Question

# Consider a street hot dog vendor that has a demand that follows a Poisson distribution with...

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

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

 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. ?

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