1)
l =1.25 mins or 48 customers per hour
u =1 min or 60 customers per hour
Wq =Ws- u = l/(u(u-l)) . = 48/60(12) = 1/15 hours = 4 mins
Ws =Ls/l = =1/(u-l) = 1/(12) =5 mins
Lq =l Wq = l^2/ u(u-l)) = 48*48/ 60 *12 = 16/5 = 3.2
Ls =l/u-l= 48/ (60-48) = 4 patients
rho = l/u =48/60= .8
b)
in minutes | in rate terms | 5% inc | 10% inc | 15% inc | 20% inc | 50%inc | 100%inc | |
l | 1.25 | 48 | 50.4 | 52.8 | 55.2 | 57.6 | 72 | 96 |
u | 1 | 60 | 60 | 60 | 60 | 60 | 60 | 60 |
base | 5% inc | 10% inc | 15% inc | 20% inc | 50%inc | 100%inc | ||
wq (mins) | 4 | 5.25 | 7.333333333 | 11.5 | 24 | -6 | -2.666666667 | |
ws (mins) | 5 | 6.25 | 8.333333333 | 12.5 | 25 | -5 | -1.666666667 | |
ls | 3.2 | 4.41 | 6.453333333 | 10.58 | 23.04 | -7.2 | -4.266666667 | |
lq | 4 | 5.25 | 7.333333333 | 11.5 | 24 | -6 | -2.666666667 | |
rho | 0.8 | 0.84 | 0.88 | 0.92 | 0.96 | 1.2 | 1.6 |
3)
1)
l =1.25 mins or 48 customers per hour
u =3 min or 20 customers per hour per server
c =3 servers
po= 1/(1+rho rho^2/2 + rho^c/ c! (1- rho/c)) = 1/(1+.8 + .8^/2 +.8^3/ 6(1- .8/3)) =.058
Wq = lq/l = .0539
Ws = wq +1/u =
Lq = rho^(c+1)/(c-1)! (c-rho)^2) Po = 2.5888
L = lWq + l/u =
rho =l/u =48/20 =2.4
utilization= l/cu =48/3*20= .8
b)
in minutes | in rate terms | 5% inc | 10% inc | 15% inc | 20% inc | 50%inc | 100%inc | |
l | 1.25 | 48 | 50.4 | 52.8 | 55.2 | 57.6 | 72 | 96 |
u | 3 | 20 | 20 | 20 | 20 | 20 | 20 | 20 |
3 | base | 5% inc | 10% inc | 15% inc | 20% inc | 50%inc | 100%inc | |
wq (mins) | 0.053932584 | 0.07431789 | 0.108608731 | 0.177637726 | 0.385570844 | -0.116546763 | -0.063681592 | |
ws (mins) | 0.103932584 | 0.12431789 | 0.158608731 | 0.227637726 | 0.435570844 | -0.066546763 | -0.013681592 | |
ls | 4.988764045 | 6.265621656 | 8.374541004 | 12.56560249 | 25.08888062 | -4.791366906 | -1.313432836 | |
lq | 2.588764045 | 3.745621656 | 5.734541004 | 9.805602492 | 22.20888062 | -8.391366906 | -6.113432836 | |
rho | 2.4 | 2.52 | 2.64 | 2.76 | 2.88 | 3.6 | 4.8 | |
p0 | 0.056179775 | 0.042799058 | 0.030599755 | 0.019466615 | 0.009297136 | -0.035971223 | -0.074626866 | |
utilization | 0.8 | 0.84 | 0.88 | 0.92 | 0.96 | 1.2 | 1.6 |
please use the formulas mention in the text I am unable to add spreadsheet to this
There is no need to develop a Simio model as part of this homework. You can...
**LOOKING FOR FORMULAS, ANSWERS PROVIDED. 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...
I specifically need help with part c and d
Problem #1 (10 points) Customers arrive to a fast food restaurant with two servers with interarrival times that follow an exponential distribution with a mean of 5 minutes. Service times (per server) also follow an exponential distribution with a mean of 8 minutes. There is enough space in the restaurant to accommodate a very large number of customers waiting to be served. (a) (3 points) What is the probability that at...
Question 1 Unless otherwise stated, assume all times reported refer to averages from exponential distributions and that we are looking at stable processes. If the average time between arrivals is 10 minutes, what is the arrival rate? a. 6 jobs per hour b. 0.1 jobs per minute c. 0.001666 jobs per second d. All of the above 1 points Question 2 For a system with a single server, if the arrival rate is six jobs per hour and the average...
Jobs arrive at a single-CPU computer facility with interarrival times the are IID exponential random variables with mean 1 minute. Each job specifies upon arrival the maximum amount of processing time it requires, and the maximum times for successive jobs are IID exponential random variable with mean 1.1 minutes. However, if m is the specified maximum processing time for a particular job, the actual processing time is distributed uniformly between 0.55m and 1.05m. The CPU will never process a job...
Consider a single server ,poison input queue with mean arrival rate of 12/hr .currently the server works according to an exponential distribution with mean service time of 6 minutes .Management has a training course which will result in an improvement in the variance of service time but a slight increase in the mean .After completion of course its estimated that the service time is now normally distributed and that the mean service time will increase to 4.5 minutes ,but the...
People's Software Company has just set up a call center to provide technical assistance on its new software package. Two technical representatives are taking the calls, where the time required by either representative to answer a customer's questions has an exponential distribution with a mean of 8 minutes. Calls are arriving according to a Poisson process at a mean rate of 10 per hour. By next year, the mean arrival rate of calls is expected to decline to 5 per...
Consider the M/G/1 queue with FIFO service (see Homework 6) Assume that (1) the arrival rate is 1 customer per minute, and (2) the service times are exponentially distributed with average service time 45 seconds. 07. Find the server utilization 88. Find the average value of the waiting time (in minutes). 9. Find the probability that an arriving customer will wait in the queue for at least 1 minute. 10. Find the probability that an arriving customer who finds the...
roblem Consider a single server queueing system where the customers arrive according to a Poisson process with a mean rate of 18 per hour, and the service time follows an exponential distribution with a mean of 3 minutes. (1). What is the probability that there are more than 3 customers in the system? (2). Compute L, Lq and L, (3). Compute W, W and W (4). Suppose that the mean arrival rate is 21 instead of 18, what is the...
The Burger Dome waiting line model studies the waiting time of customers at its fast-food restaurant. Burger Dome's single-server waiting line system has an arrival rate of 0.75 customers per minute and a service rate of 1 customer per minute. Adapt the Black Sheep Scarves spreadsheet shown below to simulate the operation of this waiting line. Make sure to use the random values for both interarrival and service times generated in the worksheet_12-23. Assuming that customer arrivals follow a Poisson...
I need matlab code for solving this problem
Clients arrive to a certain bank according to a Poisson Process. There is a single bank teller in the bank and serving to the clients. In that MIM/1 queieing system; clients arrive with A rate 8 clients per minute. The bank teller serves them with rate u 10 clients per minute. Simulate this queing system for 10, 100, 500, 1000 and 2000 clients. Find the mean waiting time in the queue and...