lec12 mm1k queueing system2

7/31/2019 Lec12 MM1k Queueing System2

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OR372-Dr.Khalid Al-Nowibet 1

4.2 M/M/1/k Queueing Model

Characteristics1. Interarrival time is exponential with rate

Arrival process is Poisson Process with rate

2. Interarrival time is exponential with rate

Number of services is Poisson Process with rate

3. Single Server

4. System size is finite = k

5. Queue Discipline : FCFS

NotationM / M / 1 / k / FCFS


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Steady-State Distribution

State of the systemsystem is in state n if there are n customers in the

system (waiting or serviced)

Let Pnbe probability that there are n customers in the

system in the steady-state. n = 0 , 1 , 2 , 3 , , k


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Steady-State Distribution

Rate Diagram:1. If system changes state, where to go?

2. How fast the system changes state?

0 1 2 k-1 k. . .


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Steady-State DistributionBalance Equations:

For each state n:

Average Rate out of state n =

Average Rate in to state n =

Average

Rate in toState n

Average

Rate out ofState n =

k n}stateinPr{systemk)n(rates

k

k}stateinPr{systemn)k(rates




Steady-State DistributionBalance Equations:

n = 0 P0 = P1n = 1 P1 + P1 = P0 + P2 (+)P1 = P0 + P1n = 2 P2 + P2 = P1 + P3 (+)P2 = P0 + P3n = 3 P3 + P3 = P2 + P4 (+)P3 = P2 + P4. . . . . . . . . . .

n = k Pk= Pk-1 Pk= Pk-1

0 1 2 k-1 k. . .

AverageRate in to

State n

AverageRate out of

State n=




Steady-State DistributionSolution of Balance Equations:

P0 = P1(+)P1 = P0 + P2(+)P2 = P1 + P3. . . . . . . . . . .

Pk= Pk-1

Eq-1 P0 = P1Eq-2 (+)P1 (/)P1 = P2 P1 = P2Eq-3 (+)P2 (/)P2 = P3 P2 = P3

. . . . . . . . . . .

Eq-k Pk-1 = Pk


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Make all equations functions of P0 only:

Eq-1 P0 = P1 P1 = (/)P0Eq-2 P1 = P2 from Eq-1 P2 = (/)

2 P0Eq-3 P2 = P3 from Eq-2 P3 = (/)

3P0

. . . . . . . . . . .

Eq-k Pk-1 = Pk from Eq-(k-1) Pk= (/)kP0


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Computing P0 :

P0 + P1 + P2 + P3 + +Pk= 1

P0 + (/)P0 + (/)2P0 + (/)

3P0 + + (/)

kP0 = 1

P0 [ 1 + (/)+ (/)2 + (/)3 + + (/)k] = 1

P0 = [1 + (/)+ (/)2

+ (/)3

+ + (/)k

]1

1Pn =n


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Computing P0 :

All Pn are functions of P0 . Then Pn > 0 if and only if P0 > 0

Then P0 > 0 for any value of and since is finite sum

( )=

=k

n

n

0

01P

( )

=

0n

n


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Pn = P0

n = 1,2,3, ,k

For any value of and

n

=

=k

n

n

n

0

nP


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Pn = P0

n = 1,2,3, ,k

for any value of ( can be > 1)

n

1n 11P + = k

n

= 10 11P + = k


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why (/) can be >1 ??

If system is full arrival rate = 0Number of customers in system does not go to


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Performance Measures

In steady state

e , , P0LB = E[busy servers] = E[#Cust. in service]Ls = Lq + LB

Ws = Wq + (1/)Ls = Ws

Lq = WqLB = WB

Know 4 measures all measures are known

System isin Steady Stead


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1. Effective Arrival Rate e:

Finite size = k

Total

Arrivals

e

b

= e + b

1

Rate of Enteringthe system

Rate of Not Enteringthe system


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1. Effective Arrival Rate e:

e = . Pr{an arrival enters the system}

= . Pr{system is not full}= . [P0 + P1 + P2 + + Pk1 ]= . [1 Pk] = Through-put Rate

b = . Pr{an arrival cant enter the system}= . Pr{system is full}= . Pk


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2. Average Customers in System Ls:

3. Average Busy servers LB:

LB

= E[busy servers] = E[#Cust. in service]

LB = 0 . P0 + 1 (P1 + P2 + P3 + ) = 1 P0

=

=k

0n

ns PnLFinite sum

11

1

1 +

= k

=


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4. Utilization of the System U:

U = Pr{ n > 0 } = P1 + P2 + P3 + + Pk= 1 P0

5. Average Customers in Queue Lq:

Lq = Ls

LBor

Lq = 0.(P0+P1) + 1.P2 + 2.P3 + + (k1)Pk

=


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6. Average Waiting time in System Ws:

Ls = e .Ws

7. Average Time Spent in Queue Wq:

Lq = e .Wq

LsWs =

e

LqWq =

e


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ExampleConsider the car-wash station in Example-2. Assume now that the itis not allowed for care to wait on the side of the road. So, station has

made some modifications so that 6 cars can wait inside the station(See diagram). Also, a driver is hired to move cars from parking tothe machine. The driver takes an average of 2 minutes to move the carto the machine.

Car-Wash

Machine

Car-Wash Station

N

o

Parking


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ExampleAssuming that the arrival rate is 9 cars per hour and the washing timeis 6 minutes. Also, assume Poisson arrivals and exponential service.

Answer the following questions in steady-state:

1. What is the average number of cares waiting in station?2. If the car wash costs 15 SR and the station works from 8:00am to

8:00 pm how much money the collects per day on average? Howmuch the station losses?

3. On average How much it takes for a customer until he leaves with

his car washed?4. The management decided to buy another machine if the oldmachine works more than 85% of the time. Will the managementbuy a new machine?


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Example = 9 cars/hourE[S]= E[driving]+E[washing] = 2 min + 6 min = 8 min

= 1/8 cars/hr = 7.5 cars/hr single machine = 9/7.5 = 1.2k =(max. # waiting) + (max. # in service) = 6+1= 7 (max. system size)

M/M/1/k queueing system

n = 0,1,2, ,7

=

= k

n

n

n

0

nP=


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Example = 9 cars/hour = 7.5 cars/hr M/M/1/k=7 system

1. Average number of cares waiting in station = Lq = Ls (1P0)

Lq = Ls (1P0) = 4.424 (10.061) = 3.485 cars

n 0 1 2 3 4 5 6 7

n 1 1.2 1.44 1.73 2.07 2.49 2.99 3.580.217

1.520

16.50Pn 0.061 0.073 0.087 0.105 0.126 0.151 0.181 1.00

nPn 0.000 0.073 0.175 0.314 0.503 0.754 1.086 4.424


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2. car wash costs = 15 SRworks hours = 12 hours

E[money collected per day] =(15SR) E[cars washed per day] (12hr)

E[cars washed per day] = e = (1 P7) = 9 (10.217) = 7.047 car E[money collected per day] =(15SR)(7.047)(12hr) = 1268.46 SR

E[money lost per day] =(15SR) E[cars not washed per day] (12hr)E[cars not washed per day] = b = . P7 = 9 (0.217) = 1.953 car E[money lost per day] =(15SR)(1.953)(12hr) = 351.54 SR


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3. E[time until customer leaves with his car washed] = WsWs = Ls /e = 4.424/7.047 = 0.6278 hrs

4. The management decided to buy another machine if the oldmachine works more than 85% of the time. Will the managementbuy a new machine?

Percentage of working time for the old machine = Pr{ n > 0 } = UU = 1 P0 = 1 0.061 = 0.939 > 0.85 Buy a new machine

lec12 mm1k queueing system2

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