Problem Set #4
1. Pedestrian Light, revisited – 3 pts
Little’s Law: L = λ.W and Lq = λ.Wq, with λ = λL + λR, W =
1
"
+ W q and L = Lq + .
u
u
T
T
" LA = ( # L + # R )
2
2
!
!
N 0 "1
N 0 "1
$ LB =
Rule B: W B =
2( #L + #R )
2
&
&
1
( "L + "R ).T0 #
1
!Rule C: W = T0 #1+
%
( ) LC =
%1+
(
C
2 $ 1+ ( "L + "R ).T0 '
2
$ 1+ ( "L + "R ).T0 '
!
2. Squaring – 4 pts
a) 2 pts
!
FY (y) = P[Y " y] = P[X 2 " y] = P # y " X " y
Rule A: W A =
[
Let FY(y) be the CDF of fY(y):
y
FY (y) =
$
f X (x).dx = FX
# y
Then, fY (y) =
d
FX
dy
!
X
[ ( y ) " F (" y )] = 2.1 y [ f ( y ) + f (" y )]
b) 2 pts
!
( y ) #F (# y )
X
X
X
!
1
f X (X) = ,"X # [$1;2]
3
Therefore, fY (y) =
!
1
2. y
[ f ( y ) + f (" y )]
X
X
% 1
'' 3. y ,#y $ [0;1]
=&
' 1 ,#y $ [1;4]
(' 6. y
]
!
3. M/M/k Queue – 6.5 pts
a) 2 pts
- # " &n
/% (
/ $ µ ' P ,)n * [0;k +1]
0
/
P(n) = . n! n
/ #"&
/ %$ µ ('
/ n+k P0 ,)n * [k;,[
0 k .k!
- $ # 'n
$ # 'n 0
/ k*1 & )
2
,
, & )
%µ(
%µ( 2
/
Since + Pn = 1 " P0 +
+ + n*k
= 1, we have: P0 =
/
n!
k .k!2
n= 0
n= 0
n= k
/
2
.
1
1
n
#"&
# " &n
k)1 % (
+ % (
µ
$µ'
* n! + * k$n)k'.k!
n= 0
n= k
This becomes:
! P0 =
1
n
n
#"&
# "&
%
(
k)1
+ % (
µ
$µ'
* n! + * k$n)k'.k!
n= 0
n= k
=
1
n
# "&
!
k
m
k)1 % (
+
# " &
$µ' # "& 1
* n! + %$ µ (' k! , *%$ µ.k ('
n= 0
m= 0
=
1
n
#"&
k
k)1 % (
1
$µ' # "& 1
* n! + %$ µ (' k! , "
n= 0
1)
µ.k
1
P0 =
n
#"&
k)1 % (
"k
$µ'
+
* n! µ k)1.(µ.k ) ")(k )1)!
n= 0
We can then deduce the exact value for each Pn.
!
b) 1.5 pt
%
& % )k 1 # & % )m
& % )k 1
& % ) k +1
P0
k.µ
Lq = $ (n " k).Pn =P0 ( +
, $ m( + = P0 ( +
,
2 =( +
2
&
' µ * k! m= 0 ' k.µ *
' µ * k! &
% ) 'µ*
% )
n= k
1"
k.k!
1"
(
+
(
+
' k.µ *
' k.µ *
1 L
1
Since W = W q + = q + ,
µ " µ
k
" .P0
1
W =
2 +
$
" ' µ
µ k +1 .k.k!&1#
)
!
% k.µ (
#
!
!
c) 1.5 pts
A simulation shows that, when fixing Wq = 1 and 1/µ = 1 min, the progression of the maximum
number of calls is faster than the progression of the number of servers required. In addition, the
value (1-ρ) tends to 0.
d) 1.5 pts
In such a case, the service quality is extremely sensitive to volume fluctuation. A mere 7%
increase in calls and the call center can no longer service customers fast enough: the expected waiting time becomes infinite.
On February 27th, trading volumes exceeded the trading capacity of the stock market, creating an
enormous queue of transactions. The value of stocks lagged far behind, creating confusion among traders.
4. M/M/k Queue with discouraged voters – 6.5 pts
a) 2 pts
n
(n " k)(n " k + 1)
Since # (i " k) =
2
i= k
.
# " &n
0
% (
$µ'
0
P0 ,)n * [0;k +1]
0
! We have: P(n) = /
n!
n
0 # "&
(n+k )(n+k +1)
0 %$ µ ('
+
20
P
,
e
,)n * [k;-[
0 n+k
1 k .k! 0
n
. $ # 'n
1
$ #'
0
3
&
)
&
)
(n*k )(n*k +1)
,
k*1
,
*
%µ(
%µ(
!
0
3 =1
20
+ Pn = 1 " P00+ n! + + k n*k .k! - e
3
n= 0
n= k
0n= 0
3
/
2
This becomes: P0 =
!
#"&
% (
µ
* $ n!'
n= 0
n
k)1
1
=
# " &n
, % (
(n)k )(n)k +1)
)
$µ'
20
+ * n)k + e
k
.k!
n= k
b) 1.5 pts
n%k
k%1
#
!#
%
"Av = $ "n .Pn = $ " .Pn + $ " .e 10 .Pn
n= 0
!
n= 0
n= k
1
n
#"&
m
% ( # &k
m(m +1)
)
" 1 ,# " &
$µ'
* n! + %$ µ (' k! + *%$ µ.k (' + e 20
n= 0
m= 0
k)1
$
L
In steady state, we get: W q = q =
"Av
$
% (n # k).P
% m.P
n= k
m= 0
n
k#1
$
% ".P + % ".e
n
n= 0
n= k
#
n#k
10
m +k
=
.Pn
$
m
& k#1
)
#
"(% Pn + % e 10 .Pm +k +
'n= 0
*
m= 0
c) 1.5 pt
No, it’s not. People would balked out do are not taken into account here. The quality of service
!
will be overestimated.
d) 1.5 pt
Percentage of people who balked: PBal = 1"
!
m
$
& k"1
)
"
#Av
= 1" (% Pn + % e 10 .Pm +k +
#
'n= 0
*
m= 0