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Main points of this past exam are: Listed Transformation, Datapath Elements, Shift Operation, Output Values, Amount Required, Listed Transformation, Hexadecimal, Shift Amount, Decimal Value, Logical Operation
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4 problems, 7 pages Exam Three 1 December 201 0 Instructions: This is a closed book, closed note exam. Calculators are not permitted. If you have a question, raise your hand and I will come to you. Please work the exam in pencil and do not separate the pages of the exam. For maximum credit, show your work. Good Luck! Your Name ( please print ) ________________________________________________ 1 2 3 4 total 18 32 25 25 100
4 problems, 7 pages Exam Three 1 December 201 0 Problem 1 ( 2 parts, 18 points) Datapath Elements Part A (9 points) Consider the following input and output values for a shift operation. Determine the shift type and amount required to achieve the listed transformation. I/Os are in hexadecimal. Input Value Output Value Shift Type Shift Amount (signed decimal value) 87654321 FFF 87654321 00000008 87654321 43218765 Part B (9 points) Consider the following input and output values for a logical operation. Determine the logical function and function code (in hexadecimal) required for the operation. X Input Y Input Output Logical Function Function Code 87654321 00FF00FF 00650021 87654321 00FF00FF 8 79A43DE 87654321 00FF00FF 789ABCDE Problem 2 (3 parts, 3 2 points) Memory Systems Part A (1 2 points) Consider a DRAM chip organized as 64 million addresses of 16 bit words. Assume both the DRAM cell and the DRAM chip is square. The column number and offset concatenate to form the memory address. Using the organization approach discussed in class, answer the following questions about the chip. Express all answers in decimal. number of columns column decoder required ( n to m ) type of mux required ( n to m ) number of muxes required number of address lines in column number number of address lines in column offset Part B (10 points) Consider a 4 Gbyte memory system with 512 million addresses of 8 byte words using DRAM chips organized as 64 million addresses by 16 bit words. word address lines for memory system chips needed in one bank banks for memory system memory decoder required ( n to m ) DRAM chips required
4 problems, 7 pages Exam Three 1 December 201 0 Problem 3 ( 5 parts, 25 points) Microcode Using the supplied datapath, write microcode fragments to accomplish the following procedures. Express all values in hexadecimal notation. Use ‘X’ when a value is don’t cared. Smile when you're happy. For maximum credit, complete the description field. Part A ( 6 points) $6 ← mem[0x 3000 ]. Use only register 6. # X Y Z rwe im en im va au en^ -a /s en^ lu lf su en st ld en en^ st^ r/ -w msel description 1 2 Part B ( 5 points) Mask all but the eight least significant bits of $8 Use only register 8. # X Y Z rwe im en im va au en^ -a /s^ lu en lf su en st (^) enld en^ st -w^ r/ msel description 1 Part C ( 4 points) Subtract 8 from register $3. Use only registers 3. # X Y Z rwe im en im va au en^ -a /s en^ lu lf su en st (^) enld en^ st -w^ r/ msel description 1 Part D ( 6 points) mem[$4] ← 0x1957. Use only registers 4 and 5. # X Y Z rwe im en im va au en^ -a /s en^ lu lf su en st (^) enld en^ st -w^ r/ msel description 1 2 Part E ( 4 points) $7 ← 32 * $7. Use only register 7. # X Y Z rwe im en im va au en^ -a /s en^ lu lf su en st (^) enld en^ st -w^ r/ msel description 1
4 problems, 7 pages Exam Three 1 December 201 0 Problem 4 (1 part, 25 points) Assembly Programming Part A (25 points) Complete this subroutine that averages an array of integers in memory. Assume $4 contains the number of integers and $5 contains the starting address of the array. Return the average in $7. Use the $8 for the input value, $7 for the sum, and $6 for the end test. label instruction comment **AvgN: # offset = 4 * # elems
jr $31 # return to caller**
4 problems, 7 pages Exam Three 1 December 201 0 MIPS Instruction Set instruction example meaning arithmetic add add $1,$2,$3 $1 = $2 + $ subtract sub $1,$2,$3 $1 = $2 - $ add immediate addi $1,$2,100 $1 = $2 + 100 add unsigned addu $1,$2,$3 $1 = $2 + $ subtract unsigned subu $1,$2,$3 $1 = $2 - $ add immediate unsigned addiu $1,$2,100 $1 = $2 + 100 set if less than slt $1, $2, $3 if ($2 < $3), $1 = 1 else $1 = 0 set if less than immediate slti $1, $2, 100 if ($2 < 100), $1 = 1 else $1 = 0 set if less than unsigned sltu $1, $2, $3 if ($2 < $3), $1 = 1 else $1 = 0 set if < immediate unsigned sltui $1, $2, 100 if ($2 < 100), $1 = 1 else $1 = 0 multiply mult $2,$3 Hi, Lo = $2 * $3, 64-bit signed product multiply unsigned multu $2,$3 Hi, Lo = $2 * $3, 64-bit unsigned product divide div $2,$3 Lo = $2 / $3, Hi = $2 mod $ divide unsigned divu $2,$3 Lo = $2 / $3, Hi = $2 mod $3, unsigned transfer move from Hi mfhi $1 $1 = Hi move from Lo mflo $1 $1 = Lo load upper immediate lui $1,100 (^) $1 = 100 x 216 logic and and $1,$2,$3 $1 = $2 & $ or or $1,$2,$3 $1 = $2 | $ and immediate andi $1,$2,100 $1 = $2 & 100 or immediate ori $1,$2,100 $1 = $2 | 100 nor nor $1,$2,$3 $1 = not($2 | $3) xor xor $1, $2, $3 (^) $1 = $2 ⊕ $ xor immediate xori $1, $2, 255 (^) $1 = $2 ⊕ 255 shift shift left logical sll $1,$2,5 $1 = $2 << 5 (logical) shift left logical variable sllv $1,$2,$3 $1 = $2 << $3 (logical), variable shift amt shift right logical srl $1,$2,5 $1 = $2 >> 5 (logical) shift right logical variable srlv $1,$2,$3 $1 = $2 >> $3 (logical), variable shift amt shift right arithmetic sra $1,$2,5 $1 = $2 >> 5 (arithmetic) shift right arithmetic variable srav $1,$2,$3 $1 = $2 >> $3 (arithmetic), variable shift amt memory load word lw $1, 1000($2) $1 = memory [$2+1000] store word sw $1, 1000($2) memory [$2+1000] = $ load byte lb $1, 1002($2) $1 = memory[$2+1002] in least sig. byte load byte unsigned lbu $1, 1002($2) $1 = memory[$2+1002] in least sig. byte store byte sb $1, 1002($2) memory[$2+1002] = $1 (byte modified only) branch branch if equal beq $1,$2,100 if ($1 = $2), PC = PC + 4 + (1004) branch if not equal bne $1,$2,100 (^) if ($1 ≠ $2), PC = PC + 4 + (1004) jump jump j 10000 PC = 10000* jump register jr $31 PC = $ jump and link jal 10000 $31 = PC + 4; PC = 10000*
