Question

Topics 1. MIPS instruction set architecture (ISA). 2. Performance. 3. MIPS datapath and control. Exercise 1 Consider the memory and register contents shown below. Registers Ox0100 FFF8 13 ($t 5) 14 ($t6) 0x0100 FFFC 0x0101 0000 Memory 0x0000 0000 0x0001 1100 0x0A00 со00 0x1234 4321 OxBAOO OOBB 15 OXAAAA 0000 0x1111 1010 0x7FFF FFFD 0x0100 FFFO 0x0101 0008 (St7) Ox0101 0004 16 ($80) 0x0101 0008 17 ($sl) Show what changes and give the new values in hexadecimal after the following instructions are executed. Consider the execution of each instruction independent of the others. (a) lw $t7, 24($s0) (b) sw $t5, -16 ($51) (c) sll $t7, $t7, 31 (d) add $t6, $t5, $s1

Answer 1

Dlw $t7,241$so) 24 in hexadecimal = Ox18 $17 - Mem Coxo 100FFF0 +0x18] - Mem Coxo101 0008) value at register $t7 win charged to OXBA 0000BB toser Backspa -b) sw $t5, -16 ($51) Mem COX1010008 -OX10 )= Mem Cox0100FFF8] = $15 Value at memory address Ox0100FFF8 is Ox AAA AOOO Ec) s11 $17, $47,31 value at Register $t7 = 0x7FFFFFFD $t7 = (0X7 FFFFFFD) <<31 - 0111 1111 1111 1122 1111 1111 1111 1101) <<31 $7 = 1000 0000 0000 0000 оооооооо 0000 0000 $t7 OX 8000000 (d) add $to $65 $1 $16= $15 + $51 $t 6 = OxAAA A0000 + OX01010008 = OXABABOOOS Value at register $+6 = OX ABA BOODS (2) execution time = (total ins buchon * CPI) (clock sate execution time = (2500555 + 1) (2:57 10^4) = 1.000222 mset

(3) Instruction per second for Computer A=(clock rate) / CPI = (3.5*100g) 11.2 = 2.9167 & long Instouchons per second for computer B = c clock soate ) ICPI = (3.0*10ng) (1 - 3* lola Computer B is faster than computer A. Speedup (3*109)1(2.9167 * 1019) = 1-02857 times faster F11