Throughput Notes

INTRODUCTION In any process, throughput can be defined as the number of elements passing through a given system in each unit of measurable time. That unit of time can vary in different cases, and can be anything from seconds to years. Throughput can be easily spotted inside any type process and with any type of unit. The process can be either a service or a manufacturing process such as: doing someone’s makeup, manufacturing a product, or producing a candy in a factory. In this picture, workers are putting the lids on the cars that are being assembled. Throughput in this particular case can be depicted as the work in process, or in this case, the number of cars being assembled that enter this process per second, minute, day, or year. Previously, throughput was touched upon in the lesson on Little’s Law. Throughput is an extremely important element of little’s law. Little’s Law: Inventory= Throughput ( R ) x Time Throughput= Inventory/Unit of time Inventory=(Inventory/Unit of time ) x Time Symbolic Representation of the Equation: I=R x T It is imperative to realize that without throughput, there would be no way to calculate the amount of time it takes for a unit to go through a whole system or specific subsystems. In the same way, without throughput there would be no way to calculate inventory within a process, regardless if it’s in the manufacturing sector or in the service sector. HOW THROUGHPUT IS CALCULATED One may ask how is something like throughput calculated? There may be periods where there may be a lot of units coming in, such as Monday mornings in a Coffee Shop, or in the contrary dead hours, around 11 pm in a coffee shop, this concept is also known as variability. Throughput is calculated in any process in three steps: 1) Witnessing the process multiple times 2) Calculating the number of elements that go into any system or subsystem per unit of time 3) Calculating the average of those calculated rates For example, let’s suppose Maria wants to know the number of customers that on average go in and out of her coffee shop. This particular experiment will be extremely helpful in deciding her staff during certain hours of the day. She collected the following data. 9 am 5 customers/min (on average) 10 am 2 customers/min (on average) 11 am 2 customers/min (on average) 12 pm 4 customer/min (on average) 1pm 2 customer/min (on average) 2 pm 1 customer/min (on average) Based on these calculations, she can calculate the average number of customers per minute that come into her store, which is 2.67 people/min. One can also describe throughput in the context of a flow system. In such cases, throughput can also be considered as the average rate in which an object flows. Likewise, we have also come across the concept of capacity. Capacity is defined as the maximum rate at which a system can handle production. Average throughput can be anywhere below capacity value. However, the maximum flow at the period of the most congestion is maximum capacity. Throughput values for that specific process cannot exceed that capacity. CYCLE TIME AND TAKT TIME Cycle time measures how long an entire process, such as baking baklava, can take place at the maximum capacity of unit inflows. Cycle time relates to capacity by focusing on the internal capabilities of the process. However, throughput, the number of inflow of units per unit of time, focuses on the external demand for the specific process. Takt time is related to throughput. Therefore, Takt time, unlike cycle time, focuses on external demand, and measures how long an entire process takes at the average throughput or demand. The formulas are as follows: Cycle Time= 1/Capacity Takt time= 1/Throughput (demand) Suppose Maria has a capacity of 50 people per hour. Her cycle time for the capacity would be: Cycle Time= 1/50 hours x60 min/hour = 1.2 min Suppose Maria has an average throughput of 28 customers per hour coming in to her coffee shop. Her Takt time would be: 1/28 hours x 60 min/hour= 2.14 min It is extremely important to know that capacity should always be greater than throughput. This fact infers that the Cycle time will always be less than the takt time. Remember, throughput is the average number of inflow and outflow units per unit of time. When applying these concepts to make effective decisions, one needs to know the purpose of such equations. As managers, if we want a smooth flow, we want to make as much units as possible in small amount of time. Therefore, our goals would be to maximize capacity and throughput. In hand, our goals will also be to lessen takt time and cycle time. TERMS AND SYMBOLS RELATING TO THROUGHPUT AND CAPACITY In the future examples, we will be using the terms Ra, which refers to throughput, and Rp, which refer to capacity. In problems we will be given resources, people or objects performing each activity or sub process of the entire process. We will see a value known as Tp, which is the unit load of a resource unit. That is, how long a resource needs to work for each flow unit produced. Using that one can, in turn, discover capacity! THE PROCESS OF EFFECTIVE CAPACITY, THROUGHPUT, AND UTILIZATION We will connect the concept of throughput, the equivalent of demand, after in depths look into the process of calculating effective capacity for individual resource units and, in turn, resource pools hourly and daily. We will then compare our demand, or throughput, with the maximum capacity our individual activities are able to withhold. That would determine our utilization. Based on our utilization, we can make effective decisions. CALCULATING EFFECTIVE CAPACITY FOR RESOURCE UNITS AND THEIR CORESPONDING POOL There are two types of capacity: Theoretical and Simple. Simple is also known as Effective Capacity. What makes them different from each other is that theoretical capacity does not take into account Capacity Waste factor, where as Simple, or effective capacity does consider Capacity Waste factor. Continuing from the previous lesson of capacity, we can expand it to find out the effective capacity of each resource unit and resource pool. We, then, use the pool capacity (claims/min) to determine how many units are capable of going through a system each day or each month. This gives us great insight into the way the business is run, where it’s going, and where it could be. EXAMPLE You’ll go through this example to measure the effective capacity of each individual resource unit, such as workers, and the resource pool, which totals how many worker units the company has per resource position. We can then use these figures and the amount of hours that the workers work to determine utilization based on a throughput value. In this example, we will review the overall process of CSUN’s Counseling Service. We will observe the number of people that enter the office and utilize the resources daily. We will also see the demand in comparison to the capacity each resource (worker) is capable of doing. By doing this, we can calculate utilization. Through this number, we will see if the counseling service office needs to do more advertising in order to reach out to more students. The counseling office consists of two kinds of workers: A receptionist and a Counselor. There is one receptionist checking in patients at the front desk. There are five counselors giving therapy to students who walk in the door. The receptionist spends a maximum of 1.5 minutes checking in each student. The counselor spends a maximum of 60 min with each student in therapy. Using this information, we want to find out the effective capacity of each resource unit. In order to do this, we need to inverse the fraction. 1.5 minutes becomes 1/1.5 minutes. And 60 minutes becomes 1/60. We are turning min/claim into claims/minute. This is to give us an easier time calculating effective capacity. We then multiply the claim per minute by the number of resources, or workers, on the scene. We only have one receptionist. However, we have 5 counselors. We do this because we need to find the capacity of the entire resource pool, so we can see what area of the process is the bottleneck of the entire process. The bottleneck of the process will be the resource with the minimum capacity. The number of people working on the job matters in how the job gets done. The first line contains the values of the receptionist; the second line contains the values of the counselors. Now that we have the effective capacity of the resource pool, in other words, claims that they can handle per minute, we can calculate how many claims, or students, they can serve in a day. We do this by multiplying these values by the number of minutes they work in their shift. Both the receptionist and Counselors work 540 minutes in their daily shift. We multiply the effective capacity of the resource pool for each position by 540 to come to the number of people they are capable of serving a day The receptionist can serve a maximum of 360 people during his shift, while the counselors can only serve a maximum of 45 people per shift. Throughput for the counseling services has been calculated. It is 35 students in one day. Using the current daily demand, we can compare it to the maximum effective capacity that each resource pool serves per day. We do so by dividing throughput by the capacity of each resource pool. We turn this into a percentage which gives us the utilization value. What does this tell us? Given the current demand, the receptionist only utilizes 10% of his capacity. Also, this table shows that given the current demand, the counselors utilize 78% of their capacity. Using these numbers, those who run the counseling services may decide to broadcast their services in order to reach out to the student population. As we all know, depression and suicide can be a major problem among young adults. Perhaps they may see a greater need to reach out to the student population more than ever before. They will have the quantitative capabilities to do so in addition to their qualitative motivations. Given the information from this specific case, we can also calculate values that we touched upon earlier. OTHER CALUCLUATIONS Rp (Capacity): 45 Cycle Time: 1/45*5*540=60 R (Throughput): 35 Takt Time: 1/35 *5*540=77 Flow Time: 60 + 1.5= 61.5 Theoretical Capacity & Capacity waste Factor As we stated before, the effective capacity of a resource unit is defined as 1/Tp. Tp in this case is unit load, which is the time of production of a certain resource including the wasted time. In order to calculate Theoretical capacity, it is important to measure the Theoretical unit load, which is the time of production of a resource unit with no wasted time included. Effective Capacity= c/Tp (Unit load) Theoretical Unit load= Tp * (1-CWF) (Capacity Waste factor) Theoretical Capacity= c/Theoretical Unit load In these types of cases, capacity waste factor represents a percentage amount of time that was wasted due to set- ups or perhaps break downs. In these cases, the capacity waste factors will be given in each problem. For example, in car body shop, on average it takes a person 25 minutes to fully paint a car. And it has been determined that there is a Capacity Waste factor of 20% because it takes the mechanic 5 minutes to set up all of the tools needed. In this case these are the calculations. Unit load= 25min Effective capacity = c/Tp = 1/25*60min per minute or 2.4 per hour Theoretical Unit load= Tp * (1-CWF) = 25* (1- .20)= 20 Theoretical capacity= c/Theoretical Tp= 1/20 or 3 per hour CONCLUSION Mastering this process is crucial to understand how to compare process capacity to demand in order to come up with a utilization percentage. With utilization numbers, you have the numbers to back up making effective decisions in order to improve sales and customer turnabout. You can see how far the business would be comfortable in increasing customer numbers without reaching the capacity and stressing workers. Or, perhaps, you can see how many customers a business needs to hire in order to find a new bottleneck and improve capacity or lessen time. At the same time, as managers, it is imperative to eliminate as many non-value added activities that account for Capacity Waste factor. By eliminating these types of activities, we will see that our Effective capacity will increase, and our goals must be to maintain that effective capacity as close as possible to the theoretical capacity which would mean that there are not as many non value added activities and which infers that our process is running smoothly. Problem 1 On average it takes a barber 35 minutes to cut someone’s hair. During those 35 minutes, seven of those minutes is wasted by the barber in order to prepare his equipment and adjust any modifications to the machines he uses. If he work eight hours per day, what is his Simple and theoretical capacity? CWF= 7/35= 0.2 or 20% Unit load (Tp): 35 min Effective Capacity= 1/Tp= 1/35*60= 1.7/hr Or 1.7/hr* 8rhs/day= 13.6/day Theoretical Unit Load (Ttp)= Tp-(1-.20)= 35min *.80=28 min Theretical Capacity: 1/TTP= 1/28*60= 2.14/hr Or 2.14/hr*8hrs/day= 17.14 This means that if the barber did not waste any time in each haircut, he would be able to cut 3 people’s hair in addition to what he already had. Now with this data, he can work something out to decrease as much as possible his non value added activities and thus increase his capacity. Problem 2 ESP Guitar Company manufactures custom guitars from well known guitars players, Kirk Hammett, and James Hetfiled from legendary band Metallica. Both guitars required to spend some time in the design department, the painting department, and the detailing department. The ESP Explorer guitar requires four hours in the design department, two hours in the painting department, and three hours in the detailing department. On the other hand, the ESP Flying V guitar requires 3 hours in the design department, 2 hours in the panting department, and 2 hours in the detailing department. Each department works for 8 hours each day, and the Capacity waste factor are as followed: Design= 25%, painting= 0%, Detailing=50%. There are two people in the design department, three machines on the painting department, and five guitar engineers taking care of the detail department. What is the effective capacity and theoretical capacity for each resource, and their corresponding pool? Unit load: Explorer design Painting Details 4 hours 2 hours 3 hours Unit load: Flying V 3 hours 2 hours 2 hours Unit Load: 60%Explorer 40% Flying V 3.6 hours 2 hours 2.6 hours No. of resources 2 3 5 Design: Unit load (Tp)= 3.6 hours Effective capacity= 1/Tp= 1/3.6hrs Theoretical Unit load (TTp)= Tp*(1 – CWF)= 3.6 hours*(1-.25)= 2.7 hours Theoretical capacity= 1/TTp= 1/2.7 hours Painting: Unit load (Tp)= 2 hours Effective capacity= 1/Tp= 1/2hrs Theoretical Unit load (TTp)= Tp*(1 – CWF)= 2 hours*(1-0)= 2hours Theoretical capacity= 1/TTp= 1/2 hours Details: Unit load (Tp)= 2.6 hours Effective capacity= 1/Tp= 1/2.6 hours Theoretical Unit load (TTp)= Tp*(1 – CWF)= 2.6 hours*(1-.50)= 1.3hours Theoretical capacity= 1/TTp=1/1.3 hours Effecti Theoreti Numbe Effecti Theoreti Effectiv Theoreti ve cal r of ve cal e Cap. cal Cap. capacit capacity resourc Capacit capacity of Rof Ry per per es y of of pool pool pool per resourc resource pool per day day e (x8hour (x8hours s) ) 1/3.6hr 1/2.7hr 2 0.56/hr 0.74/hr 4.44 5.93 Desig n Painti 1/2hr 1/2hr 3 1.5/hr 1.5/hr 12 12 ng details 1/2.6hr 1/1.3hr 5 1.92/hr 3.8/hr 15.4 30.8 In this case the bottleneck of this system is the design department because it is the one with the least capacity out of the entire pool Cycle time= 8 hours /4.44= 1.8 hours Problem 3 Health blend produces many types of multivitamins which manufactures around the whole United States. All of their multivitamins have to go through a mixing department, and encapsulating department, and a bottling department. However, every single department can only handle a certain demand for each multivitamin. There is a demand 200,000 units for the month of April; half of the well known “Gold” multivitamin, and the other half for the “Silver” multivitamin. Health blend manufactures the multivitamins by orders of 1,000 units. In order to complete a Gold multivitamin order, it need to spend three hours in the mixing department, 4 hours in the encapsulating department, and 2 hours in the bottling department. On the other hand, the Silver multivitamin, on average, it spends four hours in the mixing department, five hours in the encapsulating department, and three hours in the bottling department. The capacity waste factors are as followed: mixing= 15%, encapsulating= 25%, and bottling=10%, and the product mix What is the effective capacity and theoretical capacity for each resource, and their corresponding pool? Unit load: Gold Unit load: Silver Unit load: Number of 50%Gold Resources 50%Silver Mixing 3hrs/order 4hrs/order 3.5hrs/order 3 machines Encapsulating 4hrs/order 5hrs/order 4.5hrs/order 5 machines Bottling 2hrs/order 3hrs/order 2.5hrs/order 2 machines Mixing: Unit load (Tp)= 3.5hrs/order Effective capacity= 1/Tp= 1/3.5hr Theoretical Unit load (TTp)= Tp*(1 – CWF)= 3.5hours*(1-.15)= 2.98 hr Theoretical capacity= 1/TTp= 1/2.98 hr Encapsulating: Unit load (Tp)= 4.5hrs/order Effective capacity= 1/Tp= 1/4.5hr Theoretical Unit load (TTp)= Tp*(1 – CWF)= 4.5hours*(1-.25)= 3.37hrs Theoretical capacity= 1/TTp= 1/3.37hr Bottling: Unit load (Tp)= 2.5hrs/order Effective capacity= 1/Tp= 1/2.5hr Theoretical Unit load (TTp)= Tp*(1 – CWF)= 2.5hours*(1-.10)= 2.25hrs Theoretical capacity= 1/TTp= 1/2.25hr Mixing Encapsulati ng Bottling Effecti ve capacit y per resourc e Theoretic al capacity per resource Numbe r of resourc es 1/3.5hr 1/4.5hr 1/2.98 hr 1/3.37hr 3 5 Effecti ve Capacit y of pool resourc e 0.86/hr 1.11/hr 1/2.5hr 1/2.25hr 2 0.8/hr Cycle time= 8 hours/ 6.9= 1.16 hours Theoretic Effectiv Theoretic al e Cap. al Cap. of capacity of RR-pool of pool pool per per day day (x8hours) (x8hour s) 1.01/hr 6.9/day 8.1/day 1.48/hr 8.9/day 11.8/day 0.89/hr 6.4/day 7.1/day