Spring 1999, Computer Org, Assignment 5

Answers

1. No circuits on test
2. No circuits on test
3. (4 points) Mic-2
   • Explain in detail why INVOKEVIRTUAL is only 11 instructions in Mic-2, but 22 instructions in Mic-1. Indicate why some instructions could be combined, and why some instructions could be eliminated altogether.
     • invokevirtual1,7,10,13,21 are eliminated because IFU does all fetching and PC incrementing
     • invokevirtual2-4 combined into 1 instruction: MAR = CPP + MBRU2 due to IFU and additional bus
     • invokevirtual5 becomes OPC = PC because IFU increments PC
     • invokevirtual6 becomes PC = MDR becuase IFU does all fetching
     • invokevirtual8,9,11 combined into TOS = SP - MBRU2 due to IFU and additional bus
     • invokevirtual12 beomes TOS = MAR = H = TOS + 1 because H is needed as temp reg
     • invokevirtual14,15,16 combined into MDR = SP + MBRU2 + 1 due to IFU and additional bus
     • invokevirtual17-20 remain unchanged, but have different labels
     • invokevirtual22 becomes LV = TOS; goto (MBR1) due to merging the interpreter loop and the IFU.
   • Do the same analysis for ILOAD.
     • iload1,2 combined into MAR = MBRU1 + LV; rd due to IFU and additional bus
     • iload3 unchanged but with new label
     • iload4 is not needed dut to IFU doing all fetching and incrementing PC. wr is moved to next instruction
     • iload5 besomes TOS = MDR; wr; goto (MBR1) due to merging the interpreter loop and the IFU and moving wr from previous instruction to this instruction.
4. (2 poionts) Whenever the ALU does subtraction for unsigned numbers, it inverts the normal carry out bit to set the carry flag. How would this be implemeted in hardware?
   • AND the ENB, ENA, INVA, and INC input lines of the ALU. XOR the result with the carry out of the high bit. This will toggle the carry out whenever B-A is performed.
5. (2 points) It is not reliable to use the N bit to determine the relative order of two unsigned numbers. Give me an example of subtracting unsigned numbers as follows:
   • Perform A - B where the result is negative and A < B.
     • 7Fh - FFh = 80h
   • perform C - B where the result is positive and C < B. B should be the same number as in the first subtraction.
     • 7Eh - FFh = 7Fh
6. (4 points) Mic-3. Figure 4-33 on page 258 shows how the SWAP instruction from Mic-2 would be sequenced in Mic-3.
   • Create a similar table for ILOAD.

     iload1	iload2	iload3
     A=LV; B=MBRU1
     C = A + B	B = SP
     MAR = C; rd	C = B + 1
     MDR = mem	MAR = SP = C
     		B = MDR
     		C = B
     		TOS = C; wr

   • Create a similar table for IRETURN

     ireturn1	ireturn3	ireturn4	ireturn5	ireturn6	ireturn7	ireturn8
     B = LV
     C = B
     MAR=SP=C;rd
     MDR = mem
     	B = MDR
     	C = B
     	LV=MAR=C;rd
     	MDR = mem	B=LV
     		C=B+1	B=MDR
     		MAR=C	C=B	B=SP
     			PC=C;rd	C=B
     			MDR=mem	MAR=C
     					B=MDR
     					C=B	B=TOS
     					LV=C	C=B
     						MDR=C;wr
     						goto(MBR1)

7. (2 points) Consider the direct-mapped cache on page 267.
   • What would be the tag and line for the hex address 087F473ADh?

     8	7	F	4	7	3	A	D
     1000	0111	1111	0100	0111	0011	1010	1101
     10000111111101000111001110101101
     Tag			Line			Word	Byte
     1000011111110100			01110011101			011	01

   • If this were a full associative cache, then what would be the tag and the line for the same address?

     8	7	F	4	7	3	A	D
     1000	0111	1111	0100	0111	0011	1010	1101
     10000111111101000111001110101101
     Tag						Word	Byte
     100001111111010001110011101						011	01