Assignment 13, due Jul 31

On every assignment, write your name legibly as it appears on your University ID card! Homework is due on paper at the start of class on the day indicated (Tuesday or Thursday). Exceptions will be made only by advance arrangement (excepting "acts of God"). Late work must be turned in to the TA's mailbox (ask the CS receptionist in 14 MLH for help). Never push homework under someone's door!

1. Background: In homework 12, problem 3b, it was noted that before running the following sequence of instructions to turn on the Hawk MMU, it is necessary to load the MMU with at least one entry that maps the virtual address for the current page to somewhere sensible.

   a) Write a sequence of Hawk instructions that maps the first page of the Hawk ROM (containing the entire Hawk monitor) to frame 0 of the virtual address space, and sets the access rights appropriately. (0.5 points)

   b) Write a sequence of Hawk instructions that maps the first page of the Hawk RAM to frame 10₁₆ of the virtual address space, and makes the mapping and sets the access rights appropriately. (0.5 points)

2. Background: Here are two ways of computing the same thing, based on code lifted from the SAVEREGS section of the trap handler outlined in Chapter 13 in the Hawk manual:

   FIRST:  CPUGET  R1,TSV
           CPUGET  R3,TPC
           STORE   R1,R2,svR2
           STORE   R3,R2,svPC
           CPUGET  R1,TMA
           CPUGET  R3,PSW
           STORE   R1,R2,svMA
           STORE   R3,R2,svPSW
   
   SECOND: CPUGET  R1,TSV
           STORE   R1,R2,svR2
           CPUGET  R1,TPC
           STORE   R1,R2,svPC
           CPUGET  R1,TMA
           STORE   R1,R2,svMA
           CPUGET  R1,PSW
           STORE   R1,R2,svPSW
   

   These two pieces of code do exactly the same thing. The first solution uses two registers, the second uses only one. If we did not have pipelined computers, the two alternatives would be equally fast, but on a pipelined machine, one alternative is distinctly faster than the other.

   a) Which alternative is faster on a pipelined computer, and why? (0.5 points)

   b) How much faster? You can answer this by careful use of a pipeline diagram. The answer will be a number, a count of the number of clock cycles saved by the faster solution when compared to the slower solution. (0.5 points)

3. Background: The simplified instruction fetch stage discussed near the middle of Chapter 15 omits two important features of the Hawk computer and one important feature of the Sparrowhawk:

   a) What does this diagram imply about the word size of the data fetched from RAM for each instruction, and how does this differ from the Sparrowhawk? (0.5 points)

   b) What additional issue does the Hawk raise that would add even more complexity to this pipeline stage? (0.5 points)