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Unformatted text preview: ISyE 3232
H. Ayhan Stochastic Manufacturing and Service Systems Fall 2011 Homework 9
November 4, 2011
(Due November 9)
1. Daily demand for paint brushes at a particular store follows the demand distribution:
d
0
P (D = d) .5 1
.15 2
3
4
.
.3 .04 .01 The stock level is reviewed every evening and when warranted an order is placed at the central warehouse to augment stock. Orders arrive over night and are available to meet the demand on the morning
of the next day. The ﬁxed cost for placing an order at the end of the day is $0.2 and the per unit order
cost is $0.05, the daily per unit holding cost is is $0.01 and the per unit penalty cost for unﬁlled orders
is $0.5. Compute the long run expected daily cost under the following inventory replacement policies:
(a) If the number of brushes at the end of the day is 2 or less the management orders enough brushes
to bring the on hand inventory to 6 brushes
(b) If the number of brushes at the end of the day is 3 or less the management orders enough brushes
to bring the on hand inventory to 6 brushes
2. Consider a bank with two tellers. Teller 1 has an exponential service time with mean 3 minutes and
teller 2 has an exponential service time with mean 6 minutes. Alice, Betty, and Carol enter the bank at
the same time. Alice goes to teller 1 and Betty goes to teller 2 while Carol waits for the ﬁrst available
teller.
(a) What is the expected time that Carol spends in the bank?
(b) What is the expected time until the last of the three customers leaves the bank?
(c) What is the probability that Carol is the last one to leave?
3. Sue and Liz arrive at a beauty salon together and plan to leave together. Sue needs a perm, and Liz
a manicure. The duration of a perm is exponentially distributed with rate 5/hr; that of a manicure is
exponentially distributed with rate 10/hr. If both are served immediately, what is expected duration
one has to wait for the other? ISyE 3232 Stochastic Manufacturing and Service Systems Fall 2011 H.Ayhan
Solutions to Homework 9
1. Let {Xn : n ≥ 0} be the stock level at the evening of day n. The problem states an
(s, S ) inventory policy, i.e., if the stock level is s or less than s at the review time,
we make orders to bring the stock level up to S ; we do not order otherwise. We can
establish the relation between Xn+1 and Xn as follows:
S − D if Xn ≤ s
Xn+1 =
.
(1)
Xn − D if Xn > s
By Equation (1), {Xn } is a discretetime Markov chain. Note that we allow negative
stock levels, which indicate unﬁlled orders at review times.
(a) In this case, s = 2 and S = 6. We have the state space S = {−1, 0, 1, 2, 3, 4, 5, 6}.
The transition probability matrix can be written as, 0
0
0 .01 .04 .3 .15 .5
0
0
0 .01 .04 .3 .15 .5 0
0
0 .01 .04 .3 .15 .5 0
0
0 .01 .04 .3 .15 .5 P = .01 .04 .3 .15 .5 0
0 0 0 .01 .04 .3 .15 .5 0 0 0
0 .01 .04 .3 .15 .5 0 0
0
0 .01 .04 .3 .15 .5
By π P = π and i∈S πi = 1, we can obtain the stationary distribution as,
π = (0.0019, 0.0104, 0.0695, 0.1217, 0.1901, 0.2808, 0.1221, 0.2035) Let f (i) be the cost incurred at state i, where i = −1, 0, . . . 6. We have that,
f (−1) = 0.5 + 0.2 + 7 × 0.05 = 1.05, order 7
f (0) = 0.2 + 6 × 0.05 = 0.5, order 6
f (1) = 0.01 × 1 + 0.2 + 5 × 0.05 = 0.46, order 5
f (2) = 0.01 × 2 + 0.2 + 4 × 0.05 = 0.42, order 4
f (i) = 0.01i for i = 3, 4, 5, 6, no orders
Thus, the longrun expected daily cost, denoted by c, is,
c= 6
i = −1 1 πi f (i) = 0.1255. (b) In this case, s = 3, S = 6. The state space S = {0, 1, 2, 3, 4, 5, 6} and the transition
probability matrix is, 0
0 .01 .04 .3 .15 .5
0
0 .01 .04 .3 .15 .5 0
0 .01 .04 .3 .15 .5 0 .01 .04 .3 .15 .5 P = 0 .01 .04 .3 .15 .5 0 0 0 .01 .04 .3 .15 .5 0 0
0 .01 .04 .3 .15 .5
We have the stationary distribution, π = (0.0035, 0.0154, 0.1151, 0.1173, 0.3467, 0.1508, 0.2513).
Let f be deﬁned above. We have that,
f (0) = 0.2 + 6 × 0.05 = 0.5, order 6
f (1) = 0.01 × 1 + 0.2 + 5 × 0.05 = 0.46, order 5
f (2) = 0.01 × 2 + 0.2 + 4 × 0.05 = 0.42, order 4
f (3) = 0.01 × 3 + 0.2 + 3 × 0.05 = 0.38, order 3
f (i) = 0.01i for i = 4, 5, 6, no orders
Thus, the longrun average daily cost is c = π f = 0.1382.
2. Let T1 and T2 be the exponentially distributed service times of Sellers 1 and 2, respectively, i.e., T1 ∼ exp(λ) and T2 ∼ exp(µ), where λ = 1/3 and µ = 1/6.
(a) Let Ta be the time Carol spends in the bank. We thus have,
T1 if T1 ≤ T2
Ta = min(T1 , T2 ) +
.
T2 if T1 > T2
Thus, E(Ta ) = E(min(T1 , T2 )) + E(T1 )Pr {T1 ≤ T2 } + E(T2 )Pr {T1 > T2 } = 6 (2) (b) Let Tb be the time till the last one leaves the bank. We have Tb = min(T1 , T2 ) +
max(T1 + T2 ). Note that
− max(T1 , T2 ) = min(−T1 , −T2 )
= −T1 − T2 + T1 + T2 + min(−T1 , −T2 )
= −T1 − T2 + min(T1 + T2 − T1 , T1 + T2 − T2 )
= −T1 − T2 + min(T1 , T2 )
Thus, Tb = T1 + T2 and E(Tb ) = E[T1 ] + E[T2 ] = 3 + 6 = 9.
2 (c) By (a) and (b), we need to calculate Pr {Ta ≥ Tb }. By (b), Tb = T1 +T2 . Subtracting
Tb from Equation (2) gives,
min(T1 , T2 ) − T2 if T1 ≤ T2
Ta − Tb =
min(T1 , T2 ) − T1 if T1 > T2
Thus,
Pr {Ta − Tb ≥ 0} = Pr {T1 ≤ T2 }Pr {min(T1 , T2 ) − T2 ≥ 0}
+ Pr {T1 > T2 }Pr {min(T1 , T2 ) − T1 ≥ 0}
= Pr {T1 ≤ T2 }Pr {T2 ≤ T1 } + Pr {T1 > T2 }Pr {T1 ≤ T2 }
= 2Pr {T1 ≤ T2 }Pr {T2 ≤ T1 }
1/3
1/6
=2×
1/3 + 1 /6 1/3 + 1 /6
= 4/ 9
3. Let Ts and Tl be the time Sue and Liz will spend at the beauty salon, respectively.
Thus, Ts ∼ exp(λ) and Tl ∼ exp(µ), where λ = 5 and µ = 10. Let T be the duration
one has to wait for the other. It’s clear that T = max(Ts , Tl ) − min(Ts , Tl ). By 2(b),
T = Tl + Ts − 2 min(Ts , Tl ) and
E(T ) = E(Tl ) + E(Ts ) − 2E[min(Ts , Tl )] = 1/5 + 1/10 − 2 × (1/15) = 1/6 3 ...
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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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