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Answer:
For plastic
Year 1 97,000 units per year and 2 operators
Year 2 115,000 units per year and 3 operators
Year 3 136000 units per year and 3 operators
Year 4 141000 units per year and 3 operators
For bronze
Year 1 21000 units per year, 2 machines and 4 operators in total
Year 2 24000 units per year, 2 machines and 4 operators in total
Year 3 29000 units per year, 3 machines, 5 operators in total
Year 4 34000 units per year, 3 machines, 6 operators in total
Explanation:
for plastic,
since there is only one machine that is operated by 4 operators,
Units per operator= Machine speed/number of operator
= 200,000/4
= 50,000 units per operator
so for year 1, for 97,000 units 2 operators will be enough and for year 2, year 3 and year 4, 3 operators will be enough
for Bronze
since there are 3 machines and total number of 6 operators,
Units per operator= sum of speed of all machines/total number of operator
= 36000/6
= 6,000 units per operator
so for year 1 for 21,000 units per year, 4 operators will be required (6000×4) with 2 machines
for year 2 for 24000 units per year, 4 operators will be requried with 2 machines
for year 3 for 29,000 units per year, 3 machines will be required with 5 operators to meet the demand
for year 4 for 34,000 units, 3 machines will be requried with 6 operators
For plastic are:
- According to the first Year: 97,000 units per year and 2 operators
- According to the second Year: 115,000 units per year and 3 operators
- According to the third Year: 136000 units per year and 3 operators
- According to the fourth Year: 141000 units per year and 3 operators
For bronze are
- Year first 21000 units per year, 2 machines and 4 operators in total
- Year second 24000 units per year, 2 machines and 4 operators in total
- In a third Year 29000 units per year, 3 machines, 5 operators in total
- Year fourth 34000 units per year, 3 machines, 6 operators in total
About for plastic, is more detail below:
Thus, there is only 1 machine that is operated by 4 operators,
Units per operator= Machine speed/number of operator
= 200,000/4
= 50,000 units per operator
- Then As for year 1, for 97,000 units 2 operators will be enough and for year 2, year 3, and year 4, 3 Its operators will be also enough
About for Bronze more detail
when there are 3 machines and a total number of 6 operators,
- As per a Unit per operator= sum of the speed of all machines/total number of operator
= 36000/6
= 6,000 units per operator
- Then for a year 1 for 21,000 units per year, 4 operators will be required (6000×4) That with the 2 machines
- Also, for year 2 for 24000 units per year, 4 operators will be required with the 2 machines
- Thus, for year 3 for 29,000 units per year, 3 machines will be required with 5 and its operators to meet the demand.
- Although for year 4 for 34,000 units, 3 machines will be required with the 6 operators.
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