Question

A company produces crushed limestone for the road construction industry. The rock is crushed in a quarry and then sent by trucks to the construction site. Each truck must be weighed before exiting the quarry. There is one weight scale. The trucks arrive at the weigh station and, if necessary, line up to be weighed on a first come, first-served (FIFO) basis. Truck arrivals are Poisson distributed with an average of 12.5 trucks per hour. The time needed to weigh each truck varies but is exponentially distributed with an average of 4 minutes per truck. The entire weighing operation is at a steady state rate. The weigh station does not have a practical limit on the parking space available, hence it can be considered to have an “infinite” calling population.

A second weight scale and station, identical to the first, can be purchased and installed for an investment cost of $135,000. Cash operating costs for this second station total $45,000 per year (52 weeks). Trucks would still form a single line and follow FIFO servicing. The new scale would have the same operating characteristics, including exponentially distributed service times with an average of 4 minutes per truck.

d. Assume the operation runs 8 hours per day for 5 days per week. How many total minutes would be saved per week by adding the second scale?

e. The owner of the company estimates that the out-of-pocket cash costs for a truck and driver to sit idle total $35 per hour. How many weeks would it take to repay the investment in the second scale? (Hint: You must consider the station cash operating costs.)

Answer 1

arrival rate lambda = 12.5 trucks per hour

Service rate mu= 60/4 =15 per hour

Utilisation rho = lambda /mu = 12.5 /15 =0.833

average waiting time in line for trucks

= lambda / mu ( mu-lambda) = 12.5 / 15 x( 15-12.5) = 1/3 hour

cumulative waiting hours per hour = 12.5 x1/3 =4.166 hours

annual wait time = 52 x8x5 x4.166 = 8666.66 hours

Cost of waiting per hour with single channel = 12.5 x1/3 x35 = 145.83 per hour

annual cost of waiting = 8666.66 x35 = 303333 per year

Performance with 2 channels

P(0) = 1 / [ (1+rho) + (rho)² x mu / (2 xmu -lambda)]

= 1 / [ 1+0.833+ 0.5947] =0.411

Waiting time in line =P(0) x mu x (rho)² /( 2xmu-lambda)²

= 0.411 x 15 x0.6938 / (17.5)² = 0.0139 hours

cumulative waiting time per hour = 12.5 x0.0139 =0.1737 hour

yearly wating time = 52x8x5 x0.1737 =361.4

cost of waiting per year = 361.4 x35 =12649

cost saving per year = 303333 -12649 =290684

Payback time = [( 135000+45000) / 290684] years

= 0.619 years.