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CuikSuiqbad RekaDayaUpaya Rekayasa Daya Upaya Industry Manufacturing Engineering
5.4 AUTOMATED FLOW LINES WITH STORAGE BUFFERS One of the methods by which flow lines can be made to operate more efficiently is to add one or more parts storage buffers between workstations along the line. The buffer zones divide the line into stages. If one storage buffer is used, the line is divided into two stages. If two storage buffers are used at two different locations along the line, a three-stage line results. The upper limit on the number of storage zones is to have in-process inventory capacity between every workstation. ‘Die number of stages will then be equal to the number of stations. For an n-stage line, there will be n – l storage buffers. This, of course, does not include the inventory of raw work parts which are available for feeding onto the front of the flow line. Nor does it include the inventory of finished pans that accumulate at the end of the line. In a flow line without internal work part storage, the workstations are interdependent. When one stations breaks down, all the other stations will be affected, either immediately or by the end of a few cycles of operation. The other workstations will be forced to stop production for one of two reasons : 1. Starving of stations. If a workstation cannot continue to operate because it has no pans to work on, the station is said to be starved of parts. When a breakdown occurs at a given station in the line, the stations following the broken-down station will become starved. 2. Blocking of stations. This occurs when a station is prevented from passing its work part along to the following station. When a station breakdown occurs, the preceding or upstream stations are said to be blocked, because they are unable to transfer work parts to the station that is down. The terms “starving” and “blocking” are often used in reference to manual how lines. However, they are useful for explaining how buffer storage zones can be used to improve the efficiency of automated flow lines. When an automated flow line is divided into stages by storage buffers, the overall efficiency and production rate of the line are improved. When one of the stages in the line breaks down, the storage buffers prevent the other stages from being immediately affected. Consider a twostage transfer line with one storage buffer between the stages. If the first stage breaks down, the inventory of work parts that has accumulated in the storage buffer will allow the second stage to continue operating until the inventory has been exhausted. Similarly, if the second stage breaks down, the storage zone will accept the output of the first stage. The first stage can continue to operate until the capacity of the storage buffer is reached. The purpose served by the storage buffer can be extended to flow lines with more than two stages. The presence of these in-process inventories allows each stage to operate somewhat independently. It is clear that the extent of this independence between one stage and the next depends on the storage capacity of the buffer zone separating the stages. The two extreme cases of storage buffer effectiveness can be identified as 1. No buffer storage capacity at all 2. Storage buffers with infinite capacity NO BUFFER STORAGE CAPACITY. In this case, the flow line acts as one machine. When a station breakdown occurs, the entire line is forced down. The efficiency of the line was previously given by Eq. (5.5). We will rewrite this equation as
The subscript 0 identifies this as the efficiency of the line with no buffer storage capacity. It represents a starting point from which the line efficiency can be improved by adding provision for in-process storage. INFINITE-CAPACITY STORAGE BUFFERS. The opposite extreme is the case where buffer zones of infinite capacity are installed between each stage. If we assume that each of these buffer zones is half full (in other words, each buffer zone has an infinite supply of parts as well as the capacity to accept an infinite number of additional parts), then each stage is independent of the rest. Infinite storage buffers means that no stage will ever be blocked or starved because of the breakdown of some other stage. Of course, a flow line with infinite-capacity buffers cannot be realized in practice. However. if such a line could be approximated in real life, the overall line efficiency would be determined by the bottleneck stage. We would have to limit the production on the other stages to be consistent with that of the bottleneck stage. Otherwise, the in-process inventory upstream from the bottleneck would grow indefinitely, eventually reaching some practical maximum, and the in-process inventory in the downstream buffers would decline to zero. As a practical matter, therefore, the upper limit on the efficiency of any such a line is defined by the efficiency of the bottleneck stage. If we assume the cycle time Tc to be the same for all stages, the efficiency of early stage is given by
where the subscript k is used to identify the stage. The overall line efficiency would be given by
where the subscript oo is used to indicate the infinite storage buffers between stages. BUFFER STOCK EFFECTIVENESS IN PRACTICE. By providing one or more buffer zones on a flow line, we expect to improve the line efficiency above Eo but we cannot expect to achieve Eoo, simply because buffer zones of infinite capacity are not possible. Hence, the actual value of E will lie somewhere between these extremes :
Before embarking on the problem of evaluating E for realistic levels of buffer capacity, it seems appropriate to comment on the practical implications of Eqs. (5.l6) through (51.9) : 1. Equation (5.19) indicates that if E0 and Eoo are nearly equal in value, relatively little advantage will be gained from the addition of storage buffers to the line. If Eoo is significantly larger than Eo, storage buffers can provide a pronounced improvement in the line’s efficiency. 2. The workstations along the line should be divided into stages so as to make the efficiencies of all stages as equal as possible. In this way the maximum difference between Eo and Eoo will be achieved, and no single stage will stand out as a significant bottleneck. 3. The efficiency of an automated How line with buffer storage can be maximized under the following conditions : a. By setting the number of stages equal to the number of workstations. That is, all adjacent stations will be separated by storage buffers. b. By providing that all workstations have an equal probability of breakdown. c. By designing the storage buffers to be of large capacity. The actual capacity would be determined by the average downtime. If the average downtime (in particular, the ratio Td/Tc) is large, more buffer capacity must be provided to ensure adequate insulation between workstations. 4. Although it is not obvious from Eqs. (5.16) through (5.l9), the “law of diminishing returns” operates as the number of stages increases. The biggest improvement in line efficiency comes from adding the first storage buffer to the line. As more storage buffers are added, the efficiency improves, but at an ever-slower rate. This will be demonstrated in Example 5.6 later in this section. Analysis of a two-stage line Much of the preceding discussion is based on the work of Buzacott [2], who has pioneered the analytical research on flow lines with buffer stocks. Several of his publications on the subject are listed in the references at the end of this chapter [2,4,5,6,7]. Other researchers have also studied the problem of flow lines with storage buffers, especially the two-stage line, and the interested reader is referred to their work [12,15,16]. Our presentation of the two-stage flow line problem will closely follow Buzacott’s analysis developed in reference [2]. The two-stage line is separated by a storage buffer of capacity b. The capacity is expressed in terms of the number of work parts the storage zone can hold. Let F1 and F2 represent, respectively, the breakdown rates of stages 1 and 2. We will use the term r to define the ratio of breakdown rates as follows :
The ideal cycle time, To is the same for both stages. We assume that the downtime distributions of all stations within a stage are the same, and that the average downtimes of stages I and 2 are Td1 and Td2. Over the long run, both stages must have equal efficiencies. For example, if the efficiency of stage I would tend to be greater than the stage 2 efficiency, the inventory in the storage buffer would tend to build up until its capacity b is reached. Thereafter, stage l would eventually be blocked when it tried to outproduce stage 2. Similarly, if the efficiency of stage 2 is greater, the buffer inventory would become depleted, thus starving stage 2. Accordingly, the efficiencies of the two stages would tend to equalize over time. The overall line efficiency for the two stages can be expressed as
where Eo is the line efficiency without buffer storage. The value of Eo was given by Eq. (5.l6), but we will write it below to explicitly define the two-stage system efficiency when b = 0 :
The term D1h(b) appearing in Eq. (5.25 represents the improvement in efficiency that results from adding buffer capacity (b > 0). D1 is the proportion of total time that stage 1 is down. However, Buzacott defines D1 as
Note that the form of this equation is different from the case given by Eq. (5.6) when we were considering the How line to be an integral mechanism, uncomplicated by the presence of buffer stocks. The term h(b) is the ideal proportion of the downtime D1 (when stage 1 is down) that stage 2 could be up and operating within the limits of buffer capacity b. Buzacott [2] presents equations for evaluating h(b) using a Markov chain analysis. The equations cover several different downtime distributions and are based on the assumption that the probability of both stages being down at the same time is negligible. Four of these equations are presented in Table 5.l. In sample calculations, the author has found that Eq. (5.21) tends to overestimate line efficiency. This is because of the assumption implicit in the computation of h(b) that both stages will not be broken down at the same time. Another way of stating this assumption is this: During the downtime of stage l, stage 2 is assumed to always be operating. However, it is more realistic to believe that during the downtime of stage 1, stage 2 will be down a certain portion of the time, this downtime determined by the efficiency of stage 2. Hence, it seems more accurate to express the overall line efficiency of a two~stage line as follows, rather than by Eq. (5.21) :
where E2 represents the efficiency of stage 2 by Eq. (5.l7). In work subsequent to the research documented in reference [2], Buzacott develops a more exact solution to the two-stage problem than that provided by Eq. (5.2l). This is reported in reference 5 and seems to agree closely with the results given by Eq. (5.29). EXAMPLE 5.5 This example will illustrate the use of Eq. (5.29) and Table 5.l to calculate line efficiency for a transfer line with one storage buffer. The line has I0 workstations, each with a probability of breakdown of 0.02. The cycle time of the line is I min, and each time a breakdown occurs, it takes exactly 5 min to make repairs. The line is to be divided into two stages by a storage bank so that each stage will consist of five stations. We want to compute the efficiency of the two-stage line for various buffer capacities. Solution : First, let us compute the efficiency of the line with no buffer capacity.
By Eq. (5.16),
Next, dividing the line into two equal stages by a buffer zone of infinite capacity, each stage would have an efficiency given by Eq. (5.l7).
By Eq. (5.18), Eoo = 0.6667 represents the maximum possible : efficiency that could be achieved by using a storage buffer of infinite capacity. TABLE 5.1 Formulas for Computing h(b) for a Two-Stage Flow line under Several Downtime Situations The following definitions and assumptions apply to Eqs. (5.25) through (5.28) used lo compute h(b) : Assume that the two stages have equal repair limes and equal cycle times. That is,
and
let
when B is the largest integer satisfying the relation
and L represents me leftover units, the number by which b exceeds BTd/Tc. Enally, let r = F1/F1, as given by Eq_ (5.20). With these definitions and assumptions, we can express the relationships for two theoretical downtime distributions as developed by Buzacott [2]: 1. Constant repair distribution. Each downtime occurrence is assumed to require a constant require time Td.
2. Geometric repair distribution. This downtime distribution assumes that the probability that repairs are completed during any cycle is independent of the time since repairs began. Define the parameter K :
Then for the two cases
Now, we will see how Eq. (5.29) and the formulas in Table 5.1 are used. Let us investigate the following buffer capacities: b = 1, 10, 100, and °°. When b = 1, according to Eq. (5.24) in Table 5.1, the buffer capacity is convened to B = 0, and L = 1. We are dealing with a constant repair-time distribution and the breakdown rates are the same for both Stages, so r = 1. We will therefore use Eq. (5.26) to compute the value of h(1) :
Using Eq. (5.23) to get the proportion of total time that stage 1 is down
Now the line efficiency can be computed from Eq. (5.29)
Only a very modest improvement results from the use of a storage buffer with capacity of one work part. When b = 10, B = 2 and L = 0. The value of h(10) is again computed from Eq. (5.26) :
The resulting line efficiency is
We see a 12% increase in line efficiency over Eo from using a buffer capacity of 10 work units When b = 100, B = 20, L = 0, and
Therefore, E = 0.50 + 0.25 ( 0.952 ) ( 0.6667 ) = 0.6587 A 32% increase in line efficiency results when the buffer capacity equals 100. Comparing this to the case when b = 10, we can see the law of diminishing returns operating. When the storage capacity is infinite (b = °°),
E = 0.50 + 0.25 ( 1.0 ) ( 0.6667 ) = 0.6667 according to Eq. (5.29), which equals the result given by Eq. (5. 18). It is of instructional value to compare this result for b = 0°° given by Eq. (5.29), with the result given by Eq. (5.21), which neglects the possibility of overlapping stage downtime occurrences. According to Eq. (5.2l), the line efficiency would be calculated as E = 0.50 + 0.25 ( 1.0 ) = 0.75 which, of course, exceeds the maximum possible value given by Eq. (5. l8). The corrected formula, Eq. (5.29), provides better calculated results, since it accounts for the occurrence of overlapping breakdowns for the two stages. Flow lines with more than two stages We will not consider exact formulas for computing efficiencies for How lines consisting of more than two stages. lt is left to the interested reader to consult some of the references, in particular [1], [2], and [5]. However, the previous discussion in this section should provide a general guide for deciding on the configuration of multistage lines. Let us consider some of these decisions in the context of an example. EXAMPLE 5.6 Suppose that the flow line under consideration here has I6 stations with cycle time of 15 s (assume that all stations have roughly equal process times). When station breakdowns occur, the average downtime is 2 min. The breakdown frequencies for each station are presented in the following table. Station pt Station pt 1 0.01 9 0.03 2 0.02 10 0.01 3 0.01 11 0.02 4 0.03 12 0.02 5 0.02 13 0.02 6 0.04 14 0.31 7 0.01 15 0.33 8 0.01 16 0.31 We want to consider the relative performances when the line is separated into two, three, or four stages. Solution: First, the performance of the line can be determined for the single-stage case (no storage buffer) From the table, the frequency of downtime occurrences is given by Eq. (5.9) :
The line efficiency is, by Eq. (5.16)
To divide the line into stages, we must first decide the optimum locations for the storage buffers. The stations should be grouped into stages so as to make the efficiencies of the stages as close as possible. Then, to assess relative performances for the two-, three-, and four-stage lines, we will base the comparison on the use of storage buffers with infinite capacity. For the two-stage line, the breakdown frequency of F = 0.30 should be shared evenly between the two stages. From the foregoing table of p, values it can be determined that the storage buffer should be located between stations 8 and 9. This will yield equal F values for the two stages.
The resulting stage efficiencies an
From Eq. (5.18), E.. = 0.4545. Similarly, for the three-stage configuration, the frequency of breakdowns should be divided as equally as possible among the three stages. Accordingly, the line should be divided as follows ; Stage Stations Fk Ek 1 1-5 0.09 0.5814 2 6-10 0.10 0.5556 3 11-16 0.11 0.5319 The third stage, with the lowest efficiency, determines the overall three-stage line efficiency Eoo = 0.5319. Finally, for four stages the division would be the following : Stage Stations Fk Ek 1 1-4 0.07 0.6410 2 5-8 0.08 0.6098 3 9-12 0.08 0.6998 4 13-16 0.07 0.6410 The resulting line efficiency for the four-stage configuration would be Eoo = 0.6098 The example shows the proper approach for dividing the line into stages. lt also shows how the line efficiency continues to increase as the number of storage banks is increased. However, it can be seen that the rate of improvement in efficiency drops off as more stage buffers are added. In the foregoing calculations we assumed an infinite buffer capacity. For realistic capacities, the efficiency would be less as was illustrated in Example 5.5. The maximum possible efficiency would be achieved by using an infinite storage bank between every workstation. Station 6 would be the bottleneck stage. We leave it for the reader to figure out the line efficiency for the I6-stage case.
This entry was posted in Computer Integrated Manufacturing, part II on May 3, 2013 [https://rekadayaupaya.wordpress.com/2013/05/03/5-4-automated-flow-lines-withstorage-buffers/] .
One thought on “5.4 AUTOMATED FLOW LINES WITH STORAGE BUFFERS”
Ankush January 18, 2016 at 18:27
I am unable to locate table 5.1 in the article. Can you can please help?