Question Solved1 Answer 1 In August 2003, a car dealer is trying to determine how many 2004 models should be ordered. Each car costs the dealer $10,000. The demand for the dealer's 2004 models has the probability distribution shown in Table 4. Each car is sold for $15,000. If the demand for 2004 cars exceeds the number of cars ordered in August, the dealer must reorder at a cost of $12,000 per car. If the demand for 2004 cars falls short, the dealer may dispose of excess cars in an end- of-model-year sale for $9,000 per car. How many 2004 models should be ordered in August? TABLE4 No. of Cars Demanded Probability 20 25 30 35 40 .30 .15 .15 .20 .20

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Transcribed Image Text: 1 In August 2003, a car dealer is trying to determine how many 2004 models should be ordered. Each car costs the dealer $10,000. The demand for the dealer's 2004 models has the probability distribution shown in Table 4. Each car is sold for $15,000. If the demand for 2004 cars exceeds the number of cars ordered in August, the dealer must reorder at a cost of $12,000 per car. If the demand for 2004 cars falls short, the dealer may dispose of excess cars in an end- of-model-year sale for $9,000 per car. How many 2004 models should be ordered in August? TABLE4 No. of Cars Demanded Probability 20 25 30 35 40 .30 .15 .15 .20 .20
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Transcribed Image Text: 1 In August 2003, a car dealer is trying to determine how many 2004 models should be ordered. Each car costs the dealer $10,000. The demand for the dealer's 2004 models has the probability distribution shown in Table 4. Each car is sold for $15,000. If the demand for 2004 cars exceeds the number of cars ordered in August, the dealer must reorder at a cost of $12,000 per car. If the demand for 2004 cars falls short, the dealer may dispose of excess cars in an end- of-model-year sale for $9,000 per car. How many 2004 models should be ordered in August? TABLE4 No. of Cars Demanded Probability 20 25 30 35 40 .30 .15 .15 .20 .20
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Answer: Given Data Let , q = Number of 2004 model cars ordered in August d = Number of 2004 model cars demanded after August Now determine the smallest value of q for which&#160;E(q+1)-E(q) >= 0&#160; To calculate&#160;E(q+1)-E(q)&#160;, consider two possibilities: Case 1 : If&#160;d <= q&#160; then in this case , ordering q + 1&#160; units instead of q units which cause to be overstocked by one more unit . The probability that Case 1 will occur is simply&#160;P(D <= q)&#160;,&#160; where D is the random variable representing demand. Case 2 : If&#160;d >= q+1&#160;&#160;then in this case , ordering q + 1&#160; units instead of q units enables to be short one less unit . The probability that&#160; Case 2 will occur is&#160;P(D >= q+1)=1-P(D <= q) Now check conditions written above for the give problem. If&#160;d <= q&#160;, the costs shown in table given below are incurred t tt tttComputation of Total Cost if&#160;d <= q ttt&#160; tt tt ttt&#160; tttCost tt tt tttBuy q cars at $10000/car ttt10000q tt tt tttSell d cars at $15000/car ttt-15000d tt tt tttDispose of excess q-d cars at $9000/car ttt-9000(q-d) tt tt tttTotal cost ttt10000q-15000d-9000(q-d) tt t Hence total cost for case&#160;d <= q&#160;is , 10000 q-15000 d-9000(q-d)=10000 q-15000 d-9000 q+9000 d quad rarr(1)&#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160;=1000 q-6000 d Compare equation (1) with equation , c(d,q) =&#160;c_(0)q&#160;+ (terms not involving q) Hencew&#160;C_(0)&#160; is the per - unit cost of being overstocked . Thus get the value&#160;C_(0)&#160;= 1000 If&#160;d >= q+1&#160; , the costs shown in table given below are incurred t tt tttComputation of Total Cost if&#160;d >= q+1 ttt&#160; tt tt ttt&#160; tttCost tt tt t ... See the full answer