Work Measurement PFC & Time Study Webinar

[Audio] Hello everyone, and welcome to the global Industrial Engineering webinar series. The Global IE team is presenting a series of 7 LIVE webinars to re-enforce foundational IE practices to all engineers of all levels; and, to introduce this as new material to our new engineers as well. As a reminder, all presentations will be given in English… we kindly ask you to convey this information to those that will need it in your local languages. My name is (SAY YOUR NAME) and will be your presenter today. The experts onboard are (LIST THEIR NAMES). They will help answering any question you may have during the presentation by using the Q&A chatbox in the upper right corner of your screen. Our LIVE Q&A session will take on additional questions right after the presentation, and we will feature several questions so all may see and hear our response. This LIVE event is being recorded! The recording and powerpoint slides will be available in about a week in the IE Portal. In the unfortunate event of a network connection issue, you will still be able to download the recording and watch it. Additionally, all of the Q&A questions will be recorded as well. Today we will focus on Work Measurement, Man-Machine Analysis… thank you for joining, and we will now begin. CLICK TO NEXT SLIDE.

[Audio] In this webinar we will explain the …. Definition of a Man-Machine system and examples in Jabil Different elements and terminologies in a Man-Machine production system How to derive Cycle Time in case of a Man-Machine production system How to set Time Standard for these type of situation, and finally How to find the optimum man to machine ratio in different Man-Machine situations Just for a good understanding: It will not handle capacity modeling..

[Audio] A Man-Machine system is essentially a semi-automatic process where operators and machines are engaged together in continuous production of a job to produce a finished unit. This type of man-machine production system are also termed as 'restricted work'. (The operator is more or less depended on the machine to perform some other activities) LASERPOINTER: It contains both manual time elements (Hand time or Interactive Machine Time IMT) and automatic machine time. The operator can have control over the output only to the extent of the manual work content, but not the content of the work that must be executed by the machine. Note that the IMT is always less than or equal to the Machine time if the operator is focused on the same unit under the machine process..

[Audio] Some of the common Man-Machine combinations in Jabil are: Component preparation, In Circuit Testing Functional Verification Testing Routers, Screwing, Automated Visual Inspection, Press-Fit Soldering Or a customized Automation process, All of these often involve operators to load and unload the machine equipment, Or multiple machines could be handled by the same operator..

[Audio] If we look at a typical production line in Jabil, we will use this Webinar to zoom in at the area where there are operators interacting with machines. We will not focus on SMT lines, since that output is mainly driven by the machines itself, and operators are only supporting the machines. (for example by replenishing the reels, or during change-over). LASERPOINTER Our IE focus for this webinar is at the area after the SMT lines: starting from ICT, followed by possibly Routers, manual assembly or semi-automated processes, FVT and all the way to their backend packout stations..

[Audio] Let's have a look at a typical test operation (this could also be another automated machine process like screwing or soldering…). In this example an operator will load a tester manually by hand for 30 sec. Then the Tester runs automatically for 120 seconds. At the end of the test, the operator unload the tester manually for 20 sec. And inspects the tested unit for 50 seconds. From Time study perspective, there are various ways to represent a man-machine system. For a single operator single machine system (most simplest one) it can be shown as mentioned here: LASER POINTER + EXPLANATION DIAGRAM This figure represents one cycle: The 30 sec manual time to load the test un blue, The 120 sec machine testing time in orange, The 20 sec unload time un blue again and finally, The 50 sec of Inspection time in green..

[Audio] In reality, during continuous production, an operator will start the cycle with unloading a tested part, and loading a new part into the tester to start the automated test. The inspection will happen DURING the machine time, so the Inspection can be categorized as Interactive Machine Time IMT. So if we move the unload action to the beginning of the diagram, and the inspection in parallel with the machine time, it looks like this. LASERPOINTER Unloading > Loading > Inspection during testing. This way of representing the cycle is called a pump diagram (because it looks like an air-pump that you may have used to inflate the tires of a bicycle).

[Audio] From time period perspective, there are various elements in a man-machine production system. Let's look first at a system with 1 Man and 1 Machine. A (in orange), the machine controlled time element: represents the Period when the machine is working independently B (in green), the inside work or Interactive Manual Time (IMT): represents the Period when both the machine and the operator are engaged C1 and c2 (in blue), outside work: are the periods when the operator works but the machine is not working And finally d (in grey) is the Operator Idle Period when machine works but the operator is idle.

[Audio] Let's have a look now at a Man-Machine process when more than 1 machine is involved. LASERPOINTER You can see here that during a cycle, an operator will unload and load the 1st MACHINE, and starts inspection when the machine is running. At the end of the inspection, the operator can unload and load MACHINE 2, and do the inspection of the product while MACH 2 is testing. At the end of the inspection of the part coming from MACHINE 2, the operator can go back to MACHINE 1 and can start with unloading there again. BUT you can see that there is some time when MACHINE 1 is waiting to have the part unloaded. This is the Machine IDLE time (what is basically lost time or waste from the perspective of the machine, but not necessarily the entire line). It's our task as Engineers to try to reduce this waiting time of the machines to a minimum to optimize the usage of our assets..

[Audio] Now that we are all aware of the different elements within a Man Machine system, we can have a deeper look at the different cycle times that will define the Man Machine system and how to calculate them. Assume one operator handles N number of machines, and in this case N equals 1. The different cycletimes involved are: LASERPOINTER OPERATOR BUSY TIME = time that operator is working in a cycle = the number of machines the operator is handling in 1 cycle, multiplied with the sum of HAND + IMT cycletimes = N x c1 +b + c2 So 1 x (30+50+20) equals 100 sec. Let's call this T1.

[Audio] LASERPOINTER MACHINE BUSY CYCLETIME = time that machine is busy and can't be started before this cycle has been done (so from loading c1, followed by the automated machine time a, up to and including the unloading c2). Keep in mind, loading and unloading a machine is required for a machine to run! The sum of these times (30+120+20) equals 170 seconds, Let's call this T2 Note: The MACHINE TIME (as used in Time studies and IEDB) = time that the machine is running = MACH = 'a' only.

[Audio] The (Normal) CYCLETIME = maximum of the time the operator is busy or the machine is busy = max of T1 and T2 So the Max (100,170) = 170 sec If Output per Cycle = 1; Cycle Time / Pc. = 170 sec Remember, that is still our Normal cycle time, it is not our standard time because we have not added in our PF&D allowances yet. We will cover this in a few more slides..

[Audio] Kindly note that in our example, the operator will be idle for 70 sec in each cycle This is WASTE Because of that it could be worthwhile to have the operator starting a second machine starting from the moment the operator finished the inspection. Let's have a look at that situation in the next slide:.

[Audio] In this example one operator handles 2 identical machines, so N = 2. (Here we present the pump diagram in the format starting from unloading, but that is irrelevant to the calculation, we do it this method because it is to present and understand the cycle for our calculations. LASERPONTER (explain process steps and calculation) Same as in previous example, but the operator will unload a finished unit, load a new unit into the machine, and then start inspecting the processed 1st unit. Instead of waiting now until the machine completed it's machine time, the operator can start unloading a second machine at the end of the inspection. So for MACH 2: after the unloading, also there the operator can load a new unit into the machine and start the inspection process during when that machine is running. At the moment the unit from MACH 2 has been inspected, the operator can go back to MACH 1 and start the 'unload, load, inspect' cycle there again. Let's look at the calculations for the cycletimes now: Operator busy time T1… Machine busy time T2… Cycletime… Output per cycle… So Cycle time per Piece… Note that in this scenario the operator does not have any idle time. The 70 seconds IDLE time the operator had in the past, is completely reduced, But now the machine has some waiting time which is called Machine IDLE time; Please note that all the calculations above are based on the Normal Time for each of the elements in the operation. Let us see how to establish the Standard Time by applying the applicable PF&D allowance..

[Audio] Quick reminder: The Allowance is necessary to get the Standard Time which a normal operator can achieve on regular basis under normal operational conditions.​ An Allowance, or PF&D percentage, is given to take care of the following elements representing non-working time: P for P-ersonal needs (for example toilet break…) F for F-atique (Rest pauses to reduce tiredness) And D for D-elays (because of small unavoidable interruptions).

[Audio] How about PF&D Allowances in a Man-Machine environment? LASERPOINTER: For MACHine times only P&D are applicable. BUT no F, because a machine (unlike a human) will not suffer from Fatigue (not getting tired) HAND, or manual work, can be like Loading and Unloading to/ from machine during when the machine is not running; Here we need to apply P (for personal breaks), F (for Fatigue) and D (for small Delays) IMT is like doing inspection of an unloaded part when the machine is processing the next loaded part Also here, P, F and D should be applied because it is the work of the operator. So P&D are applicable for both Manual and Machine Times BUT F is applicable only for Manual Times (Hand + IMT), For Machine Times: no Fatigue factor needed. *Remark: Most commonly P for Mach is applied because Machine stops while operator is away during break BUT in case the machine will continue running whiole the operator is away during a break P for Mach could be 0. So in a Timestudy the applicable PF&D% for a machine operation is mostly only the 1 or 2% for minor Delays..

[Audio] Remember that the Interactive Machine Time IMT internal to the Machine Time should always be smaller or equal to the Machine time. BUT if the work content of the Interactive Machine Time IMT increases due to adding in PF&D%, it could become greater than machine time. To solve this, and store that time in the correct type, we should increase the MACHine time OR part of IMT moves to HAND time LASERPPOINTER TO EXPLAIN.

[Audio] So now we also know how to add PF&D% to our normal times in a man-machine system, we can now define the correct standard times: Remember: in JABIL we add the Applicable PF&D% to our Normal Times to get the standard times (to upload in IEDB for example), we do not include P: So if we assume PF = 11%, with a P of 5% (as derived in the way mentioned during the PF&D Webinar) This gives us a F of 6%. And we assume a D of 1%. So the total PF&D% is 12% and thus, the Applicable PF&D is 7% because 12 – 5 = 7. LASERPOINTER: The HAND time= 30 + 7% + 20 + 7% = 53.5 sec THE IMT = 50 + 7% = also 53.5 sec And the MACH time equals 120 + 1% or 121.5 sec. The operator busy time is 1 machine multiplied with the sum of hand and IMT time., 107 sec in total The machine busy time is the sum of HAND and MACH time, 174.7 sec on total. Therefore the Standard cycle time = MAX of (operator busy time and Machine busy time), so 174.7 seconds for 1 piece..

[Audio] We can actually add these Applicable Allowances immediately into a TimeStudy summary sheet as well: So for each Work Element we need to add the correct Allowance: The first element (loading the tester), THE HAND standard time is the normal Time added with an applicable PF&D of 7% (this was the 6% for F + the 1% for D) The 2nd element (tester running), is a MACHine type and therefore we only need to add 1% for D to the Normal TIme 3rd element is the Inspection. For this Interactive Machine Time we need to add again the applicable PF&D, so 7%. And finally the 4th element, again a HAND type for unloading the tester with an applicable allowance of 7%. This will give us a MACH time of 121.2 sec IMT of 53.5 And HAND time of 53.5 as well. Resulting in a Standard time of 174.7 seconds per unit..

[Audio] Let's calculate an example together in case we have 1 operator and 2 machines. (Also notice that in this example we will reduce the IMT to 25sec, instead of 50, to make sure there is no IDLE time for the machine) LASERPOINTER: The operator busy time is 2 machine multiplied with the sum of hand and IMT time., 160.5 sec in total The machine busy time is the sum of HAND and MACH time, 174.7 sec on total. Therefore the Standard cycletime = MAX of (operator busy time and Machine busy time) or 174.7 seconds, and since we get 2 pieces each cycle this gives us a CT of 87.35 sec per piece..

[Audio] In case the IMT is bigger, let's say again 50sec, there is Machine idle time because when MACH 1 is ready, it has to wait for the start of the next cycle until the operator is available after finishing the inspection at MACH 2. Calculating the cycletime in the same way as before, will increase the maximum cycletime to 214 seconds or 107 seconds per piece because there are two machine. This will have an impact on how many machines 1 operator can handle. For that, we can look at the Load Factor in the following slides:.

[Audio] A way to indicate how optimized the utilization of the process is, is to look at the Load Factor. This represents the ratio of how busy the operator is compared to the total cycletime. We can calculate this by dividing the manual time (HAND+ IMT) by the total cycletime (HAND + MACH).: In the example here it gives us 59[break]% The opposite of the load factor will give us the optimum number of machines one operator can handle: 1 divided by the load factor equals 1.7. So one operator could only handle 1 machine. (since 1.7 is less than 2, we round down) If this number is closer to 2 than 1, we should try to improve the number of machines an operator can handle, in our case here, that means trying to have an operator handling 2 machines, Therefore we have to find ways to reduce the HAND or IMT times in our cycle..

[Audio] For the optimum utilization of an Operator, still it's possible that one operator handles 2 machines, even when the load factor is less than 2. In this case the operator is 100% utilized BUT the utilization of machine gets reduced due to "Machine IDLE" time. Total Cycletime increases to 200 seconds (instead of 170) So the optimum number of Man-Machine will be based on what we are trying to optimize. A good utilization of the operators, OR a good utilization of the machines… it depends on the priorities. You need to consider the full line balance, upstream and downstream process, and the available of both operator and machine resources to meet our customer's demand for this..

[Audio] To summarize: We have seen how a Man-Machine system can be represented by a pump diagram, and the different elements HAND, IMT and MACH..

[Audio] To calculate the Standard Time, each element has it's own allowance to be added. With a special remark that P for MACH could be 0 when the machine continues to run during a break. When 2 or more machines are involved, there is the possibility that a machine will have to wait in an IDLE state before it will be used again. We explained how to define the optimal number of machines an operator can handle by taking 1 divided by the load factor. And finally that to optimize we have to prioritize if we want to have the best 'men-utilization' of 'machine-utilization'.

[Audio] Before leaving, we would like to thank you for your participation, and remind you that all the webinars will become available on the IE Portal. More detailed IE training modules will be available through Workday..