hmwk2sol - ISyE 3232 Stochastic Manufacturing and Service...

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Unformatted text preview: ISyE 3232 Stochastic Manufacturing and Service Systems Fall 2011 H. Ayhan Solutions to Homework 2 1. a) To calculate the pmf for min(D, 7), we should crete the following table, D 5 6 7 8 9 7 7 7 7 7 7 min(D, 7) 5 6 7 7 7 P(D = x) 1/10 2/10 2/5 2/10 1/10 From here, we can find the pmf, 1/10 if k = 5 2/10 if k = 6 P(min(D, 7) = k ) = 7/10 if k = 7 0 otherwise Hence, E [min(D, 7)] = 7 ￿ dP(min(D, 7) = d) = d=5 1 2 7 ∗5+ ∗6+ ∗ 7 = 6.6 10 10 10 b) By constructing the same type of table as in part (a.), we can construct the pmf of (7 − D)+ by 1/10 if k = 2 2/10 if k = 1 + P ((7 − D) = k ) = 7/10 if k = 0 0 otherwise Hence, E [(7 − D)+ ] = 2. 1 2 ∗2+ ∗ 1 = 0.4 10 10 a) To assist calculating the expected value, we can construct the following table. D d≤8 d>8 8 8 8 1 max(D, 8) 8 d With the help of the table the expected value can be calculated. ￿ 10 E [max(D, 8)] = max(s, 8) ds = ￿ 5 8 5 8 ds + 5 ￿ 10 8 = 8 ∗ P (D ≤ 8) + s ds 5 ￿ 10 ￿ 2 ￿￿10 3 s￿ ￿ =8∗ + 5 10 ￿8 24 36 = + 5 10 = 8.4 8 s ds 5 b) This value can be calculated using a method similar to Problem 1.b., but first we need to change the form using the following fact. min(f (x)) = − max(−f (x)) To see that this transformation works, try drawing f (x) = x2 + 2. Then we have the following. − (D − 8) = min(D − 8, 0) = − max(− (D − 8) , 0) = − max(8 − D, 0) = − (8 − D) + + The following table shows the values for (8 − D) , much like in part (a.). D d≤8 d>8 8−D 8−d≥0 8−d<0 Which allows the following calculation. ￿ ￿ − E (D − 8) = −E [(8 − D)+ ] ￿ 10 =− max(8 − s, 0) ds =− ￿ ￿ 5 8 max(8 − s, 0) ds − 5 8 8−s ds − 0 5 5 ￿ 2 ￿￿ 8 8∗3 s￿ ￿ =− + 5 10 ￿ =− 5 24 39 =− + 5 10 = −0.9 2 max(8 − D, 0) 8−d 0 0 0 0 ￿ 10 8 max(8 − s, 0) ds The zero follows from the table above. 3. p = 40, cv = 10, sv = 5. From the newsvendor class notes Profit(7) = (p − cv ) min{D, 7} − (cv − sv )(7 − D)+ Hence E [Profit(7)] = 30E [min{D, 7}] − 5E [(7 − D)+ ] = 196 using the values found in question 1. 4. Similar to the previous question, but first because demand ranges from 5 to 15, the expected values in Problem 2. must be recalculated. First observe that a continuous uniform distribution function from 5 to 15 has the following density function, 1 if 5 ≤ s ≤ 15 f (s) = 10 0 otherwise a) E [min(D, 7)] = 7 ∗ P (D > 7) + ￿ 7 5 s ds = 7 ∗ 7/10 + 10 + ￿ ￿￿ 7 s2 ￿ ￿ = 49 + 6 = 6.1 20 ￿5 10 5 b) From the definition, (7 − D) can be calculated as follows. ￿ 0 if D > 7 + (7 − D) = max{7 − D, 0} = 7 − D if D ≤ 7 From this definition it is possible to calculate the necessary expected value. ￿∞ + + E [(7 − D) ] = (7 − s) f (s) ds = ￿ 0 5 0 + (7 − s) f (s) ds + ￿ 15 ￿ 15 5 + (7 − s) f (s) ds + ￿ ∞ 15 + (7 − s) f (s) ds 1 s ds + 0 The zeros come from the p.d.f. 10 5 ￿7 ￿ 15 +1 +1 = (7 − s) ds + (7 − s) ds 10 10 5 7 ￿7 1 + = (7 − s) ds + 0 The zero comes from the definition of (7 − s) 10 5 ￿7 ￿7 7 s = ds − ds 10 10 5 5 ￿7 ￿7 7￿ 1 s2 ￿ = s￿ − ·￿ 10 ￿ 10 2 ￿ =0+ 5 (7 − s) + 5 21 24 = − 10 20 = 0.9 With these values the remaining calculations are the same as the previous question. E [Profit(7)] = 30E [min{D, 7}] − 5E [(7 − D)+ ] = 30 · 6.1 − 5 · 0.9 = 178.5 3 5. If D has an exponential distribution with mean 7 then the pdf of D is given by fD ( x) = 1 −1x e7. 7 As in the last 2 questions we need to find E [min{D, 7}] and E [(7 − D)+ ]. E [min(D, 7)] = 7 ∗ P (D ≥ 7) + + E [(7 − D) ] = ￿ 7 0 ￿ 7 0 s −s/7 e ds = 7e−1 + (7 − 14e−1 ) ≈ 4.43 7 7 − s −s/7 e ds = 7 − 7 ￿ 7 0 s −s/7 e ds = 7 − (7 − 7e−1 ) ≈ 2.575 7 E [Profit(7)] = 30E [min{D, 7}] − 5E [(7 − D)+ ] = 120.25. 4 ...
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This note was uploaded on 03/08/2012 for the course ISYE 3232 taught by Professor Billings during the Fall '07 term at Georgia Institute of Technology.

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