Respuesta :
Answer:
7.313%
Explanation:
In the problem above, the total cost for last year production is equivalent to the addition of labor cost, energy cost, capital cost, and material cost. Thus,
Total cost (last year) = (300*12) + (3000*0.6) + (9000*0.02) + (48*6) = 3600 + 1800 + 180 + 288 = 5868.
Last year productivity = unit produced/total cost = 1000/5868 = 0.17042
Total cost (this year) = (275*12) + (2850*0.60) + (10000*0.02) + (43*6) = 3300 + 1710 + 200 + 258 = 5468
This year productivity = 1000/5468 = 0.18288
%change in productivity = [(0.18288 - 0.17042)/0.17042]*100% = 7.313%
The percent change in productivity for one month last year versus one month this year on a multifactor basis with dollars as the common denominator is 7.313%
- The calculation is as follows;
Total cost (last year) is
= (300 × 12) + (3000 × 0.6) + (9000 × 0.02) + (48 × 6)
= 3600 + 1800 + 180 + 288
= 5868.
Now
Last year productivity = unit produced ÷ total cost
= 1000 ÷ 5868
= 0.17042
Now
Total cost (this year) = (275 ×12) + (2850 ×0.60) + (10000 × 0.02) + (43 × 6)
= 3300 + 1710 + 200 + 258
= 5468
Now
This year productivity = 1000 ÷ 5468
= 0.18288
Now
%change in productivity = [(0.18288 - 0.17042) ÷ 0.17042]
= 7.313%
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