Perform operations in excel .... (and shows the formulas) Company Z has the following historical information about sales for one of its hydraulic pump models for the last 4 years: (PICTURE OF THE...


Perform operations in excel ....  (and shows the formulas)



Company Z has the following historical information about sales for one of its hydraulic pump models for the last 4 years:



(PICTURE OF THE PERIOD AND YEARS)




On the other hand, the manufacturing process used by Company Z is sequential* and has the following characteristics:



(THE OTHER IMAGE)


* Sequential; in the sense that the output of Mi is the input of Mi+1.
** In any order; in the sense that they can be interchanged, e.g. MA -> MB or MB -> MA.




The company's operating costs are as follows:



Cost per final unit produced "defect free": $100



Cost per final unit produced "with defects": $150



Cost per excess inventory (units "free of defects"):$50



Cost per inventory shortage ("defect free" units): $200



Revenue per final "defect free" unit produced and sold: $250




We have the following operating conditions:




- Any sequence of 3 machines can be used.
- Only one machine can be used once, within the same sequence of machines.
- You can work UP TO 720 hours per month.
- A "level production" type production strategy is followed.


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Establish the 1st and 2nd best machine sequence that allows company Z to obtain higher profits (profit = total revenue - total costs) for a hypothetical 5th year.





In both cases PROVE that both sequences meet the requirements of being the 1st and 2nd best sequences.


Machines THAT CAN be used in any order**.<br>Production rate per machine un/hr<br>Machine efficiency %<br>Machine reliability%<br>M1<br>M2<br>M3<br>М4<br>4.75 4.25 6.50<br>1.00 0.85<br>1.00 0.80<br>7.00<br>0.90<br>0.95<br>0.70<br>0.95<br>

Extracted text: Machines THAT CAN be used in any order**. Production rate per machine un/hr Machine efficiency % Machine reliability% M1 M2 M3 М4 4.75 4.25 6.50 1.00 0.85 1.00 0.80 7.00 0.90 0.95 0.70 0.95
period<br>year<br>1<br>2<br>3<br>4<br>1<br>1800<br>1950<br>2050<br>1900<br>

Extracted text: period year 1 2 3 4 1 1800 1950 2050 1900
Jun 04, 2022
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