University:
California State University, Northridge
Course:
MATH 103 | Mathematical Methods for Business
Academic year:

2024
Views:

494

Pages:

7
Author:

Vinnaneumma00c

3.7 a) How long must the plant operate on peak days? From 7am-6pm, oranges come in at a rate of 10,000kg/hr and are processed, and thus leave the plant, at 8000kg/hr. Because inflows exceed outflows, inventory will build up at a rate of ∆R = 10,000-8,000kg/hr = +2,000 kg/hr. Because we cannot have oranges stored overnight, we start with an empty plant so that inventory at 7am is zero: I (7 am) = 0. Because inventory builds up linearly at 2,000kg/hr, the inventory at 6pm is I(6pm) = 2,000 kg/hr * 11 hr = 22,000kg. 3.7 After 6pm, no more oranges come in, yet processing continues at 8000 kg/hr until the plant is empty. Thus, inflows is less than outflows so that inventory is depleted at a rate of ∆R = 0 - 8,000 kg/hr = - 8,000 kg/hr. Thus, since we have that I(6pm) = 22,000kg, we know that inventory depletes linearly from that level at a rate of -8,000 kg/hr. Thus, to empty the plant, inventory must reach zero and this will take an amount of time Dt where: 22,000 kg = 8,000 kg/hr ∆t or ∆t = 22,000/8,000 hr = 2.75 hr = 2 hr 45min. Thus, the plant must operate until 6pm + 2hr 45min = 8:45pm. This can all be graphically summarized in the inventory build up diagram shown below. Inventory I (t) 22,000 kg Bin capacity 6,000 kg 7 am Bin is full: First truck waits 6 pm 8:45 pm Last truck leaves 3.7 b) At what point during the day must a truck first wait before unloading into the storage bin? Inventory builds up in the bins. When the bin is full, then the trucks must wait. This happens at: 2,000 kg/hr ∆t = 6,000kg, The first truck will wait after ∆t = 6,000/2,000 hr = 3 hr, which is at 10am. c) What is the maximum amount of time that a truck must wait? The last truck that arrives (at 6pm) joins the longest queue, and thus will wait the longest. It will be able to start dumping its contents in the bins when the bins start depleting. This is at 22,000 kg - 8,000 kg/hr ∆t = 6,000. ∆t = (22,000-6,000)/8,000 hr = 2 hr, after 6pm. The maximum truck waiting time is therefore 2 hours, and that is 8pm. 3.7 e) Among trucks that wait, how long is the average wait? Procedure1: The first truck among those that wait, will wait for 0 hour. The last truck among those that wait, will wait for 2 hours. On average a truck waits for (0+2)/2 = 1 hour Procedure2: The maximum inventory is 22,000. But 6000 of it is in the bin. So the waiting inventory on trucks has a maximum of 22,000 – 6,000 = 16,000 Therefore, the average inventory for those trucks that wait is 8,000. This inventory is depleted at the rate of 8000/hrs I = RT 8,000 = 8000T T = 8000/8000 = 1 hour 3.7 d) How long a truck wait on the average? The average wait among trucks that wait is 1 hour. But this is for the last 8 hours of the day. In the first 3 hours no truck waits. Therefore the average wait is 0 hour. Therefore, the average wait among all trucks is (3×0 + 8×1)/11 = 0.727 hour or 43.6 minutes Truck Inventory 16 Bin capacity 6,000 kg 7 am 10 am 6 pm 8 pm