Modify the problem so that aggregation is performed on the integer values sent to the Aggregate$()$ function, and the float average of the values $($the aggregation$)$ broadcast back to the processes.

Use the attached C$*$ template. Pay close attention to the comments in the template.

The output should be approximately:

finished $3$ loops

SEQUENTIAL EXECUTION TIME: $4715$

PARALLEL EXECUTION TIME: $2663$

SPEEDUP: $1\mathrm{.}77$

NUMBER OF PROCESSORS USED: $12$

$-----$below is the template$--------$

$/*$ Assign$6*/$

$/*$ Compute, Aggregate, and Broadcast using streams $*/$

#define LOOPS $3$

#define PROCESSES $10$

typedef struct $\{$

int addr;

int valu;

float aggr;

$\}$ aggmsg$\_$t;

float verify$[$LOOPS$]=\{5\mathrm{.}5,15\mathrm{.}5,25\mathrm{.}5\}$;

$/*$

$*$ define a stream array of the correct type and size named 'aga'

$*/$

$/*$

$*$ Aggregate receives a process count parameter. Aggregate must initially block on a receive,

$*$ waiting for its input. It counts its input, and performs steps needed for aggregation.

$*$ When it has received all messages, it completes the aggregation and broadcasts the result.

$*$ It leaves itself in a state so that it can be reused, until receiving an endmarker message,

$*$ one with an addr field of $0.$

$*/$

void Aggregate$($int prct$)$

$\{$

$\}$

$/*$

$*$ process$()$ tests the Aggregation$()$ function.

$*/$

void process$($int i$,$ int j$)$

$\{$

aggmsg$\_$t pmsg;

pmsg$.$addr $=$ i;

pmsg$.$valu $=$ j $*$ PROCESSES $+$ i;

pmsg$.$aggr $=0\mathrm{.}0$;

send$($aga$[0],$ pmsg$)$;

recv$($aga$[$i$],$ pmsg$)$;

if $($pmsg$.$aggr $!=$ verify$[$j$])$

cout $<<$ "process $"<<$ myid $<<"$ aggr $"<<$ pmsg$.$aggr $<<"$ not equal to expect value $"<<$ verify$[$j$]<<$ endl;

$\}$

$/*$

$*$ initialization for Aggregate if needed

$*/$

void initAgg$()$

$\{$

return;

$\}$

int main$()$

$\{$

int i$,$ j;

aggmsg$\_$t emsg;

$/*$

$*$ initialization code

$*/$

for $($j $=0$; j $<$ LOOPS; $++$j$)$

$\{$

forall i $=1$ to PROCESSES do

process$($i$,$ j$)$;

$\}$

$/*$

$*$ termination code

$*/$

cout $<<$ "finished $"<<$ j $<<"$ loops" $<<$ endl;

return $0$;

$\}$