deterministic performance estimation

Deterministic Performance Estimation

Prof. dr. K.J. Roodbergen

Prof. dr. I.F.A. Vis

Prof. dr. M.B.M. de Koster


1

Contents

1 Learning objectives .................................................................................... 2

2 Flow diagram .............................................................................................. 3 Incorporating data into a flow diagram ........................................................... 3

Other tools ....................................................................................................... 4

Exercises .......................................................................................................... 5

3 Throughput times ....................................................................................... 6 Throughput times with multiple paths............................................................. 6

Throughput time estimates for different types of products in one system ...... 7

Difference between actual and estimated throughput times ............................ 8

Exercises .......................................................................................................... 9

4 Design and effective capacity .................................................................. 10 Sufficiency of design capacity ...................................................................... 10

Effective capacity .......................................................................................... 11

Exercises ........................................................................................................ 11

5 Bottleneck ................................................................................................. 13 Exercises ........................................................................................................ 14

6 Departure rate .......................................................................................... 16

7 Utilisation and efficiency ......................................................................... 18 Utilisation rate for n identical parallel machines........................................... 18

Difficulties in calculating utilisation rates..................................................... 19

Productive utilisation rate .............................................................................. 20

Relation between utilisation rate and throughput time .................................. 20

Exercises ........................................................................................................ 21

8 Work-in-progress ..................................................................................... 23 Exercises ........................................................................................................ 23

9 Solving bottleneck problems ................................................................... 25 Exercises ........................................................................................................ 26

10 Large exercises ...................................................................................... 27

Answers to exercises ....................................................................................... 30 Flow diagrams ............................................................................................... 30

Throughput times .......................................................................................... 31

Design and Effective Capacity ...................................................................... 31

Bottlenecks .................................................................................................... 32

Departure rate ................................................................................................ 33

Utilisation and efficiency .............................................................................. 33

Work-in-Progress .......................................................................................... 34

Solving bottleneck problems ......................................................................... 34

Large exercises .............................................................................................. 34


2

1 Learning objectives The learning objectives related to this subject are:

• You must be able to draw a flow diagram for a given situation description.

• For any given description of a production/service facility, using the

assumption that all values are deterministic, you must be able to determine

the following performance criteria:

- throughput time,

- design capacity

- location and maximum design capacity of the bottleneck,

- throughput (= departure rate),

- (productive) utilisation

- work-in-progress,

• You must understand the following concepts and be able to estimate the

effects of these concepts on the mentioned performance criteria:

- varying arrival rates and operating times,

- impact of stochasticity on various performance criteria,

- bottleneck processes,

- product batching and machine setup times (or switchover times),

- using serial and parallel production machines/servers,

- using additional machines/servers.

• You must be able to calculate the minimum batch size for a process such

that a given process will not be a bottleneck.


3

2 Flow diagram Flow diagrams (see also Heizer and Render, Chapter 7) illustrate the

movement of materials, information or people through all processes required to

transform resources into products or services. Here we will just introduce a

very rudimentary approach, using only square blocks and arrows. Actually,

there exists a formal guideline for making such drawings, which we will ignore

here. Our main purpose is to show how you can transform a certain amount of

information into a structured drawing. Blocks will be used to represent a

production or service process. With arrows you can indicate the direction of

flow from one activity to another (e.g. transport of products or walking of

customers).

Flow diagrams can be used to analyse and record activities in an objective and

structured way. For example, a manager might use a flow diagram to determine

the time that passes between the moment the customer/product is taken into

“production” and the moment the customer/product is ready (i.e. throughput

time). In the next sections, we will discuss several quantitative performance

measures, which can be used to analyse activities with flow diagrams. Example

2.1 presents a flow diagram, illustrating the production process of a CD.

Example 2.1 Empty CDs arrive in the production system. In this production system a CD will be

automatically wrapped in a box. The CD will be transported to the labelling machine and

thereafter a label will be attached to the box. Next, the CD will be transported to a checking

point where the quality of the wrapping will be checked. It is decided if the product can be

transported to the warehouse or should be transported to an employee to be unwrapped and

wrapped and labelled again. Thereafter the CD will also be transported to the warehouse. The

related flow diagram is given in Figure 1 .

Incorporating data into a flow diagram

To study the performance of the production system we need to use quantitative

data related to the various production processes. We can incorporate

interarrival times, arrival rates, production times, travel times or storage times

into a flow diagram.

The interarrival time can be defined as the time between two subsequent

arrivals of products at their entrance in the process. The arrival rate indicates

wrapping

CD

labeling

warehouse

checking

unwrapping

wrapping

and

labeling

wrapping

CD

labeling

warehouse

checking

unwrapping

wrapping

and

labeling

wrapping

CD

labelinglabeling

warehousewarehouse

checkingchecking

unwrappingunwrapping

wrapping

and

labeling

wrapping

and

labeling

Figure 1:

Flow diagram

related to

example 2.1


4

the number of products that arrive per time unit (e.g. number of products that

arrive per hour). The interarrival time can be translated into the arrival rate. For

example, if the average interarrival rate is 10 minutes than 6 products will

arrive per hour. Clearly, the arrival rate thus equals 6 products per hour.

Both deterministic and stochastic values can be used to indicate production,

travel and storage times. Deterministic values are constant and do not react to

changes in the system. Stochastic values change over time and are usually

expressed in terms of a probability distribution. For example, the time required

for a certain production process follows a normal distribution with a mean of 6

minutes and a standard deviation of 30 seconds. Example 2.2 illustrates the

drawing of a flow diagram with data.

Example 2.2 Every 10 minutes an empty CD arrives in the production system. In this production system a

CD will automatically be wrapped in a box in 660 seconds. The CD will be transported to the

labelling machine in 1 minute and thereafter a label will be attached to the box. The time to

attach a label follows a normal distribution with a mean of 6 minutes and a standard deviation

of 1 minute. Next, the CD will be transported to a checking point in 1 minute where the quality

of the wrapping is checked in 2 minutes. On average 90% of the CDs can be directly

transported to the warehouse in 5 minutes. The remaining CDs are transported in 1 minute to

an employee to be unwrapped in 2 minutes. Next, the CD needs to be packed and labelled

again by an employee. The time to wrap and label again follows a normal distribution with a

mean of 8 minutes and a standard deviation of 2 minutes. Thereafter, the CD will be

transported in 5 minutes to the warehouse. The related flow diagram with important data is

given in Figure 2.

Other tools

Except for flow diagrams, the following tools exist to illustrate the movement

of people or products through a system (see also Heizer and Render, Chapter

7). Namely,

• time-function mapping: flow diagram in which the time is added on a

horizontal axis

• process charts: analysing the movement of people or material by using

symbols, time and distance

• Service blueprinting: used to focus on the customer and the interaction with

the customer

packingCD

labeling

warehouse

Quality check

unpacking Pack+labelOne CD

arrives

every 10

minutes

660 sec.Normal (6,1)

min2 min.

2 min.Normal(8,2)

min1 min.

1 min.

5 min.5 min.

1 min.1 min.

10%

90%

packingCD

packingCD

labeling

warehouse

Quality check

unpacking Pack+label

labeling

warehouse

Quality check

unpacking Pack+labelOne CD

arrives

every 10

minutes

660 sec.Normal (6,1)

min2 min.

2 min.Normal(8,2)

min1 min.

1 min.

5 min.5 min.

1 min.1 min.

1 min.

1 min.

5 min.5 min.

1 min.1 min.

10%

90%

10%

90%Figure 2:

Flow diagram

related to

example 2.2

Rate a in hourly rate . Indicating this is expressed in hour.


5

Exercises

Exercise 2.1 Consider economy passengers at an airport. Every minute a new passenger arrives. A

passenger walks in approximately 4 minutes to the 3 check-in counters. The passengers are

waiting in a single line for the counters. The time to check in follows a normal distribution

with a mean of 12 minutes and a standard deviation of 4 minutes. After the check-in procedure

the passengers go, with a travel time of 2 minutes, to the customs. The time required at the

customs equals 5 minutes. There is one customs officer. Thereafter, the passengers walk with

luggage trolleys to their gates in 10 minutes. Draw a flow diagram to indicate the movement of

the passengers. Incorporate all available data.

Exercise 2.2 Consider the production process, where 12 products arrive per hour. Products are randomly

assigned to process 1a or 1b. From multiple observations it is known that on average 60% of

the products is assigned to process 1a and 40% to process 1b. One operator is available at

process 1a. The operation time for process 1a follows a normal distribution with a mean of 3

minutes and a standard deviation of 1/2 minute. Conveyor A, capable of transporting a large

number of products simultaneously, transports a product in 9 minutes to process 2. Also one

operator is available for process 1b. The operation time for process 1b takes exactly 6 minutes.

Conveyor B, capable of transporting a large number of products simultaneously, transports a

product in 5 minutes to process 2. 4 operators are working in parallel at process 2. The

operation time per product equals exactly 30 minutes. Thereafter, product A leaves the system.

Draw a flow diagram to indicate the movement of the products. Incorporate all available data.

Exercise 2.3

From 9.00 am to 5.00 pm one customer arrives every 2 minutes. 60% of the customers need to

withdraw money from one automated teller machine. The time to withdraw money follows a

normal distribution with a mean of 2 minutes and a standard deviation of 30 seconds. 50% of

these customers want a receipt. It takes another 45 seconds to wait for the receipt to be printed.

After receiving the money and, if applicable, a receipt the customers leave the bank. The

second type of customers (30%) are customers that want to be helped by one clerk to withdraw

and/or deposit Euros. 45% need to withdraw Euros, 35% need to deposit Euros and 20% want

both. The time to withdraw Euros takes on average 4 minutes and the time to deposit takes on

average 5 minutes. After receiving the money the customer leaves the bank. The third type of

customers wants to withdraw or deposit foreign money. There is another counter with one clerk

where these types of customers are helped. The time to order foreign money takes 30 seconds.

Thereafter, the customer and the employee need to wait for 1 minute until the time-lock of the

safe is opened. Thereafter, the employee counts the foreign money and prepares a receipt for

the customer. The time to perform these two actions follows a normal distribution with a mean

of 2 minutes and a standard deviation of 15 seconds. Thereafter, the customer leaves the bank.

Draw a flow diagram to indicate the movement of the customers. Incorporate all available data.


6

3 Throughput times The throughput time (also called “manufacturing cycle time”, see Heizer and

Render, Chapter 16) can be defined as the time that passes between the

moment at which the product/customer enters the system and the moment at

which the product/customer is ready. For example, the time required for a

patient from entering the emergency room of a hospital until being treated and

having left the hospital.

We can estimate deterministic throughput times by simply adding up the

expected processing times of the different processes. In estimating the

deterministic throughput time we ignore any stochastic effects, such as

probability distributions for processing times and just use the average value

(also called mean or expected value). Furthermore, we ignore waiting line

effects. The calculated throughput time is, therefore lower than or equal to the

actual throughput time.

Example 3.1 Every 5 minutes a product arrives to be processed. The time required at machine 1 equals

exactly 4 minutes. With a conveyor, the product will be transported to process 2 in 10 minutes.

The time required at process 2 follows a normal distribution with a mean of 3 minutes and a

standard deviation of 1 minute. Thereafter, the product leaves the system. This production

process is illustrated in Figure 3.

We calculate the throughput time by just adding up the various average processing times. Thus,

the throughput time equals 4 + 10 + 3 = 17 minutes.

Note, that you should not include the interarrival time in your calculations, because until a

product has arrived, it is not yet in the system, which we intend to measure.

Throughput times with multiple paths

At the start of the system it might be “uncertain” which path will be followed

by the product or customer through the various processes in the system. For

example, the CD in the production system of Figure 1 may or may not need to

be unwrapped and rewrapped before being transported to the warehouse. For

path the probability is given that products/customers will follow this specific

path. In estimating throughput times you need to use these probabilities.

Follow the next steps in calculating throughput times with multiple paths:

1. Estimate the throughput time for one of the possible paths in the production

process

2. Multiply the resulting throughput time with the probability that this path

will be followed by a product/customer

3. Repeat steps 1 and 2 until all paths have been handled.

4. Estimate the throughput time by adding up the results of step 2 for all

paths.

In Out

Interarrival

time

5 minutes

Process 1

Time required:

Exactly 4

minutes

Process 2

Time required:

Normal (3,1)

minutes

Transport

(on conveyor)

requires

on average 10

minutes

In Out

Interarrival

time

5 minutes

Process 1

Time required:

Exactly 4

minutes

Process 2

Time required:

Normal (3,1)

minutes

Transport

(on conveyor)

requires

on average 10

minutes

In Out

Interarrival

time

5 minutes

Process 1

Time required:

Exactly 4

minutes

Process 2

Time required:

Normal (3,1)

minutes

Transport

(on conveyor)

requires

on average 10

minutes

Figure 3

Flow diagram

related to

example 3.1


7

Example 3.2 We calculate the throughput time for the production system with multiple paths as illustrated in

Figure 4.

Path 1: takes 4+10 + 3 minutes = 17 minutes

Path 2: takes 6 + 10 + 3 minutes = 19 minutes

Throughput time: 0.8*17 + 0.2 * 19 = 17.4 minutes

Throughput time estimates for different types of products in one system

If different types of products/customers follow processes in the same

production/service system, you might be asked to estimate a throughput time

for each type of product/customer.

Example 3.3 Both DVDs and CDs are handled in the same system. DVDs are wrapped in production

process 1a, whereas CDs are wrapped at production process 1b. Both CDs and DVDs are

transported by conveyor to production process 2 where they are labelled. All important data are

illustrated in the flow diagram in Figure 5.

The throughput time for a DVD equals 4 + 10 + 3 = 17 minutes.

The throughput time for a CD equals 6 + 10 + 3 = 19 minutes.

In Out

Interarrival

time

5 minutes

Process 1a

Time required:

Exactly 4

minutes

Process 2

Time required:

Normal (3,1)

minutes

Transport

(on conveyor)

requires

on average 10

minutes

Process 1b

Time required:

Exactly 6

minutes

80%

20%

In Out

Interarrival

time

5 minutes

Interarrival

time

5 minutes

Process 1a

Time required:

Exactly 4

minutes

Time required:

Exactly 4

minutes

Process 2

Time required:

Normal (3,1)

minutes

Time required:

Normal (3,1)

minutes

Transport

(on conveyor)

requires

on average 10

minutes

Transport

(on conveyor)

requires

on average 10

minutes

Process 1b

Time required:

Exactly 6

minutes

Time required:

Exactly 6

minutes

80%

20%

In Out

Interarrival

time

5 minutes

Process 1a

Time required:

Exactly 4

minutes

Process 2

Time required:

Normal (3,1)

minutes

Transport

(on conveyor)

requires

on average 10

minutes

Process 1b

Time required:

Exactly 6

minutes

DVD

CD

In Out

Interarrival

time

5 minutes

Interarrival

time

5 minutes

Process 1a

Time required:

Exactly 4

minutes

Time required:

Exactly 4

minutes

Process 2

Time required:

Normal (3,1)

minutes

Time required:

Normal (3,1)

minutes

Transport

(on conveyor)

requires

on average 10

minutes

Transport

(on conveyor)

requires

on average 10

minutes

Process 1b

Time required:

Exactly 6

minutes

Time required:

Exactly 6

minutes

DVD

CD

Figure 4:

Production

system with

multiple

paths

Figure 5:

Production

system

related to

example 3.3


8

Difference between actual and estimated throughput times

It is important to realise that there may be a large difference between a

deterministic estimate of a throughput time and the actual throughput time.

This has mainly to do with the fact that we have excluded the effect of waiting

lines (also called buffers) on the throughput time. So let us momentarily put

aside our assumption that all processing times are deterministic. We will use a

triangle (upside down) to depict a waiting line. Consider the following

example.

Example 3.4 Consider the production system depicted in Figure 6. Products arrive at this process at a rate of

7 product per hour (Poisson distributed). First, give a deterministic estimate of the throughput

time. Then determine the average time products spend in the buffer before process 1.

Out

In Process 1

1 machine

Time required per product

On average 7.5 min.

Neg. exponentially distributed

3 parallel operators


Normal (23,4) minutes



on average 14 min.

Process 3

Process 3

A deterministic estimate for throughput time means we ignore any waiting line effects (as we

did before), so: 7.5+23+14 = 44.5 minutes

The time product spend in the waiting line can be estimated with an M/M/1 waiting line model

(see e.g. Heizer and Render, Module D). λ = 7 products/hour and µ = 60/7.5 = 8 products per

hour. Then the time in the waiting line can be calculated as:

)( λµµ

λ

−=qW = 7/8=0.875 hours = 52.5 minutes

Example 3.4 clearly shows that the deterministic estimate of 44.5 minutes

underestimates the true throughput time significantly: already in the first buffer

products spend 52.5 minutes, so real throughput time is at least 97 minutes. In

general, a deterministic estimate of throughput time will always underestimate

true throughput time (i.e. the deterministic estimate is always lower than the

true throughput time). Deterministic throughput time estimates are useful

though for mutual comparisons, and simple because they provide information

that is very simple to obtain. Information on buffers, is generally much more

difficult to obtain. In example 3.4 we had the luxury of meeting the exact

requirements of an M/M/1 waiting line model for the first buffer, however, it is

far from straightforward to estimate the time products spend in the buffer

before process 2 (an M/G/S model may be helpful, but that is outside the scope

of this course). And it is even more problematic to estimate the time in the third

buffer, because we do not even know the distribution function for process 3.

Figure 6:

Production

system

related to

example 3.4


9

In practice, simulation is often used to be able to estimate performance criteria

for complex systems, including the effect of waiting lines (sse Heizer and

render, Module F).

Exercises

Exercise 3.1 Give a deterministic estimate for the throughput time (in minutes) of an economy passenger for

the situation as described in Exercise 2.1.

Exercise 3.2 Give a deterministic estimate for the throughput time (in minutes) of a product for the situation

as described in Exercise 2.2.

Exercise 3.3 Give a deterministic estimate for the throughput time (in minutes) for each of the three types of

customers (withdraw money at automated machine, withdraw/deposit Euros at clerk and

withdraw/deposit foreign money) for the situation as described in Exercise 2.3.


10

4 Design and effective capacity Capacity refers to the upper limit a process can handle. The goal of capacity

planning is to match supply and demand. There might be a gap between

achieved capacity and desired capacity. Overcapacity results in non-working

machines and employees and high operational costs. On the other hand,

undercapacity might result in loss of customers and strained machines and

staff. Also refer to Supplement 7 of Heizer and Render.

The design capacity of a production/service process indicates the theoretical

maximum output of the process in a given period of time. The design capacity

will be calculated for a process in an ideal situation. Thus, without waiting

times and by using deterministic processing times. In production processes,

design capacity is usually expressed in the number of units that can be

produced. In service processes capacity can be measured in varying terms such

as number of beds, number of patients treated and number of meals served. To

calculate the design capacity of a process in a given time period one can follow

the following steps:

1. determine the deterministic processing time of the process

2. translate the processing time into a production rate, which indicates the

number of units produced per time unit. If there are multiple

operators/machines working in parallel at a process you need to multiply

the production rate of a single operator with the number of

operators/machines to obtain the total production rate of that process.

(Note: use the same time unit in all steps)

Example 4.1 Consider a production process where the process time follows a normal distribution with a

mean of 3 minutes and a standard deviation of 1 minute. Determine the maximum number of

products that can be handled within 24 hours at this production process 2.

1. deterministic processing time = 3 minutes

2. production rate = 60/3 = 20 products per hour

3. design capacity per 24 hours = 24 * 20 = 480 products

What will be the design capacity if there are two operators in parallel working at this process?

1. deterministic processing time = 3 minutes

2. production rate = 60/3 = 20 products per hour; 2 operators in parallel results in a total

production rate of 40 products per hour

3. design capacity per 24 hours = 24 * 40 = 960 products per 24 hours.

Sufficiency of design capacity

For each process in a system we can determine the design capacity. An

important question for managers is if the design capacity of a process is

(theoretically) sufficient to handle the expected number of products that need

to be handled at that process. To answer this question, we need to compare the

number of arriving products at the process with its design capacity.

Follow the next steps to determine whether the design capacity is sufficient for

a process:


11

1. calculate the design capacity for the process

2. calculate the expected number of products arriving at that process (Note: if

products follow different paths, you need to take the related percentage into

account when calculating arrival rates for a process)

3. compare the answers to steps 1 and 2.

Example 4.2 Examine the production system with multiple paths in Figure 4. Is the design capacity of

process 1a sufficient?

Step 1:

Design capacity of process 1a: 60/4 = 15 products per hour

Step 2:

Every 5 minutes a product arrives: 60/5 = 12 products per hour.

80% follows process 1a: 9.6 products per hour.

Step 3:

Design capacity of process 1a: 9.6 products arrive per hour, while the machine can handle 15

products per hour: sufficient

Effective capacity

The effective capacity is the capacity that can be expected given the current

constraints. The difference between effective and design capacity may be

caused by, for example, the fact that a process was designed for a (slightly)

different product or has a known machine failure which impacts the

performance of the process. By subtracting such known productivity loss from

the calculated design capacity, we obtain the effective capacity.

Example 4.3 Consider a machine for which it is known that it is in repair for 10% of the time. The

machine’s design capacity is 150 products per hour. What is the effective capacity?

Effective capacity = 0.9 * 150 = 135 products per hour

Example 4.4 Consider a machine for which the following information is available. The machine’s design

capacity is 300 products per hour. However, due to the machine’s age, it is necessary to adjust

its settings after every 130 products. This adjustment requires 5 minutes. What is the effective

capacity of this machine?

During the adjustment, theoretically, 5/60*300 = 25 products could have been made. So after

making 130 products we miss production of 25 products. Thus, we make only 130 products out

of a theoretical amount of 155 products. The effective capacity is, therefore, 130/155 * 300 =

251.6 products per hour.

Exercises

Exercise 4.1


12

Determine the design capacity per hour of the processes “check-in” and “customs” at the

airport (expressed in number of economy passengers of the airport) as described in Exercise

2.1.

Exercise 4.2 Consider a gas station at which customers arrive by car between 9am and 9pm. There are three

types of customers, namely Euro 95, Diesel and LPG customers. 70% of the customers want

Euro 95, 20% want diesel and the rest want LPG. Every 0.75 minutes a customer arrives. 3 fuel

pumps contain both Euro 95 and diesel. Customers are waiting in one line for these 3 fuel

pumps. A fourth fuel pomp only contains LPG. Filling up a car follows a normal distribution

with a mean of 2 minutes and a standard deviation of 20 seconds irrespective of the type of

fuel. After filling up, the customer walks in 30 seconds to the shop. All customers need to pay

at one counter. 80% of the customers pay electronically, which takes 45 seconds. The rest of

the customers pay cash in 30 seconds. Thereafter, they walk back to their cars in 30 seconds

and leave by car the fuel pomp. After a customer has left, the next customer can start filling up.

a. During opening hours, how many customers can theoretically be handled by the Euro

95/Diesel pumps, how many customers can theoretically be helped by the LPG pomp and

how many customers can theoretically be helped by the counter?

b. Can you say anything about the number of customers that can be served by the system as

a whole? (if not, return to this question after you have read sections 6 and 7).

Exercise 4.3 Determine if the design capacity of the automated teller machine, the “Euro clerk” and the

“foreign money clerk” are, each individually, sufficient to handle the arriving customers at the

bank as described in Exercise 2.3.


13

5 Bottleneck The bottleneck of a system limits the output of the production/server system.

Thus, the bottleneck is the slowest process in the system and has less design

capacity than the prior and following processes. If none of the processes in the

system is the bottleneck, then we say that the arrival process is the bottleneck.

To determine the bottleneck in a system with a single path, you can follow the

next steps:

1. Calculate the design capacity of each process

2. Calculate the expected number of products arriving at the system

3.

a. if the design capacity of all processes is sufficient you can conclude

that the bottleneck of the system is the arrival process

b. if one or more processes have insufficient design capacity, select the

process with the smallest design capacity. Then, this process forms the

bottleneck in the system

To determine the bottleneck in a system with multiple paths, you can follow

the next steps:

1. Calculate the design capacity of each process,

2. Calculate for each process the expected number of products arriving at that

process,

3.

a. if the design capacity of all processes is sufficient, you can conclude

that the bottleneck of the system is the arrival process

b. Check for each path if one or multiple processes on that path have

insufficient design capacity; select the process with the smallest

design capacity. Then, this process forms the bottleneck of this path.

Example 5.1 Examine the production system in Figure 7. Every 4 minutes a product arrives at the

production system. Process 1 only starts processing products if three products have arrived at

the process (batch size = 3). The time to collect these three products before the processing can

start equals 5 minutes (setup time). The time per product in the batch equals 5 minutes.

Thereafter, products are transported on a conveyor to production process 2. Two operators are

working in parallel at this process. The time to handle a single product equals 10 minutes.

Thereafter, the product leaves the system.

Which of the processes is the bottleneck in the system?


14

System with a single path:

Step 1 Calculate design capacity:

Design capacity of process 1: processing time per batch = 3*5 + 5 = 20 minutes

60/20 = 3 batches per hour = 9 products per hour

Design capacity of process 2: 60/10 * 2 operators = 12 products per hour

Step 2 Calculate arrival rate

Arrival rate: 60/4=15 products per hour

Step 3: Check sufficiency

Both the design capacity of process 1 and 2 are insufficient.

process 1 has smallest capacity: bottleneck

Exercises

Exercise 5.1 Consider the system depicted in Figure 8:. Which of the processes in the system is the

bottleneck?

Exercise 5.2 Consider the system depicted in Figure 9:. Which of the processes in the system is the

bottleneck?

Out

Interarrival

time (of

individual

products):

7 minutes

Process 1

Time required:

4 minutes per product

Batch size: 2

Setup time per batch: 5 minutes

Process 2

Time required:


Batch size: 4


Tranport

(on conveyor)

requires

5 minutesOut

Interarrival

time (of

individual

products):

7 minutes

Interarrival

time (of

individual

products):

7 minutes

Process 1

Time required:


Batch size: 2


Time required:


Batch size: 2


Process 2

Time required:


Batch size: 4


Time required:


Batch size: 4


Tranport

(on conveyor)

requires

5 minutes

Tranport

(on conveyor)

requires

5 minutes

Out

Interarrival

time

5 minutes

Process 1


Time required

per product:

Normal(17, ½)

minutes

Process 3


Time required per

product:

exactly 12 minutes

Process 2

1 machine

Time required:

exactly 3 minutes

Out

Interarrival

time

5 minutes

Interarrival

time

5 minutes

Process 1


Process 1


Time required

per product:

Normal(17, ½)

minutes

Time required

per product:

Normal(17, ½)

minutes

Process 3


Process 3


Time required per

product:

exactly 12 minutes

Time required per

product:

exactly 12 minutes

Process 2

1 machine

Process 2

1 machine

Time required:

exactly 3 minutes

Time required:

exactly 3 minutes

In Out

InterarrivalTime per

individual

product

4 minutes

Process 1

Time required:5 minutes

per product

Process 2

2 operators

in parallel

Time required:

Exactly 10

minutes

Transport (on conveyor)

requires

on average 10

minutes

Batch size: 3 Setup time per batch:

5 minutes

5 minutesTime required:

minutes

on average 10

5 minutes

Figure 7: Production

system related

to

example 5.1

Figure 8:

Flow diagram

related to

exercise 5.1

Figure 9:

Flow diagram

related to

exercise 5.2


15

Exercise 5.3 Consider a bank during opening hours with three types of customers. Each two minutes a

customer arrives from 9.00 am to 5.00 pm. 60% of the customers need to withdraw money

from one automated teller machine. The time to withdraw money follows a normal distribution

with a mean of 2 minutes and a standard deviation of 30 seconds. 50% of these customers want

a receipt. It takes another 45 seconds to wait for the receipt to be printed. After receiving the

money and, if applicable, a receipt the customers leave the bank. The second type of customers

(30%) are customers that want to be helped by one clerk to withdraw and/or deposit Euros.

45% need to withdraw Euros, 35% need to deposit Euros and 20% want both. The time to

withdraw Euros takes on average 4 minutes and the time to deposit takes on average 5 minutes.

After receiving the money the customer leaves the bank. The third type of customers wants to

withdraw or deposit foreign money. There is another counter with one clerk where these types

of customers are helped. The time to order foreign money takes 30 seconds. Thereafter, the

customer and the employee need to wait for 1 minute until the time-lock of the safe is opened.

Thereafter, the employee counts the foreign money and prepares a receipt for the customer.

The time to perform these two actions follows a normal distribution with a mean of 2 minutes

and a standard deviation of 15 seconds. Thereafter, the customer leaves the bank. Which of the

processes in the system is a bottleneck?


16

6 Departure rate The departure rate (i.e. throughput) of a system indicates the number of

products/customers that leave the system per time unit. Clearly, the departure

rate is determined by the output of the bottleneck in the system. As a result, the

value of the departure rate only equals the value of the arrival rate if the arrival

process is the bottleneck in the system.

Example 6.1 Examine the production system in Figure 7.

a. What are the arrival and departure rates of the system per hour?

The departure rate of this production system is determined by the output of process 1 (see

example 5.1). Thus, the departure rate equals the output of process 1 per hour. The departure

rate is 9 products per hour and the arrival rate is 15 products per hour.

b. A new automated machine has been purchased for process 1 and its processing time is

exactly 3 minutes per product. Products will not be batched anymore but be treated

individually. What will be the new departure rate?

Design capacity of process 1 is 60/3 = 20 products per hour

Process 2 forms bottleneck of the system.

Departure rate = 12 products per hour

c. The management decides that a third operator will be used at process 2. What will be the

new departure rate? Design capacity of process 2: 60/10 *3 = 18 products per hour

Design capacity of process 1 is still 20 products per hour

Arrival rate is 15 products per hour

Arrival process forms bottleneck of the system.

Arrival rate = departure rate = 15 products per hour.

Exercises

Exercise 6.1

Consider the production process above. What is the expected departure rate per hour?

Exercise 6.2 Consider an airport during lunch time (12.00-13.00). There are two types of passengers,

namely business class (20%) and economy class (80%) passengers. Each minute a new

passenger arrives. A passenger walks in approximately 4 minutes to the check-in counter.

Economy class passengers are waiting in one line for the counters. 3 of the 5 counters are open

during lunch time. The time to check in follows a normal distribution with a mean of 12

minutes and a standard deviation of 4 minutes. At one of the counters new employees are

trained during lunch time. As a result, the check-in time of this specific counter is on average 2

times as high as at the other counters. For business class passengers there is one counter open

Out

Interarrival time:

4 minutes

Process 1

1operator

Required time

Normal(3 , ½)

minutes

Proces

(2 operators parallel)

Required time per product:

Exactly 10 minutes

Proces

1 machine

Required time:

Exactly 6 minutes

Out

Interarrival time:

4 minutes

Process 1

1operator

Process 1

1operator

Required time

Normal(3 , ½)

minutes

Required time

Normal(3 , ½)

minutes

Proces


Proces


Required time per product:

Exactly 10 minutes

Proces

1 machine

Proces

1 machine

Required time:

Exactly 6 minutes


17

during lunch time with higher skilled staff. The check-in time follows a normal distribution

with a mean of 10 minutes and a standard deviation of 2 minutes. After the check-in procedure

both types of passengers go, with a travel time of 2 minutes, to the customs. The time required

at the customs equals 5 minutes. There are separate counters for business and economy

passengers. Economy class passengers walk with luggage trolleys to their gates. Business class

passengers are transferred with a vehicle. Each vehicle has a capacity of 1 passenger. The total

time for a vehicle to travel with a passenger to a gate and to return empty to the customs equals

on average 13 minutes. Calculate the total number of business class passengers, which need a

vehicle during lunch time.


18

7 Utilisation and efficiency Utilisation can be defined as the fraction of total time in which a

machine/service-unit is used for production/service. More mathematically, we

define the utilisation rate for a single machine/operator as:

(7.1)

For example, consider an operator, who is busy for 70% of its time and idle for

30% of its time. The utilisation rate of this operator is 0.7.

Example 7.1 Consider a production process where every 5 minutes a product arrives. The production time is

4 minutes per product. The utilisation rate can be calculated according to the two options in

equation (7.1). Thus,

utilisation rate = 4/5 = 0.8

or

arrival rate = 12 products per hour; production rate = 15 products per hour;

utilisation = 12/15 = 0.8

In equation (7.1) we divide the total time in operation by the total time. It is

important to define these terms properly. In calculating operating time, we

need to include setup times of machines (e.g. time required to switch between

two colours).

Efficiency

We defined utilisation as actual output divided by design capacity. It is,

however, often already known in advance that there are certain additional

constraints, such as planned lunch breaks and machine failures. These

constraints limit the utilisation. To take these factors explicitly into

consideration, we define efficiency. Efficiency is a measure for the achieved

output compared to the effective capacity. If there are no additional constraints,

then efficiency equals utilisation (because then effective capacity equals design

capacity).

Example 7.2 Consider a machine, which is busy for 70% of its time and idle for 20% of its time. The

machine is in repair for 10% of the time.

The utilisation of this machine is 0.7

The efficiency of this machine is: 0.7 / (0.7 + 0.2) = 0.778

Utilisation rate for n identical parallel machines

We can also calculate the utilisation rate for several machines/operators

working in parallel at the same process. In this case, the utilisation rate depends

on the number of products that arrive at the process (arrival rate) and the

capacitydesign

output actual erator machine/op single a of raten utilisatio

time total

operationin timetotal erator machine/op single a of raten utilisatio

=

=

or equivalantly as:


19

number of units handled per machine/operator (production rate). The utilisation

rate ρ for a process consisting of n identical machines/operators can be

expressed as follows:

, (7.2)

where as λ = arrival rate and µ = production rate per machine/operator.

Note: use the same time-unit for λ and µ.

Example 7.3 Consider an airport with three check-in counters. 20 passengers arrive per hour at these

counters. Each employee can serve up to 9 passengers per hour. The utilisation rate of the

check-in counter equals:

Difficulties in calculating utilisation rates

Make sure that you read carefully whether the arrival rate or interarrival time

(similarly: production time versus production rate) has been given. Never

combine an arrival rate with a production time or an interarrival time with a

production rate in calculating utilisation rates.

Example 7.4

In calculating utilisation rates for serial processes, you need to locate the

bottleneck first. In this way, you can calculate the number of products arriving

at each of the processes.

µ

λρ

n=

74.09*3

20===

µ

λρ

n

Process 1arrival rate:

15 products per hourproduction rate:

20 products per hour

Process 1every 15 minutes

a product arrivesproduction rate:



utilisation rate = 4/20 = 0.2 or 3/15 = 0.2



















utilisation rate = 4/20 = 0.2 or 3/15 = 0.2Process 1every 15 minutes









20

Example 7.5 Consider the production process in Figure 10.

Per hour, 15 products arrive at the system, but due to the bottleneck at process 1 only 5

products arrive at process 2 per hour.

Utilisation of process 2 = 5/12 = 0.42

Productive utilisation rate

In calculating productive utilisation rates we exclude setup times from

operating times. Setup times occur, for example, if the paint cartridge of a

painting machine must be replaced.


Production time per batch: 4*3 + 1 = 13 minutes

Time elapsed until next batch is completed: 4*4 = 16 minutes

utilisation = 13/16 = 0.81

productive utilisation rate: 12/16 = 0.75

Relation between utilisation rate and throughput time

As we noted in Section 3, there may be a large difference between the

deterministic estimate of the throughput time and the actual throughput time.

This large difference is mainly caused by high utilisation rates. Consider, for

example, the following three systems consisting of a single machine with a

waiting line. Arrivals follow a Poisson distribution and processing times are

negative exponentially distributed (so we can use a M/M/1 waiting line model).

a. λ = 10 product per hour, µ = 30 products per hour

b. λ = 10 product per hour, µ = 15 products per hour

c. λ = 10 product per hour, µ = 11 products per hour

Process 1 Process 2

production rate:



arrival rate:


Process 1 Process 2

production rate:



arrival rate:


Process 1 Process 2

production rate:



arrival rate:


Process 1

time required: 3 minutes per product

production starts when 4 products have

arrived (batch size is 4)

setup time per batch: 1 minute

Interarrival time:

4 minutes

Process 1





Process 1





Interarrival time:

4 minutes

Figure 10:

Flow diagram

related to

example 7.5

Figure 11: Flow diagram

related to

example 7.6


21

The time products spend in the waiting line can be calculated as (see Heizer

and Render, Module D) WS = 1/(µ – λ). So the time in the system is:

a. WS = 0.05 hour = 3 minutes

b. WS = 0.2 hour = 12 minutes

c. WS = 1 hour = 60 minutes

As you can see, the waiting time increases rapidly if µ approaches λ. In the

extreme, if λ > µ, then in theory the waiting line would continue to grow

indefinitely. In practice, this will generally mean that processes upstream are

halted until there is sufficient space again in the buffer. This is the situation

where we speak of a bottleneck (refer to Section 5).

The example, we sketched here, holds in general. If the utilisation approaches 1

then the waiting time approached infinity. Note that “the utilisation approaches

1 is equivalent to the notion “µ approaches λ” we used before, because

utilisation = λ/µ. This relation is given graphically in Figure 12.

Exercises

Exercise 7.1 Consider a machine, which is busy 25% of its time, idle for 65% of its time and in repair for

10% of its time. What is efficiency for this machine?

Exercise 7.2

Consider the service process depicted above. What is the utilisation rate for the process?

Exercise 7.3 Consider a machine where each 10 minutes a single product arrives. The batch size equals 4

products. The set-up time per batch equals 6 minutes. The total processing time for processing

4 products equals 16 minutes. What are the utilisation rate and the productive utilisation rate

for this machine?

Arrival rate:

6 customers

per hour

Workstation with 3

parallel employees.

One employee can serve

a customer in exactly 8

minutes

Waiting time

utilisation

0 % 50 % 100 %

Figure 12:

Relation

between

utilisation and

waiting time


22


23

8 Work-in-progress The work-in-progress (WIP) indicates the number of products/customers that

have been taken into production/service but have not yet been finished.

Generally, we can not compute the WIP exactly. We distinguish between two

options to estimate the WIP.

Firstly, we can use Little’s equation:

L = λ * W (8.1)

where,

L = average WIP

λ = average number of products arriving per time-unit

W = throughput time

Note: use the same time-units for λ and W.


A product arrives each 5 minutes. λ = 1/5 product per minute.

Throughput time = 17 minutes (see Example 3.1)

WIP = λ * W = 1/5 * 17 = 3.4 products

Secondly, we can use equation (8.2).

(8.2)

where,

ρi = utilisation of machine/service-unit i

Xi = batch size on machine/service-unit i

n = number of machines/service-units

Example 8.2 Consider the production process in Figure 13:.

WIP = 2/4*1 + 13/16*4 = 3.75 products (see example 7.6 for utilisation of process 2)

Exercises

Exercise 8.1

∑=

n

i

ii X1

*ρ

Out

Interarrival time

(of individual products):

4 minutes

Process 1 Process 2

Required time:

Exactly 2 minutes

Required time:


Batch size = 4

Set-up time = 1 minute

Out

Interarrival time


4 minutes

Process 1 Process 2

Required time:

Exactly 2 minutes

Required time:


Batch size = 4


Out

Interarrival time


4 minutes

Process 1 Process 2

Required time:

Exactly 2 minutes

Required time:


Batch size = 4


Figure 13: Flow diagram

related to

example 8.2


24

Consider the process depicted above. Determine the average Work-In-Progress with Little’s

equation?

Exercise 8.2 What is the average number of economy passengers (WIP) at the airport as described in

Exercise 2.1?

Out

Interarrival time


7 minutes

Process 1 Process 22 operators in parallel

Required time:



On average 11 minutes


Requires on average 8 minutes

Out

Interarrival time


7 minutes

Process 1 Process 22 operators in parallel

Required time:



On average 11 minutes


Requires on average 8 minutes


25

9 Solving bottleneck problems If capacity exceeds demand, we may stimulate demand by, for example,

reducing production or service tariffs. If demand exceeds capacity we may

want to increase capacity by making staffing changes or by adjusting

equipment and by redesigning processes and products.

Solving bottleneck problems can be viewed in different time horizons. On a

long term planning we can adjust capacity by adding facilities and equipment

and by adjusting the product design. On an intermediate planning level setup

times can be decreased or personnel can be trained to achieve a higher output.

To modify capacity on the short run already existing capacity can be used more

wisely by scheduling staff and jobs in a different way (for example, by using

over-time and working in two groups), or by using larger batch sizes or by

outsourcing activities to other companies .

In this section we treat two examples for managers to solve capacity problems.


How many operators (servers) are required at process 2 to ensure a departure rate of at

least 20 products per hour?

Production rate of process 2 with one operator = 4 products per hour

To achieve a production rate of 20 products per hour, at least 20/4 = 5 operators need to work

in parallel. However, if we use 5 operators, the utilisation will be exactly 1.0. If there is only a

small amount of uncertainty in the system (quite realistic), 5 operators will be insufficient.

Therefore, we need a minimum of 6 operators. Did you notice an additional problem? The

arrival rate is not high enough to achieve the required departure rate.

In Out

Average

interarrival

time:

8 minutes

Process 1

Time required:

Exactly 3 minutes

Process 2

(x operators

in parallel)Transport

(on conveyor)

requires

on average 10

minutes

Time required

per product

exactly 15 minutes

In Out

Average

interarrival

time:

8 minutes

Process 1

Time required:

Exactly 3 minutes

Process 2

(x operators

in parallel)Transport

(on conveyor)

requires

on average 10

minutes

Time required

per product

exactly 15 minutes

Figure 14:

Flow diagram

related to

example 9.1


26


What is the minimum batch size to prevent process 1 from being a bottleneck?

Process 1 is not a bottleneck if the utilisation rate is smaller than 1.

Utilisation = (3*X + 10)/4*X < 1

If X = 10 then both arrival and process 1 are bottleneck. So, the minimum batch size is 11.

Exercises

Exercise 9.1

Consider the process depicted above. What is the minimum batch size to prevent process 1

from being a bottleneck?

Process 1


batch size = X

set-up time per batch = 10 minutes

Interarrival time:


Process 1


batch size = X


Process 1


batch size = X


Interarrival time:


In Out

Arrivals:

10 products

per hour

Process 1

Time required: 4.2 minutes per product

Production batch size is x

Set-up time per batch: 25 minutes

Figure 15:

Flow diagram

related to

example 9.2


27

10 Large exercises

Exercise 10.1A A manufacturer of car audio equipment is planning to produce and to sell a new type of car

radio next year. It will be a simple model that will be offered for a very fair price. To restrict

the costs of these radios, they will be assembled from modules that are also used for other

types. For the production of the radios only one assembly line will be needed.

The above figure shows a schematic overview of the assembly line, as it has been designed for

the producer by an engineering consultant. Assembly will take place in three steps. In the first

step (assembly 1) the supplied empty casing will be provided with a printed circuit board

(PCB) and the mechanics for a CD player (the lining or interior). In the second step (assembly

2), this printed circuit board is soldered to the casing, after which the whole is mounted with a

cover. Two parallel workstations are available for this operation. The radio is tested in the third

step and a label with the type and serial number will be attached (assembly 3).

The following facts are known about the assembly process:

• An empty casing arrives at the assembly line on average every 150 seconds

• The assembly of the interior work takes 120 seconds

• Soldering the PCB and mounting the cover takes 260 seconds

• Testing and labelling on the radio takes 140 seconds

• Printed circuit boards and covers are always sufficiently available

Use “Deterministic Performance Estimation” techniques to determine:

• The number of radios produced in 4 weeks (1 week consists of 5 working days; each

working day consists of 8 working hours)

• The throughput time of radios

• The utilization rates of the three different operations

• The average amount of work in progress (WIP)

Exercise 10.1B One of your colleagues has built a simulation model of the radio assembly as described in

question A. In this simulation, she could do what you could not when doing these calculations

by hand: include stochasticity (i.e. include uncertainty by specifying a probability distribution

of the arrival times and the production times). Some of the results from her simulation model

are as follows:

Total production: 3834 radios

Throughput time: 973 seconds

Utilization rate interior work: 0.79

Utilization rate soldering: 0.85

Utilization rate testing: 0.94

WIP: 5.2 products

Can you explain the differences with the values that you calculated?

Car-radio assembly

buffer 1 buffer 2

Interior work

Cover plates

assembly 1 assembly 3 buffer

assembly 2

assembly 2


28

Exercise 10.2A The figure below shows a scheme of the production line of scissors. The basic material for

scissors is stainless steel, which is offered in plates. A punching machine can make 10 half

pairs of scissors out of one plate at once. Two punching machines are used, one for either half.

Afterwards, the half pairs of scissors are deburred. The half pairs of scissors are then grinded

on a grinding machine, after which they are ready for assembly. Hereto, two workstations are

available where both half pairs are being assembled to become one pair of scissors. Then, the

scissors are put on a conveyor to be transported to an inspection station where they will be

tested thoroughly. The certified scissors will go to a packaging machine, which will pack them

per 10 in a box. The unapproved scissors are thrown into a garbage bin, which will be emptied

when 4 disapproved scissors are in it.

The following is known of the production process:

• On average every 2.5 minutes a plate is being delivered.

• The rejection percentage is 5% on average.

• A batch consists of 10 half pairs of scissors. All operations take place batchwise. In the

assembly, a batch is built up of 2x10 half pairs of scissors = 10 scissors. The assembly of

20 half pairs of scissors takes place on one of the machines and lasts 380 sec on average.

• The capacity of the buffers between the operation steps is 40 scissor-halves (4 batches) per

buffer.

Furthermore, the following times are known:

Parameter Distribution Mean Std. deviation

Interarrival time (of plates) Exponential 150 150

Punching per plate Normal 160 5

Deburring per batch Normal 75 15

Grinding per batch Normal 130 30

Assembly per batch Normal 380 120

Inspection per pair of scissors Normal 19 5

Packaging per 10 scissors Normal 150 30

B

B

B

B

B

In

Out

Punching

Punching

Deburring Grinding

Inspection Packing

Assembly

Assembly waste

Scissors factory

B

B


29

Calculate, with “Deterministic Performance Estimation” techniques the following performance

indicators

• The daily production of the scissors factory (1 day consists of 8 working hours)

• The throughput time of scissors in the production process.

• The utilization rates of the 6 different operation steps.

• The total amount of Work in Progress (WIP).

Exercise 10.2B The board of directors of the factory is considering decreasing the prices of the scissors by

15%, since a market survey showed that by this decrease, the market share could be increased

by 50%. Certainly production will have to be raised to achieve this. The board proposes to

simply decrease the average interarrival time of plate material from 150 to 100 seconds. Will

this measure increase the output by the desired 50%?

Hint: determine the bottleneck and calculate the design capacity of this bottleneck.


30

Answers to exercises

Flow diagrams

OutArrival process:arrival rate:

12 per hour

Time required:

Normal(3, ½)

minutes


exactly 30 minutes

Tranport

(on conveyor A)

requires

on average 9

minutes60%

Tranport

(on conveyor B)requires

on average 5

minutes

40%

Process 1a

1 operator

Process 1b

1 operator

Time required:

6 minutes

Process 2

(4 operators in parallel)

Answer

exercise 2.1

10 min

Answer

exercise 2.2


31

Throughput times

Throughput time of economy passenger =

4 + 12 + 2 +5 +10 = 33 minutes

Throughput time of a product:

step 1 and 2:

path 1: 0.60*(3+9+30) = 25.2

path 2: 0.40*(6+5+30) = 16.4

step 3:

25.2 + 16.4 = 41.6 minutes

Withdraw at automated teller: 2 + ½*¾ = 2.375 min

Withdraw/Deposit Euro: 0.45*4 + 0.35*5 + 0.20*9 = 5.35 min

Withdraw/Deposit foreign: ½ + 1 + 2 = 3.5 min

Design and Effective Capacity

Design capacity check-in:

step 1: deterministic processing time = 12 min

step 2: production rate = 5 passengers per hour per counter

production rate = 15 passengers per hour in total

step 3: design capacity = 15 passengers per hour

Design capacity customs:

step 1: deterministic processing time = 5 min

step 2: production rate = 12 per hour

step 3: design capacity = 12 passengers per hour

arrivalinterarrival time: 2 minutes

between 9am and 5 pm

leave

withdraw

automated tellerNorm (2,1/2) min

50% 50%

receipt 3/4 min

60%

withdraw/deposit

clerk

45%

35%

20%

With

draw

4 m

in

dep

osit

5 m

in

both

9 m

in

foreign clerk

1/2 + 1 + Norm(2,1/4) min

30%

10%

Answer

exercise 3.1

Answer

exercise 2.3

Answer

exercise 3.2

Answer

exercise 3.3

Answer

exercise 4.1


32

a. Design capacity Euro95/diesel:

step 1: processing time of car at fuel pomp = time to fill up + time to walk to shop + time to

pay + time to walk to car = 2 + 0.5 + 0.80*0.75 + 0.20*0.5 + 0.5 = 3.7 min

step 2: production rate = 60/3.7* 3 fuel pumps = 48.6 cars

step 3: design capacity during opening hours = 12 * 48.6 = 583 cars

Design capacity LPG:

step 1: processing time of car at fuel pomp = time to fill up + time to walk to shop + time to

pay + time to walk to car = 2 + 0.5 + 0.80*0.75 + 0.20*0.5 + 0.5 = 3.7 min

step 2: production rate = 60/3.7* 1 fuel pumps = 16.2 cars

step 3: design capacity during opening hours = 12 * 16.2 = 194 cars Design capacity counter:

step 1: processing time = 0.80*0.75 + 0.20*0.5 = 0.7 min

step 2: production rate = 60/0.7* 1 counter = 85.7 cars

step 3: design capacity during opening hours = 12 * 85.7 = 1028 cars

b. Total number of customers arriving during opening hours = 960.

90% wants Euro95/diesel = 864 customers. However, only 583 can be handled.

10% wants LP = 96 customers. All LPG customers can be served.

Total number of customers that get to the counter during opening hours = 583 + 96 = 679

customers. This is less than the design capacity, thus they can all be served at the counter.

Total number of customers served during opening hours = 583 + 96 = 679 customers.

Step 1:

• Deterministic processing time automated teller machine: 2 + ½*3/4 = 2.375 min per

customer; design capacity = 202 customers per 8 hour

• Deterministic processing time Euro clerk: 0.45*4 + 0.35*5 + 0.20*9 = 5.35 min; design

capacity = 89 customers per 8 hour

• Deterministic processing time foreign clerk: ½ + 1 + 2 = 3.5 min; design capacity = 137

customers per 8 hour

Step 2:

• Arrival process = 60/2 = 30 customers per hour * 8 hours = 240 customers

• Automated teller machine = 0.60 * 240 = 144 customers per day

• Withdraw/Deposit Euro clerk = 0.30 * 240 = 72 customers per day

• Withdraw/Deposit foreign clerk = 0.10 * 240 = 24 customers per day

Step 3:

• Automated teller machine: sufficient

• Withdraw/deposit Euro clerk: sufficient

• Withdraw/deposit foreign clerk: sufficient

Bottlenecks

System with single path:

Step 1:

Design capacity process 1:

deterministic processing time for 2 products = 4*2 + 5 = 13 minutes

production rate = 4.6 batches per hour

design capacity = 9.2 products per hour


deterministic processing time = 4*5 + 9 = 29 min

production rate = 2.07 batches per hour


Step 2: Arrival rate:

60/7 = 8.6 products per hour

Step 3: Thus, design capacity process 1 is sufficient and design capacity process 2 is

insufficient. Process 2 is bottleneck.

Answer

Exercise 4.2

Answer

exercise 4.3

Answer

exercise 5.1


33

System with single path:

Step 1:


deterministic processing time = 17 minutes

production rate = 60/17* 3 operators = 10.6 per hour



deterministic processing time = 3 min

production rate = 20 products per hour

design capacity = 20 products per hour


deterministic processing time = 12 min

production rate = 60/12 * 2 operators = 10 products per hour

design capacity = 10 products per hour

Step 2: Arrival rate:

60/5 = 12 products per hour

Step 3:

Thus, design capacity process 1 is insufficient, design capacity process 2 is sufficient and

design capacity process 3 is insufficient. Design capacity process 3 is smaller than design

capacity process 1, process 3 is bottleneck.

System with multiple paths:

Steps 1, 2 (see answer exercise 4.3):

• Automated teller machine: sufficient

• Withdraw/deposit Euro clerk: sufficient

• Withdraw/deposit foreign clerk: sufficient

Step 3:

Design capacity all processes is sufficient: Arrival process is bottleneck

Departure rate

design capacity process 1: 20 per hour



arrival rate: 15 per hour

bottleneck: process 2

departure rate = 10 products per hour

arrival rate: 60 passengers per hour of which 20% is a business class passenger = 12 per hour

design capacity check-in: 6 passengers per hour

design capacity customs: 12 passengers per hour

bottleneck: check-in, consequently departure rate = 6 business class passengers which require a

vehicle per hour

Utilisation and efficiency

efficiency = 0.25/(0.25 + 0.65) = 0.28

utilisation rate =

27.0

)8

60(*3

6

*==

µ

λ

n

Answer

exercise 5.3

Answer

exercise 5.2

Answer

exercise 6.1

Answer

exercise 6.2

Answer

exercise 7.1

Answer

exercise 7.2


34

Production time per batch = 16 + 6 = 22 minutes

Time elapsed = 4*10 = 40 minutes


productive utilisation rate = 16/40 = 0.4

Work-in-Progress

W = 3 + 8 + 11 = 22 minutes

λ = 1/7 per minute

WIP = 1/7 * 22 = 3.14 products

W = 33 minutes (see exercise 3.1)

λ = 1 passenger per minute

WIP = 33 passengers

Solving bottleneck problems

Process 1 is not a bottleneck if the utilisation rate is smaller than 1.

Utilisation = (4.2x+25) / 6x

1 > (4.2x+25) / 6x

6x > 4.2x+25

1.8x > 25

x > 25/1.8

x > 13.9

So, minimum batch size is 14.

Large exercises

indicator expected tot. radios 4 weeks=576000 sec. prod. rate: 1 per 150 sec → 3840

throughput time 120+260+140 = 520 sec

ut.rate interior work 120/150 = 0.8

ut.rate soldering 130/150 = 0.87

ut.rate testing 140/150 = 0.93

WIP λ (average # prod/sec) = 1/150 S=520, L=λW= 3.47

ΣρiXi = Σut.rates*batchsize = 0.8*1+0.87*1+0.87*1+0.93*1= 3.47

To explain the effect that occurs, imagine the following. There is a process that requires

exactly 5 minutes per product. Products arrive at this process every 6 minutes. How long do

you expect the queue to be? The queue length will be zero, because every time the process is

ready, it will need to wait 1 minute until the next product arrives. Now assume that the arrival

times are uncertain. The average is still 6 minutes, but the time between two arrivals may

vary. Suppose 3 products arrive shortly after each other. Then 1 product is taken into

production, and the other 2 products must wait. So now there is a queue! On the other hand,

you may ask “But what happens if the time between arrivals is temporarily larger?” In our

initial situation, the queue length was zero. If even fewer products arrive, then the queue length

will still be zero (it clearly cannot become negative). These quiet periods can only be used to

reduce the length of the waiting line that appeared due to the first effect. In total we, therefore,

see an increase in the queue length. If the queue length increases, this means that products are

waiting, which will increase the throughput time and the WIP. The effects in the situation

described in this exercise are similar, only now we have the assembled effect of many

uncertainties. The basic line of reasoning is, however, the same. Uncertainty usually increases

Answer

exercise 7.3

Answer

exercise 8.2

Answer

exercise 8.1

Answer

exercise 9.1

Answer

exercise 10.1A

Answer

exercise 10.1B


35

WIP and throughput times. Note that in this case the simulated throughputtime is 87% higher

than the deterministic throughput time.

Then why are the utilization and the total production almost equal between the simulation and

the calculation (if you would run the simulation for a couple of years, then the numbers will be

even closer to each other)? Let’s take the same example as before. The process requires exactly

5 minutes per product. One product arrives every 6 minutes. The utilization is 5/6 = 0.83. Now

assume that there is uncertainty in the production time, 50% of the time it is 2 minutes and

50% of the time it is 8 minutes. What is the utilization? Simply (0.5*2+0.5*8)/6 = 0.83. The

same holds if you introduce other probability distributions (e.g. a normal distribution).

Sometimes the process is faster, sometimes slower, but the average remains the same.

indicator expected daily production 91.2 boxes

throughput time 914 (=160 + 75 + 130 + 380 + 19 + 150)

ut.rate punching 0.533

ut.rate deburring 0.50

ut.rate grinding 0.867

ut.rate assembly 380/(4∗150) = 0.633

ut.rate inspection 0.633

ut.rate packaging 0.475

WIP 30.5 scissors (30.22 via Σi ρiXi)

Explanation:

• Daily production. The bottleneck is the interarrival rate, 150 seconds. Each 150 seconds, 1

plate is being delivered. 1 plate = 10 half pairs of scissors = 5 scissors. Of these, 5% is

rejected. So, on average only 0.95 * 5 = 4.75 scissors can be produced per 150 seconds.

Thus, the daily production is 3600 ∗ 8 ∗ 4.75 / 150 = 912 scissors = 91.2 boxes. Note that

production is always measured in terms of the end products, which are boxes of 10 scissors

in this case.

• Utilization. When computing the utilization rate for packaging, you have to take the

rejection percentage into account as well as the fact that the arrival time for packaging is

300 seconds, because 2 batches are needed and the arrival time of batches is 150 seconds.

So the utilization rate for packaging is (150*0.95)/300 = 0.475.

• WIP via Little’s formula (L=λW): W = 914, λ = 5 products/150 seconds → L = 1/30*914 =

30.5 products. It is important to express both W and λ in the same units (we used seconds).

• WIP = Σi ρiXi = 0.533 ∗ 5 + 0.533*5 + 0.5 ∗ 5 + 0.867 ∗ 5 + 0.633 ∗ 10 + 0.633 ∗ 10 +

0.633*1 + 0.475 ∗ 10 = 30.22.

The new bottleneck is grinding with an operating time of 130 seconds. So the daily production

is now 3600*8*5 (number of scissors per plate) * 0.95 (percentage of good scissors) / (130 *

10 (10 scissors per box)) = 105.2 boxes. Note that the daily production is exactly the

production of the bottleneck, grinding. Increasing the daily production by 50% is therefore not

possible without adapting the current machinery (more specifically, at least one extra grinding-

machine would be needed to increase the production).

Answer

exercise 10.2A

Answer

exercise 10.2B

deterministic performance estimation

Documents

large exercises

productive utilisation

estimated throughput

utilisation rates

throughput time estimates

departure rate

effective capacity

flow diagram