Question

Consider a street hot dog vendor that has a demand that follows a Poisson distribution with mean of 0.75 customer/minute while the time to serve customer follows a negative exponential distribution with the mean of 1 min. 1. Utilization of the vendor 2. The probability that a new customer arrival has to wait 3. The probability that there is no customer in the system 4. What is the probability that the system will have 5 or more customers? 5. Average number of customers in the system 6. Average number of customers in the waiting line 7. Average time a customer spends in system 8. Average waiting time of a typical customer in line 9. If service time becomes an constant of 1 min, what is the new “Average waiting time of a typical customer in line”? (M/D/1)

Consider the street hot dog vendor (M/M1 example in slide 31)

•

= 0.75 customer/min

•

= 1 customer/min

•

Customer arrival rate now has increased from 0.75 customer/min to 2.2

customers/min. Answer below questions:

A.

What will be the new system utilization

B.

What is the minimum number of new helpers he should hire to manage this new

demand?

C.

On average, how much time a customer spends in the queue if the vendor hires 2 extra

help?

I ONLY NEED HELP WITH PART 2 A,B,C

Answer 1

street hot dog vendor
Mean arrival rate, ?	2.2	customers/min
Mean service rate, ?	1	customers/min
A.
The new system utilization, ?=?/?
	220%
B.
the minimum number of new helpers needed will make the utilization as 100% (or 1)
1 = ?/(m?)
1 = 2.2/(m*1)
Solve for m:
m	2.20	helpers
=	3	(rounding off to the next whole number)
C. time a customer spends in the queue if the vendor hires 2 extra help :
Time in the queue, Wq=?/(?*(?-?))
here, ? = 2* 1 = 2 customers/hour
i.e.	?<?
hence, this form of queue is not feasible, thus the average time customer spends in the queue cannot be calculated.