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Logistical and Transportation Planning Methods
Problem Set #4
Due: November 8, 2006
Problem 1
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Logistical and Transportation Planning Methods
4.
We use the bounds discussed in class: B −
λ (σ X2 + σ S2 )
1+ ρ
≤ Wq ≤ B with B =
.
2λ
2(1 − ρ )
1 25
1
=
and σ S2 =
, therefore
2
9
12
λ
3 25 1
( + )
5
9 12 = 8.583 and 1 + ρ = 1 + 0.9 =1.583
B=
3
2(1 − 0.9)
2λ
2⋅
5
Thus, the upper and lower bounds are 7 ≤ Wq ≤ 8.583.
σ X2 =
By Little’s Law, 4.2 ≤ Lq ≤ 5.150 .
5.
6.
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Logistical and Transportation Planning Methods
7.
8.
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Logistical and Transportation Planning Methods
9.
10.
We now have two classes of customers.
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Logistical and Transportation Planning Methods
E[S1 ] = 1.2 min

For Class A customers: λ1 = 14.4 per hour and  2
. Thus
0.4 2
=
min 2
σ
 S1
12

14.4
ρ1 =
= 0.288 .
60 1.2
E[S1 ] = 1.7 min

For Class B customers: λ 2 = 21.6 per hour and  2
. Therefore,
0.6 2
min 2
σ S1 =
12

21.6
ρ2 =
= 0.612 .
60 1.7
From Equation 4.107 of the L+O book, we have:
Wo
Wo
and Wq 2 =
1 − 0.288
(1 − 0.288)(1 − 0.288 − 0.612)
0.04 

2
14.41.2 2 +
 + 21.6 1.7 + 0.03
3 

Wo is given by equation 4.102: Wo =
.
2 ⋅ 60
Finally we have Wq = 0.4 ⋅ W q1 + 0.6 ⋅ W q 2 = 6.29 min.
Wq1 =
(
)
In this new case, Wq is lower than in part 1 and part 9 because we are now giving the priority to
the customers with the shortest expected time.
A corollary takes advantage of this phenomenon ( p239 of the textbook):
To minimize the average waiting time for all users in the system, assign priorities according to
the expected service times for each user class: the shorter the expected service time, the higher
the priority of the class.
11.
This time, Class B customers have the priority. They have the greatest expected service time.
Therefore, the overall Wq increases.
Numerical answer: Wq = 8.28
Problem 2
We define the states (i, j, k)
where:
- i describes System 1 and i = 0, 1 or b ( for blocked)
- j describes System 2 and j = 0, 1 or b
- k describes Sytem 3 and k = 0 or 1
There are 13 states.
The corresponding state transition diagram is:
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Logistical and Transportation Planning Methods
Problem 3
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Logistical and Transportation Planning Methods
3λ
000
100
0.4 µ
µ
λ
2λ
101
102
0.4 µ
0.6 µ
µ
010
0.4 µ
0.6 µ
0.6 µ
µ
011
2λ
012
λ
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Logistical and Transportation Planning Methods
Problem 4
(a) A state is defined by (a,b,c) where:
a = the type of the customer(s) (0, 1, or 2) currently in service (you cannot have
one server occupied by a Type 1 customer and the other server occupied by a
Type 2 customer at the same time),
b = the number of Type 2 customers (0, 1, or 2) in the system,
c = the number of Type 1 customers (0, 1, 2, 3 or 4) in the system.
The corresponding state transition diagram for the system is:
2,2,0
µ2
λ2
λ1
µ1
2,1,0
λ2
µ2
1,1,1
λ2
λ1
0,0,0
1,1,2
2µ1
λ1
1,0,1
µ1
2µ1
2,1,1
λ1
λ1
1,0,2
µ2
λ1
λ2
1,0,3
2µ1
1,0,4
2µ1
µ2
λ1
2,1,1
(b) The event of interest can happen only by having a set of three consecutive transitions from
state (1,0,4) to (1,0,3) to (1,0,2) to (1,1,2). We have:
P=
2µ1
λ2
⋅
2µ1 + λ1 λ1 + λ 2 + 2 µ1
Problem 5
(c) A state is defined by (a,b) where a describes the first facility and b the second one.
a = 0 if there are no customers
1 if there is one customer
2 if there are 2 customers
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Logistical and Transportation Planning Methods
b1 if there is one customer and he/she is blocked
b2 if there are two customers and one is blocked
b3 if there are two customers and both are blocked.
b = 0 if there are no customers
1 if there is one customer.
The corresponding state transition diagram for the system is:
λ
0,0
µ1
µ2
µ2
λ
b1,1
µ2
λ
1,1
µ1
µ2
2,0
2µ1
µ2
λ
0,1
λ
1,0
b3,1
2µ1
b2,1
µ1
µ2
2,1