Answer to Question #237050 in Operations Research for opr

Question #237050

A company is involved in the production of two items (X and Y). The resources need to produce X and Y are twofold, namely machine time for automatic processing and craftsman time for hand finishing. The table below gives the number of minutes required for each item:

Machine time Craftsman time Item

X 13 20

Y 19 29

The company has 40 hours of machine time available in the next working week but only 35 hours of craftsman time. Machine time is costed at Β£10 per hour worked and craftsman time is costed at Β£2 per hour worked. Both machine and craftsman idle times incur no costs. The revenue received for each item produced (all production is sold) is Β£20 for X and Β£30 for Y. The company has a specific contract to produce 10 items of X per week for a particular customer. Formulate the problem of deciding how much to produce per week as a linear program. Solve this linear program graphically.


1
Expert's answer
2021-09-22T00:09:30-0400



Let π‘₯ be the number of items of 𝑋, 𝑦 be the number of items of π‘Œ. Then the LP is maximise

20π‘₯ + 30𝑦 βˆ’ 10(π‘šπ‘Žπ‘β„Žπ‘–π‘›π‘’ π‘‘π‘–π‘šπ‘’ π‘€π‘œπ‘Ÿπ‘˜π‘’π‘‘) βˆ’ 2(π‘π‘Ÿπ‘Žπ‘“π‘‘π‘ π‘šπ‘Žπ‘› π‘‘π‘–π‘šπ‘’ π‘€π‘œπ‘Ÿπ‘˜π‘’π‘‘)

subject to:

13π‘₯ + 19𝑦≀ 40(60) π‘šπ‘Žπ‘β„Žπ‘–π‘›π‘’ π‘‘π‘–π‘šπ‘’

20π‘₯ + 29𝑦≀ 35(60) π‘π‘Ÿπ‘Žπ‘“π‘‘π‘ π‘šπ‘Žπ‘› π‘‘π‘–π‘šπ‘’

π‘₯ β‰₯ 10 π‘π‘œπ‘›π‘‘π‘Ÿπ‘Žπ‘π‘‘

π‘₯,𝑦β‰₯ 0

so that the objective function becomes maximise

20π‘₯ + 30𝑦 βˆ’10(13π‘₯ + 19𝑦)

60 βˆ’2(20π‘₯ + 29𝑦)

60

i.e. maximise

17.1667π‘₯ + 25.8667𝑦

subject to:

13π‘₯ + 19𝑦≀ 2400

20π‘₯ + 29𝑦≀ 2100

π‘₯ β‰₯ 10

π‘₯,𝑦β‰₯ 0

It is plain from the diagram below that the maximum occurs at the intersection of π‘₯ =10 and

20π‘₯ + 29𝑦≀ 2100.

Solving simultaneously, rather than by reading values off the graph, we have that π‘₯ =10 and 𝑦=65.52

with the value of the objective function being Β£1866.5.





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