Calculate cumulative performance
$begingroup$
Suppose a man produces $15$ nos. of item $A$ having cycle time of $8.99 sec$ in $6 min$.
Performance I:
Cycle Time in seconds $= 8.99$
Total Units Produced $= 15$
Total Runtime in seconds $= 6times60 = 360$
$$P1 = ((8.99times15) / 360)times100=37.48%$$
The same man on same machine produces $218$ nos. of item $B$ having cycle time of $12 sec$ in $78 min$.
Performance II:
Cycle Time in seconds $= 12$
Total Units Produced $= 218$
Total Runtime in seconds $= 78times60 = 4680$
$$P2 = ((12times218) / 4680)times100=55.89%$$
So how can get his overall performance of complete $6 + 78 = 84 min$?
algebra-precalculus
edited Jan 10 at 16:10
Shubham Johri
5,560818
asked Jan 10 at 15:51
Aniket BhansaliAniket Bhansali
1084
$endgroup$
add a comment |
$begingroup$
Suppose a man produces $15$ nos. of item $A$ having cycle time of $8.99 sec$ in $6 min$.
Performance I:
Cycle Time in seconds $= 8.99$
Total Units Produced $= 15$
Total Runtime in seconds $= 6times60 = 360$
$$P1 = ((8.99times15) / 360)times100=37.48%$$
The same man on same machine produces $218$ nos. of item $B$ having cycle time of $12 sec$ in $78 min$.
Performance II:
Cycle Time in seconds $= 12$
Total Units Produced $= 218$
Total Runtime in seconds $= 78times60 = 4680$
$$P2 = ((12times218) / 4680)times100=55.89%$$
So how can get his overall performance of complete $6 + 78 = 84 min$?
algebra-precalculus
edited Jan 10 at 16:10
Shubham Johri
5,560818
asked Jan 10 at 15:51
Aniket BhansaliAniket Bhansali
1084
$endgroup$
add a comment |
$begingroup$
Suppose a man produces $15$ nos. of item $A$ having cycle time of $8.99 sec$ in $6 min$.
Performance I:
Cycle Time in seconds $= 8.99$
Total Units Produced $= 15$
Total Runtime in seconds $= 6times60 = 360$
$$P1 = ((8.99times15) / 360)times100=37.48%$$
The same man on same machine produces $218$ nos. of item $B$ having cycle time of $12 sec$ in $78 min$.
Performance II:
Cycle Time in seconds $= 12$
Total Units Produced $= 218$
Total Runtime in seconds $= 78times60 = 4680$
$$P2 = ((12times218) / 4680)times100=55.89%$$
So how can get his overall performance of complete $6 + 78 = 84 min$?
algebra-precalculus
edited Jan 10 at 16:10
Shubham Johri
5,560818
asked Jan 10 at 15:51
Aniket BhansaliAniket Bhansali
1084
$endgroup$
Suppose a man produces $15$ nos. of item $A$ having cycle time of $8.99 sec$ in $6 min$.
Performance I:
Cycle Time in seconds $= 8.99$
Total Units Produced $= 15$
Total Runtime in seconds $= 6times60 = 360$
$$P1 = ((8.99times15) / 360)times100=37.48%$$
The same man on same machine produces $218$ nos. of item $B$ having cycle time of $12 sec$ in $78 min$.
Performance II:
Cycle Time in seconds $= 12$
Total Units Produced $= 218$
Total Runtime in seconds $= 78times60 = 4680$
$$P2 = ((12times218) / 4680)times100=55.89%$$
So how can get his overall performance of complete $6 + 78 = 84 min$?
algebra-precalculus
algebra-precalculus
edited Jan 10 at 16:10
Shubham Johri
5,560818
asked Jan 10 at 15:51
Aniket BhansaliAniket Bhansali
1084
edited Jan 10 at 16:10
Shubham Johri
5,560818
asked Jan 10 at 15:51
Aniket BhansaliAniket Bhansali
1084
edited Jan 10 at 16:10
Shubham Johri
5,560818
edited Jan 10 at 16:10
Shubham Johri
5,560818
edited Jan 10 at 16:10
Shubham Johri
5,560818
5,560818
asked Jan 10 at 15:51
Aniket BhansaliAniket Bhansali
1084
asked Jan 10 at 15:51
Aniket BhansaliAniket Bhansali
1084
asked Jan 10 at 15:51
Aniket BhansaliAniket Bhansali
1084
1084
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Just add up the total working time and divide by the total time
$$frac {8.99cdot 15+12cdot 218}{60cdot 84}approx 54.58%$$
This is close to the second phase because the second phase is most of the total time.
answered Jan 10 at 16:01
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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$begingroup$
Just add up the total working time and divide by the total time
$$frac {8.99cdot 15+12cdot 218}{60cdot 84}approx 54.58%$$
This is close to the second phase because the second phase is most of the total time.
answered Jan 10 at 16:01
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Just add up the total working time and divide by the total time
$$frac {8.99cdot 15+12cdot 218}{60cdot 84}approx 54.58%$$
This is close to the second phase because the second phase is most of the total time.
answered Jan 10 at 16:01
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Just add up the total working time and divide by the total time
$$frac {8.99cdot 15+12cdot 218}{60cdot 84}approx 54.58%$$
This is close to the second phase because the second phase is most of the total time.
answered Jan 10 at 16:01
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Just add up the total working time and divide by the total time
$$frac {8.99cdot 15+12cdot 218}{60cdot 84}approx 54.58%$$
This is close to the second phase because the second phase is most of the total time.
answered Jan 10 at 16:01
Ross MillikanRoss Millikan
301k24200375
answered Jan 10 at 16:01
Ross MillikanRoss Millikan
301k24200375
answered Jan 10 at 16:01
Ross MillikanRoss Millikan
301k24200375
answered Jan 10 at 16:01
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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