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Sergeu [11.5K]
3 years ago
5

A company producing custom-made teddy bears is considering several options for expanding their existing capacity. There are 3 po

ssibilities. The first is a low-end machine, which would take 10 minutes/bear for the (machined) manufacturing process. In addition, an average of 30 minutes of detail work would have to be done by hand (per bear). The second option is a high-end machine, which would take 8 minutes/bear, and reduce the amount of hand detail work to 25 minutes/bear. The final option is to subcontract out the bears. The subcontractor is willing to provide up to 400 bears per year for a flat fee of $2,000. Additional bears would cost $8 each. There is no difference in bear quality between the 3 options. It costs $10,000 to buy the low-end machine. Yearly maintenance is $1,000. The purchase price for the high-end machine is $15,000, while maintenance is $2,200. Management estimates the cost for running a machine at $6/hour. Labor costs are $15/hour. Assume the factory runs 350 days/year for 8 hours/day.a) If you expect a yearly demand of 12,000 bears, which option is the cheapest over a 3-year time horizon?b) If the service times on both machines are Exponentially distributed, and the job arrivals have a Poisson distribution with a rate as specified in part a), what is the expected time between the job's arrivals at the factory to the time it is complete and can leave? (Assume that there are more than enough workers to cover the hand detail workc) At what arrival rates (demand levels) would the different options make sense, given a 3-year time horizon?Please write down the answer step by step

Business
1 answer:
kupik [55]3 years ago
5 0

Answer:

Answer for the question:

A company producing custom-made teddy bears is considering several options for expanding their existing capacity. There are 3 possibilities. The first is a low-end machine, which would take 10 minutes/bear for the (machined) manufacturing process. In addition, an average of 30 minutes of detail work would have to be done by hand (per bear). The second option is a high-end machine, which would take 8 minutes/bear, and reduce the amount of hand detail work to 25 minutes/bear. The final option is to subcontract out the bears. The subcontractor is willing to provide up to 400 bears per year for a flat fee of $2,000. Additional bears would cost $8 each. There is no difference in bear quality between the 3 options. It costs $10,000 to buy the low-end machine. Yearly maintenance is $1,000. The purchase price for the high-end machine is $15,000, while maintenance is $2,200. Management estimates the cost for running a machine at $6/hour. Labor costs are $15/hour. Assume the factory runs 350 days/year for 8 hours/day.a) If you expect a yearly demand of 12,000 bears, which option is the cheapest over a 3-year time horizon?b) If the service times on both machines are Exponentially distributed, and the job arrivals have a Poisson distribution with a rate as specified in part a), what is the expected time between the job's arrivals at the factory to the time it is complete and can leave? (Assume that there are more than enough workers to cover the hand detail workc) At what arrival rates (demand levels) would the different options make sense, given a 3-year time horizon?Please write down the answer step by step

is given in the attachment.

Explanation:

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What is the expected value when a $1 lottery ticket is bought in which the purchaser wins exactly $10 million if the ticket cont
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We expect to lose $0.37 per lottery ticket

<u>Explanation:</u>

six winning numbers from = { 1, 2, 3, ....., 50}

So, the probability of winning:

P(win) = \frac{ no of favorable outcomes}{no of possible outcomes}

P(win) = \frac{1}{^5^0C_6} \\\\P (win) = \frac{6! X (50 - 6)!}{50!} \\\\P(win) = \frac{6! X 44!}{50!} \\\\P(win) = \frac{1}{15,890,700}

The probability of losing would be:

P(loss) = 1 - P(win)

P(loss) = 1 - \frac{1}{15,890,700} \\\\P(loss) = \frac{15,890,699}{15,890,700}

According to the question,

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The expected value is the sum of the product of each possibility x with its probability P(x):

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