Tutorial 5: Queuing
Question1
M/M/1
ƛ=4/hour
µ=6/hour
ƍ= ƛ/ µ=2/3=0.67
1. U= ƍ=0.67=67%
2. Probability of waiting = probability of busy wash bay = 67%
3. ∑
= (1- ƍ) ∑
ƍ) = (1- ƍ) ((1- ƍ8)/ (1- ƍ)) = 1- ƍ8 = 1-0.678 = 0.96 = 96%
4. 1-0.67x = 90% ; -0.67x = -0.1 ; ln(0.1)=xln(0.67) ; x=ln(0.1)/ln(0.67) ; x= -2.3/-0.4 =
5.75 = 6
5. M/M/1/5
Number of cars lost per 24h: ƛ P5 = 4 P5
P5 = (1- ƍ)/ (1- ƍ6) ƍ5= (1-0.67) / (1-0.676)* 0.675 = 0.05 = 5%
6. P0 = (1- ƍ)/ (1- ƍ6) = 0.37 = 37%
7. W = L/ ƛe = L/ ƛ(1-P5)
L = (ƍ/ 1-ƍ) - (6ƍ6/ 1-ƍ6) = 1.4
W = 0.37 hour
8. 4 - Lq = 4 - [L - (1 - P0)] = 3.23
9. P5 = 0.05 = 5%
Question 2
M/M/1
µ=3.33/hour
λ=5/hour
ƍ= λ/ µ=1.5
ƍ>1.5: no steady state will develop. Therefore, he can't manage to serve all his
patients.
µ=3.33: In order to have a steady state, λe must be < 3.33. Therefore, λe=3 is a
good choice.
λe=λ(1-pN)
3=5(1-pN)
1-pN=3/5
PN=2/5=0.4
PN=
* ƍn=0.4
=
Let X=1.5N:
=
0.1X=0.4
X=4
1.5N=4
Nln(1.5)=LN(4)
0.405N=1.39
N=1.39/0.405=3.43 =4
The queuing capacity must be of 2.43. Therefore, we may accept 3 chairs in the
waiting room.
Lq=L-(1-p0)
L=
-
p0 =
= 3.59
= 0.07
Lq = 2.66 patients
p0 = 0.07=7%hour=4.2minutes
λe=λ(1-p5) =5*[1-(p0* ƍ5)]=3.25
3.25*40*10=1300TD
Question3
M/M/1
λ=3/minute
µ=5/minute
ƍ= λ/ µ=3/5
1. U= ƍ=3/5
2. W= ; L=
=3/2 ; W=
=0.5minute
3. Lq=L- ƍ=0.9car/minute
4. p0=1- ƍ=0.4 minute ; 0.4*60=24minutes/hour
M/M/2
1) U= = =
2) W = Wq+
= = 0.3
; Wq=
; Lq = p0*
]-1=0.54 Lq = 0.06 ; Wq =
3) Lq = 0.06
*
; p0 = [∑
+
= 0.02 ; W = Wq+ = 0.22 minute
*
4) p0 = 0.05
Question 4
c0 : operationg cost
cp : penalty cost
CT : c0 + cp L
λ= 4
o Copier 1:
µ1 =
= 0.18 jib/hour = 4.32 jobs/day
CT1 = (15*24)
+[80*(
)] = 1360TD
o Copier 2:
µ2 = 5.18
CT2=751,186TD
o Copier 3:
µ3 = 7.2
CT3=676TD
o Copier 4:
µ4 = 9.5
CT4=706,18TD
We have to choose copier 3.
Question 5
1. Cp: penalty cost
Cµ: speed cost
C(µ)= Cµ*µ+ CpL = Cµ*µ+ Cp( ƛ/ µ- )
2. C’(µ)= Cµ - ƛ Cp/ (µ- ƛ)2 = Cµ*µ+ Cp( ƛ/ µ- ƛ)
C’(µ)=0 Cµ = ƛ Cp/ (µ- ƛ)2; or equivalently (µ- ƛ)2= ƛ
Therefore, µ= ƛ+√
⁄
3. We have ƛ=3; Cp=100; Cµ=0.1*500=50
Therefore, µ=3+√ is the optimal speed to order.