Question

All trucks traveling on Freeway 99 between Sacramento and Fresno are required to stop at a weigh station. Trucks arrive at the weigh station at a rate of 165 per 8-hour day, and the station can weigh, on the average, 180 trucks per 8-hour day.

1. Determine the average time spent waiting and being weighed at the weigh station by each truck.
2. Determine the average waiting time before being weighed for each truck.
3. Determine the average number of trucks waiting
4. Determine the average number of trucks waiting and being weighed
5. Determine the portion of time the weigh station is busy.
6. What would happen if the arrival rate became higher than the service rate?
7. If the truck drivers find out they must remain at the weigh station longer than 20 minutes on the average, they will start taking a different route or traveling at night, thus depriving the state of taxes. The state of California estimates it loses $15,000 in taxes per year for each extra minute that trucks must remain at the weigh station. A new set of scales would have the same service capacity as the present set of scales, and it is assumed that arriving trucks would line up equally behind the two sets of scales. It would cost $100,000 per year to operate the new scales. Should the state install the new set of scales? What is the net saving per year?

       (Remember that this is not a multiple server problem)

8. Suppose passing truck drivers look to see how many trucks are waiting to be weighed at the weigh station. If they see more than five trucks in line, they will pass by the station and risk being caught and ticketed. What is the probability that a truck will pass by the station?

Answer 1

Answer:-

Mean Arrival rate (λ) = 165/8 = 20.625 per hr

Mean Service rate (μ)= 180/8 = 22.5 per hr

(an) Average time spent pausing and being weighed at weight station by each truck (Ws)=1/(μ-λ) = 1/(22.5-20.625 ) = 1/1.875‬ =0.533 hr

(b) the normal gauging time before being weighed by each truck (Wq)=λ/μ(μ-λ) = 20.625/22.5(22.5-20.625 )=0.4888 hr

(c) Average number of trucks pausing (Lq)=λ2/μ(μ-λ)= (20.625)2/22.5(22.5-20.625 )=10.082 trucks

(d) Average number of trucks pausing and being gauged =(Ls)=λ/(μ-λ) = 20.625/(22.5-20.625 ) = 11 trucks

(e) Portion of time checkpoint is occupied (λ/µ) = 20.625/22.5 =0.9166 = 91.66%

(f) If the appearance rate is higher than the administration First thing is the use of the checkpoint will be 100%, however the weighting time for the trucks in the line and the framework will be expanded.

(A) Probability =(λ/μ)5+1 =(20.625/22.5)5+1 =0.6470

***Please please like this answer***