question archive A manufacturing company regularly consumes a special type of glue purchased from a foreign supplier
A manufacturing company regularly consumes a special type of glue purchased from a foreign supplier
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A manufacturing company regularly consumes a special type of glue purchased from a foreign supplier. Because the supplier is foreign, the time gap between placing an order and receiving the shipment against that order is long and uncertain. This time gap is called “lead time.” From past experience, the materials manager notes that the company’s demand for glue during the uncertain lead time is normally distributed with a mean of 187.6 gallons and a standard deviation of 12.4 gallons. The company follows a policy of placing an order when the glue stock falls to a predetermined value called the “reorder point.” Note that if the reorder point is x gallons and the demand during lead time exceeds x gallons, the glue would go “stock-out” and the production process would have to stop. Stock-out conditions are therefore serious.
a. If the reorder point is kept at 187.6 gallons (equal to the mean demand during lead time) what is the probability that a stock-out condition would occur?
b. If the reorder point is kept at 200 gallons, what is the probability that a stockout condition would occur?
c. If the company wants to be 95% confident that the stock-out condition will not occur, what should be the reorder point? The reorder point minus the mean demand during lead time is known as the “safety stock.” What is the safety stock in this case?
d. If the company wants to be 99% confident that the stock-out condition will not occur, what should be the reorder point? What is the safety stock in this case?
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Answer:
a) calculate p (x>187.6)
Given We use standardized normal distributionz=σx−μμ=187.6σ=12.4x=187.6 the probability can be calculated asp(x>187.6)=p(σx−μ>12.4187.6−187.6)=p(z>0)=1−p(z≤0)=1−0.5=0.5Hence, the required the probabilityp(x>187.6)=0.5
b) calculate p (x> 200)
Given We use standardized normal distribution
z=σx−μμ=187.6
σ=12.4x=200the probability can be calculated asp(x>200)=p(σx−μ>12.4200−187.6)
=p(z>1)=1−p(z≤1)=1−0.8413=0.1587Hence, the required the probabilityp(x>200)=0.1587
c) calculate the 5th percentile, p (X ≥ x) = 0.95
given, μx=187.6σ=12.4P(x≥x1)=0.95 first,we computed zz=−1.64 second, we can calculate x using the formula
z=σx−μthird, x can be calculated as follows:
X1=Zσ+μfourth, putting the given values, we have
x1=−1.64∗12.4+187.6=167.264fifth,hence, the required x is167.3
d) calculate percentile, 1% p (X ≥ x) = 0.99
given, μx=187.6σ=12.4P(x≥x1)=0.99 first,we computed zz=−2.33second, we can calculate x using the formula z=σx−μthird, x can be calculated as follows: X1=Zσ+μfourth, putting the given values, we have
x1=−2.33∗12.4+187.6=158.708fifth,hence, the required x is158.7
