Unformatted text preview: ISyE 3232 Stochastic Manufacturing and Service Systems Fall 2011 H. Ayhan
Solutions to Homework 7
1. You must be careful in this problem because λ is not the arrival rate in this problem. Because
λ is being used in the service rate distribution, I will use η (read as “eta”) to denote the arrival
rate. Because the arrival rate is given in terms of hours, I will ﬁrst convert it into minutes to
match the service distribution’s units. (You can also convert the service rate information into
hours; both methods will work, so long as the units agree.)
30 customers
1 hour
1 customer
=
η=
1 hour
60 minutes
2 minutes
Because the arrival distribution is exponential, the squared coeﬃcient of variation is 1, so
c2 = 1, which is always true for the exponential distribution. Now that the arrival information
A
is calculated, I need to determine the mean and variance for the arrival distribution (which if
you look carefully should see is not an exponential distribution). The ﬁrst way to do this is
to calculate the mean and variance directly from the deﬁnitions. Let S be a random variable
representing the service time.
∞
∞
1
3
E (S ) =
sf (s) ds =
s4λ2 se−2λs ds = = minutes
λ
2
0
0
∞
2
∞
2
Var(S ) = E S 2 − (E (S )) =
s2 f (s) ds −
sf (s) ds
= ∞ 0 s2 4λ2 se−2λs ds − 0 ∞ s4λ2 se−2λs ds 0 2 0 = 3
1
1
9
− 2= 2=
2λ 2
λ
2λ
8 An alternative method for calculating the mean and variance is possible that avoids the necessity of calculating the above integrals but requires knowledge of the Erlangk distribution.
The p.d.f. for an Erlangk distribution with parameter α is, α k s k e − αs for s ≥ 0
(k − 1)!
g (s) = 0
otherwise By setting α = 2λ and k = 2, and considering only s ≥ 0, I can rewrite the given p.d.f. as
follows.
2
(2λ) se−(2λ)s
α2 se−αs
f (s) = 4λ2 se−2λs =
=
(2 − 1)!
(k − 1)! I have now shown that the service times have an Erlang2 distribution with parameter α = 2λ.
What is useful about this result is that an Erlangk distribution with parameter α corresponds
to the distribution of the sum of k i.i.d. exponential random variables with rate α. So let
S = X1 + X2 where X1 and X2 are independent exponential random variables with rate α,
allowing the mean and variance to be calculated as
E ( S ) = E ( X1 + X2 ) = E ( X1 ) + E ( X2 ) =
Var(S ) = Var(X1 + X2 ) = Var(X1 ) + Var(X2 ) =
1 1
2
1
1
+==
αα
α
λ
1
1
2
1
+ 2= 2= 2
2
α
α
α
2λ Having calculated the mean and variance of the service times, I can now calculate the service
rate µ, and the squared coeﬃcient of variation for the service distribution c2 .
S
µ=
c2 =
S 1
2
=
E (S )
3
Var(S ) (E (S )) 2 = 1
2 The last piece of information I need to calculate is the traﬃc intensity,
ρ= η
3
=
µ
4 (a) Using Kingman’s formula, the average waiting time for each customer (in the queue) is
Wq = c2 + c2
a
s
2 ρ
=
µ−η 1+
2 1
2 3
4
2
3 − 1
2 = 3.375 minutes (b) Because the arrival rate is less than the service rate, the throughput of the system is η
Using Little’s Law, the average number of customers in the queue is
Lq = η Wq = 1
· 3.375 = 1.6875 customers
2 (c) Since the mean service time is E (S ) = 3/2 minutes, the average time spent at the site by
a customer is
W = Wq + m = 3.375 + 3/2 = 4.875 minutes
By using Little’s Law again, the average number of customers at the site is
L = ηW = 1
· 4.875 = 2.4375 customers
2 2. (a) To determine the state space for Xn , you should consider how the system moves from
state to state. If the inventory drops below 3 units during a particular day, an order is
placed to return the inventory to 6 units by the next morning, therefore the day can never
start with fewer than 3 units in stock. The stock is only ever replenished to at most 6
units, so the day will never start with more than 6 units. From these observations, the
state space can be written as
S = {3, 4, 5, 6}
The initial state is deterministic, which means the initial distribution is given by a probability distribution with only a single nonzero value,
1 for i = 5
P(X0 = i) =
0 otherwise
or equivalently as a vector,
aT = 0 0 Next, we ﬁnd the transition matrix. Note that 1 0 P(Xn+1 = 3Xn = 3) = P(Xn+1 = 4Xn = 3) = P(Xn+1 = 5Xn = 3) = 0
2 because whenever the inventory level goes below 3 we order up to 6. Therefore,
P (Xn+1 = 6Xn = 3) = 1
Now suppose the inventory at the beginning of the day is 4 units. If the demand during
the current day (day n) is 1, we end up with 3 units of inventory, and, because we do
not order, we will have 3 units of inventory at the beginning of the next day (day n + 1).
Because the demand equals 1 with probability 1/6, the probability of transitioning from
state 4 to state 3 is 1/6.
P(Xn+1 = 3Xn = 4) = P(D = 1) = 1
6 If the demand is 2 or 3, the inventory level drops below 3, so we order, hence
P (Xn+1 = 6Xn = 4) = P(D = 2) + P(D = 3) = 5
6 Proceeding by the same logic, row by row, the transition matrix is given by, 0001
1 0 0 5 6
P = 6 1 3 6 0 2
6
6
2
3
1
0
6
6
6
where Pij = P(Xn+1 = j + 2Xn = i + 2). (b) Let Yn be the number of units of inventory in stock at the end of day n. The inventory
at the end of a day can be any value from 0 to 5. It cannot be 6 because at the beginning
of a day the maximum number of items we can have in inventory is 6 and the demand is
strictly greater than zero with probability 1. So the state space in this case is
S = {0, 1, 2, 3, 4, 5}.
If it still seems odd to exclude 6, include 6 in the transition matrix and see what happens.
You should notice in the construction involving state 6, that state 6 cannot be reached
from any initial state, including state 6, for this reason we do not include it in this
formulation.
The initial state is deterministic and the initial distribution is given by
1 if i = 2
P(Y0 = 2) =
0 otherwise
Observe that
P (Yn+1 = 5Yn = 0) = P (D = 1) = 1/6
Similarly
P (Yn+1 = 4Yn = 0) = P (D = 2) = 3/6
P (Yn+1 = 3Yn = 0) = P (D = 3) = 2/6 3 Continuing in this fashion, we arrive at 0
0 0
P = 2 6 0
0 the following transition matrix 00231
6
6
6
0 0 2 3 1
6
6
6 0 0 2 3 1
6
6
6
3
1
0 0 0 6
6 2
3
1
0 0
6
6
6
0 2
6 3
6 1
6 0 where Pij = P (Yn+1 = j − 1Yn = i − 1). 3. Xn represents the number of consecutive days without injury on the morning of day n. If today
the number of consecutive days without injury is m then it is possible (with probability 99/100)
to have one more day without injury and begin the next morning with m + 1 consecutive days
without injury. Because this logic holds for every nonnegative integer, the state space is equal
to the nonnegative integers,
S = Z+ = {0, 1, 2, . . .} To gain some intuition for how the transition probabilities are calculated, consider the following
table. The ﬁrst row gives the number of consecutive days without injury on the morning of
day n. The second row indicates whether an injury occurred later during the day but before
the next morning. The last row shows the number of consecutive days without injury on the
following morning, which is a result of the previous day’s events.
n
Morning of day n (Xn )
Injury Today?
Morning of day n+1 (Xn+1 ) 1
0
No
1 2
1
No
2 3
2
Yes
0 4
0
No
1 5
1
No
2 6
2
No
3 7
3
Yes
0 8
0
Yes
0 This example serves to illustrate the progression of states for the Markov chain in one particular
instance. In general, if at the start of the day there have been m ≥ 0 consecutive days without
injury, then one of two outcomes can occur. If no injury occurs then the number of consecutive
days without injury has increased by 1, so on the morning of the next day there will have been
m + 1 consecutive days without injury. From this logic the ﬁrst transition probability can be
determined.
99
P(Xn+1 = m + 1Xn = m) = P(No Injury) =
100
If an injury occurs, then the streak has been broken and the number of consecutive days
without injury resets to 0. This allows one more transition probability to be calculated.
1
100
Notice that the sum of these two probabilities is 1, and they correspond to the same row of
the transition matrix. Because the sum of the elements of each row of the transition matrix
must equal 1, all other entries must be 0. Also, because m was generic, this formula and logic
applies to all m ≥ 0. The probabilities can be summarized by writing as follows. 99 if j = i + 1 100 1
Pi,j = P(Xn+1 = j Xn = i) = 100 if j = 0 0
otherwise
P(Xn+1 = 0Xn = m) = P(Injury) = 4 By applying the transition probability formula row by row, we can calculate the ﬁrst several
rows of the matrix until a pattern develops
1 99
0
0
0
0 ···
100
100
99
1
0
0
0
0 · · · 100 100
1 99 100
0
0
0
0 · · ·
100 99
1
0
0
0
0 · · · 100 100
1 99 100
0
0
0
0
· · ·
100 .
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.
..
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The initial distribution is deterministic, and so it can be expressed as a probability distribution
taking one nonzero value,
1 if i = 0
P(X0 = i) =
0 otherwise
or in vector notation
a= 1 0 5 0 0 ··· ...
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 Fall '07
 Billings
 Probability theory

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