A certain program consists of a sequential part (Part1) that takes 10 millisecon

Question

A certain program consists of a sequential part (Part1) that takes 10 milliseconds to scan a file directory to locate 16 files each or which requires the same processing. Part1 then calls a function (Part2), that takes 48 milliseconds per file to process. All files require the same 48 ms to process. The program is run on two different systems, the first system has a single processor and the second system has a total of 32 cores. When the program is run on the 32-core system the program spawns multiple processes each of which executes Part2 to process one file in parallel with the other processes. If the total time required to spawn all of the required processes is 4 milliseconds, what speedup would be provided by running the program on the 32-core system compared to running it on the single processor system?  

Explanation / Answer

void metric linear unit (int c[][], int a[][], int b[][])

Here's my unit code:

.text
main:
j mm

mm:
la $a3, array_A # base address for array_A loaded into $a3
la $a1, array_B # base address for array_B loaded into $a1
la $a2, array_C # base address for array_C loaded into $a2

li $t1, four # $t1 = four (row-size and loop end)
li $s0, zero # i = 0; initialize first for loop
loop1:
li $s1, zero # j = 0; restart second for loop
loop2:
li $s2, zero # k = 0; restart third for loop
sll $t2, $s0, two # $t2 = i * four (size of row of c)
addu $t2, $t2, $s1 # $t2 = i * size(row) + j
sll $t2, $t2, two # $t2 = computer memory unit offset of [i][j]
addu $t2, $a2, $t2 # $t2 = computer memory unit offset of [i][j]
lw $t4, 0($t2) # $t4 = two bytes of c[i][j]
loop3:
sll $t0, $s2, two # $t0 = k * four (size of row of b)
addu $t0, $t0, $s1 # $t0 = k * size(row) + j
sll $t0, $t0, two # $t0 = computer memory unit offset off [k][j]
addu $t0, $a1, $t0 # $t0 = computer memory unit address of b[k][j]
lw $t5, 0($t0) # $t5 = two bytes of b[k][j]
sll $t0, $s0, two # $t0 = i * four (size of row of a)
addu $t0, $t0, $s2 # $t0 = i * size(row) + k
sll $t0, $t0, two # $t0 = computer memory unit offset of [i][k]
addu $t0, $a3, $t0 # $t0 = computer memory unit address of a[i][k]
lw $t6, 0($t0) # $t6 = two bytes of a[i][k]
mul $t5, $t6, $t5 # $t5 = a[i][k] * b[k][j]
add $t4, $t4, $t5 # $t4 = c[i][j] + a[i][k] * b[k][j]
addiu $s2, $s2, one # $k = k + one
bne $s2, $t1, loop3 #if (k != 4) head to loop3
sw $t4, 0($a2) # c[i][j] = $t4
#----------TEST-------------
li $v0, 1

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