Next count the items contributing to the FP:$-$

A$=$ the number of external inputs $=2($temperature signal, volume signal$)$

B$=$ the number of external outputs $=4($alarm report, status of the system,

control signal, readings of the sensor$)$

C$=$ number of inquiries $=0$

D$=$ number of external files $=1($archive file$)$

E$=$ number of internal files $=1($alarm control file$)$

With average complexity for each item:$-$

UFC$=4$A$+5$B$+4$C$+10$D$+7$E$=8+20+0+10+7=45$

Now treat functions as simple:$-$

UFC$=3$A$+4$B$+3$C$+7$D$+5$E$=6+16+0+7+5=34$

Finally treat functions as complex:$-$

UFC$=6$A$+7$B$+6$C$+15$D$+10$E$=12+28+0+15+10=65$Now if we consider Adjusted FP Count,

o find the range of possible values for

o Average case, simple case and complex case.

$$ If$,$ on average, an FP takes $2$ person$-$days of effort to

implement,

o then find total effort in

o Average case, simple case and complex case