learning outcomes

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Learning Outcomes

• Mahasiswa akan dapat menghitung penyelesaian model antrian tunggal dan ganda dalam berbagai contoh aplikasi..

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Outline Materi:

• Model Antrian Ganda M/M/C• Jaringan Antrian• Contoh Penerapan

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• Type: Multiple servers; single-phase.

• Input source: Infinite; no balks, no reneging.

• Queue: Unlimited; multiple lines; FIFO (FCFS).

• Arrival distribution: Poisson.

• Service distribution: Negative exponential.

M/M/S Model

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M/M/S Equations

λMμMμ

μλ

M!1

μλ

n!1

1P

M1M

0n

n0

Probability of zero people or units in the system:

PM!ML

M

sAverage number of people or units in the system:

PM!MW

M

sAverage time a unit spends in the system:

Note: M = number of servers in these equations

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M/M/S Equations

sq

sq

WW

LLAverage number of people or units waiting for service:

Average time a person or unit spends in the queue:

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M/M/2 Model Equations

(2 + )(2 -)

Ws = 4 42 - 2

Lq = Wq

Wq = 2

Ls = Ws

P0 = 2 - 2 +

Average time in system:

Average time in queue:

Average # of customers in queue:

Average # of customers in system:

Probability the system is empty:

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M/M/2 Example

Average arrival rate is 10 per hour. Average service time is 5 minutes for each of 2 servers.

= 10/hr, = 12/hr, and S=2Q1: What is the average wait in the system?

Ws = 412 4(12)2 -(10)2

= 0.1008 hours = 6.05 minutes

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M/M/2 Example

= 10/hr, = 12/hr, and S=2Q2: What is the average wait in line?

Also note:

so

Ws =1 Wq +

Wq =1 Ws - = 0.1008 - 0.0833 =0.0175 hrs

Wq = (10)2

12 (212 + 10)(212 - 10)= 0.0175 hrs = 1.05 minutes

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M/M/2 Example

= 10/hr, = 12/hr, and S=2Q3: What is the average number of customers in line and in the system?

L q = Wq = 10/hr 0.0175 hr = 0.175 customers

L s = Ws = 10/hr 0.1008 hr = 1.008 customers

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M/M/2 Example

= 10/hr and = 12/hrQ4: What is the fraction of time the system is empty (server is idle)?

= 41.2% of the timeP0 = 212 - 10 212 + 10

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M/M/1, M/M/2 and M/M/3

1 server 2 servers 3 servers

Wq 25 min. 1.05 min. 0.1333 min. (8 sec.)

0.417 hr 0.0175 hr 0.00222 hr

WS 30 min. 6.05 min. 5.1333 min.

Lq 4.167 cust. 0.175 cust. 0.0222 cust.

LS 5 cust. 1.01 cust. 0.855 cust.

P0 16.7% 41.2% 43.2%

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Service Cost per Day

= 10/hr and = 12/hr Suppose servers are paid $7/hr and work 8 hours/day and the marginal cost to serve each customer is $0.50.

M/M/1 Service cost per day = $7/hr x 8 hr/day + $0.5/cust x 10 cust/hr x 8 hr/day = $96/day

M/M/2 Service cost per day = 2 x $7/hr x 8 hr/day + $0.5/cust x 10 cust/hr x 8 hr/day = $152/day

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Customer Waiting Cost per Day

= 10/hr and = 12/hr Suppose customer waiting cost is $10/hr.M/M/1 Waiting cost per day = $10/hr x 0.417 hr/cust x 10 cust/hr x 8 hr/day = $333.33/dayM/M/1 total cost = 96 + 333.33 = $429.33/day

M/M/2 Waiting cost per day = $10/hr x 0.0175 hr/cust x 10 cust/hr x 8 hr/day =$14/dayM/M/2 total cost = 152 + 14 = $166/day

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Unknown Waiting Cost

Suppose customer waiting cost is not known = C.

M/M/1 Waiting cost per day = Cx 0.417 hr/cust x 10 cust/hr x 8 hr/day = 33.33C $/dayM/M/1 total cost = 96 + 33.33CM/M/2 Waiting cost per day = Cx 0.0175 hr/cust x 10 cust/hr x 8 hr/day =1.4C $/dayM/M/2 total cost = 152 + 1.4C M/M/2 is preferred when 152 + 1.4C < 96 + 33.33C or

C > $1.754/hr

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M/M/2 and M/M/3

Q: How large must customer waiting cost be for M/M/3 to be preferred over M/M/2?M/M/2 total cost = 152 + 1.4C M/M/3 Waiting cost per day = Cx 0.00222 hr/cust x 10 cust/hr x 8 hr/day = 0.1776C $/dayM/M/3 total cost = 208 + 0.1776C

M/M/3 is preferred over M/M/2 when 208 + 0.1776C < 152 + 1.4C

C > $45.81/hr

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= Mean number of arrivals per time period.– Example: 3 units/hour.

= Mean number of arrivals served per time period.– Example: 4 units/hour.

• 1/ = 15 minutes/unit.

Remember: & Are Rates

If average service time is 15 minutes, then μ is 4 customers/hour

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• M/D/S

– Constant service time; Every service time is the same.

– Random (Poisson) arrivals.

• Limited population. – Probability of arrival depends on number in service.

• Limited queue length.

– Limited space for waiting.

• Many others...

Other Queuing Models

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