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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.
 Fall '07
 Billings

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