CHAPTER 3: PROCESS FLOW MEASURES
3.1 Objective
The objective of this chapter is to identify key operational measures that may be used to study process
flows. They are linked together using Little's law. We then present a series of examples that show how
process flow analysis may be used to study performance. The objective is to study current performance as
well as identify target areas for improvement. We also link the operational measures of performance to
financial measures.
In a class of 100 minutes we start by discussing the importance of building a time based capability in
today's competitive environment. We then establish Little's law to set up other operational measures namely inventory and throughput that impact flow time. Several examples from the chapter are discussed
to make this relationship clear. We then link these operational measures to financial measures to identify
what form improvements may take. We then discuss the Kellogg CRU Rental case to demonstrate how
such an analysis may be used to identify key areas for improvement.
3.2 Additional Suggested Readings
We assign a short case as supplemental reading for the analysis of process flows. The case is used to do a
thorough analysis of flows and identify key drivers of cost and revenue in a process. This understanding
is then used to identify actions that improve performance.
 “CRU Computer Rentals”. Kellogg Case. Author: Sunil Chopra. Available from:
http://www.kellogg.northwestern.edu/cases/index.htm. Suggested assignment questions are contained
in the case.
3.3 Solutions to the Chapter Questions
Discussion Question 3.1
The 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 3.2
In 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).
13
14
Chapter 3
Discussion Question 3.3
First, 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 3.4
For 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 3.5
If GM and Toyota have same turns, and we know that
turns = 1/flow time = 1/T,
it follows that their average flow times are the same. We also know that Toyota's throughput is twice that
of GM. Thus, from
I=RT
it follows that Toyota has twice the inventory of GM. Thus, the statements are inconsistent, both
companies have the same flowtime but Toyota has higher inventory than GM.
Discussion Question 3.6
Yes, low inventories means few flow units are held in the buffer. In contrast, fast inventory turns means
short flow times; i.e., flow units do not spend a long time in the process. As such, one can have high
turns with high or low inventories (it all depends on what the throughput is).
Discussion Question 3.7
A 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., lower
working 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 have
Chapter 3 15
Average 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 have
Average 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 have
Average inventory I = average balance = $3,000
Turns = 6 per year.
Average flow time T = 1 / turns = 1/6 year = 2 months.
Thus
Throughput R = I / T = 3,000/2 = $1,500 / month.
Exercise 3.4 (ER)
First draw the flowchart with all the data given:
Potential admits
10%
Buffer 1
Registration
Buffer 2
Doctor
R =55/hr
I=7
I = 34
T = 2 min
T1 = 5min
T2 = 30min
90%
Simple Prescription
We assume a stable system. This implies that average inflow equals average outflow at every stage. In
this case 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:
16
Chapter 3
(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
flow rate 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 minutes
4b: prescription folks: T = 5 minutes
OK, 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.
The answer to the other questions is found as follows:
1. On average, how long does a patient spend in the emergency room?
We know the flow time of each flow unit. The average flow time over all flow units is the weighted
average: 10% of total flow units spend 76.7 minutes while 90% spend 51.7 minutes. Thus, the grand
average is:
T = 10% * 76.7 + 90%*51.7 = 54.2 minutes.
2. On average, how many patients are being examined by a doctor?
This question asks for the average inventory at the doctor's activity. Again, first calculate inventory of
each type of flow unit:
(a) Potential admits: R = 5.5 patients/hr, T = 30 min = 0.5 hr, thus, I = RT = 5.5/hr*0.5 hr = 2.75
patients
(b) Simple prescription: R = 49.5 patients/hr, T = 5 min = (5/60) hr, thus I = RT = 49.5*(5/60) =
4.125 patients
Thus, total inventory at the doctor is 2.75 + 4.125 = 6.865 patients.
3. On average, how many patients are in the ER?
This question asks for total inventory in ER = inventory in buffer 1 + inventory in registration + inventory
in buffer 2 + inventory with doctors = 7 + 1.83 + 34 + 6.865 = 49.695 patients.
Chapter 3 17
Exercise 3.5 (ER, triage)
The process flow map with the triage system is as follows:
Buffer 3
Potential admits
5.5/hr
Buffer 1
Registration
Buffer 2
Triage Nurse
55/hr
49.5/h
r
Buffer 4
Simple Prescription
The inventory, and time spent in various locations are as follows. In each case the calculated quantity is
italicized.
Exercise 3.5
Location
Inventory
Buffer 1
20
Registration
1.83
Buffer 2
5
Triage Nurse
0.92
Buffer 3
1
Potential Admits
2.75
Buffer 4
15
Simple Prescriptions
4.13
Total
50.63
Avg time spent in ER
Avg time for patients admitted
Avg time for patients subscription
Throughput
(per hr)
55
55
55
55
5.5
5.5
49.5
49.5
Throughput
(per min)
0.9167
0.9167
0.9167
0.9167
0.0917
0.0917
0.8250
0.8250
Flow Time (min)
21.82
2
5.45
1
10.91
30
18.18
5
55.23
71.18
53.45
Throughput through ER, R = 55 patients / hour = .9167/min.
Average inventory in emergency room, I = sum of inventory in all stages = 50.63 patients
Average time spent in the emergency room = I/R = 50.63/.9167 = 55.23 minutes.
For patients that are eventually admitted, average time spent in the emergency room = time in
buffer 1 + registration + buffer 2 + triage nurse + buffer 3 + doctor (potential admit) = 71.18
minutes.
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Chapter 3
Exercise 3.6 (ER, triage with misclassification)
In this case the process flow map is altered somewhat since there are some patients sent from simple
prescriptions to buffer 3.
Buffer 3
Potential admits
4.95/hr
Buffer 1
Registration
Buffer 2
Triage Nurse
55/hr
0.55/hr
50.05/hr
Buffer 4
Simple Prescription
(We will assume that the doctor "instantaneously" recognizes misclassification so that a misclassified
patient does not spend 5 minutes with the doctor. However, if you assume such person also spends 5
minutes, the entire methodology below follows, only increase the relevant flow time by 5 minutes.)
The inventories, throughputs and flow times are as follows:
Location
Inventory
Throughput
(per hr)
Buffer 1
5
55
Registration
1.83
55
Buffer 2
5
55
Triage Nurse
2.75
55
Buffer 3
1
5.5
Potential Admits
2.75
5.5
Buffer 4
15
50.05
Simple Prescriptions
4.13
49.5
Total
37.46
Avg time spent in ER
Avg time for patients admitted (correctly classified)
Avg time for patients admitted (first misclassified)
Avg time for patients subscription
Throughput
(per min)
Flow Time (min)
0.9167
0.9167
0.9167
0.9167
0.0917
0.0917
0.8342
0.8250
5.45
2
5.45
3
10.91
30
17.98
5
Avg time for patients admitted
Grand average
Throughput through ER, R = 55 patients / hour
Average inventory in emergency room, I = sum of inventories in all stages = 37.46
Average time spent in the emergency room T = I/R = 37.46/.9167 = 40.86 minutes.
To calculate flow times, we should distinguish three types of flow units:
40.86
56.82
74.80
38.89
58.62
40.86
Chapter 3 19
(1) those that are correctly identified as potential admits the first time: flow rate = 55*9% =
4.95/hr. Average flow time = time in buffer 1 + registration + buffer 2 + triage nurse + buffer
3 + doctor (potential admit) = 56.82 minutes
(2) those that are first mis-identified as simple prescription and later corrected and redirected to
potential admits: flow rate = 55*1% = 0.55/hr. Average flow time = time in buffer 1 +
registration + buffer 2 + triage nurse + buffer 4 + buffer 3 + doctor (potential admit) = 74.80
minutes
(3) those that are correctly identified to get a simple prescription the first time: flow rate =
55*90% = 49.5/hr. Average flow time = time in buffer 1 + registration + buffer 2 + triage
nurse + buffer 4 + buffer 3 + doctor (simple prescription) = 38.89 minutes
For patients that are eventually admitted, that is types (1) and (2), average time spent in the
emergency room is the weighted average of their flow times. Type (1) is fraction
4.95/(4.95+0.55) = 90% of those admitted and type (2) is 0.55/(4.95+0.55) = 10% of those
admitted. Thus, average flow time for patients that are eventually admitted is 90%*56.82min +
10%*74.80.8min = 58.62 min.
Note that the overall average flow time over all patients is: 9%*56.82min + 1%*74.80min +
90%*38.89.8min = 40.86 min, in agreement with the number derived directly from Little's Law
above.
Exercise 3.7 (Orange Juice Inc)
First let us notice that there are two periods in the day:
1. 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.
Thus, 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.
20
Chapter 3
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
2. 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 t where:
22,000 kg - 8,000 kg/hr t = 0,
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 above.
3. Truck dynamics: for this the inventory diagram is really useful. Notice that we have taken a total
process view of the plant, including the truck waiting queue. Thus, inventory is total inventory in the bins
+ inventory in the trucks (if any are waiting). So, let's draw the thick line on the inventory build-up
diagram, representing the bin storage capacity. First 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,
Chapter 3 21
so that the first truck will wait after t = 6,000/2,000 hr = 3 hr, which is at 10am. Now, the last truck that
arrives (at 6pm) joins the longest queue, and thus will wait the longest. That "unfortunate" truck 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,
or after t = (22,000-6,000)/8,000 hr = 2 hr, after 6pm. Thus, the last truck departs at 8pm and the
maximum truck waiting time is therefore 2 hours.
Now, among all the trucks that do wait (i.e., those arriving after 10am), the first truck waits practically
zero minutes, and the last truck waits 2 hours, culminating in an average of (0 + 2)hrs/2 = 1 hour.
Notice that the trucks arriving before 10am do not wait. Thus, the overall average truck waiting time is
(# trucks arriving before 10am * 0 + # trucks arriving after 10am * 1hr)/(total # of trucks).
Because input rate is 10,000kg/hr and each truck carries 1,000 kg/truck, the truck input rate is 10
trucks/hr, so that the overall average truck waiting time is:
(10 trucks/hr * 3hrs * 0 + 10 trucks/hr * 8hrs * 1hr)/(10 trucks/hr * 11 hrs) = 8/11 hr = 43.63min.
Average waiting time can also be calculated by noticing that the area of the upper triangle in the build-up
diagram represents the total amount of hours waited by all trucks:
Area = (22,000 - 6,000)kg * (8pm - 10 am) /2 = 16,000 kg * 10 waiting hr /2
= 80,000 kg waiting hrs = 80,000 kg waiting hrs / (1,000 kg/truck) = 80 truck waiting hrs.
Now, we just calculated that there are 80 trucks that do wait, hence the average waiting time among those
trucks that do wait is 80 truck waiting hrs/ 80 trucks = 1 hour.
Exercise 3.8 (Jasper Valley Motors)
Part a.
TURNStotal = 1/Ttotal so Ttotal = 1/8 years = 1.5 months
Itotal = RtotalTtotal = 160 vehicles/month * 1.5 months = 240 vehicles, which is the answer.
Typical errors: wrong units and stating that “I = 160*(1/8) = 20 vehicles.”
Part b.
Similar to part a, we have Tnew = 1/7.2 years = 1.667 months and Tused = 1/9.6 years = 1.25 months.
Inew = 0.6 * 160 vehicles/month * 1.667 months = 160 new vehicles
22
Chapter 3
Iused = 0.4 * 160 vehicles/month * 1.25 months = 80 new vehicles
Total monthly financing costs then 160*$175 + 80*$145 = 28,000 + 11,600 = $39,600/month.
Cost per vehicle are then $39,600/month (160+80) = $165 per vehicle per month, which is the answer.
Typical errors:
1. Not realizing that the cost driver is inventory, not throughput. (Taking a throughput-weighted
average would yield 60%* $175 + 40%*$145 = 163, instead of the correct inventory-weighted.)
2. Not taking a weighted average. Clearly, the answer must fall between $145 and $175.
3. Giving total monthly costs instead of per vehicle.
Part c.
From Little’s Law, cutting time 20% while holding R unchanged will reduce inventory by 20%. From
part b, average monthly financing costs for new vehicles is 160*$175 = $28,000/month. A 20% drop
gives $5,600 per month, which is the answer.
Typical errors:
1. Assuming the service works also on used cars, leading to 20%$39,600/month= $7920/mo.
2. Only stating the value per car per month: We reduce Tnew from Tnew = 1/7.2 years = 1.667 months by
20%*1.667mo = 1/3 mo. This saves 1/3mo * $175/new car,mo = $58.33/new car. (Multiplying by
96new cars/mo would have yield the correct $5600/mo.)
3. Reducing the flow time by 20% does not mean that turnover is increased by 20%. (On the contrary,
actually, turnover increases here from 7.2 to 9, which is 25%.)
Exercise 3.9 (Cheapest Car Rentals)
Part a.
Throughput
Inventory
Customers
300/week
2000.5 + 1002
= 300 cars
1 week
Clean area
240/week
100 cars
Repair area
60/week
120 cars
0.42 weeks
2 weeks
Flow Time
Part b:
Number of cars owned by Cheapest = 300 + 100 + 120 = 520
Number of cars on rent = 300
Revenue per week
= (#of cars on rent for short term) x $200/week + (#of cars on rent for LT) x $120/week
= I_ST x $200/wk + I_LT x $120/wk
Recall from [5a] that I_ST = R_ST x T_ST = 200/wk x 0.5 wks = 100 and I_LT = 100x2=200
Chapter 3 23
= 100$200 + 200$120 = $44,000/week
Cleaning cost per week = 240$5 = $1,200/week
Repair cost per week = 60$150 = $9,000/week
Depreciation per week = 520$40 = $20,800/week
Profit per week = 44,000 - (1,200 + 9,000 + 20,800) = $13,000/week
Part c:
Decreasing flow time in repairs by 1 week will lower the inventory in repairs from 120 to 60. This
will reduce the number of cars required by 60 and thus weekly depreciation by $2,400. All other
revenues and costs will be unchanged.
Decreasing repair cost per car by $30 lowers weekly repair cost by 6030 = $1,800.
Decreasing the flow time in repairs is thus more effective since it results in larger savings.
Exercise 3.9 (The Evanstonian)
Part a.
Answer: __162 of each type__
For Leisure Travelers we have:
RL = 135/3 = 45 guests/night and TL = 3.6 nights → IL = RL TL = 45 × 3.6 = 162 guests
For Business Travelers we have RL = 135 × 2/3 = 90 guests/night and TL = 1.8 nights → IL = RL TL = 90
× 1.8 = 162 guests
Part b:
ANSWER: __12.5/MONTH___
The total inventory is then IT = 324. We then have Turns = RT/IT = 135/324 = 0.4167 turns per
day = 12.5 turns per month.
An alternative approach: The average stay is TT = (1/3) × 3.6 + (2/3) × 1.8 = 2.4 days.
Turns = 1/TT = 1/2.4 = 0.4167 turns per day = 12.5 turns per month.
Part c:
ANSWER: $____230/DAY______
24
Chapter 3
From the analysis above on the average night, 324 rooms are occupied with half of the guests
being leisure travelers. Thus the average rate that The Evanstonian receives per occupied room
is:
½ × 250 + ½ × 210 = $230.
Exercise 3.11 (ABC Corporation)
2004
Factory flow time = I/R = 20,880/97,380 = 0.2144 years
AR flow time = 21,596/99,621 = 0.2168 years
(must use COGS dollars)
(must use sales dollars)
2005
Factory flow time = I/R = 25,200/98,350 = 0.2562 years
AR flow time = 22,872/110,644 = 0.2067 years
Performance in the factory has worsened while that in AR has improved.
Overall, net income has increased so that at the aggregate corporate level we have an
improvement.
Chapter 3 25
3.4 Test Questions
3.4.1
Customers walk into a bank and are served at an average rate of 20 per hour. One average there
are 5 people waiting in queue to be served. How long does the average customer wait in queue?
Answer to 3.4.1:
Throughput R = 20 per hour,
Inventory I = 5
Waiting time in queue T = I/R = 5/20 = 0.25 hours = 15 minutes.
3.4.2
At a hospital emergency room, patients come in at the rate of 20 per hour and on average, are
processed at the rate as well. Patients wait in queue till they are called in for registration. After
registration, patients are taken to an inner area where they are assigned a bed where they wait till
an intern is ready to see them. Once an intern sees them, she either identifies them to be a routine
case, prescribes medication and releases them, or she admits them to the hospital for further tests
and treatment. On average, it is found that there are 30 patients waiting for registration and
another 20 waiting to be seen by an intern.
A study has determined that 75% of the patients simply require routine treatment. A new system
separates the emergency room into two treatment areas, one for routine care and another for
emergency cases. A triage nurse who assigns the patient to one of the two tracks sees a patient
after registration. Patients for routine care are seen by an intern who either prescribes medication
and releases them or determines that the triage nurse has made a mistake and the patient really
needs emergency care. In this case the patient is moved to the emergency track. Patients for
emergency care are seen by an intern who either admits them for further treatment or determines
that they really are routine cases, prescribes medication and releases them. It is determined that
there are, on average, 20 people waiting to be registered, 5 waiting to be seen by the triage nurse,
10 waiting to be seen by an intern in the emergency area, and 10 waiting to be seen by an intern
in the routine care area. One average, 75% of the patients are classified by the triage nurse as
needing routine care.
(a)
On average, how long did patients spend in the emergency room before the triage system (either
before being released or admitted)?
(b)
On average, how long do patients that are finally admitted to the hospital spend in the emergency
room after the introduction of the triage system? Assume that the triage nurse makes no mistake
in identifying a patient as needing routine or emergency care.
(c)
On average, how long do patients spend in the emergency room after the triage system is
introduced (either before being released or admitted)? Assume that the triage nurse makes no
mistake in identifying a patient as needing routine or emergency care.
26
Chapter 3
Answer to 3.4.2:
(a)
Pre Triage process:
Register
Intern
Throughput = 20 per hour
Average Inventory = 20 + 30 = 50
Average flow time = Average inventory / Throughput = 50 / 20 = 2.5 hours.
(b)
Triage Process:
Admit
Register
Triage
Routine
Registration:
Average Inventory = 20
Throughput = 20 per hour
Average flow time = 20 / 20 = 1 hour
Triage:
Average Inventory = 5
Throughput = 20 per hour
Average flow time = 5 / 20 = 0.25 hour
Admit:
Average Inventory = 10
Throughput = 5 per hour
Average flow time = 10 /5 = 2 hours
Average time spent by people admitted = 1 + 0.25 + 2 = 3.25 hours
(c)
In this case we have
Average Inventory = 20 + 5 + 10 + 10 = 45
Throughput = 20 per hour
Average flow time = 45 / 20 = 2.25 hours.
Chapter 3 27
3.4.3
A car repair shop has two hoists where cars can be lifted for repair work. Currently customers
come in at the rate of 4 per hour and are processed at a similar rate. On average 8 cars are waiting
to be processed, 4 needing routine repairs and 4 needing major repairs. People are served on a
first come first serve basis.
(a)
How long do customers wait on average before being processed?
(b)
The repair shop owner feels that he is losing many customers needing routine repair because of
the long wait. He dedicates one hoist for routine repair and one for major repairs. A study
indicates that routine repairs are processed at the rate of 3 per hour and major repairs at the rate of
1 per hour. There are now 5 people waiting on average for routine repairs and 3 waiting on
average for major repairs. How long does each type of customer now wait before being served?
(c)
With the new system, what is the average waiting time over all customers?
Answers to 3.4.2:
(a)
For the current process we have
Inventory, I = 8 cars,
Throughput, R = 4 cars per hour.
Thus,
Flow time, T = I/R = 8/4 = 2 hours.
Cars wait an average of 2 hours before being served.
(b)
At the routine repair hoist, we have
Inventory, I = 5 cars,
Throughput, R = 3 cars per hour.
Thus,
Flow time, T = I/R = 5/3 = 1.67 hours.
Cars wait an average of 1.67 hours before being served at routine repairs.
At the major repairs hoist we have
Inventory, I = 3 cars,
Throughput, R = 1 car per hour.
Thus,
Flow time, T = I/R = 3/1 = 3 hours.
Cars wait an average of 3 hours before being served at major repairs.
(c)
With the new system, we have
Inventory, I = 5 + 3 = 8 cars,
Throughput, R = 3 + 1 = 4 cars per hour.
Thus,
Flow time, T = I/R = 8/4 = 2 hours.
Cars wait an average of 2 hours before being served even in the new system.
28
3.4.4
Chapter 3
A product sold at a supermarket has a shelf life of 21 days. If daily demand averages 100 units,
what is the maximum inventory that the supermarket should carry.
Solution to 3.4.4:
In this case, throughput R = 100 /day.
The maximum flow time cannot exceed 21 days (ideally should be much lower). Thus the supermarket
should carry a maximum of 21100 = 2,100 units in inventory. Ideally, the inventory carried should be
significantly lower.
3.4.5
Accounts Receivable (AR) receives an average of 1,000 payments per week. There is an average
of 500 checks waiting to be deposited.
(a)
What is the average time spent by each check waiting to be deposited?
(b)
Assume that the checks coming in are either small or large. Small checks are for an average of
$500 and large checks are for an average of $5,000. 20% of the checks coming in are large with
the rest being small. Currently no distinction is made between the large and small checks. As a
result 20% of the checks waiting to be processed are large. Is it worth reallocating resources so
that large checks wait an average of 0.3 weeks while small checks wait an average of 0.8 week
before being processed?
Answer to 3.4.5:
(a)
Throughput R = 1,000 per week, Inventory I = 500. Thus flow time T = I/R = 500/1000 = 0.5
weeks.
(b)
Currently AR carries an inventory of 500 checks worth 500 0.2 5,000 + 500 0.8 500 =
$70,000 waiting to be deposited. Now consider the case where resources are reallocated. Flow
time of large checks = 0.3 weeks, throughput R = 200/week. Thus average inventory of large
checks = 0.3 200 = 60 checks = 60 5,000 = $30,000. Flow time of small checks = 0.8 week,
throughput = 800/week. Thus, inventory of small checks = 0.8 800 checks = 640 checks = 640
500 =$32,000. Total inventory in this case = $30,000 + $32,000 = $62,000. Since $62,000 is less
than $70,000, the change will reduce the dollar value of checks waiting to be deposited and is
beneficial. This is in spite of the fact that the average inventory of checks will increase from 500
to 860.
3.4.6
A firm has seen its demand quadruple over a 2 year period. During the same period, the number
of turns has doubled. What has happened to inventories over the same period?
Answer to 3.4.6:
Since turns have doubled, flow time is half of what it was. Throughput (demand) meanwhile has increased
by a factor of 4. Thus inventories should have increased by a factor of 0.5 4 = 2 over the same period.
Inventories have thus doubled.