Presentation on theme: "3. 7 MBPF. Orange Juice Inc. produces and markets fruit juice"— Presentation transcript:
1 3. 7 MBPF. Orange Juice Inc. produces and markets fruit juice 3.7 MBPF. Orange Juice Inc. produces and markets fruit juice. During the orange harvest season, trucks bring oranges from the fields to the processing plant during a workday that runs from 7 a.m. to 6 p.m. On peak days, approximately 10,000 kilograms or oranges are trucked in per hour. Trucks dump their contents in a holding bin with a storage capacity of 6,000 kilograms. When the bin is full, incoming trucks must wait until it has sufficient available space. A conveyor moves oranges from the bins to the processing plant. The plant is configured to deal with an average harvesting day, and maximum throughput (flow rate) is 8,000 kilograms per hour.Assuming that oranges arrive continuously over time, construct an inventory buildup diagram for Orange Juice Inc. In order to process all the oranges delivered during the day, how long must the plant operate on peak days? (Assume, too, that because Orange Juice Inc. makes fresh juice, it cannot store oranges.) Assuming, finally, that each truck holds about 1,000 kilograms of oranges, at what point during the day must a truck first wait before unloading into the storage bin? What is the maximum amount of time that a truck must wait? How long will trucks wait on average? Among trucks that do wait, how long is the average wait?
2 Directions o) Assume a truck as your flow unit. a) How long must the plant operate on peak days?b) At what point during the day must a truck first wait before unloading into the storage bin?c) What is the maximum amount of time that a truck must wait?d) Among trucks that wait, how long is the average wait?e) How long a truck wait on the average?
3 3.7 (a) How long must the plant operate on peak days? From 7am-6pmoranges come in at a rate of 10,000kg/hrare processedleave the plant, at 8000kg/hrBecause inflows exceed outflows, inventory will build up at a rate ofDR = 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 isI(6pm) = 2,000 kg/hr * 11 hr = 22,000kg.
4 3.7 (a) How long must the plant operate on peak days? After 6pm, no more oranges come inYet 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 ofDR = 0 - 8,000 kg/hr = - 8,000 kg/hr.We have I(6pm) = 22,000kgInventory depletes linearly from that level at a rate of -8,000 kg/hr.To empty the plant,22,000 kg = 8,000 kg/hr DtorDt = 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.
5 3.7 (a) Inventory built-up diagram Bin capacity6,000 kgInventoryI(t)22,000 kg6pm8:457amBin is full:First truck waitsLast truck leaves
6 3.7 (b) and (c)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 Dt = 6,000kg,The first truck will wait after Dt = 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 at22,000 kg - 8,000 kg/hr Dt = 6,000.Dt = (22,000-6,000)/8,000 hr = 2 hr, after 6pm.The maximum truck waiting time is therefore 2 hours, and that is 8pm.
8 3.7 (d)d) 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 hourProcedure2: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 = 16,000Therefore, the average inventory for those trucks that wait is 8,000.This inventory is depleted at the rate of 8000/hrsI = RT8,000 = 8000TT = 8000/8000 = 1 hour
9 3.7 (e)e) 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 = hour or 43.6 minutes