Number three please I got 12337sec for a B 880012337 C 08890

Number three please
I got .12337sec for a
B. 8800/..12337
C .088902 sec
D 98454.39
Review view Tell me what you want to do need to edit, it\'s safer to stay in Protected View. Enable Editing Question 3 (20pts: Assume that there are 5 hops between two devices, A and B. The processing delay of interconnecting devices is 5usec. All links have a speed of 1Mbits/second. The speed of light in the link is 3x108 m/sec, and the length of each link is 1 km. With packet switched networking, the packet size is 1050Bytes, which includes 50 Bytes header. There are 10 packets that need to be transmitted from A to B. The queuing delay at each interconnecting devices is 5 msec a. What is the total delay for A to send the 10 packets to B using packet switching? (5pts) b. What is the throughput of packet switching? hat is the efficiency? (5pts c. Assume same amount of data needs to be transmitted over a circuit switching network. Let the total time needed for both setting up and terminating the connection be 7ousec. What is the total delay in this case? (5pts) d. What is the throughput of a circuit switching network? What is the efficiency? Compare the results to (b). (5pts) Question 4: UNESCI 1070 students only):

Solution

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

.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 ordinal 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) attend loop3
sw $t4, 0($a2) # c[i][j] = $t4
#----------TEST-------------
li $v0, 1
lw $a0, ($a2)
syscall
li $v0, 4
la $a0, new_row
syscall
#----------TEST-------------

addiu $s1, $s1, one # $j = j + one
addi $a2, $a2, 4
bne $s1, $t1, loop2 # if (j != 4) attend loop2

addiu $s0, $s0, one # $i = i + one
bne $s0, $t1, loop1 # if (i != 32) attend L1

Exit:
li $v0, 10 #exits
syscall

.data
array_A: .word 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
array_B: .word 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
array_C: .word 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
output_row_string_C: .asciiz \"Matrix C Output Row \"
colon_string: .asciiz \":
space_string: .asciiz \" \"
new_row: .asciiz \"\ \"
char_space: .space 2

Number three please I got .12337sec for a B. 8800/..12337 C .088902 sec D 98454.39 Review view Tell me what you want to do need to edit, it\'s safer to stay in
Number three please I got .12337sec for a B. 8800/..12337 C .088902 sec D 98454.39 Review view Tell me what you want to do need to edit, it\'s safer to stay in

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