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Sloan [31]
2 years ago
14

An airline ticket counter forecasts that 220 people per hour will need to check in. It takes an average of 2 minutes to service

a customer. Assume that arrivals are poisson distributed and service times exponentially distributed and that customers wait in a single queue for the first available agent.
a) If we want the average time a customer spends in the queue and in service to be 10 minutes or less, how many ticket agents should be on duty?
b) There airline wants to minimize the cost of idle ticket agents and the cost of customers waiting in the queue. The salary of a ticket agent is £12 per hour and the cost of a customer waiting in the queue is £5 per customer per hour. How many ticket agents should be on duty?​​
Engineering
1 answer:
Nadusha1986 [10]2 years ago
6 0

A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents

B) The number of agents required on duty to reduce cost = 9 agents

<u>Given data : </u>

Arrival rate of customers ( β ) = 220 per hour

Service rate ( mu ) = 60 minutes / 2 minutes = 30 customer per hour

utilization ( rho ) = 220 / 30 ≈ 7

at least 8 server personnel are required for stability of the queue

A<u>) Determine the number of agents required to achieve a wait time of 10 minutes or less per customer</u>

waiting time = 10 - 2 = 8 minutes

number of customers waiting ( ∝ ) = 7 and required server = 8

assuming   Lq = 5.2266

Hence the waiting time in line = Lq / arrival rate

                                                  = 5.2266 / 220 = 0.0238 hour

                                                  = 0.0238 * 60 = 1.428 minutes

Since the waiting time ( 1.428 minutes ) is less than the original waiting time ( 2 minutes ) the number of agents that will achieve a wait time of 10 minutes or less is = 8 agents

<u>B) Determine the number of</u><u> ticket agents</u><u> that should be on duty to minimize cost </u>

salary of ticket agent = £12 per hour

cost of customer waiting in queue = £5 per hour per customer

<em> </em><u>i) When 8 agents are used </u>

waiting time of customers = 0.0238 * 220 = 5.236

waiting cost for customers = 5.236 * 5 = £26.18

employee cost = 8 * 12 = £96

∴ Total cost = 96 + 26.18

                    = £ 122.18

<u>ii) When 9 agents are used </u>

waiting time for customers = 0.0074 * 220 = 1.628

Wq = 1.6367 / 220 = 0.0074

waiting cost for customers = 1.6367 * 5 = £ 8.1835

assuming Lq = 1.6367

employee cost = 9 * 12 = £ 108

∴ Total cost = 108 + 8.1835 = £ 116.18

From the calculations in ( i ) and ( ii ) the Ideal number of ticket agents that should be on duty to minimize cost should be 9 agents.

Hence we can conclude that A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents and The number of agents required on duty to reduce cost = 9 agents.

Learn more about cost reduction : brainly.com/question/14115944

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A wooden pallet carrying 540kg rests on a wooden floor. (a) a forklift driver decides to push it without lifting it.what force m
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Answer:

The appropriate solution is "1481.76 N".

Explanation:

According to the question,

Mass,

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Coefficient of static friction,

\mu_s = 0.28

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The applied force will be:

⇒ F=\mu_s mg

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8 0
2 years ago
Using the results of the Arrhenius analysis (Ea=93.1kJ/molEa=93.1kJ/mol and A=4.36×1011M⋅s−1A=4.36×1011M⋅s−1), predict the rate
uysha [10]

Answer:

k = 4.21 * 10⁻³(L/(mol.s))

Explanation:

We know that

k = Ae^{-E/RT} ------------------- euqation (1)

K= rate constant;

A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;

E = activation energy = 93.1kJ/mol;

R= ideal gas constant = 8.314 J/mol.K;

T= temperature = 332 K;

Put values in equation 1.

k = 4.36*10¹¹(M⁻¹s⁻¹)e^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}

k = 4.2154 * 10⁻³(M⁻¹s⁻¹)

here M =mol/L

k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)

 or

k = 4.21 * 10⁻³((L/mol)s⁻¹)

or

k = 4.21 * 10⁻³(L/(mol.s))

3 0
3 years ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and
k0ka [10]

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

z1= \frac{\left(150-137\right)}{27.7}=0.4693

z2=\frac{\left(201-137\right)}{27.7}=2.3105

P(0.4693

B.)

z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289

\\P(2.9309

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

3 0
3 years ago
Read 2 more answers
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