each file you listed. CourseNana.COM

The printout of the report script gives us the command line that was used, then tells us how fast the simulation was going (Exe Speed), how much real time elapsed while the simulator was running (Exe Time), and how much simulated time on the simulated machine it took to run the benchmark (Sim Time). Note that we have simulated an execution that takes much less than a millisecond, but it probably took many milliseconds for the simulator to simulate that execution. Because simulation executes on one machine but is trying to model another, when reporting results there can be some confusion about which one we are referring to. To avoid this confusion, the simulation itself is called “native” execution and the simulated execution is sometimes also called “target” execution. CourseNana.COM

A)In this simulation, the simulated processor needed 0.462 milliseconds of simulated time (“Sim Time” in the report) to run the application, but the simulator itself took 0.580 seconds of native execution time to do this simulation (“Exe Time” in the report). CourseNana.COM

The printout of the report script also tells us how accurate the simulated processor’s branch predictor was overall (the number under Total). The next two numbers are in a parenthesis under the RAS label, and they report the accuracy of the return address stack (RAS) and what percentage of branches actually used the RAS. Next, under the BPRed label we have the accuracy for branches that didn’t use the RAS (i.e. they used the direction predictor and possibly the BTB and/or the BTAC), and then we have statistics for the BTB and BTAC, which are less interesting for now. CourseNana.COM

B)The overall accuracy of the simulated branch predictor in this simulation was 91.07 percent. The RAS tends to always be highly accurate but it only took care of 11.34 percent of all dynamic branch instructions (i.e. branches as they occur at runtime). That means that the direction predictor handles the remaining 88.66 percent of dynamic branch instructions and achieves 89.94 percent accuracy. CourseNana.COM

Next, the report tells us how many dynamic instructions were completed (nInst) and the breakdown of these instructions, i.e. the percentage of nInst that comes from control flow instructions like branches, jumps, calls, returns, etc. (the number under the BJ label), from loads, from stores, from integer ALU, and from floating point ALU operations. The rest of that line (LD Forward and beyond) is less interesting for now. CourseNana.COM

C)The number of completed dynamic instructions in this run of lu on the simulated processor is 323668 , and control flow instructions represent 9.49 percent of these instruction. CourseNana.COM

Next the report tells us the overall IPC achieved by the simulated processor on this application, how many cycles the processor spent trying to execute instructions, and what those cycles were spent on. This last part is broken down into issuing useful instructions (Busy), waiting for a load queue entry (LDQ) to become free, then the same for store queue (STQ), reservation stations (IWin), ROB entries, and physical registers (Regs, the simulator uses a Pentium 4-like separate physical register file for register renaming). Some of the cycles are wasted on TLB misses (TLB), on issuing wrong-path instructions due to branch mispredictions (MisBr), and some other more obscure reasons (ports, maxBr, Br4Clk, Other). Note that each of these “how the cycles were spent” items is a percentage although they do not have the percentage sign. Also note that the simulator models a multiple-issue processor, so the breakdown of where the cycles went is actually attributing partial cycles when needed. For example, in a four-issue processor we may have a cycle where one useful instruction is issued and then the processor could not issue another one because the LDW became full. In this cycle, 0.25 cycles would be counted as “Busy” and 0.75 cycles toward “LDQ”. CourseNana.COM

D)Given that branches represent 9.49 percent of all dynamically completed instructions, and that only 8.93 [Rounded to the nearest hundredth] percent of all completed branch instructions are mispredicted, we can conclude that only .85 [Rounded to the nearest hundreth] percent of all dynamically completed instructions are mispredicted branches. But now we see that branch mispredictions result in wasting 45.2 percent of the processor’s performance potential. Why is this last number (percentage of performance wasted) so high compared to the previous one (percentage of instruction that are mispredicted branches)? CourseNana.COM

According to the numbers found by running the first simulation it takes 45.2% of all cycles to deal with mispredictions while mispredictions only account for .85% of all dynamically completed instructions. The reason the last number is so high compared to the previous one is because the processor needs additional steps to account for the mispredictions. A branch instruction is a control dependence and must force the pipeline to flush the previous instructions if it is mispredicted causing this waste in the processor’s performance. We can calculate the penalty caused by a misprediction like this: 1 + 0.85 * x = 1.43. Isolating the x we get x = 0.5 which equals our penalty for misprediction. CourseNana.COM

Now run two additional simulations, this time using 64 and then 256 as the matrix size for lu (note the changes in –f and –n parameters): ~/sesc/sesc.opt -fn64.rpt -c ~/sesc/confs/cmp4-noc.conf -olu.out -elu.err lu.mipseb -n64 -p1 ~/sesc/sesc.opt -fn256.rpt -c ~/sesc/confs/cmp4-noc.conf -olu.out -elu.err lu.mipseb -n256 -p1 CourseNana.COM

Note: remember to check lu.err and lu.out after each simulation to make sure that the simulated run has completed correctly! CourseNana.COM

E)As we increase the matrix size (-n parameter) from 32 to 64, and then to 256, the overall branch predictor improves from 91.07 percent to 94.74 percent and then to 94.72 percent. Why does this accuracy improve as we increase the –n parameter in lu? CourseNana.COM

The reason the accuracy improves is due to improvement of the branch predictors overtime. As the matrix becomes larger, a lot of branches are executed more often than before. This allows the predictors to get a better history of more difficult to predict patterns of taken vs not taken branches. For example in a n=32 matrix we may not see enough of a certain branch to be able to follow it’s pattern to begin predicting it with a very high accuracy. But with the n=64 and n=256, we have a better chance to find the pattern with more costly predictors such as PShare or GShare predictors. Also, as the branches are executed more often because the matrix gets larger, the initial mispredictions it takes to find the difficult to predict patterns cause less of an impact. CourseNana.COM

Note: Predictor accuracy is essentially the overall probability that an occurrence of branch instruction will be predicted correctly, i.e. the fact that we complete more instructions or that we complete more branches does not by itself explain why each branch occurrence is now more likely to be predicted correctly. Note: We are not asking for guesswork - the answer to this question should include either conclusive evidence or excellent reasoning. CourseNana.COM

F)Submit all three simulator-generated report files ( sesc_lu.mipseb.n32.rpt, sesc_lu.mipseb.n64.rpt, and sesc_lu.mipseb.n256.rpt ). Remember that each of these should be a result of only one simulation! CourseNana.COM

Part 2 [40 points]: Compiling and simulating a small application

Now let’s see how to compile our own code so we can run it within the simulator. For that purpose, modify the hello.c in the ~/sesc/tests/ directory to print a greeting with your first and last name. The hello.c should be exactly as follows (e.g. do not omit the \n at the end of the string in the printf call, do not add extra spaces or other characters to the string, etc.), except that the name (the part in bold and cursive letters) should be your first and last name as show in Canvas: CourseNana.COM

#include <stdio.h>
int main(int argc, char *argv[]){
  printf("Hello! My name is ** \n");
  return 0;

Now you should try to compile and run your code natively, to see if it’s working: CourseNana.COM

gcc -o hello hello.c

To compile our hello.c application, we can use the cross-compiler directly: CourseNana.COM

/mipsroot/cross-tools/bin/mips-unknown-linux-gnu-gcc –O0 -g
-static -mabi=32 -fno-delayed-branch -fno-optimize-sibling-
calls -msplit-addresses -march=mips4 -o hello.mipseb hello.c

We disable optimization (-O0), include debugging information in the executable file (-g), link with static libraries (-static) to avoid dynamic linking which requires additional setting up in the simulator, tell the compiler to avoid using branch delay slots (-fno- delayed-branch), regular call/return for all procedure calls (-fno-optimize-sibling-calls), use some optimizations that allow faster addressing (-msplit-addresses), and use a 32-bit implementation of the MIPS IV ISA (-mabi=32 –march=mips4). CourseNana.COM

After the compiler generates the MIPS executable file hello.mipseb for us, we can execute it in the simulator. Run this command line exactly as shown below – you should not change the command line options, run it from a different directory, or specify the path for hello.mipseb – that would change the simulated execution and prevent us from checking if you did it correctly and if everything is working. CourseNana.COM

~/sesc/sesc.opt –fhello0.rpt -c ~/sesc/confs/cmp4-noc.conf
-ohello.out hello.mipseb

To complete this part of the project, get the simulation report summary (using report.pl) and fill in the blanks in the following statements: CourseNana.COM

G)In this simulation, the simulated processor executed 4866 (“nInst” in the report) instructions. CourseNana.COM

H)Why does it take so many instructions to execute even this small program? CourseNana.COM

Note: Again, answers based on guesswork will not earn points here! CourseNana.COM

After researching the helpful notes below, I noticed some important things. First from executing “mips-unknown-linux-gnu-objdump” with “-d and -f” flags, I noticed that the entry point is not

it is <__start> which then calls CourseNana.COM

<__libc_start_main>. Already we have a lot of instructions that have not executed any instructions specific to the work completed in our main function in hello.c. Then after noticing this I researched __libc_start_main which executes instructions to accomplish initializing libc and its symbols. Finally, after researching “mips-unknown-linux-gnu-strip” I found that strip discards all symbols from the object file significantly reducing the number of mips instructions originally found from the objdump. My conclusion from this research is that a statically linked c program must load all libraries and symbols before it can execute the actual program that is why when we performed the strip to discard symbols the instructions were reduced. Helpful Note #1: You can use mips-unknown-linux-gnu-objdump with a “-d” option to see the assembler listing of the hello.mipseb executable. In the listing produced by the disassembler, the code for the C “main” can be found by searching for “
”. Helpful Note #2: When an executable file is loaded, it does not begin execution at main(). For what really happens, you can look up the documentation for the __glibc_start_main function which is part of the GNU standard C library (GLIBC). Helpful Note #3: Not all of the code you see in the listing from Hint 1 is actually executed. For example, you can use mips-unknown-linux-gnu-strip on the hello.mipseb executable to dramatically reduce its code size without breaking it (you can easily find online what the “strip” utility does). Even among the code that is not removed by “strip”, a lot of the code is only executed under special circumstances (failure to allocate memory, etc.) and is not reached in the specific run we are simulating. Note: No guesswork or vague answers! The answer should be very specific, i.e. to answer this you need to point out specific behavior in specific code. Helpful Note: You can use objdump to see what main is really doing (it’s not calling printf!) then find the source code for that function online instead of trying to figure out what it does by looking at the (barely human-readable) disassembly. J)Submit your last hello.c source code (the one with the exclamation mark that made is execute fewer instructions than with one less exclamation), the original report file (sesc_hello.mipseb.hello0.rpt ), and the one from the last simulation (the one that resulted in a decrease in completed instructions, e.g. it may be sesc_hello.mipseb.hello3.rpt ).

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