Unformatted text preview: IE:3610 Stochastic Modeling (Fall 2014)
Homework 9 (due Thu, Nov 6)
Solutions
HW9 (Due Thu, Nov 6): 17.63; 17.610 (a,c,e); 17.611; 17.612; 17.615. 17.63
Solution:
M/M/1 model
Proportion of time no one is waiting to be served =
= P (number of customers in the system ≤ number of servers) =
= P (n ≤ s)
λ = 10, µ = 15 =⇒ ρ = 2/3
P0 = 1 − ρ = 1/3, P1 = (1 − ρ) ∗ ρ = 1/3 ∗ 2/3 = 2/9
P (n ≤ s) = 1/3 + 2/9 = 5/9 = 0.5556
17.610(a,c,e)
Solution:
(a) M/M/1 model with λ = 30 and µ = 40
λ
30
L = µ−λ
= 40−30
= 3 customers,
1
1
W = µ−λ = 40−30 = 0.1 hours,
30
λ
= 40(40−30)
= 0.075 hours,
Wq = µ(µ−λ)
Lq = λWq = 30 ∗ 0.075 = 2.25 customers,
P0 = 1 − ρ = 1 − 3/4 = 1/4 = 0.25,
P1 = (1 − ρ)ρ = 0.1875,
P2 = ρ2 (1 − ρ) = 0.140625
P (more than 2 customers at the checkout stand) = P (n > 2) =
= 1 − (P0 + P1 + P2 )
P (more than 2 customers at the checkout stand) = 1 − 0.578125 =
0.421875 ≈ 0.423
(c) M/M/1 model with λ = 30 and µ = 60
λ
30
L = µ−λ
= 60−30
= 1 customer,
1
1
W = µ−λ = 60−30 = 0.033 hours,
λ
30
Wq = µ(µ−λ)
= 60(60−30)
= 0.017 hours,
Lq = λWq = 30 ∗ 0.017 = 0.51 customers,
P0 = 1 − ρ = 1 − 0.5 = 0.5,
P1 = (1 − ρ)ρ = 0.25,
P2 = ρ2 (1 − ρ) = 0.125
P (more than 2 customers at the checkout stand) = P (n > 2) =
1 = 1 − (P0 + P1 + P2 )
P (more than 2 customers at the checkout stand) = 1 − 0.875 = 0.125
(e) The manager should hire another person to help the cashier by
bagging the groceries
17.611
Solution:
The three criteria are:
1) the average number of customers in the queue should not exceed 1:
Lq ≤ 1
2) with 95% probability the number of customers in the queue (waiting for clearance) should not exceed 4, which means the total number
of
customers in the system should not exceed 5 with probability 95%:
P4+s
i=0 Pi ≥ 0.95
3) P (Wq ≤ 30 mins) ≥ 0.99 or P (Wq ≤ 0.5 hrs) ≥ 0.99
17.611.(a) M/M/s model with λ = 10, µ = 20, s = 1
Allcriteria
the criteria
are currently
satisfied
(a) All the
are currently
satisfied. s= Data
10
20
1 Results
(mean arrival rate)
(mean service rate)
(# servers) Pr(W > t) = 0.006738
when t = L=
Lq = 1
0.5 W=
Wq = 0.05 0.1 0.5
0.5 Prob(W q > t) = 0.003369
when t = 0.5 n
0
1
2
3
4
5 model
with λ = 15, µ = 20, s = 1
(b) None(b)
of M/M/s
the criteria
are satisfied.
None of the criteria
are currently satisfied
Data
s= 15
20
1 (mean arrival rate)
(mean service rate)
(# servers) Pr(W > t) = 0.082085
when t = Pn cumulative 0.5
0.25
0.125
0.0625
0.03125
0.015625 0.5
0.75
0.875
0.9375
0.96875
0.984375 Results L=
Lq = 3
2.25 W=
Wq = 0.15 0.2 0.5
0.75 Prob(W q > t) = 0.061564
when t = 0.5 n 2 Pn (c) The first and third criteria are satisfied, but the second is not.
Data
25 (mean arrival rate) cumulative 0
0.25
0.25
1
0.1875
0.4375
2
0.140625
0.578125
3 0.10546875 0.68359375
4 0.079101563 0.76269531
5 0.059326172 0.82202148 Results
L = 2.051282051 Wq = Pr(W > t) = 0.006738
when t =
0.5
when t =
0.5 0.5
0.5 Prob(W q > t) = 0.003369
Prob(W q > t) = 0.003369
when t =
0.5
when t =
0.5 s=
s= (mean arrival rate)
(mean
(mean arrival
servicerate)
rate)
(mean service rate)
(# servers)
(# servers) Pr(W > t) = 0.082085
Pr(W > t) = 0.082085
when t =
0.5
when t =
0.5 Pn
Pn n
n
0
01
12
23
34
45
5 0.5
0.5
0.25
0.25
0.125
0.125
0.0625
0.0625
0.03125
0.03125
0.015625
0.015625 L=
LLq ==
Lq = Results
Results 3
3
2.25
2.25 W=
W ==
W
q
Wq = 0.2
0.2
0.15
0.15 (b) None of the criteria are satisfied.
(b) None of the criteria
are satisfied.
Data
Data
15
15
20
20
1
1 0.05 cumulative
cumulative
0.5
0.5
0.75
0.75
0.875
0.875
0.9375
0.9375
0.96875
0.96875
0.984375
0.984375 0.75
0.75 Prob(W q > t) = 0.061564
Prob(W q > t) = 0.061564
when t =
0.5
when t =
0.5 Pn
Pn n
n
0
01
12
23
34
45
5 0.25
0.25
0.1875
0.1875
0.140625
0.140625
0.10546875
0.10546875
0.079101563
0.079101563
0.059326172
0.059326172 L=
LLq ==
Lq = Results
2.051282051
2.051282051
0.801282051
0.801282051 W=
W=
Wq =
Wq = 0.082051282
0.082051282
0.032051282
0.032051282 cumulative
cumulative
0.25
0.25
0.4375
0.4375
0.578125
0.578125
0.68359375
0.68359375
0.76269531
0.76269531
0.82202148
0.82202148 (c) M/M/s model with λ = 25, µ = 20, s = 2
(c) The first and third criteria are satisfied, but the second is not.
firstthird
andData
third criteria
are satisfied,
but theisResults
second
(c) The The
first and
criteria
are satisfied,
but the second
not. is not
s=
s= Data
25
25
20
20
2
2 (mean arrival rate)
(mean
(mean arrival
servicerate)
rate)
(mean service rate)
(# servers)
(# servers) Pr(W > t) = 0.001022
Pr(W > t) = 0.001022
when t =
0.5
when t =
0.5 0.625
0.625 Prob(W q > t) = 0.000266
Prob(W q > t) = 0.000266
when t =
0.5
when t =
0.5 n
n
0
01
12
23
34
45
5 Pn
cumulative
Pn
cumulative
0.230769231
0.23076923
0.230769231
0.288461538 0.23076923
0.51923077
0.288461538
0.180288462 0.51923077
0.69951923
0.180288462
0.112680288 0.69951923
0.81219952
0.112680288
0.07042518 0.81219952
0.8826247
0.07042518 0.92664044
0.8826247
0.044015738
0.044015738 0.92664044 Criteria: 1) Lq = 0.8013 ≤ 1 is satisfied;
P
P6
2) 4+s
i=0 Pi =
i=0 Pi = 0.027509836 + 0.926640437 = 0.954150273 ≥
0.95 is satisfied;
3)P (Wq ≤ 0.5hr) = 1 − 0.0002661716
≥ 0.99 is satisfied
1716
17.612
Solution:
Guidlines (M/M/4 model):
1)The average number of customers waiting in line to begin service
should not exceed 1: Lq ≤ 1 3 2)At least 95 percent of the time, the number of customers waiting in
line should not exceed 5: P (n − s ≤ 5) ≥ 0.95 or P (n ≤ 9) ≥ 0.95
3)For at least 95 percent of the customers, the time spent in line waiting
17.612.to begin service should not exceed 5 minutes: P (Wq ≤ 5 mins) ≥ 0.95 (a) All the guidelines
currently
(a)M/M/4are
model.
All met.
the guidelines are currently met. s= Data
2
1
4 (mean arrival rate)
(mean service rate)
(# servers) W = 1.086956522
W q = 0.086956522 Pr(W > t) = 0.007902
when t = Results
L = 2.173913043
Lq = 0.173913043 5
0.5 Prob(W q > t) =
when t = 7.9E06
5 n Pn 0
1
2
3
4
5
6
7
8
9 0.130434783
0.260869565
0.260869565
0.173913043
0.086956522
0.043478261
0.02173913
0.010869565
0.005434783
0.002717391 cumulative
0.1304
0.3913
0.6522
0.8261
0.9130
0.9565
0.9783
0.9891
0.9946
0.9973 (b)The first two guidelines will not be satisfied in a year, but the third
(b) The will
firstbe.
two guidelines will not be satisfied in a year, but the third will be. s= Data
3
1
4 (mean arrival rate)
(mean service rate)
(# servers) W = 1.509433962
W q = 0.509433962 Pr(W > t) = 0.023901
when t = Results
L = 4.528301887
Lq = 1.528301887 5
0.75 Prob(W q > t) = 0.003433
when t = 5 4 (c) Five tellers are needed in a year. n Pn 0
1
2
3
4
5
6
7
8
9 0.037735849
0.113207547
0.169811321
0.169811321
0.127358491
0.095518868
0.071639151
0.053729363
0.040297022
0.030222767 cumulative
0.0377
0.1509
0.3208
0.4906
0.6179
0.7134
0.7851
0.8388
0.8791
0.9093 1
2
3
4
5
6
7
8
9 0.260869565
0.260869565
0.173913043
0.086956522
0.043478261
0.02173913
0.010869565
0.005434783
0.002717391 0.3913
0.6522
0.8261
0.9130
0.9565
0.9783
0.9891
0.9946
0.9973 (b) The first two guidelines will not be satisfied in a year, but the third will be. s= Data
3
1
4 (mean arrival rate)
(mean service rate)
(# servers) W = 1.509433962
W q = 0.509433962 Pr(W > t) = 0.023901
when t = Results
L = 4.528301887
Lq = 1.528301887 5
0.75 Prob(W q > t) = 0.003433
when t = 5 n Pn 0
1
2
3
4
5
6
7
8
9 0.037735849
0.113207547
0.169811321
0.169811321
0.127358491
0.095518868
0.071639151
0.053729363
0.040297022
0.030222767 cumulative
0.0377
0.1509
0.3208
0.4906
0.6179
0.7134
0.7851
0.8388
0.8791
0.9093 Fiveare
tellers
areinneeded
(c) Five(c)
tellers
needed
a year. in a year: 1) Lq = 0.354227405 ≤ 1 is satisfied
2) P (n − s ≤ 5) = P (n ≤ 10) = 0.988982111 ≥ 0.95 is satisfied
3) P (Wq ≤ 5 mins) = 1 − (1.07213E − 05) ≥ 0.95 is satisfied
17.615
Solution:
M/M/1 model with λ = 20 per hr,
µ = 1/2 per min = 30 per hr
1717
P ( An arriving customer does not have to wait before service) = P0 =
1 − λ/µ = 1/3
Expected price of gasoline per gallon: 4 ∗ 1/3 + 3.5 ∗ 2/3 = $3.667 5 ...
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 Spring '08
 KROKHMAL
 Ode, mean arrival rate, mean service rate

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