EE 4720 Computer Architecture - HW 7 Solution


Problem 1

The solution below is partly repeated. The first table shows the entire solution. (You might need to maximize your browser window.) The second table, for convenience, shows the solution starting at cycle 11.

Some points:

Time 0    1    2    3    4    5    6    7    8    9    10   11   12   13   14   15   16   17   18   19   20   21   22   23   24   25   26   27   28
muld IF   ID   5:A1 5:A2 5:A5 5:A4 5:WB
addd      IF   ID   3:RS 3:RS 3:RS 3:RS 3:A1 3:A2 3:A3 3:A4 3:WB
muld           IF   ID   6:RS 6:RS 6:RS 6:M1 6:M2 6:M3 6:M4 6:WB 6:WB
subd                IF   ID   4:RS 4:RS 4:RS 4:A1 4:A2 4:A3 4:A4 4:WB 4:WB
loop:
ld                       IF   ID   2:EX 2:MI 2:WI                                                                       IF   ID
subd                          IF   ID                            3:RS 3:RS 3:A1 3:A2 3:A3 3:A4 3:WB
ld                                 IF                            ID   1:EX 1:MI 1:WB
subd                                                             IF   ID   4:RS 4:RS                4:A1 4:A2 4:A3 4:A4 4:WB
ld                                                                    IF   ID   2:EX 2:MI 2:WB
subd                                                                       IF   ID                  3:RS                     3:A1 3:A2 3:A3 3:A4 3:WB
subi                                                                            IF                  ID   1:EX 1:MI 1:WB
addi                                                                                                IF   ID   2:EX 2:MI 2:WB
bnez                                                                                                     IF   ID        1:EX 1:MI 1:WB
Time 11   12   13   14   15   16   17   18   19   20   21   22   23   24   25   26   27   28
muld 
addd 3:WB
muld 6:WB 6:WB
subd 4:A4 4:WB 4:WB
loop:
ld                                                               IF   ID
subd      3:RS 3:RS 3:A1 3:A2 3:A3 3:A4 3:WB
ld        ID   1:EX 1:MI 1:WB
subd      IF   ID   4:RS 4:RS                4:A1 4:A2 4:A3 4:A4 4:WB
ld             IF   ID   2:EX 2:MI 2:WB
subd                IF   ID                  3:RS                     3:A1 3:A2 3:A3 3:A4 3:WB
subi                     IF                  ID   1:EX 1:MI 1:WB
addi                                         IF   ID   2:EX 2:MI 2:WB
bnez                                              IF   ID        1:EX 1:MI 1:WB

Problem 2

The modifications to the pipeline are shown below. This solution works in most situations. (See end of description.)

The BHT has a separate read and write port (so that one entry can be read while another is updated). The BHT address is the concatenation of h-m PC bits and m bits from the shift register. The BHT entry and the BHT entry address are clocked into the IF/ID latch.

The prediction is made in the ID stage simply by looking at the most significant bit of the BHT entry. An updated BHT entry is generated by adding or subtracting 1, based on the branch outcome. (In any real branch prediction scheme we would not know the outcome until several cycles later.) Using the address in the IF/ID latch, the updated entry is written to the BHT (at the end of ID) using the address from the latch. (It's important to write the entry to the same address it was read from, of course.)

The shift register is updated at the end of ID. The register shifts only if the "enable" line is asserted.

The solution below would have received full credit, but it does have a problem: If there are two consecutive branches, the shift register value used for the second branch will not be correct. How could this problem be fixed? Also, how would the design be changed if the "taken" signal were available only after ID? Solution to Problem 2


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David M. Koppelman - koppel@ee.lsu.edu
Modified 28 Apr 1997 18:50 (23:50 UTC)