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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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Determine the speed of sound in air at 400 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3
Reika [66]

Answer:

\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s

Ma= \frac{310 m/s}{400.899 m/s}= 0.773

Explanation:

For this case we have given the following data:

T= 400 K represent the temperature for the air

v = 310 m/s represent the velocity of the air

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R = 0.287 KJ/ Kg K represent the gas constant  for the air

And we want to find the velocity of the air under these conditions.

We can calculate the spped of the sound with the Newton-Laplace Equation given by this equation:

\alpha = \sqrt{\frac{K}{\rho}}=\sqrt{k RT}

Where K = is the Bulk Modulus of air, k is the adiabatic index of air= 1.4, R = the gas constant  for the air, \rho the density of the air and T the temperature in K

So on this case we can replace and we got:

\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s

The Mach number by definition is "a dimensionless quantity representing the ratio of flow velocity past a boundary to the local speed of sound" and is defined as:

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Q4 a.
dedylja [7]

The Java program that accepts a matrix of M × N order and then interchanges diagonals of the matrix is given below:

<h3>Steps:  </h3>
  • 1. We can only interchange diagonals for a square matrix.
  • 2. Therefore, we would have to create a square matrix of size [M × M].
  • 3. We would check whether the matrix is a square matrix or not. If the matrix is square then follow step 3 else terminate the program.
  • 4. Apply logic for interchange diagonal of the matrix some logic is given below.

<h3>Java Code</h3>

//  Java Program to Accept a Matrix of Order M x N &

//  Interchange the Diagonals

import java.util.Scanner;

public class InterchangeDiagonals {

   public static void main(String[] args)

   {

       // declare variable

       int m, n, i, j, temp;

       // create a object of scanner class

       Scanner sc = new Scanner(System.in);

       System.out.print("Enter number of rows ");

       // take number of rows

       m = sc.nextInt();

       System.out.print("Enter number of columns ");

       // take number of columns

       n = sc.nextInt();

       // declare a mxn order array

       int a[][] = new int[m][n];

       // if block it's execute when m is equals to n

       if (m == n) {

           System.out.println(

               "Enter all the values of matrix ");

           // take the matrix inputs

           for (i = 0; i < m; i++) {

               for (j = 0; j < n; j++) {

                   a[i][j] = sc.nextInt();

               }

           }

           System.out.println("original Matrix:");

           // print the original matrix

           for (i = 0; i < m; i++) {

               for (j = 0; j < n; j++) {

                   System.out.print(a[i][j] + " ");

               }

               System.out.println("");

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           for (j = 0; j < m; j++) {

               temp = a[j][j];

               a[j][j] = a[j][n - 1 - j];

               a[j][n - 1 - j] = temp;

           }

           System.out.println(

               " after interchanging diagonals of matrix ");

           // print interchanged matrix

           for (i = 0; i < m; i++) {

               for (j = 0; j < n; j++) {

                   System.out.print(a[i][j] + " ");

               }

               System.out.println("");

           }

       }

       // else block it's only execute when m is not equals

       // to n

       else {

           System.out.println("Rows not equal to columns");

       }

   }

}

Read more about java programming here:

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larisa86 [58]

Answer:

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Explanation:

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