Computer Engineering - Computer Engineering - Exam, Exams of Computer Science

Main points of this exam paper are: Computer Engineering, Memory Array, Gigabyte Memory, Gigabyte Memory System, Decoder Required, Chips Required, Bit Words, Memory Decoder, Byte Addressing, Big-Endian

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ECE 2030 Computer Engineering Spring 1999
4 problems, 6 pages Exam Three 28 May 1999
1
Name (print) _____________________________________________________
1234 total
30 10 30 30 100
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4 problems, 6 pages Exam Three 28 May 1999

Name (print) _____________________________________________________

1 2 3 4 total

4 problems, 6 pages Exam Three 28 May 1999

Problem 1 (3 parts, 30 points) The Memory System Problem

Imagine a one gigabyte memory array organized as 256 million address of 32 bit words. The

following three parts consider how the memory system could be built using three different DRAM

memory chips.

Part 1.A (10 points) Suppose the gigabyte memory system is built using today’s 256 megabit

DRAM organized as 32 million addresses of eight bit words.

number of chips needed in 32M word bank

number of banks for 256 million addresses

memory decoder required ( n to m )

number of DRAM chips required

Part 1.B (10 points) Suppose the gigabyte memory system is built using the original IBM PC’s 16

killobit DRAM organized as 16 thousand addresses of one bit words.

number of chips needed in 16K word bank

number of banks for 256 million addresses

memory decoder required ( n to m )

number of DRAM chips required

Part 1.C (10 points) Suppose the gigabyte memory system is built using tomorrow’s one gigabit

DRAM organized as 64 million addresses of sixteen bit words.

number of chips needed in 64M word bank

number of banks for 256 million addresses

memory decoder required ( n to m )

number of DRAM chips required

Problem 2 (2 parts, 10 points) Short Answer

Briefly explain what “byte addressing” means.

Explain the difference between big-endian and little-endian.

MIPS Assembly Language

move from Hi mfhi $1 $1 = Hi

  • 4 problems, 6 pages Exam Three 28 May
    • subtract sub $1,$2,$3 $1 = $2 - $ add instruction add $1,$2,$3 example $1 = $2 + $3 meaning
    • add immediate addi $1,$2,100 $1 = $2 +
    • 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 +
    • multiply mult $2,$3 hi, lo = $2 * $ move from coproc. register mfc0 $1,$epc $1 = $epc
    • multiply unsigned multu $2,$3 hi, lo = $2 * $
    • divide div $2,$3 lo = $2 / $3; hi $2 mod $
    • divide unsigned divu $2,$3 lo = $2 / $3; hi $2 mod $
    • and and $1,$2,$3 $1 = $2 & $ move from Lo mflo $1 $1 = Lo
    • or or $1,$2,$3 $1 = $2 | $
    • and immediate andi $1,$2,100 $1 = $2 &
    • or immediate ori $1,$2,100 $1 = $2 |
    • shift left logical sll $1,$2,5 $1 = $2 <<
    • shift right logical srl $1,$2,5 $1 = $2 >>
    • store word sw $1,100($2) memory [$2 + 100] = $ load word lw $1,100($2) $1 = memory [$2 + 100]
    • load upper immediate lui $1,100 $1 = 100 x
    • branch if equal beq $1,$2,100 if ($1 = $2), PC = PC + 4 +
    • branch if not equal bne $1,$2,100 if ($1 ≠ $2), PC = PC + 4 +
    • set if less than slt $1, $2, $3 if ($2 < $3), $1 = 1 else $1 =
    • set if less than immediate slti $1, $2, 100 if ($2 < 100), $1 = 1 else $1 =
    • set if less than unsigned sltu $1, $2, $3 if ($2 < $3), $1 = 1 else $1 =
    • set if less than imm. uns. sltiu $1, $2, 100 if ($2 < 100), $1 = 1 else $1 =
    • jump j 10000 PC =
    • jump register jr $31 PC = $
    • jump and link jal 10000 $31 = PC + 4; PC =

4 problems, 6 pages Exam Three 28 May 1999

Problem 4 (1 part, 30 points) The Microcode Problem

Using the datapath on the next page, write a microcode fragment that computes the equation ( 100 )= ( 5 R 5 − 9 R 6 ) 64. The result is

stored at memory location 0x100. Express all values in hexadecimal notation. Use ‘X’ when a value is don’t cared. For maximum

credit, complete the description field. Use only R5, R6, and R1. R5 and R6 can be modified after their initial value has been used.

cycle X Y Z rwe im en im va au en -a/s lu en lf su en st ld en st en r/-w msel description 1 2 3 4 5 6 7 8 9