Queuing Theory

1

MISRIMAL NAVAJEE MUNOTH JAIN ENGINEERING COLLEGEDEPARTMENT OF MATHEMATICS

PROBABILITY AND QUEUING THEORY (MA 2262)SEMESTER-IV

QUESTION BANK – IVUNIT– IV – QUEUEING THEORY

PART- A

Problem 1 For / /1 : /M M FIFO model, write down the Little’s Formula.

Solution:i) S SL W

ii) q qL W

iii)1

S qW W

iv) S qL L

Problem 2 For / / : /M M C N FIFO model, write down the Formula for

a) Average number of customers in the queue.b) Average waiting time in the system.

Solution:

a)

0 1 1! 1

n c n cqL n c

C

when

c

b)1

1S SW L

1

1

0

i

nn

C c n

1

S qL L

Problem 3 What is the probability that a customer has to wait more than 15 minutes to

get his service completed in a / /1M M queuing system, if 6 per hour 10 per

hour?

Solution:Probability that the waiting time in the system exceeds t is

W t

t

e dW e

Queuing Theory

2

1

10 6 .14

150.3679

60SP W e e

Problem 4 What is the probability that an arrival to an infinite capacity 3 server Poisson

queueing system with 2

and 0

1

9P entries the service without waiting.

Solution:

P[Without waiting] 0 1 23P N P P P

0

1

!

n

nP Pn

when 3n c .

21 2 1 1 53 2

9 9 2 9 9P N .

Problem 5 In a given / /1 : /M M FCFS queue, 0.6 , What is the probability

that the queue contain 5 or more students?

Solution:

555 0,6 0.0467P X

Problem 6 What is the effective arrival rate for / /1 : 4 /M M FCFS queuing model

when 2 and 5

Solution:

1 1

0 1 5

1 / 1 2 / 55 1

1 / 1 2 / 5K

5

3/ 5 0.65 1 5 1

1 0.010241 2 / 5

0.6

5 1 5 1 0.6070.989

5 0.393 1.965 2

Problem 7 a) In / /1 : /M M K FIFO write down the expression for P0.

b) In / /1 : / , 3/ , 4 / , 7M M K FIFO hr hr K Calculate P0

Solution:

a) 0 1

1

1

KP of

1

1of

K

Queuing Theory

3

b) 0 11

1

KP of

8

3

41 0.27783

14

Problem 8 In / / : / , 10 / , 15 / , 2M M S FIFO hr hr S Calculate P0

Solution:1

1

00

1 1

! !

n sS

n

sP

n S s

121 10 1 10 30

11! 15 2! 15 30 10

1

2

Problem 9 In a two server system with infinite capacity and 20 / hr and 11/ hr ,

find P0.

Solution:1

1

00

1 1

! !

n sS

n

sP

n S s

1220 1 20 11 2

111 2 11 22 20

1

21

Problem 10 Write the steady states equations in / /1 : /M M K FIFO where

0,1,2... 1n for n K

0 for n k

1, 2,3nand n

Solution:

1 1 1 1n n nP P P n K

1 0 0P P n

Queuing Theory

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Problem 11 In / /1 : / , 3/ , 5 / , 7M M K FIFO hr hr K , what is the

probability that the server will be idle.

Solution:The probabilities that the server will be idle

0 1 8

3 21 10.45 5 0.407

1 0.0168 0.983311

5

KP

Problem 12 In / / : / ,M M S K FIFO write the expression for P0.

Solution:

11

00

1 1/ /

! !

n ss k

n s

n n s

Pn s s

Problem 13 What is effective arrival rate in / / : /M M S K FIFO

Solution:Effective arrival rate

1

1

0

s

nn

s s n

Problem 14 / / : /In M M S K FIFO with

3.76 / 4, 2, 7,hr and s k compute 5P where 0 0.3617P .

Solution:

5 0

1/

!

n

n sP P

s s

5

5 2

1 3.760.3617

2!2 4

0.0166

PART-B

Problem15 A repairman is to be hired to repair machines which break down at anaverage rate of 3 per hour. The breakdown follows Poisson distribution. Non- productivetime of machine is considered to cost Rs.16/hr. Two repairmen have been invited. One isslow but cheap while the other is fast and expensive. The slow repairman charges Rs.8per hour and he services machines at rate of 4 per hour. The fast repairman demandsRs.10 per hour and service at the average rate of 6 per hour. Which repair man should behired?

Queuing Theory

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Solution:

For the slow man: Model / /1 : / 3 / , 4 /M M FIFO Hr hr

Expected waiting time of a machine1 1

14 3

hr

Non productive cost =16 / hrNon productive cost = 3 16 =Rs.48.Charges paid to the repairman 8 3 24 Total cost =48+24=72For fast man: M

3/ , 6 / .hr hr

Expected waiting time1 1

3

hour

Non productive cost1

3 16 163

Charge paid to repairman 10 1 10 Total cost =16+10=26Fast repairman can be hired.

Problem 16 A two person barber shop has 5 chairs to accommodate waiting customers.Customers who arrive when all 5 chairs are full leave without entering barber shop.Customers arrive at the average rate of 4 per hour and spend an average of 12 minutes in

the barber’s chair. Compute 0 1,P P and average number of customers in the queue

Solution: This is model / / : / 0 4 / , 5 / , 2, 2 5 7M M C K FIF hr hr C N

1 7

0 20 2

12 5

1 1 1/ /

! ! 2

4 8 2 2 21 1 ...

5 25 5 5 5

0.4287

in n

nn n

Pn C

0

1/ , 0

!n n c

P P nC C

7

7 7 2

14 / 5 0.4287

2 2!P

0.0014.

0 21 1

! 1

C

N C N Cq

PCL P N C

C

Where / C

2 5 5

2

0.40.4287. 0.8 1 0.4 5 0.6 0.4

2! 1 0.4

0.15 Customers

Queuing Theory

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Problem 17 Derive the formula fori) Average number qL of customers in the queue.

ii) Average waiting time of a customer in the queue for

/ /1 : /M M FIFO MODEL.

Solution:

1 1 1

1q n n nn n n

L n P nP P

0 01 0 0

1n n nn n n

nP P P nP P

01sL P

2

1 1

ii) Average waiting time of a customer is the queue qW

0

w

qW W e dw

where / e w

is the probability density

function of the waiting time distribution of the queue.

0

. ww e dw

00

w we e dwW

1

Problem 18 On average 96 patients over 24 hour day require the service of anemergency classic. Also on average a patient requires 10 minutes of active attention.Assume that the facility can handle any one emergency at a time. Suppose that it costs theclinic Rs.100 per patient treated to obtain an average servicing time of 10 minutes andthat each minute of decrease in this average time would cost Rs.10 per patient treated.How much would have to be budgeted by the clinic to decrease the average size of the

queue from1

13

patient to1

2patient?

Solution: This is / /1 : /M M FIFO model96

4 /24

hr .

6 / hr

4 1. 1

3 3qL

Now1

13

customers to be reduced to1

2, then we have to find 1 such that

Queuing Theory

7

2 114 32 0

2ie

1 1 18 4 0 8 Patient /hr

Average time required to each patient'1 15

8 2hr

Decrease in the time required to attend a patient15 5

102 2

min.

Decrease in each minute cost Rs.10 per patient.

Cost5

10 252

.

Budget required for each patient= 100+25=125 to decrease the size of the queue.

Problem 19 Obtain the expressions for steady state probability of a M/M/C queuingsystem

Solution:

1n n n C

C n c

It can be treated as a birth death process with n for all n and n

is given by (1)

In a birth death process

1

01

ni

n

i

P Pi

and

1

1

01

1

1

ai

n i

n

P

i

for M/M/C.

0 110

1 2 1

.... nn

n

P P

0... 1

1 2

nP

n if n C

0

1n

nof n c

ni P

1 .......... 1

1 2 1

n

n

c cP c if n c

n c

1.... 1 1

1 2

nn c

c c

1

!

nn

n

C of n cC

Queuing Theory

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1

00 ! !

n nC

n c nn n cn

Pn c c

Problem 20 Arrivals at a telephone booth are considered to be Poisson with an averagetime of 12 mins. The length of a phone call is assumed to be distributed exponentiallywith mean 4 min. Find the average number of persons waiting in the system. What is theprobability that a person arriving at the booth will have to wait in the queue? Alsoestimate the traction of the day. Where phone will be in use

Solution: This is model / /1 : /M M FIFO

112, 12

per min

1 14,

4

per min

1

12 0.5.1 1

4 12

SL

0 1 0 1S sP L P t P [No customer in the system]

0

11 1 1 / / .

3P

P Phone will be idle = 0

21

3P

P Phone will be in use =1

3Problem 21 There are three typists in an office. Each typist can type an average of 6letters per hour. If letters arrive for being typed at the rate of 15 letters per hour, whatfraction of time all the typists per hour will be busy? What is the average number ofletters waiting to be types?

Solution: This is / / : /M M C FIFO

3, 15 / , 6 /C hr hr

P [all the three typists buy] 3P N 3

0

3! 13

P

Queuing Theory

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3

0 1

0

1

1 1

!

! 1

r

n

P

nn

cc

32 2.5

1

2.5 11 2.5

152!3! 1

6 3

0.0449.

3 0.7016P N 1

02

. ! 1

c

qL P

c c

4

2.50.0449 3.5078

253 6 1 2

3

Problem 22 A bank has two tellers working are savings account. The first teller handleswith drawals only the second teller handles deposits only. It has been found that theservice time distributions for both deposits and withdrawals are exponential with meanservice time of 3 minutes per customer. Depositors are found to arrive in a poissonfashion through out the day with mean arrival rate of 16 per hour withdrawals also arrivein a poisson fashion with mean arrival rate of 14 per hour. What would be the effect onthe average waiting time for the customers if each teller could handle both withdrawalsand deposits?

Solution:1. When the tellers handle separately withdrawals and deposits. This is Model

/ /1 : /M M FIFO .

For depositors 16 / , 20 /hr hr

1

12min5

qW hr

For withdrawals 14 / , 20 /hr hr

7 minqW

ii) If both tellers do both service, one it will be a / /1 : /M M FIFO model.

2 20 / 16 14 30 /C hr hr

Queuing Theory

10

11

00

1 1

! !

n ci

n

cP

n c c

1230 1 30 40 1

120 2 20 40 30 7i

2

02 2

30

1 1 1 1 120

. ! 20 2.21 73011

40

c

qW Pc c

c

0.0642 3.857hr or

Problem 23 In a heavy machine shop, the overhead crane is 75% utilized. Time studyobservations gave the average slinging time as 10.5 minutes with a SD of 8.8 minutes.What is the average calling rate for the service of the crane and what is the average delayin getting service? If the average service time is cut to 8.0 minutes, with a standarddeviation of 6.0 minutes, how much reduction will occur on average in the delay ofgetting service?

Solution: This is model / /1 : /M G FIFO

Average delay is getting service

2 21/

2 1p

0.75, 5.71/ , 8.8.hr

220.75 8.8 60

1 5.712 1 0.75 60 5.71

Average calling rate 0.75 5.71 4.28 per hr. If the service time is cut to

8 minutes, then60 4.29

7.5 / , 0.5718 7.5

hr

utilization of the crane reduced to 57.1 percent and average delay in getting service

220.571 600 60

1 7.52 1 0571 60 7.5

qW

8.3 minutes.

a reduction of 18.5 minutes approximately 70%

Problem 24 A Super market has two girls ringing up sales at the customers. If the servicetime for each customer is exponential with mean 4 minutes and if the people arrive in aPoisson fashion at the rate of 10 per hour

a) What is the probability of having to wait for service?b) What is the expected percentage if idle time for each girl?

Queuing Theory

11

Solution:

This is model / / : /M M C FIFO

1 1 1, ; 2 .

6 4 3C

C

121

00

1 4 1 4 2.1/ 4 1

! 6 2! 6 2.1/ 4 1/ 6 2

nc

n

Pn

1 0

2 1 1.

3 2 3P P

a) 2

1 1 12 1

2 3 6c

P C Pc

b) Probability of any girl being idleExpected number of idle girls

Total number of girls .

0 1

1 12 1

2 1 42 3 0.672 6

P P

C

Hence expected percentage of idle time each girl is 67%.

Problem 25 At a one man barber shop, the customers arrive following poisson process atan average rate of 5 per hour and they are serviced according to exponential distributionwith an average service. Rate of 10 minutes assuming that only 5 seats are available forwaiting customers find the average time a customer spends in the system.

Solution:

This is model / /1 : /M M N FIFO

5 / .hr 1

60 6 510

per hour N

5

6P

0 61

51

1 0.16666 0.2505.1 0.66515

16

NP

SS

LW where

1 6

6

51 6

5 6

6 5 51 1 1

6

n

S

n

L

n

Queuing Theory

12

6.0335 2.015 5 198

1 0.335 0.665

'01 P

6 1 0.2505 4.497

'

1 1.980.441

4.497s SW L hrs

Problem 26 At a railway station, only one train is handled at a time. The railway yard issufficient only for two trains to wait while the other is given signal to leave the station.Trains arrive at the station at an average of 12 per hour. Assuming poisson arrivals andexponential service distribution, find the steady state probabilities for the number oftrains in the system also find the average waiting time of a new train coming into theyard. If the handling rate is reduced to half, what is the effect of the above results?

Solution:

This is Model / /1 : /M M K FIFO

Hence 6 / , 12 / , 3hr hr k

Steady state probability for the number of trains in the system 1 2 3,P P and P .

0 0 1

1

1

n

n KP P when P

4 4

61

0.5 0.512 0.53330.93751 0.56

112

1

1 0 0.2667P P

2

2 0P P

= 0.1334

3

3 0 0.0667P P

Average waiting time of a new train coming in the yard is1

s SW L

where

1

1

1

1

K

s K

K

L

Queuing Theory

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4

4

4 0.560.7333.

12 6 1 0.5

1 1 12 1.05333 5.6004c

0.73330.1309

5.6004sW hrs

Case II

If the handling rate is reduced to half, then 6 / , 6 / 3hr hr and k

Hence1 1

1 4nP

k

1 2 3

1

4P P P

Average waiting time of a new train coring into the yard in

1s

s

LW

3

1.52 2

s

KL Since train

10

11 6 1 4.5

4P

1

1

1.50.333 ( 20 )

4.5s

s

LW hrs or

Problem 27 At a port there are 6 unloading berths and 4 unloading crews. When all theberths are full, arriving ships are diverted to an overflow facility 20 kms down the river.Tankers arrive according to a poisson process with a mean of 1 for every 2 hrs. If it takesfor an unloading crew on the average, 10 hrs to unload a tanker, the unloading timefollows an exponential distribution Determine.

a) How many tankers one of the port on the average?b) How long does a tanker spend at the port as the average?

Solution:1

1 0 0.0361P P

2

2 0

10.0901

2!P P

3

3 0

10.1502

3!P P

4.8214 4 0.00721 3 0.0361 2 0.0901 0.1502 4.3539sL tankers.

b) Time spend by a tanker at the port

Queuing Theory

14

1

1

10

1 c

S S nn

W L Where C c n

1 0 1 2 34 3 2S P P P P

1

4 4 0.00721 3 0.0361 2 0.0901 0.150210

0.35324.3539

12.32500.3532

SW

Problem 28 A petrol pump has 2 pumps. The service time follows the exponentialdistribution with a mean of 4 minutes and ears arrive for service in a poisson process atthe rate of 10 cars per hour. Find the probability that a customer has to want for service.What proportion of time the pumps remain idle?

Solution:

This is model / / : /M M C FIFO

1 410 / 15 / 2

60hr is hr C

P[ a customer has to wait ] P n c

0

! 1

c

c

P

C

2

0

10

152

102! 1

30

P n P where

1

1

00

1 1

! !

n cc

n

cP

n c c p

121 10 1 10 30

11! 15 2! 15 30 10

12 1 1

13 3 2

Probability that a customer has to want in

22

1 132 . 0.1667

2 2 62.3

P n

Queuing Theory

15

The traction of time when the pumps are busy =traffic intensity10 1

30 3c

Hence the pumps remain idle 1

1 21 67%

3 3

Problem 29 A supermarket has two girls serving at the customers. The customers arrivein a Poisson fashion at the rate of 12 per hour. The service time for each customer isexponential with mean 6 minutes. Find

i) The probability that an arriving customer has to want for service.ii) The average number of customers in the system, andiii)The average time spent by a customer in the supermarket.

Solution: This is / / : /M M C FIFO . Here

1 612 / , 10 / , 2

60X hr hr c

i) Probability that an arrival customer has to wait is

0

1

C

P

P n c

Cic

2 1

0

12 1 12 20 1

10 2! 10 20 12 4P

212 1

10 42 0.45

122! 1

20

P n

The probability that an arrival customer has to want=0.45

ii) Average number of customer is the system

1

02

1

. !1

c

sL Pc c

c

3

2

12

1 1 12101.875 2

2.2! 4 10121

20

customer

iii) Average time spent by a customer in the supermarket =Average waiting timeof a customer in the system

Queuing Theory

16

1 1.8750.1563 9.375

12s sW L hours

minutes.

Problem 30 Four counters are being run on the frontier of a country to check thepassports of the tourists. The tourists choose a counter at random. If the arrival at the

frontier is poisson at the rate and the service time is exponential with parameter2

,

find the steady average queue at each counter.

Solution:

This is / / : /M M C FIFO

Average queue length

02

1

. !1

qL Pc c

c

1

3

00

1 1

!

n c

n

cP

n Ci c

12 342 2 1 2

1 2 22 6 4! 2

3

23

2

1 25 3 40.1739

4.4! 23 231

2

qL

Problem 31 In a given M/M/1 queuing system the average arrivals is 4 customers perminutes 0.7 what are

1) Mean number of customers sL in the system

2) Mean number of customers qL in the queue.

3) Probability that the service is idle.4) Mean waiting time sW in the system.

Solution:0.7 0.7

2.33.1 1 0.7 0.3

sL

2.333 0.7 1.633.q sL L

0 1 1.07 0.3P

Queuing Theory

17

2.3330.583

4s

s

LW

Problem 32 A TV repairman finds the time spend on his job has an exponentialdistribution with mean 30’.If he repairs sets in the model in which they come in and ofthe arrival of sets is approximately poisson with an average rate of 10 sets per 8 hoursday, what is the repairman’s expected idle time each day ? How many jobs ahead of theaverage set just brought in?

Solution:

We are given10 5

8 4 sets per hour

160 2

30 sets per hour

Then5

8

The probability of no unit in the queue is 0

5 31 1

8 8P

Hence the idle time for repair man in an 8-hours day3

8 38

hrs

524 1

5 324

sL

jobs = 2 TV Sets (approx)

Problem 33 If for a period of 2 hours in the day trains arrive at the yard every 20minutes but the service time continuous to remain 36 minutes. Calculate the following forthe above period.

i) The Probability that the yard is emptyii) The average number of trains on the assumption that the line capacity of the yard

is limited to 4 trains only.

Solution:

This model is / /1 : /M M K FIFO Hence1

20 per min;

1

36 per min ,

4K .i) Probability that the yard is empty

0 1

1

,

1

KP

5

361

20 0.044736

120

Queuing Theory

18

ii) Average number of trains

1

1

1

1

K

S K

K

L of

5

5

1 365

20 2030 27

1 1 36136 20 20

=3 trains.

Problem 34 Suppose there are 3 typists is a typing pool. Each typist can type an averageof 6 letters / hr. If letters arrive to be typed at the rate of 15 letters/hr

a) What is the probabilities that there is only one letter in the systemb) What is the average number of letters waiting to be typed?c) What is the average time a letter spends in the system?

Solution:

This is / / : /M M S FIFO1

1

00

1 1

! !

n ss

n

sP

n s s

132

0

1 15 1 15 18

! 6 3! 6 18 5

n

n n

11 18

1 2.5 3.125 15.6256 3

0.04494.

a) 1 0

1

!

n

P P Since n sn

1

1 150.04494 0.11235

1! 6

b) Average numbers of letters waiting to be typed

1

02

1

!1

S

qL Pss

s

Queuing Theory

19

4

2

15

1 60.04494

3.3! 151

18

3.51094 .Letters

C)1

s sW L

1qL

1 153.51094 0.40073

15 6hr

24.0438 .

Problem 35 An automobile inspection station has 3 inspection stalls. Assume that canswait in such a way that when a stall becomes vacancy the car at the head of the time pullsup to it. The Station can accommodate at most 4 cars waiting at one time. The arrivalpattern is poisson with a mean of I car every minute during the peak hours. The servicetime is exponential with mean 6 min. Find the average number of customers is the queueduring peak hours and the average waiting time in the queue.

Solution:

This is / / : /M M C K FIFO Model.

3, 7.c k

60 / 10 / .hr hr 1

1

00

1 1

! !

m C m ci k

m m v

Pm C c

132 7

3

0 3

1 1 66 6

! 3! 3

mm

m mm

1

2 3 2 3 41 11 6 6 6 1 2 2 2 2

2! 3!

7 1

7 18 36 31 1141 0.000.8764

i) Average no of customer is the queue.

0 2

11 1

! 1

c

K C K CqL P K C

c

62

3Where

c

Queuing Theory

20

3 4 4

2

2 16 .0.0008764 1 2 4 1 2 2

3! 1 2

3.09ii) Average waiting time in the queue.

1

'

'0

1 c

q q mm

W L where c c m P

2

1

0

10 3 3 mm

m P

0 1 210 3 3 2P P P

10 3 3 0.00087 2 0.005256 0.015768

29.76 30 3.09

0.103 .30

qL hrs

0

1; 3

!

n

nHence P P n cn

1

1

16 0.0008764

1!P 0.005256

2

2

16 0.00087

2!P 0.015768

Problem 36 Patients arrive at a clinic according to Poisson distribution at a rate of 60patients per hour. The waiting room does not accommodate more than 14 patientsInvestigation time per patient in exponential with mean rate of 40 per hour.

a) Determine the effective arrival rate at the clinic.b) What is the probability that an arriving patient will not wait?c) What is the expected time (waiting) until the patient is discharged from the

clinic?

Solution: This is / / : /M M I K FIFO model hence 60 patients / hr,

40 / . 14 1 15hr K

a) Effective arrival rate

101 P

Where 0 1

1

1

KP of

6

601

40 0.000762460

140

Queuing Theory

21

1 40 1 0.0007624 39.9695 / hr

b) An arrived will rut want =there is no patient is the clinic.

P[ a patient will not wait ] 0 0.0007624P

c) Effective waiting time

1S

S

LW

1

1

1

1

K

S K

K

When L

16

16

6016

60 40

40 60 601

40

2 16.00 14 Patient

14

0.3503 21.01839.9695

sW hr or

Problem 37 A group of users in a computer browsing centre has 2 terminals. Theaverage computing job requires 20 min of terminal time and each user requires somecomputation about once every half an hour .Assume that the arrival rate is poisson andservice rate is exponential and the group contains 6 users. Calculate

a) The average number of users waiting to use one of the terminals and in thecomputing job.

b) The total time lost by all users per day when the centre is opened 12 hrs/day.

Solution:

This is / / : /M M S K FIFO Hence1 30 1 1 20 1

;60 2 60 3

2 / , 3 / 2, 6hr hr s k

a) The average number of users waiting to use terminals and in theComputing job is nothing but me average number of uses in the system.

1

0

S

S q nn

L L S S n P

1

1

00

1 1

! !

n s n sS k

n n s

Pn S s

12 2 3 4

2 1 2 1 1 1 11 1

3 2! 3 3 3 3 3

0.5004

Queuing Theory

22

0 2

1 1! 1

s

K S k sqL P y K S Where y

sS

2 4 4

2

12 1 2 130.5004 . 1 43 3 3 32

2!3

0.0796

1

0

s

S q nn

L L S S n P becomes

0 12qL S P P

0 02qL S P P

2

0.0796 2 2 0.5004 0.50043

0.7452 .

b) Average waiting time of a user in the queue qW

1

11

0

0 1

1

1

22 3 2 2 0.5004 0.5004

3

1.9968

1 0.07960.0399

1.9968

q q

S

nn

q q

W L where

S S n P

S P P

W L hr

Problem 38 The railway marshalling yard is sufficient only for trains (there are 11lines one of which is earmarked for the shutting engine to reverse itself from the crest ofthe hump to the rate of the train). Train arrive at the rate of 25 trains per day, inter-arrivaltime and service time follow exponential distribution with an average of 30 minutesDetermine.

a) The probability that the yard is empty.b) The average queue length.

Solution:

This is model / / : /M M K FIFO

Hence 25 trains /day,1

30 0.0333 train /min

0.0174

Queuing Theory

23

10 11 1K

a) Probability that the yard is empty.

0 1

1

1

KP if

11 11

0.01741

1 0.52250.0333 0.47791 0.52250.0174

10.0333

c) Average queue length.

1

1

1

1

K

S K

K

L of

11

11

0.017411

0.0174 0.0333

0.0333 0.0174 0.01741

0.0333

0.0087211.09434 .

0.9992

1.0856