EECS 370 – Introduction to Computer Organization – Exam 1 w

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9. Single Cycle Datapath [12 pts]After graduation you join a start-up company that specializes in single-cycle LC2K processors(similar to those studied in class). The company wants you to do some verification of theirnewest processor to ensure correctness, thus you decide to develop some single-instruction testsprograms to check that the control ROM is working properly.Below is a table of possible bugs that you are checking for. For instance the first row considersthe possibility that control signal D (as shown in the picture) is forced to always be 0 (in jargon,it is said to be ‘stuck-at 0’). Which LC2K instructions could be used to reveal this bug? For eachbug, circle all instructions that could be used to expose the bug by sending them to the processorand then checking the results (by inspecting memory and register file).D stuck-at 0 add nand sw lw beq jalr noop haltC stuck-at 1 add nand sw lw beq jalr noop haltB stuck-at 0 add nand sw lw beq jalr noop haltA stuck-at 1 add nand sw lw beq jalr noop haltPage 10 of 12
10. ISA Design [14 pts]Your EECS 370 instructional staff has come up with a ridiculously fast ISA that just embarrassesopposing ISAs. This new ISA, ShoeLace, is byte-addressable, stack-based and it contains twoaccumulators (ACC1 and ACC2) to assist with computations. There are three instructions typesthat vary in length and are detailed below.R-type instructionsbits 7-4 bits 3-0unused opcodeInstruction Opcode Actionmov1 0001 Pop the top value of the stack and store it into ACC1mov2 0011 Pop the top value of the stack and store it into ACC2pop 0000 Pop the top value of the stackhalt 1000 Stop the programI-type instructions(immediate is a 12-bit valuein 2’s complement form)bits 15-4 bits 3-0immediateopcodeInstruction Opcode Actionpushi 0010 Push immediate onto the stackload 0110 Load MEM[ACC1+immediate] onto the top of the stackstore 0101 Store the top value of the stack into MEM[ACC1+immediate]bgz 0100If (ACC1 ≥ 0) PC = PC+2+immediate, else PC=PC+2; ACC1=0If ACC1 is greater than zero, branch. Then always set ACC1 tozero. Note that PC increments by 2, since this instruction takes 2bytes.U-type instructionsbits 15-14 bits 13-4 bits 3-0dest code unused opcodeInstruction Opcode Dest code Mnemonic Actionneg. 100100 neg.acc1 ACC1 = (-1)*ACC111 neg.acc2 ACC2 = (-1)*ACC2add. 101100 add.acc1 ACC1 = ACC1+ACC211 add.acc2 ACC2 = ACC1+ACC2Page 11 of 12
Page 1 and 2: The University of Michigan - Depart
Page 3 and 4: 3. MIPS Assembly [10 pts]Given the
Page 5 and 6: 5. Memory Alignment [10 pts]You are
Page 7 and 8: 7. Symbol Table and Relocation Tabl
Page 9: foo()CallersavedCalleesavedbar()Cal

EECS 370 – Introduction to Computer Organization – Exam 1 w

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