Chapter 3: Process flow measures3.3 Solutions to the Chapter QuestionsDiscussion Question 3.1Discussion Question 3.2Discussion Question 3.3Discussion Question 3.4Discussion Question 3.5Discussion Question 3.6Discussion Question 3.7Exercise 3.1 (Bank)Exercise 3.2 (Fast-Food)Exercise 3.3 (Checking Accounts)Exercise 3.4 (ER)Exercise 3.5 (ER, triage)Exercise 3.5 (ER, triage)Exercise 3.6 (ER, triage with misclassification)Exercise 3.7 (Orange Juice Inc)Exercise 3.8 (Jasper Valley Motors)Exercise 3.9 (Cheapest Car Rentals)Flow TimeExercise 3.9 (The Evanstonian)Answer: __12.5/month___Answer: $____230/day______Exercise 3.11 (ABC Corporation)Chapter 3 13CHAPTER 3: PROCESS FLOW MEASURES3.3 Solutions to the Chapter QuestionsDiscussion Question 3.1The opposite of looking at average is looking at a specific flow unit’s flow time, and the inventory status and instantaneous flow rate at a specific point in time. Because flow times change from flow unit to flow unit, it is better to look at the average over all flow units during a period of time. Similar for inventory and throughput.Discussion Question 3.2In practice, one often tracks inventory status periodically (each day, week, or month). Flow rate is typically also tracked periodically (even more frequently than inventory status because it directly relates to sales). It then is easy to calculate the average of those numbers to obtain average inventory and throughput during a period.In contrast, few companies track the flow time of each flow unit, which must be done to calculate the average flow time (over all flow units during a given period).Discussion Question 3.3First, draw a process flow chart.Second, calculate all operational flows: throughput, inventory, and flow time for each activity.Third, calculate the financial flow associated with each activity. If the activity incurs a cost (or earns a revenue), the cost or revenue rate is simply the throughput times the unit cost or revenue. If the inventory incurs a holding cost, the inventory cost rate is simply the average inventory times the unit holding cost.Fourth, summing all revenue rates and deducting all cost rates yields the profit rate, directly broken down in terms of the relevant throughputs and inventory numbers. The latter thus are the minimal set of operational measures to predict financial performance.Discussion Question 3.4For the department of tax regulations we have Average inventory I = 588 projects, Throughput R = 300 projects/yr (we assume a stable system). Thus, Average flow time T = I / R = 588 / 300 = 1.96 yr. This is larger than six months. So we should disagree with the department head's statement. Discussion Question 3.5If GM and Toyota have same turns, and we know that14 Chapter 3turns = 1/flow time = 1/T, it follows that their average flow times are the same. We also know that Toyota's throughput is twice thatof GM. Thus, from I=RT it follows that Toyota has twice the inventory of GM. Thus, the statements are inconsistent, bothcompanies have the same flowtime but Toyota has higher inventory than GM.Discussion Question 3.6Yes, low inventories means few flow units are held in the buffer. In contrast, fast inventory turns meansshort flow times; i.e., flow units do not spend a long time in the process. As such, one can have high turnswith high or low inventories (it all depends on what the throughput is).Discussion Question 3.7A short cost-to-cash cycle means that it does not take long to convert an input into a sold output. Clearly,this is good because we do not need to finance the input for a long time before it earns revenue (i.e., lowerworking capital requirements).Short cost-to-cash cycle requires short flow times, which imply low inventories (for a given throughput),or high throughput (for a given inventory).Exercise 3.1 (Bank)For the bank we haveAverage inventory I = 10 people, Throughput R = 2 people/min (we assume a stable system). Thus, Average flow time T = I / R = 10/2 min = 5 min. Exercise 3.2 (Fast-Food)For the fast food outlet we haveAverage inventory I = 10 cars. Throughput evaluation is as follows: Cars attempt to enter the drive through area at a rate of 2 cars/min. However 25% of cars leave when they see a long queue. Thus, cars enter the drive through at a flow rate R = 75% * 2 cars/min = 1.5 cars/min. Thus Average flow time T = I / R = 10/1.5 min = 6.67 min. Exercise 3.3 (Checking Accounts)For a checking account we haveAverage inventory I = average balance = $3,000Turns = 6 per year.Average flow time T = 1 / turns = 1/6 year = 2 months.ThusT1 = 5min T2 = 30min I = 34 I = 7 Buffer 1 Registration Buffer 2 Doctor Potential admits Simple Prescription R =55/hr 10% 90% T = 2 min Chapter 3 15Throughput R = I / T = 3,000/2 = $1,500 / month.Exercise 3.4 (ER)First draw the flowchart with all the data given:We assume a stable system. This implies that average inflow equals average outflow at every stage. In thiscase you are given inventory numbers I and flow rate R = 55 patients/hr. There are two flow units:(1) Those that are potential admits: flow rate = 55*10% = 5.5/hr. (2) Those that get a simple prescription: flow rate = 55*90% = 49.5/hr.To find the average flow times, we use Little's law at each activity for which the flow time is unknown:(1) Buffer 1: R = 55/hr (both flow units go through there), I = 7, so that waiting time in buffer 1 =T = I/R = 7/55 hr = 0.127 hours = 7.6 minutes.(2) Registration: flow time T = 2 min = 2/60 hr. All flow units flow through this stage. Thus flowrate through this stage is R = 55 / hr. Average inventory at registration is given by I = RT = 55*2/60 = 1.83 patients.(3) Buffer 2: R = 55/hr (both flow units go through there), I = 34, so that waiting time in buffer 2 = T = I/R = 34/55 hr = 0.62 hours = 37.1 minutes.(4) Doctor time: depends on the flow unit:4a: potential admits: T = 30 minutes4b: prescription folks: T = 5 minutesOK, now we have everything to find the total average flow times: find the critical path for each flow unit. In this case, each flow unit only has one path, so that is the critical path. We find its flow time by adding the activity times on the path:(a) For a potential admit, average flow time (buffer 1 + registration + buffer 2 + doctor) = 7.6 + 2 + 37.1 +30 = 76.7 minutes(b) For a person ending up with a prescription, average flow time (buffer 1 + registration + buffer 2 + doctor) = 7.6 + 2 + 37.1 + 5 = 51.7 minutes.Buffer 1 Registration Buffer 2