Question

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Problem-1:

At a single-phase, multiple-channel service facility, customers arrive randomly. Statistical analysis of past data shows that the interarrival time has a mean of 20 minutes and a standard deviation of 4 minutes. The service time per customer has a mean of 15 minutes and a standard deviation of 5 minutes. The waiting cost is $200 per customer per hour. The server cost is $25 per server per hour. Assume general probability distribution and no buffer capacity restriction.

a. Find the optimal number of servers to be employed to minimize the total of waiting and server costs. (Ans: Cost per hour with one server=$59.00; Cost with 2 servers = $52.19; Cost with 3 servers = $75.40: So two servers are optimal.)

b. Find the average waiting time and the average total time through the system for the optimal case. (waiting: 0.2188 min; Total: 15.2118 min)

c. Find the cost per hour, average waiting time, and average flow time for one server if the probability distributions for the interarrival time and service time are assumed to be exponential and the mean values remain the same. The cost data remain the same. Use manual calculations. (Ans: Cost per hour=$475.00, Waiting time=45.00 min, Flow time=60.00 min).

Problem-2:

At a single-phase, multiple-channel service facility, customers arrive randomly. Statistical analysis of past data shows that the interarrival time has a mean of 12 minutes and a standard deviation of 6 minutes. The service time per customer has a mean of 10 minutes and a standard deviation of 4 minutes. The waiting cost is $100 per customer per hour. The server cost is $20 per server per hour. Assume general probability distribution and no buffer capacity restriction.

a. Find the optimal number of servers to be employed to minimize the total of waiting and server costs. (Ans: Cost per hour with one server=$105.42; Cost with 2 servers = $44.12; Cost with 3 servers = $60.76: So two servers are optimal.)

b. Find the average waiting time and the average total time through the system for the optimal case. (waiting: 0.4940 min; Total: 10.4940 min)

c. Find the cost per hour, average waiting time, and average flow time for one server if the probability distributions for the interarrival time and service time are assumed to be exponential and the mean values remain the same. The cost data remain the same. Use manual calculations. (Ans: Cost per hour=$436.67, Waiting time=50.00 min, Flow time=60.00 min).

Problem-3:

At a single-phase, multiple-channel service facility, customers arrive randomly. Statistical analysis of past data shows that the interarrival time has a mean of 5 minutes and is exponentially distributed. The service time per customer has a mean of 4 minutes and is exponentially distributed. The arrival buffer capacity is 5 customers. The lost customer cost due to blocking is $200 per customer. The waiting cost is $100 per customer per hour. The server cost is $20 per server per hour. Use Performance.xls spreadsheet for queue analysis and manual calculation for costs.

a. Find the optimal number of servers to be employed to minimize the total of lost customer cost, waiting cost, and server cost. (Ans: Cost per hour with one server=$318.77; Cost with 2 servers = $58.00; Cost with 3 servers = $62.00: So two servers are optimal.)

b. Find the average overall waiting time and the average total flow time through the system for the optimal case. (Ans: Waiting time: 0.73 min; Flow time: 4.73 min)

Problem-4:

At a single-phase, multiple-channel service facility, customers arrive randomly. Statistical analysis of past data shows that the interarrival time has a mean of 10 minutes and is exponentially distributed. The service time per customer has a mean of 8 minutes and is exponentially distributed. The arrival buffer capacity is 8 customers. The lost customer cost due to blocking is $500 per customer. The waiting cost is $200 per customer per hour. The server cost is $40 per server per hour. Use Performance.xls spreadsheet for queue analysis and manual calculation for costs.

a. Find the optimal number of servers to be employed to minimize the total of lost customer cost, waiting cost, and server cost. (Ans: Cost per hour with one server=$534.45; Cost with 2 servers = $110.63; Cost with 3 servers = $123.79: So two servers are optimal.)

b. Find the average overall waiting time and the average total flow time through the system for the optimal case. (Ans: Waiting time: 1.52 min; Flow time: 9.52 min

Answer 1

Problem 1:

(a):

To start with, you need to figure the issue proclamation as a Linear Programming model – then it can be understood utilizing any of the 3 accessible techniques viz. Simples, Big M, or Random. On the off chance that if the quantity of factors is precisely 2, at that point we can utilize graphical technique = the less demanding than the over 3 strategies.

Variables: $x_1,x_2,x_3,....,x_n$

Constraints:

Min z = x₁ + x₂ - x₃ + x₄ - x₅ is same as maximize z = -x₁ – x₂ + x₃ – x₄ + x₅

in other words min z is the same as max –z

Miu1 = Arrival Mean = x bar = Arrival Miu = 20

SD1 = Arrival Standard Deviation = Arrival SD = 4

Miu2 = Service Miu = 15

SD2 = Service SD = 5

Server Cost = $25 / server / customer

Waiting Cost = $200 / customer / hour

minimize Z1 =

Let n₁ = 150,

n₂ = 100

where n₁ and n₂ are the number of customers arriving and being served respectively per units of time – say per 4 hours or half a day – which means around 150 customers would arrive between 7am to 11 am , and 100 of them would have been already served and the remaining 50 customers will be served after 11 am. The arrival rate after 11 am to 3 pm may be less than 150 so that no customer would be asked comes the next day;

square root = sqrt

Z Calc = Calculated Value for Z

Z Crit = Critical Value for Z obtained from the table or excel formula

Formula:

Z Calc = Miu1 – Miu2 / (sqrt (    SD1^2/n1 + SD2^2/n2 ) )

= (20 – 15 ) / ( sqrt ( 4*4 / 150 + 5*5 / 100 ) )

= 5 / sqrt (16/150 + 25 / 100 )

= 5 / sqrt (0.10667 + 0.25)

= 5/0.35667

Z Calc = 8.375

This Z Calc value will be used to calculate the other costs as shown below:

Waiting cost = $200 / customer / hour

Arrival mean = 20 minutes

that means when there 150 customers arriving in 4 hours, then 150/4 = 37.5 = rounded = 38 customers per hour

So the average arrival rate = 38 customers per hour

Average service rate = 100 / 4 = 25 customers per hour

Hence average number of waiting customers = 38-25 = 13 customers

so 13 customer will be waiting and will cost us 13 * 200 = $2600 = the total weighted weighting cost – this must be divided by the factor to get the un weighted weighting cost.

(The multiplication factors 9.07, 85.0542, and 776.12 are obtained from the students look up table, and are used to calculate the division factor by multiplying the multiplication factor with Z Calc)

Division factor = multiplication factor * Z Calc

Where multiplication factor is obtained from the students table;

With 1 server, the division factor = 9.07 * Z Calc = 9.07 * 8.375 = 76

Waiting cost = Total waited waiting cost / division factor = 2600 / 76 = 34

With 2 servers, the division factor = 85.0542 * Z Calc = 85.0542 * 8.375 = 712.33

Waiting cost = Total waited waiting cost / division factor = 2600/712.33 = 3.65

With 3 servers, the division factor = 776.12 * Z Calc = 776.12 * 8.375 = 6500

Waiting cost = Total waited waiting cost / division factor = 2600/6500 = 0.4

Similarly, waiting cost for 4 servers = 1.6, and 5 servers = 2.2 ( same calculation method)

Presently arrange every one of them to get the examination:

Total cost total server cost + other costs Number of Servers Cost per hour Total Server cost Other costs 25 25 59 25 50 3.65 53.65 25 75 0.4 75.4 4 25 100 1.6 101.6 25 125 2.2 127.2

(b):

Average number of waiting customers = 38 - 25 = 13 customers

Average waiting time = 20 – 15 = 5 minutes (worst case) but on the optimal case it is = 1.82/5 = 0.364

Total number of customers in optimal case = 42.20879

Total = Total number of customers in optimal case waiting time =  

(c):

The gap between the number of arriving customers and served customers = 150 – 100 = 50

other costs = 0.4,

hence total cost = gap + other costs = 50+0.4=50.4

total cost = 75.4 * 50.4

Cost per hour = total cost / number of hours = 3800/8 = $475

Average number of waiting customers = 38-25 = 13 customers

Average wait time = total wait time / number of customers

where total wait time = 2 * (arriving + served) + 83.44

=

Hence,

Average wait time = 583.44 /13

= 44.88 min

Average time for the smooth flow of customers = Average wait time + 15 = 44..88 + 15 = 59.88 min

where 15 came from rounded wait time / 3 = 45 /3 = 15 (44.88 rounded to 45)