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artcher [175]
2 years ago
13

American Vending (AV) supplies vended food to a large university. Because students often kick the machines out of anger and frus

tration, management has a constant repair problem. The machines break down on an average of three per hour, and the breakdowns are distributed in a Poisson manner. Downtime costs the company $25 per hour per machine, and each maintenance worker gets $16 per hour. One worker can service machines at an average rate of five per hour distributed exponentially. What is the optimal maintenance crew size for servicing the machines
Mathematics
1 answer:
aev [14]2 years ago
3 0

Answer:

case 2 with two workers is the optimal decision.

Step-by-step explanation:

Case 1—One worker:A= 3/hour Poisson, ¡x =5/hour exponential The average number of machines in the system isL = - 3. = 4 = lJr machines' ix-A 5 - 3 2 2Downtime cost is $25 X 1.5 = $37.50 per hour; repair cost is $4.00 per hour; and total cost per hour for 1worker is $37.50 + $4.00

= $41.50.Downtime (1.5 X $25) = $37.50 Labor (1 worker X $4) = 4.00

$41.50

Case 2—Two workers: K = 3, pl= 7L= r= = 0.75 machine1 p. -A 7 - 3Downtime (0.75 X $25) = S J 8.75Labor (2 workers X S4.00) = 8.00S26.75Case III—Three workers:A= 3, p= 8L= ——r = 5- ^= § = 0.60 machinepi -A 8 - 3 5Downtime (0.60 X $25) = $15.00 Labor (3 workers X $4) = 12.00 $27.00

Comparing the costs for one, two, three workers, we see that case 2 with two workers is the optimal decision.

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3 years ago
Typing errors in a text are either nonword errors (as when "the" is typed as "teh") or word errors that result in a real but inc
nordsb [41]

Answer:

a) X is binomial with n = 10 and p = 0.3

Y is binomial with n = 10 and p = 0.7

b) The mean number of errors caught is 7.

The mean number of errors missed is 3.

c) The standard deviation of the number of errors caught is 1.4491.

The standard deviation of the number of errors missed is 1.4491.

Step-by-step explanation:

For each typing error, there are only two possible outcomes. Either it is caught, or it is not. The probability of a typing error being caught is independent of other errors. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

10 word errors.

This means that n = 10

(a) If X is the number of word errors missed, what is the distribution of X ?

Human proofreaders catch 70 % of word errors. This means that they miss 30% of errors.

So for X, p = 0.3.

The answer is:

X is binomial with n = 10 and p = 0.3.

If Y is the number of word errors caught, what is the distribution of Y ?

Human proofreaders catch 70 % of word errors.

So for Y, p = 0.7.

The answer is:

Y is binomial with n = 10 and p = 0.7

(b) What is the mean number of errors caught?

E(Y) = np = 10*0.7 = 7

The mean number of errors caught is 7.

What is the mean number of errors missed?

E(X) = np = 10*0.3 = 3

The mean number of errors missed is 3.

(c) What is the standard deviation of the number of errors caught?

\sqrt{V(Y)} = \sqrt{np(1-p)} = \sqrt{10*0.7*0.3} = 1.4491

The standard deviation of the number of errors caught is 1.4491.

What is the standard deviation of the number of errors missed?

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.3*0.7} = 1.4491

The standard deviation of the number of errors missed is 1.4491.

6 0
2 years ago
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