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Sloan [31]
3 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]3 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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An asphalt concrete mixture includes 94% aggregates by weight. The specific gravity of aggregate and asphalt are 2.7 and 1.0, re
riadik2000 [5.3K]

Answer:

The correct solution is "5.74%".

Explanation:

The given values are:

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G_{agg}=2.7

Gravity of asphalt,

G_{asp}=1.0

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W_{agg}=0.94 \ W_m

We know that,

W_{asp}=W_m-W_{agg}

        =0.06 \ W_m

Now,

The theoretical specific gravity of mix,

⇒ G_t=\frac{W_{agg}+W_{asp}}{\frac{W_{agg}}{G_{agg}} +\frac{W_{asp}}{G_{asp}} }

By putting the values, we get

         =\frac{0.94 \ Wm+0.06 \ Wm}{\frac{0.94 \ Wm}{2.7} +\frac{0.06 \ Wm}{1} }

         =2.45

hence,

The percentage of voids will be:

⇒  %V = \frac{G_t-G_m}{G_t}\times 100

           = \frac{2.45-2.317}{2.45}\times 100

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8 0
3 years ago
forty gal/min of a hydrocarbon fuel having a spesific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fue
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Answer: 131.75minutes

Explanation:

First if all, we've to find the density of liquid which will be:

= Specific gravity × Density to pure water

= 0.91 × 8.34lb/gallon

= 7.59lb/gallon

Then, the volume that's required to fill the tank will be:

= Load limit/Density of fluid

= 40000/7.59

= 5270.1gallon

Now, the time taken will be:

= V/F

= 5270.1/40

= 131.75min

It'll take 131.75 minutes to fill the tank in the truck.

5 0
3 years ago
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devlian [24]

Answer:

The phasor technique can't be applied directly in the following cases:

a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)

b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)

c)  -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)

d)  75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)

Explanation:

For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.

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This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function e^{j(wt+\alpha)}), so it is not possible to combine with functions expressed as a sine, even though both  have  the same frequency.

If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:

sin (ωt) = cos (ωt-90º)

This means that in order to be able to represent a sine function  as a cosine, we need to rotate it 90º in the plane clockwise.

This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).

8 0
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