analysis of an operation

8/13/2019 Analysis of an Operation

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ANALYSIS OF AN OPERATION

LEONARD N. STERN SCHOOL OF BUSINESS


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complete each process step also depends on the number of units that are to be processed.Thus, it is important to write down the unit of work. For example, if the unit of work isone cookie then the time will be given to mix or bake one (a single) cookie. Otherwise ifthe unit of work is a dozen cookies, the time will be specified at each process step for 12cookies1.

Capacity and cycle time: The flow chart of a process is a static picture. To be able toanalyze some of the dynamics of a process we need to define capacity and cycle time.The capacity of a process is the maximum rateat which output can be created given aninfinite supply of inputs and orders. Cycle time can be defined both for a process and aprocess step. The cycle time for a process is defined to be time between successive ordercompletions when the process is operating at capacity. Thus,

cycle time of a process = 1/capacity of the process.The cycle time for a process step is the time to process a unit of work.

We begin with some simple examples. In these examples there is a single resource and asingle process step. This will therefore b the constraining resource and hence called thebottleneck resource. For example consider the capacity of a process called heatingwater to make tea. Assume that a kettle can hold one gallon of water and that the time ittakes to heat water to the appropriate temperature is five minutes. We say that the cycle

time for this process is five minutes. The capacity of heating is 1 gallon per fiveminutes and in an hour we can do 12 cycles (60/5). The resultant capacity of this task is12 gallons per hour. Another example is packing a computer. If it takes on the average 12minutes to pack a computer, the capacity of the process is 1 computer in 12 minutes or 5computers per hour. The cycle time is 12 minutes.Remember that Capacity is a rate itis the maximum rate at which work can be done. We did not explicitly consider theresources involved in these examples. In the first, the resources could be the cookingrange and the kettle. As both are occupied all the time (if we are working to capacity) thetwo are considered to be bottleneck resources. In the second example, assume thepacking is done by one person. The resource is labor. The labor is the bottleneck resourceas it is the single resource.

It is quite important to understand that the cycle time is the average time necessary to

carry out a process for a unit of work For example, the cycle time is five minutes pergallon and 12 minutes per computer in the two examples. If we place 50 cookies all atonce in an oven to bake and the oven completes the process in 50 minutes, the cycle timeis one minute per cookie. (To see this notice that the oven bakes 50 cookies in 50 minutesor one cookie per minute. That is the ovens capacity. Cycle time = 1/capacity = 1 /(1cookie/minute) = 1 minute per cookie.) However, if in the last example the unit foranalysis is 50 cookies then the cycle time is 50 minutes (that is it takes 50 minutes tobake 50 cookies). Obviously, cycle time and capacity depend upon the unit of work(sometimes also called unit of analysis). Usually, capacity is easier to determine but notalways. The two concepts are related as already explained. The cycle time is defined to bethe inverse of capacity. It has time as its unit of measurement.

How do we measure cycle time of a process? One method is to ensure an infinite supply

of inputs and orders, then to stand at the output end of the process and record the timebetween orders (parts, cars, satisfied customers, etc.) exiting the process. For example, ifwe observe that a car comes off the end of an assembly line roughly one every oneminute, then the cycle time = 1 minute per car. The capacity is 1 car per minute. Or, thecapacity of the assembly line is 60 cars per hour or about 240,000 cars per year for two 8hours shift operation and 250 working days per year. There are issues of quality involved

1Sometimes the unit of analysis will vary from process step to process step. This might happen if several orders arecombined to exploit economy of scale at some process step.


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in these definitions, should we count the good parts or all parts? What is the correctdefinition? For our purposes we shall not count bad or defective parts as output.

Cycle Time: When we say that the cycle time of a machine is 10 minutes per part, wemean that if the machine were never starved for inputs it would produce on the averageone good part every ten minutes. The notions of on the average andgood partare verycritical in this definition. Cycle time is the average time between parts exiting the processor task (if you were to stand and observe it over a long period of time and the process

were producing steadily at the same rate.)

Before we discuss how to determine capacity when there are several process steps orresources involved, we will formally define and measure capacity and cycle time. Oneapproach is to consider the system you wish to observe as a black box. Give it anunlimited supply of inputs. LetN(t)be the number ofgood parts produced by the systemin time t. Then the ratio,N(t)/t is a measure of the capacity, when t is sufficiently large.Why do we need the caveat that t be large? The reason is we want to obtain the averagerate as t goes to infinity. To clarify this point, say a machine were given an unlimitedsupply of inputs. It is then observed to produce 60 parts all at once every 60 minutes.Unless we observe for 60 minutes we would say that the machine is not working! Checkthat the capacity for this example is 60 parts per hour or one part every minute and that

the cycle time is one minute per part. As cycle time is the inverse of the capacity,t/N(t) isthe measure of cycle time for sufficiently large t.

In these definitions, we have assumed that carrying out a process does not interfere withother processes. In some operations there may be interference, i.e., if a person is assignedto perform two processes, the same person often cannot attend to these two tasks at thesame time. We will discuss this point in more detail in our section on bottlenecks. Thedefinition of capacity has to be modified in such circumstances. When multipleconstraining resources such as space, labor and machine are involved, the calculation ofcapacity is no trivial matter.

The notion that subdivision of processes may be necessary for capacity analysis leads tothe definitions of work area and system. In this note we shall also see how setup time,

product mix and quality affect the capacity of a system.

Work Area: A grouping of similar machines or equipment is called a work area. Forexample, a collection of drilling machines represents the drilling work area at a Fordplant. A collection of tellers comprises the customer banking work area at a Citibankbranch. A collection of order takers comprises the customer order call center at Amazon,Dell, Lands End, etc.

Flow Time:The flow time is the time spent by a typical part, job, (order), or customer inthe system. For example, if customers are processed by a teller in batches of five (oneafter another but no one can leave until all five are attended to!) and it takes the teller fiveminutes to serve a customer, the cycle time is five minutes per customer, the capacity is12 customers per hour and the flow time is 25 minutes for an order of five customers. If

they are processed one by one and there is no waiting (customers are nice enough toarrive evenly spaced) then the capacity and cycle time are unchanged but the flow time isfive minutes. (It is harder to determine flow time when all five customers arrive at thesame time and there is only one teller. The average flow time, (remember that all but thefirst customer encounters waiting time) is given by (5+10+15+20+25)/5 = 15 minutes.)

How do you measure flow time? A practical approach is to tag jobs or customers as theyenter the system. Write down the time when they entered. Record when they leave. Flowtimes for cars in an auto plant can vary from 12 to 24 hours often depending on how longthey stay in the paint shop. Understanding the source of the variability is a good starting


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point to understanding the dynamics of the system. It is a good exercise to constructexamples where the flow time is not equal to the cycle time. One example is when jobs orcustomers are processed in batches. Another example is when there is a queue and acustomer or order has to wait until served. This waiting time becomes part of the flowtime. It is clearly possible that each order going through a process could have differentflow times. Thus flow time is not a unique number but a distribution. In this case to set aservice guarantee we must decide the percent of the customer we want to satisfy withinthe given time guarantee in a make to order system. We call this percent the customerservice level. In a make to stock or assemble to order system the service level will bedetermined by the flow time distribution as well as the level of finished goods orcomponent inventories carried. We cover these topics when we discuss inventorymanagement.

Bottleneck: The resource with the smallest capacity in the system is called thebottleneck. The astute reader will note that we use the term resource and not processstep in this definition. The reason is that the same resource may be required for carryingout two processes. For example, consider the example in which, one person verifiessignatures on checks and then subsequently the same person verifies the balance in thecustomer's account. There are two process steps, but the bottleneck is the single person.Clearly having two people do the two jobs will increase the capacity. We shall assume

unless stated that each process is carried out using a separate resource (or separate set ofresources if there are more than one identical machines or workers attending to the sameprocess).

However we briefly describe how to analyze a process when the same resource (or set ofidentical resources) is used to carry out several processes. When the same resource isused to perform different processes do not compute the capacity of a process step butinstead make a list of all resources. For each resource list the processes that have to beperformed by that resource. Then determine the work that has to be done (in units oftime). Finally, determine the capacity using the formula, capacity = time availabledivided by time required (work) per order. A sample table is provided below:

Making breakfast at a Deli. Unit of work = 1 customer order

Resource NumberAvailable

Process stepswhere needed

Time requiredper unit of work

(mins)

Capacity(units per

hour)

Personmakingcoffee

1 Make coffee 2 mins. 30

1 Make toast 2.5 8.57

Make eggs 4

Personmaking hotbreakfast Wrap 0.5

Range2 1 Make eggs 6 10

The approach suggested above is not difficult but attention to detail is important.

Particularly, for each process step we must know what resources are needed and howmuch of time is needed on each resource. It is also important to write down the unit ofwork. In fact, as we shall see in our examples, the bottlenecks mayshiftdepending on thedemand pattern (product-mix) and the lot size. Improper scheduling also can createtemporary and shifting bottlenecks.

2The range is needed for more time than the person cooking the eggs.


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System Capacity:The capacity of a system is the capacity of the bottleneck. The cycletime of the system is the cycle time of the bottleneck. Caution the capacity of thesystem is not the sum of the capacities of all resources but the capacity of the mostlimiting resource. In the table above, the limiting resource is the person making hotbreakfast.

System Flow Time:The total time spent in the system by a typical order is the systemflow time. It is notnecessarily equal to the sum of flow times of all process tasks in thesystem for one simple reason: processes can be done in parallel. For example, considerthe processes for attending to your car comprising: filling the gas tank, checking thewater in the radiator and topping it up if necessary and checking the water in the batteryand topping it up if necessary. These processes can be done sequentially (serially-oneafter another) or simultaneously (in parallel- all at the same). If these processes are donesequentially and each takes five minutes, a customer can be served in 15 minutes. If,however, they are all done simultaneously, a customer can be served in five minutes.What are the labor implications of the two approaches?

Idle Time: The idle time of a resource is the time during which work is not being doneby a resource.

Transfer Time: This is the time during which the order is being moved for one workarea to another. Typically no work is being done and no value is being added.

EXAMPLE I

In example I, parts are produced every five minutes. The cycle time is five minutes perpart for the system since Machine A is the bottleneck operation, the flow time is sevenminutes in the system and the capacity is 12 parts per hour. This system is not balancedbecause the two machines A and B have different cycle times. It is indeed rare to see aperfectly balanced operating system. This is why the notion of a bottleneck is useful. Ifthe system is fully utilized, then machine B is idle for three minutes in every fiveminutes, and we say that machine Bs capacity utilizationis 2/5 or 40%.

A B

Fi n i s he d Goods

C y c l e

T i m e = 5

C y c l e

T i me = 2

Machine Machine


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EXAMPLE II

A

A

B

Finished

G o o d s

M a c h in e C y c le

T i m e = 5

W o r k A r e a C y c l e T i m e = 2 . 5

C y c l e

T i m e = 2

M a c h in e

M a c h in e

M a c h in e

In example II, the cycle time for the work area with two machines of type A is 2.5minutes. This work area is the bottleneck (note that the whole area is the bottleneck not just one of the two machines denoted as A). The capacity of the system is 60/2.5 = 24parts per hour. (Conversely, the capacity of each machine A is 12 parts per hour. So thecapacity of the work area is 24 parts/hour. The cycle time = 60/24 = 2.5 minutes.) If the

two machines A actually produced 20 parts per hour yesterday, we will say that thecapacity utilization was 20/24 = 83.33% yesterday. We can now define capacityutilization.

Capacity Utilization:The actual rate at which the system delivers outputs divided by itscapacity is called capacity utilization.This is equivalent to Capacity Utilization =(Timeused) divided by (Time Available).

The flow time in Example II is still seven minutes, because assuming that we cleverlystagger inputs to the two machines of type A, we can ensure that there is no waiting atmachine B. In that case we have managed to double capacity while maintaining the timespent by a typical part equal to seven minutes. However, if we can double the rate of

machine A (cycle time equal 2.5) then we get the same capacity but reduce flow time to4.5 minutes. This seems to be good time to introduce the notion of Gantt charts toclarify the flow time calculations. Gantt charts are used to pictorially depict the flow of atypical job(s) through the system. The X-axis of the chart is time, the Y-axis will haveseveral bars. The bars correspond to each of the processes and each of the (key)resources. In the above example the Gantt chart for two typical jobs will be as follows(not to scale):


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Time

(min.)

PROCESS A

PROCESS B

MACHINE A-1

MACHINE A-2

MACHINE B

1 2 3 4 5 6 7 8 9 10 11 12

PART 1

PART 2

PART 2

IDLE

PART 3

PART 1

PART 1 PART 3

PART 2

PART 1 PART 2 PART 3

The points to note are as follows:1. Each process has a bar. We have two bars for work area A to show that the process isbeing done in parallel.2. Each resource has a bar, and we have named the machines A-1 and A-2 to distinguishbetween the identical type A machines.3. The idle time of machine B stands out as a result of the analysis.Extend the chart for a few more parts to get a feel of why it is that we termed the workarea with the two machines as the bottleneck. You should be able to observe that: Themachines in work area A will always be busy. Therefore the work area is the bottleneck.If the system is working to full capacity then the utilization of the work areas also can beread from the chart. Finally, the chart can be used to schedule tasks so that they do notinterfere with one another, thus to determine the due dates for each order as they arereceived and accepted into the system)

EXAMPLE: Computing Utilization in Example IIWe are given that the cycle time of the bottleneck is 2.5 minutes. In the month of January1999 we find that the system produced 3100 units. What was the capacity utilizationof the bottleneck?Assume that there are 20 working days in the month and 7 working hours per day.

Basic Approach: Strategy -- Reduce everything to common units e.g., time.Time available per machine = 20 x 7 x 60 = 8400 minutesProcess A:

Time available = 2 machines x 8400 = 16,800 minutes.Time used = 3100 units = 5 x 3100 units = 15,500 minutesCapacity utilization = Time used/Time Available = 15,500/16,800 = 92.3%

Process B:Time available = 8,400 minutesTime used to produce the 3100 units = 3100 x 2 = 6,200 minutesCapacity utilization = 6,200/8,400 = 73.8%

Labor Content and Labor CostThe labor content of a task is the actual labor hours spent on doing the task. It is notnecessarily equal to the flow time and it does not include the idle time. In both previous


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examples the labor content of the process is 5 min. + 2 min. = 7 minutes. In our makingTea example, it might take just 30 seconds to put the Tea bag in the cup but the tearequires a full five minutes to steep including the insertion time of the tea bag. Thus, thelabor content of this task is only 30 seconds and the labor is free to do other tasks whilethe tea finishes steeping.

In most situations the labor cost includes idle time since most organizations would not layworkers off while they are idle for short periods of time. How should the cost of labor perpart produced be computed? A little thought reveals that it should equal the

(cycle time of the system) x(the number of direct workers in the system) x (the labor rateexpressed in $ per unit time).

For example if the labor rate is 12 dollars per hour, then in example I, the labor cost = 5min. times 2 persons (one each for the two machines) times $(12/60)/minute = $2 perpart. (Even though the worker on machine B is idle 60% of the time most organizationspay by the shift and do not deduct idle time from compensation.) In example II, the laborcost per part is 2.5 min. x 3 persons x $12/60/minute = $1.50 per part.

Productivity:Productivity is defined as output divided by input. Labor productivity is

output divided by labor hours (that is, total labor hours including idle time if any).

Efficiency: This is defined as the actual output divided by the standard output. Forexample, despite the availability of inputs, if the worker operating machine A producedonly 10 parts per hour, the efficiency of the worker is 10/12 = 83.33%. In more complexsituations efficiency is measured by standard hours of output divided by the hoursworked. (What is a standard hour? It is the time which the industrial engineer says thatmust be taken to perform the operation. The methods used to establish the standard timeare called work-study techniques.)

Available Capacity: Available Capacity is defined to be equal to Time Available timesCapacity Utilization times Efficiency.

Effectiveness: Effectiveness is sometimes defined as useful outputs divided by actualoutput. For example if none of the parts produced could be sold, effectiveness is zero. Itcould also be defined as doing the right thing.


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Connection with productivity(output/labor hours):

Example CycleTime(min.)

Outputper hour

(units)

TotalLaborUsed(min.)

LaborContentper unit

(min.)

LaborCost per

unit(min.)

Productivity(parts/labor

min)

I 5.0 12 120 7 10.0 12/120 = 0.1II 2.5 24 180 7 7.5 24/180 = 0.133

Direct Labor Utilization: Labor utilization is defined similar to capacity utilization. It isequal to:

Labor Content of what was produced divided by the Labor hours paid for.

Example Labor Contentof 1 hours

output (min.)

Labor Used inone hour

(min.)

IdleTime(min.)

Direct LaborUtilization (%)

I 12x7 = 84 120 120-84= 36 70.00

II 24x7=168 180 180-168= 12 93.33

You might like to ask yourselves whether capacity utilization, idle time andproductivity are useful measures for a firm? How do these measures reflect competitiveadvantage? Can these measures be used to understand or be related to a firmsperformance? For example, is capacity utilization of 85% good for a process industry? Aheavy machinery industry? A bakery?

Flow Time (once again): It is the length of time spent by a part or customer order in aprocess. You should be able to answer why is:

Flow Time Cycle Time?

Flow Time Labor Content ?

Production Rate: This is the rate at which a machine, work area or system producesgoods or services. It need not be the maximum rate. For example a machine may have the

capacity to produce at the rate of 15 parts per hour (note this is how a rate is expressed in parts or orders per unit time), but the manager may choose to produce at a rate of only2 parts per hour because the demand rateis 2 parts per hour. We shall denote productionrate as Th (for throughput rate).

What is the connection between, Flow Time and Capacity? Assume that asystem is producing orders at a particular production rate (which may not equal thecapacity). Let the average inventory in the system over a sufficiently long period beINV.Then:

INV = Production Rate (Th) x Flow Time.

Note that this relationship has apparently nothing to do with capacity. In queuing theorywe shall study how capacity affects various performance parameters of the system and

the ideas in the last paragraph will become clear. This equation is called Littles law. Itcan be applied to determine how many seats are needed in the bar in a restaurant. Someexamples of this law are given below.


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Process Flow Time Examples

CustomerFlow: Taco Bell processes on average 1,500 customers per day (15 hours). Onaverage there are 75 customers in the restaurant (waiting to place the order, waiting forthe order to arrive, eating etc.). How long does an average customer spend at Taco Bell?

Answer: Throughput (Th)= 1500/day = 1500/15 per hour. Inventory (I) = 75.

Time = ?

I = Th x Time => Time = I/Th = 75/100 hours = 45 min.

Job Flow: The Travelers Insurance Company processes 10,000 claims per year. Theaverage processing time is 3 weeks. Assuming 50 weeks in a year, what is the averagenumber of claims in process.?

Answer: Th = 10000/year = (10000/50)/week = 200/week. Time = 3 weeks.I = Th x Time = 200 x 3 = 600.

MaterialFlow: Wendys processes an average of 5,000 lb. of hamburgers per week. Thetypical inventory of raw meat is 2,500 lb. What is the average hamburgers flow time?

Answer: Th = 5000 lb/week, I = 2500 lb, Time = ?Time = I/Th = 2500/5000 = 0.5 week

Cash Flow: Motorola sells $300 million worth of cellular equipment per year. Theaverage accounts receivable in the cellular group is $45 million. What is the averagebilling to collection process flow time?

Answer: I = $45 million, Th = $300 million/year, Time = ?

Time = I/Th = 45/300 years = 3/20 year = 1.8 months.

Question: A general manager at Baxter states that her inventory turns three times a year.She also states that everything that Baxter buys gets processed and leaves the dockswithin six weeks. Are these statements consistent?

Inventory turns = Th/I = 3 in a year. I = Th x Time=> Time = I/Th = 1/(Th/I) = 1/Turns = 1/3 years.

Six weeks is not equal to 1/3 year. Not an accurate statement by the general manager.

Set Up Time: This is the time needed to prepare for doing an operation. Examples aremixing the dough to make pancakes, cleaning the paint nozzles prior to changing thecolor of an automated painting machine, switching on and activating a database beforeaccessing records from it, and signing on before buying things from Amazon. There maybe a significant time expended in preparation. For example, we may have to not only dothe physical set up but also undertake some trial and error production before we get theoperation correct. An example of set up time is the time required to learn a new topic likecapacity. After having spent so much time getting it correct it becomes easier when wemust use it some place!

Setup time can be either internalor external. Internal setups take up time on the machine,i.e., the machine must be stopped to do the setup. External setups are done offline.

Example: We are painting different colors on cars. If we have two paint booths whileone color is being used, we can clean the other one for use with another color. While foodis cooking, we can dice up onions needed for the next step. In other words, doing externalsetups is doing things in parallel, it saves time. The technique of reducing set up time hasbeen refined to a science, and made popular by Shingeo Shingo in his book on SingleMinute Exchange of Dies (SMED). The lower the setup time the more output variety aprocess can create per unit time. This can be a key source of strategic advantage formany firms (as it was for Dell in the 1990s).


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Lot Size:Whenever there is a significant amount of time spent setting up and preparingto do an operation, we would like to process several parts or carry out the operationsseveral times. We do this to spread our setup time investment over as many parts aspossible. If it takes 36 hours to set up a lathe to produce a spindle, and it takes 2 secondsto make a spindle, we may wish to make a few thousand having set up the lathe.Similarly given that it takes 20 minutes to mix the dough or switch on and activate adatabase, we would like to make several pancakes or process several records once the setup has been accomplished. The lot size may also be dictated by other considerations. Forexample if an oven has a capacity to bake 2 dozen cookies at a time, then we may choosea reasonable number of cookies to bake at a time. The consideration here is that the costof baking (reflected in time spent in the oven) is the same for 1 cookie to two-dozencookies. The number of units being processed as a batch is referred to as lot size. Inprocesses that can produce multiple products, lot size refers to the number of units of anyone product that are produced together before beginning the production of any otherproduct. Each product may have a different lot size. The lot size is a decision parameterchosen by the manager.

Run Time: This is the time required to produce a part or carry out an operation once theset up has been accomplished. In the lathe example it is 2 seconds. In the oven example itis the baking time. The former is a situation when parts are made one at a time and the

latter an example of batch processing. In general most tasks can be defined as one at atime, batch or continuous.

One at a Time Processing:In this process only one unit is worked on from beginning toend before the next unit can be started. An ATM is an example of such a process

Batch Processing:In this process, several units of a product are worked on at exactly thesame time. Typing with carbon paper or sending out an e-mail message to a list of peopleare examples of such a process. Moving the batch from step to step together is a methodof production known as batch processing

Continuous Process. In this process a new unit can be started as soon as the previousunit has entered the system. An automated car wash is an example of such a process. A

conveyor oven is another example.

Transfer Size:Transfer lot size is less than or equal to the lot size. It defines the numberof units that can be moved to the next operation before the complete lot has finishedprocessing at the prior operation. It reflects the fact that we may wait till the whole lot isprocessed to transfer the parts to the next operation, or transfer to the next operations in atransfer lot size that is smaller than the lot size, see Example III below. Ideally we wishto transfer parts as they are produced on a continuous basis. But this leads to excessivetransportation or material handling costs. On the other hand, if we wait for the whole lotto be processed we may create idle time in the system. The effect of small transfer sizes isto make the flow in a system more continuous. An example is check clearing in banks. Ifchecks are scrutinized for the signature by one person and passed on to another to verifythe balance in the account, will it not be advisable to pass on the checks for verifying the

balance as soon as the signature has been verified? Imagine 20,000 checks, 10 second percheck for signature verification and 10 seconds for verifying the checking balance, andtwo persons doing this task. What is the flow time for the 20,000 checks if the transferlot size is one versus 20,000?


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Capacity Determination when There is Setup Time:

The following table illustrates how to calculate capacity when there is a set up involved.The idea is that the capacity is computed in terms of the number of lots that can beproduced per unit time. Then to multiply by the lot size to obtain the capacity in units.

Set upTime

RunTime

Number ofmachines atwork center

Lot size Time perLot

Capacity (units)

(A) (B) (C) (D) (E) (F)

X Y Z L X + YL Z*L/(X+YL)

The run and set up time have to defined correctly. As an example consider the process ofbaking bread in an oven. There exists a small counter top apartment oven capable ofbaking one loaf at a time with a setup time of one minute and a baking time of 29 minutes(run time). A normal house oven cooks loaves in a batch of two with a one-minute setuptime (one loaf in each hand) and a 29-minute bake time. A small industrial strengthbaking oven processes bread continuously with a load size of two loaves, a load time ofone minute (both hands again) and a 29 minute bake time. Using the chart the results areas follows.

Per Hour Capacity

Type ofTask

(A) (B) (C) (D) (E) (F)capacity

per hour

Countertop One At ATime

1 29 1 1 30 2

House Oven Batch 1 14.5 1 2 30 4

Bakery Oven Continuous 1 0 1 2 1 120

A look at the previous chart is a powerful illustration of how different machinetypes, even with the same setup and run times, can produce vastly different outputs. Butyou would correctly argue that the cost of these machines also varies dramatically andyou would be correct. Thus, there is often a trade off between fixed cost and capacityand between capacity and variety. (High fixed cost often implies high capacity but lowvariety - low fixed cost often implies low capacity but high variety).

Of course, these capacities should also be confirmed intuitively. They must be answers towhat we think the ovens are capable of producing per hour.

Capacity Determination with setup saving: Clearly for one at a time and batchmachines we would like to spread the setup time over as many units as possible. Iftechnically feasible we would like to increase the number of units we can process before

we have to setup again.

Capacity Determination For Multiple Product Production Processes.

Few firms make only one product so we need to be able to determine capacitybased on a firm's product mix. In the best of all worlds, if every product requires thesame setup and run time, (e.g., in cookie baking the oven set up and run time do not varywhether chocolate chip or oatmeal raisin cookies are produced) then product mixcomplexity does not affect capacity. However, if setup times vary or both setup and run


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times vary with the product type then capacity will be a function of the percent of the mixand the lot size of each product produced in the mix.

Assume a firm has two products A and B and PA, PB are the percents of each productsnormally sold. We restrict the analysis to a single stage process that has one-at-a-timemachines. To determine the capacity of such a product mix we must first determine thecapacity of the process to produce each product. If we assume LA, LB are the lot sizesfor each product then the time per lot can be computed as in the previous paragraphsubstituting the individual lot sizes for Q. Then the capacity determination table(modified for set up saving) can be used to compute product capacity based on the type ofthe bottleneck machine. As an example for A the number of units produced per unit of

time would beA

a A A

L

X Y Lwhere XA and YAare the setup and run time of product A.

We must be sure that the unit of time is great enough to allow for the completion of thelot size (LA) or else we will over estimate capacity. We compute in this manner thecapacity for producing each product separately. Let CAand CBbe the capacity per unittime of products A and B when they are produced separately. Let x units of the product-mix be produced. Then x PA and x PB are produced of products A and B respectively.The time needed to produce product A is: x PA/ CAand time needed to produce productB is:x PB/ CB. The sum total of these should not exceed one (unity) thus:

BA

BA

C

PB

C

PAxCxPBCxPA

1

1// .

Or, the capacity of the process is given by:

BA C

PB

C

PA

1.

The above definition is easily extended to more than two products. When there areseveral resources, the capacity is computed for each resource and the minimum of thesebecomes the capacity of the system. Remember that the above formula works only whenthe product mix is known. When the product mix itself unknown or has to be determined,

optimization techniques can be brought into play, such as linear and mixed integerprogrammingmethods.

A final word about process choice: As depicted in the picture belo, process selectiondepends on volume. It also depends on variety. Finally, it can also be linked to the stageof the product on the product life cycle. Thus process choice must depend on thestrategic direction of the enterprise.


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Total

Cost

Volume

Large

Medium

Small

Entry Growth Mature

Minimum cost envelope

Total Cost vs. Volume Depending on Average Order Size

CAPACITY CALCULATIONS EXAMPLES & SOLUTIONS

Find the cycle time, capacity and flow time for the system in each of the following:

1. Bank tellers: assume that it takes five minutes to attend to one customer and there are10 (ten) counters (operating in parallel i.e., a customer can go to any one of thetellers and get full service there).

2. ATM machines: on the average it takes 4 minutes for one person to complete alltransactions at an ATM and there are 8 (eight) ATM machines.

3. Car assembly: there are 50 stations (in series i.e., one after the other) on the line;each station involves an operation on the car that takes 1.2 minutes. After the carassembly, the car is tested. This takes 15 minutes per car and there are five testingstations, any one of the stations can do the entire testing (i.e., the testing stations areidentical and work in parallel).

4. Four lane freeway: the freeway is 100 mile long, cars travel at 65 miles per hour andmust maintain a separation of 100 feet between cars. (One mile = 5280 feet, 65 milesper hour = 5720 feet per minute.) What is the capacity of the freeway? This one ishard; try finding the cycle time, i.e. the rate at which cars enter a lane first.

5. University: Each course is 40 hours of instruction. To prepare for each course it takes80 hours of work by an instructor. Each student has to take 25 courses (on theaverage) to obtain a degree. The average size of a classroom is 40 students. There are200 instructors, who work 300 days per year and forty hours per week. What is the

capacity of the university? This one is also hard. Hint: Compute how many classesan instructor can teach in a year.


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ANSWERS1. Bank tellers:

Capacity of one teller = 60/5 = 12 customers per hourCapacity of 10 tellers = 10x12 = 120 customers per hour for the systemCycle time = 1/capacity = 1/120 hours/customer = 0.5 minutes per customer for the

systemFlow time = 5 minutes for the system

2. ATM machines:

Capacity of one ATM = 60/4 = 15 customers per hourCapacity of 8 ATMs = 8x15 = 120 customers per hour for the system for the systemCycle time = 1/capacity = 1/120 hours/customer = 0.5 minutes per customer for thesystemFlow time = 4 minutes for the system

3. Car assembly:

Each station is a bottleneck (bottleneck is the station with the smallest capacity).Capacity of line (system) = capacity of bottleneck station = 60/1.2 = 50 cars per hourCycle time = 1/capacity = 1/50 hours/car = 60/50 minutes/car = 1.2 minutes per car (forthe system)Flow Time = 50x1.2 = 60 minutes = 1 hour for the system

After the car assembly, the car is tested. This takes 15 minutes per car and there are fivetesting stations.

Capacity of testing station = 5 (stations) x (60/15 cars per hour) = 20 cars per hourCapacity of combined system = minimum of capacity of assembly line and testing station=

= Minimum of 20 & 50 = 20 cars per hourCycle time of Combined system = 1/capacity = 1/20 hours/car = 60/20 min./car = 3

minutes/carFlow time = 60 min. (assembly) + 15 min. (testing) = 75 minutes.

4. Four lane freeway: the freeway is 100 mile long, cars travel at 65 miles per hour andmust maintain a separation of 100 feet between cars. (One mile = 5280 feet, 65 miles perhour = 5720 feet per minute.)

Cars can enter only with a spacing of 100 feet. So the time between cars entering thefreeway is the time taken to travel 100 feet = 100/(65x5280 feet/hour) = 0.0175 minutesTherefore cycle time = 0.0175 minutes! For one lane

Capacity of one lane = 1/cycletime = 57.2 cars per minute per laneCapacity of freeway = capacity of 4 lanes = 4x57.2 = 228.8 cars per minute

Flow time = time taken to travel 100 miles = 100/65 = 1.538 hoursCycle time of Freeway = 1/228.8 = 0.00437 minutes = 0.262 secondsThis means if the freeway is being used to capacity, cars will come out (or enter) every0.262 seconds.

5. University: Each course is 40 hours of instruction. To prepare for each course it takes80 hours of work. Each student has to take 25 courses (on the average) to obtain a degree.The average size of a classroom is 40 students. There are 200 instructors, who work 300days per year and forty hours per week.


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Let us assume that 300 days = 52 weeks.

To teach one class (this is the key) an instructor puts in 120 hours (40 hours in class plus80 hours preparing for the class). So the total number of classes that can be taught =200x52x40/120 = 3467 classes per year (i.e. 200 instructors work 52 weeks per year, 40hours a week so the available instructor hours = 200x52x40, each class takes 120 hours ofthe instructors time). That is the capacity in terms of classes (sections offered).

3467 classes = 3467/25 (classes per degree) = 139 degrees (for classes of size 40) peryear. But each class has 40 students. Thus the capacity = 139x40 = 5546 graduates peryear.


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Some of these examples were created by Wedad Elmaghraby (throughput Time is thesame as Flowtime)


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Toyota/Ford Hub Cap Mfg. - Make to stock - Customer Contact -

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Lands End/Amazon.com - Assemble/Pack to Order - Customer Contact -

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Car Wash/ATM - Make to Order - Customer Contact - Continuous

Raw Material

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SOME PRACTICE PROBLEMS

1. Is set up time to be included in computing capacity measures and capacity utilization?Give an example to illustrate your answer.2. Given the total number of units produced during the day can we compute cycle timeand flow time? For a given operating system, what is the minimum information needed to

compute capacity, capacity utilization, idle time, cycle time, flow time, labor content,labor cost, and to identify bottlenecks? Try to answer in a tabular form.3. What is the capacity and flow time of the system shown below (transfer lot sizeexample)?

EXAMPLE IIILOT SIZE AND TRANSFER SIZE

A B

Finished

GoodsRun

Time = 5

Run

Time = 2

Lot Size = 4

Transfer Size = 2

Lot Size = 2

Transfer Size = 2

4 at

a time

2 at

a time

2 at

a time

4. We visit an office that processes insurance claims. The manager asserts that it takesjust two days on the average to process a claim. We examine the past records and see thaton the average 200 claims are processed to completion every week. A quick estimate ofthe pending claims in the office reveals that there are 1500 claims waiting to be processedat different desks (stages of processing) in the office.

i) What is the flow rate (Th) in this example?ii) Can we say anything at all about the capacity of the office forprocessing claims?iii) What is a good estimate for the flow time in the office?iv) How can we explain the finding in (iii) to the manager in very simpleterms?

analysis of an operation

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