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Operations Research- Queuing Model

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QUEUING THEORY

HISTORY
• Queuing theory had its beginning in the research work of a Danish engineer named A.K. Erlang. • In 1909, Erlang experimented with fluctuating demand in telephonic traffic.

• 8 years later, he published a report addressing the delays in automatic dialing equipment.
• At the end of World War II, Erlang’s early work was extended to more general problems and to business applications of waiting lines.

M/M/1 SINGLE - CHANNEL WITH POISSON
Azenith Cayetano

THE M/M/1 NOTATION REPRESENTS:
Arrival distribution Service time distribution M = Poisson M = Exponential

No. of service channels open m = 1

QUEUING EQUATIONS:

λ = mean number of arrivals per time period (for example, per hour) μ = mean number of people or items served per time period

SAMPLE PROBLEM 1

Angie is the Branch Manager of Citibank Lagos and she wants to improve the service of the bank by reducing the average waiting time of the bank’s clients. She was able to determine the average arrival and the average number of clients serviced per hour.

 How many clients are in the bank at any given time? How much time does a client spend in the bank? How many clients are waiting to be served? How much time does a client spend waiting?  What is the probability that the teller is busy? What is the probability that there are no clients?

DATA TABLE
Given Description Value

m λ μ

Number of tellers
Arrivals per hour Serviced per hour

1
11 12

1. Compute the average number of clients in the system (L) at any given time: L 11 = λ / (μ - λ) = 11 / (12 – 11)

= 11 clients are in the bank on the average
2. Compute the average number of hours a client spends in the system (W): W 1 = 1 / (μ - λ) = 1 / (12 – 11) = 1 hour is the average time a client is inside the bank

3. Compute the average number of clients waiting in line (Lq):
Lq 10.0833 = λ2 / μ (μ - λ) = 112 / 12(12 – 11) = 10.08 clients are waiting in line on the average

COMPUTATION:
4. Compute the average number of hours a client waits in line (Wq): Wq 0.9167 = λ / μ (μ - λ) = 11 / 12(12 – 11)

= 0.92 hour is the average time a client is in line
5. Compute the probability that the teller is busy (ρ): (ρ) 0.9167 =λ/μ = 11 / 12 = 92% is the probability that the teller is busy 6. Compute the probability that there are no clients in the bank (Po): Po 0.0833 = 1 - (λ / μ) = 1 – (11 / 12) = 8% is the probability that the teller is idle

STATISTICS TABLE
Equations L W Description Numbers of clients in the system Average hours client spends in the system Value 11 1

Lq Wq
(ρ) Po

Average clients waiting in line Average waiting hours per client
Probability that the teller is busy Probability that there are no clients

10.0833 0.9167
0.9167 0.0833

SAMPLE PROBLEM 2

Angie, the Branch Manager of Citibank Lagos in Example 1, knows that there is also an opportunity cost for clients who are idle while waiting in line. She was able to determine the teller’s labor cost as well as the average opportunity cost of clients who are waiting.

 How much is the total cost per shift?

DATA TABLE
Given Description Value

Cs

Service cost per teller ($/hour)
Cost of waiting ($/hour) Working hours per shift

5
6 8

COMPUTATION:
1. Compute the total service cost (Ts): Ts / hour = mCs = 1(5) Ts / shift = Ts / hour(h) = 5(8) 2. Compute for the total waiting cost (Tw): Tw / hour = λWqCw = 11(0.9167)(6) Tw / shift = Tw / hour(h) = 60.50 (8) 3. Compute for the total cost (Tc): Tc / hour Tc / shift = Ts / hour + Tw / hour = Ts / shift + Tw / shift

Cw h

=5 = 40

= 60.50 = 484.00

= 5.00 + 60.50 = 40.00 + 484.00

= 65.50 = 524.00

COST TABLE
Cost Total service cost ($) Total waiting cost ($) Per Hour 5.00 60.50 Per Shift 40.00 484.00

Total cost ($)

65.50

524.00

While total service cost is only $40.00 per shift, Angie should expect a much larger total cost of $524.00 because of the opportunity loss of $484.00.

M/M/m SYSTEM
John Sy

The M/M/m notation represents:
Arrival distribution Service time distribution No. of service channels open M = Poisson M = Exponential m=2

Angie, the branch manager of Citibank Lagos wants to improve the service of the bank by hiring an additional teller. She observed that the average arrival and average number of clients serviced per teller per hour remains the same. The figures are summarized below.

 What is the probability that there are no clients? How many clients are in the bank at given time?  How much time does a client spend in the bank? How many clients are waiting to be served? How much time does a client spend waiting? What is the probability that a teller is busy?

DATA TABLE
1. Determine the probability that there were no clients in the bank (Po) from appendix C1: Po = Po (m, λ, µ) 0.3714 = Po (2, 11, 12) Given m λ μ

Description
Number of tellers Arrivals per hour Serviced per hour per teller

Value
2 11 12

= 37% is the probability that both tellers are idle

COMPUTATION
2. Compute the average number of clients in the system (L) at any given time: L = λµ (λ/µ) 2 Po /(2µ - λ) 2 + λ/ µ 1.1604 = 11(12) (11/12) 2 (0.3714)/(2(12)- 11)2 + 11/12

= 1.16 clients are in the bank on the average
3. Compute the average hours a client spends in the system (W): W = L/λ

0.1055 = 1.1604/11
= 0.11 hour is the average time a client is inside the bank

COMPUTATION
4. Compute the number of clients waiting in line (Lq): Lq = L - λ /µ 0.2438 = 1.1604 - 11/12
= 0.24 clients are waiting in line on the average

5. Compute the average number of hours a client waits in line (Wq):
Wq = Lq/λ 0.0222 = 0.2438/11 = 0.02 hour is the average time a client is in line

6. Compute the probability that the teller is busy (ρ): ρ= λ /mµ
0.4583 = 11/2 (12) 46% is the probability that a teller is busy

STATISTICS TABLE

M/D/1 CONSTANT SERVICE TIME MODEL
Ma. Clarissa Bondal

M/D/1
Some service systems have constant service times instead of exponentially distributed times. When customers or equipment are processed according to fixed cycle, as in the case of an automated teller machine, automatic car wash or an amusement park ride, constant service rate are appropriate. Since the constant rates are certain, the values for ave. length of the queue (Lq), ave. waiting time in the queue (Wq), no. of customers in the system (L), and ave. time in the system (W) are always less than they would be in the other models, which have variable service times. In the constant service rate model, both the ave. queue length and average waiting time in the queue are halved.

THE M/D/I NOTATION REPRESENTS:
Arrival distribution Service time distribution No. of service channels open M = Poisson D = Constant m=1

1. Ave. length of the queue : Lq = _____λ²____ 2μ(μ – λ)

2. Ave. waiting time in the queue: Wq = _____λ_____ 2μ(μ – λ)

M/D/1
3. Ave. number of customers in the system : L = Lq + _λ_ μ 4. Ave. time in the system :
W = Wq + _1_ μ

SAMPLE PROBLEM 1

Angie, the Branch manager of Citibank Lagos, wants to improve the service of the bank replacing the teller with an ATM. She observed that the average arrival and the average number of clients serviced per hour remain the same. Because an ATM processes according to a fixed cycle, she presumed that the service rate distribution is now constant. The figures are summarized in the data table.  How many clients are waiting to be served? How much time does a client spend waiting? How many clients are at the ATM at any given time? How much time does a client spend at the ATM?  What is the probability that the ATM is occupied? What is the probability that there are no clients?

DATA TABLE
Given
m λ μ

Description
Number of ATMs Arrivals per hour Serviced per hour

Value
1 11 12

Computation:
1) Compute the average number of clients waiting in line (Lq): Lq = _____λ²____ 2μ(μ – λ) = 5.0417 clients are waiting in the line on the average 2) Compute the average number of hours a client waits in line (Wq): Wq = _____λ_____ 2μ(μ – λ) = 0.45836 or .46 hour is the average time a client is in line

3 ) Compute the average number of clients in the system (L) at any given time: L = Lq + = 5.0417 + 11/12 = 5.96 clients are the ATM on the average. 4) Compute the average number of hours a client spends in the system (W) W = Wq + = .4583 + 1/12 = 0.5417 or .54 hour is the average time a client is at the ATM

5) Compute the probability that the ATM is busy (ρ): ρ= = 11/12 = 92% is the probability that the ATM is busy 6) Compute the probability that there are no clients at the ATM (Po): Po = 1 – ρ = 1 - .92 = 8% is the probability that the ATM is idle

SAMPLE PROBLEM 2 (BOARD WORK)

M/M/1 WITH FINITE SOURCE
Eunice de Belen

VS.

What is the difference?
There is now a dependent relationship between the length of the queue and the arrival rate. FINITE CALLING POPULATION MODEL with the assumptions: Arrival distribution Service time distribution No. of service channels open Additional assumption M = Poisson M = Exponential m=1 N= finite size of population

10 hours?

Angie, the branch manager of Citibank Lagos wants to improve the service of the bank’s 5 existing ATMs. She observed that on the average, an ATM breaks down every 100 hours and takes 10 hours to repair by the bank’s lone technician.  What is the breakdown rate of the ATM? What is the service rate of the technician? What is the probability that there are no repairs going on? How many ATMs are waiting to be repaired?  How many ATMs are at the technician’s schedule at any given time? How much time does an ATM spend waiting to be repaired? How much time does an ATM spend at the technician’s schedule? What is the effective breakdown rate? What is the probability that the technician is busy?

DATA TABLE
Given m N x y Description Number of technicians Number of ATMs Hours of use per breakdown Repair hours per breakdown Value 1 5 100 10

COMPUTATION:
1. Compute the average number of breakdowns per hour: λ = 1/x 0.01 = 1/100 = 0.01 ATM is breaking down per hour on the average 2. Compute the average number of ATMs repaired per hour (μ): μ = 1/y 0.1 = 1/10 = 0.1 ATM is repaired per hour on the average

3. Determine the probability that there are no repairs being done (P₀) from Appendix C2: P₀ 0.564 = P₀ (m, N, λ, μ) = P₀ (1, 5, 0.01, 0.1) = 56% is the probability that the technician is idle

COMPUTATION
4. Compute the average number of ATMs waiting to be repaired (Lq): Lq 0.204 = N – [(λ + μ)/ λ] (1 – P₀) = 5 – [(0.01 + 0.1)/0.01] (1 – 0.564) = 0.20 ATM is waiting to be repaired on the average

5. Compute the average number of ATM in the technician’s schedule (L) at any given time: L = Lq + (1- P₀) 0.64 = 0.204 + (1 – 0.564) = 0.64 ATM is the schedule of the technician on the average

6. Compute the average number of hours an ATM wants to be repaired (Wq):

Wq = Lq/(N – L)λ 4.6789 = 0,204/(5 – 0.64)0.01 =4.68 hours is the average time an ATM waits to be repaired
7. Compute the average number of hours an ATM in the technician’s schedule (W) W = Wq + 1/ μ 14.6789 = 4.6789 + 1/0.1 = 14.68 hours is the average time an ATM is in the technician’s schedule

8. Compute the effective breakdown rate (r): r = λ ( N – L) 0.0436 = 0.01 (5 – 0.64) = 4% is the effective breakdown rate of an ATM 9. Compute the probability that the technician is busy (ρ): ρ 0.436 = r/μ = 0.0436/0.1 = 44% is the probability that the technician is busy

Statistics Table
Equations
λ μ

Description
Breakdown per hour ATMs repaired per hour

Value
0.01 0.1

P₀
Lq L Wq W r ρ

Probability that there are no repairs
Average ATM waiting for repair Number of ATMs in the system Average waiting hours per ATM Average hours ATM spends in system Effective breakdown rate Probability that the technician is bury

0.564
0.204 0.64 4.6789 14.6789 0.0436 0.436

SAMPLE PROBLEM 2

Angie, the branch manager of Citibank Lagos knows that the opportunity cost of an ATM which is not functioning is high. She was able to determine the labor cost of the technician as well as the opportunity cost of an ATM while waiting to be repaired.

 How much is the total cost per shift?

DATA TABLE
Given
Cs

Description
Service cost per technician ($/hour) Cost of downtime ($/hour) Working hours per shift

Value
5 60 8

COMPUTATION:
1. Compute the total cost (Ts): Ts/hour Ts/shift = MCs = Ts/hour(h)

Cw h

= 1(5) = 5(8)

=5 = 40

2. Compute the total waiting cost (Tw): Tw/hour Tw/shift = WqL = Tw/hour(h) = 60(0.64) = 38.40(8) = 38.40 = 307.20

3. Compute the total cost (Tc): Tc/hour Tc/shift = Ts/hour + Tw/hour = Ts/shift + Tw/shift = 5.00 + 38.40 =40.00 + 307.20 = 43.20 = 347.20

COST TABLE
Costs Total Service Cost ($) Total Waiting Cost ($) Total Cost ($) Per Hour 5.00 38.40 43.40 Per Shift 40.00 307.20 347.20

While the total service cost is low at $40 per shift, Angie should expect a higher total cost of $347.20 because of the opportunity loss of $307.20.

THANK YOU

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...1. Executive summary The main purpose of this report is to research and analyse the role that the utilisation of information and communication technology plays in the business affairs of Argos Ltd (Britain’s biggest retailer of home-goods). This report provides the reader with information on such aspects as: • the way that Argos conducts its business • the role that ICT plays in the everyday business affairs of Argos • background information about Argos, its allies and competitors • facts and figures on Argos`s sales, revenues and customers` buying habits After understanding the background and position on the market of Argos, the report evaluates the description of the organisation`s arrangement of operations, by adopting a customer perspective using the `customer perspective framework` while highlighting the role that technology plays within it. The last part of this report is dedicated to the competitive analysis of the organisation, in means of its relative success, by applying the framework of SWOT analysis and Porter`s 5 Forces model. Comparison of the organisation is also carried out by using the examples that are widely used in the sphere of ICT. Suggestions for improvement were given where relevant. Various credible informational resources were used it this report, including academic textbooks and journal articles, research reports and statistical data and numerous online sources. To assess and understand the current trends and developments...

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Warehouse Design

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