Bayonne Packaging

OPERATIONS MANAGEMENT 2012/2013 1. Introduction to BAYONNE PACKAGING, Inc BAYONNE PACKAGING, Inc is a $43 million printer and paper converter company that produces customized paper-based packaging, for industrial customers, for promotional materials software, luxury beverages, gift good and gift candy. Presently, the business is leaded by Dave Rand and the board of the company is constituted by family members, a local banker and outside counsel. This company is implemented in the paper packaging industry that was featured by the quickly growth between 1980? s and early 1990? . Due to organizations that chase the desire to make a greater impact with their promotional materials or moved their promotional budget from print media and broadcast forms to the package itself at the moment of purchase. From the combination of this growth and Bayonne? s aspiration to improve, the company is divided in six major stages: Composition, Sheeting, Printing, Die-Cut, Fold&Glue, Finishing and Shipping dock. In the first one, the goal is to develop the printing and the package design, printing plates are made and die-cutting dies are ordered.Then in the Sheet Department, the paper is sheeted from the roll stock and stacked on skids to be printed. The following one is the Printing department that is where the artwork is printed on 4- and 6-colour presses. In the Die-Cut department, the sheets are printed into â€Å"blanks†1. After that, the Fold&Glue department consists in turning the die-cut blanks into the finished product. Here there are two options, if it is a large order it goes to the â€Å"Royal/Queen† which is an high speed machine. If it is a low volume order is goes to â€Å"Staudes† machines.After this, a selection is made to decide which products are sent to the 3A machines2. According to the analysis of this company, there are three major problems that have a huge impact on the way the organization functions which are: quality, delivery time and c ost of the final product. Taking to account the quality control problems we realized that the main issue was concentrated in Fold & Glue department where there was either a lack in the glue lines or excess of glue. Consequently, 6% of products were defective and more than 1% of final product were rejected by he customers, for example boxes popping open because it with not enough glue or no glue at all. Besides the quality control problems, there is a significant inefficiency concerning delivery on time of the customized packages- they were late more than 20% of the time. Initially, customers learned not to entirely trust in the delivery time of the product. They also realized that some other component of the marketing project might become available earlier than anticipated and this gave a sense of hope that all would be delivered on time to both of the parties.In addition, customers often were demanded by their own clients to anticipated the due date scheduled by the marketing proje ct, but on the other hand Bayonne did not have the means to respond promptly. Therefore, they want to â€Å"move up† or expedite a due date to receive product sooner than they had originally been promised. 2. Capacity utilization in work centers Machine/Work Center| Total hours per machine| Capacity utilization| Composition| 255| 73,50%| Jagenburg sheeter| 279| 80,40%| Heidelberg press| 348| 100%| Bobst die-cut| 272| 78,39%| Int. Royal/Queen F&G| 156| 44,96%| Int.Staude F&G| 179| 51,59%| Int. 3A window/patch| 145| 41,79%| In order to organize the information provided, we built a capacity utilization? s table of each work center where we, according to the provided data and by calculating capacity utilization of each work center, we measure how much the process actually does produce relative to how much it can produce if it were running at full speed (Process Capacity). Capacity Utilization = Flow rateCapacity Given the information in Exhibit 2, October of 2011 had 347 schedule d work hours and this means the maximum number of hours in each work center that an be reached. Considering that the bottleneck is the resource with the lowest capacity and that the flow rate of the resources is identical, the bottleneck is the resource with the highest capacity utilization. After computing these values above, we conclude that only one work center operates at full capacity which is the two machines of Heidelberg press (bottleneck). Here are the main reasons that could justify why the other work centers didn’t reach 100% of capacity utilization: * If demand is less than supply, the process will not run at full capacity, but only produce at the rate of demand. If there is insufficient supply of the input of the process, the process will not be able to operate that capacity. * If there is one or several processes that only have a limited availability (e. g. maintenance and breakdowns) , the process might operate at full capacity while it is running, but then go into periods of not producing any output while it is not running. This is the case in the Die-Cut work center where production is stopped in order to change dies. 3. Capacity in pieces per day for the Die-Cut centerTaking into consideration the present question, we were asked to find the capacity in pieces per day for the Die-Cut work center, specifically the Bobst Die-Cut. We also have to assume that one order is 30000 pieces. a) None of the orders can be ganged When none of the orders can be ganged, meaning that each time we process an order we will have a single setup time. In order to know the number of orders of the process per month (Q), we have to match the total time available per month to the number of orders multiplying by the total time to produce one order. Time available per month = Q x Setup time + Q x Run timeWe compute the total time available per month taking into account that â€Å"October 2011 had 347 scheduled work hours net of breaks† and that the Die-Cut department has two machines. Time available per month = 347 hrs x 60 min x 2 machines Time available per month = 41640 min Although the Standard setup time is 30 min/sheet, in reality the setup time (time to change dies) is 2/3 hours. So we assumed, an average of 2,5 hours which corresponds to 150 minutes. Setup Time per job = 150 min To compute the run time per order we have to multiply the run time per sheet by the number of sheets that compose the order.Assuming that sheets averaged 3 pieces, each order has 10000 sheets. Run time/order = Run time/sheet x N? of sheets/order Run Time/order = 0, 0075 min x 10000 sheets Run Time/order = 75 min After calculating the values above, we conclude that total time to produce one order which is divided by setup time and run time per order is 225 minutes. Time to produce one order = Setup time + Run Time Time to produce one order = 150 min + 75 min Time to produce one order = 225 min Now, we are in conditions to find Q – number of orde rs per month – which is 185,06667. Time available per month = Q x Setup time + Q x Run timeTime available per month = Q x (Setup Time + Run Time) Time available per month = Q x Time to produce one order Q = Time available per monthTime to produce one order Q = 41 640 min 225 min Q= 185,06667 orders/month Taking into consideration that 1 order equals to 30 000 pieces and consequently, equals to 10 000 sheets we can convert Q – capacity per order per month – into capacity per sheets and also per pieces both per month. Value| Calculations| Capacity/order/ month| 185,06667| Made above| Capacity/sheets/month| 1850666,667| 185,06667 x 10 000 sheets| Capacity/pieces/month| 5552000| 185,06667 x 30 000 pieces|It was required to get the capacity per pieces per day. So as we know that per day there are two shifts of 7, 5 hours each with a meal break of 30 minutes for every worker, in the end of one day there is 15 hours of work time. Knowing that the time available for mon th is 347 hours, each month has 23, 1(3) days. Therefore, after the conversion of months into days, we conclude that capacity per pieces per day equals 240000, meaning the maximum amount the resource, in this case the machines from the Die-Cut department, can produce per unit of time (per day, in this question). b) Pairs of orders can be gangedFacing this new situation, where pairs of orders can be ganged, the setup time must be allocated in a different way. Now, we are going to have one setup for every two orders. The time available per month will be the same, 41640 minutes – calculations on the sub question above. Time available per month = 41640 min In order to know the number of orders (pairs of orders, in this case) per month ( Q ), we have to match the total time available per month to the number of pairs of orders multiplying by the total time to produce one pair of orders. Time available per month = Q x Setup time + Q x Run timeTime available per month = Q x (Setup ti me + Run time) Time available per month = Q x (Time to produce a pair of orders) The setup time per job (calculations in the sub question above) is 150 minutes, and each time we process 2 orders 150 minutes will be spent to change dies. Setup Time per job = 150 min To compute the run time per pair of orders we have to multiply the run time per sheet by the number of sheets that compose a pair of orders. As we are assuming that orders averaged 10 000 sheets, we will have that each pair of orders has 20 000 sheets (2 orders x 10 000 sheets).Run time/ pair of order = Run time/sheet x N? of sheets/ pair of order Run Time/ pair of order = 0, 0075 min x 20000 sheets Run Time/order = 150 min So, the total time to produce a pair of orders which is composed of setup time and run time both per pair of orders will be 300 minutes Time to produce one order = Setup time + Run Time Time to produce one order = 150 min + 150 min Time to produce one order = 300 min We are able to compute Q – n umber of orders (pair) per month – which equals 138, 8 orders per month Time available per month = Q x Time to produce one orderQ = Time available per monthTime to produce one order Q = 41 640 min 300 min Q= 138, 8 orders/month Keeping in mind that 1 order = 30 000 pieces =10 000 sheets we can convert Q – capacity per order per month – into capacity per sheets and also per pieces both per month. Value| Calculations| Capacity/ pair of orders/ month| 138,8| Made above| Capacity/sheets/month| 2776000| 138,8 x 20 000 sheets| Capacity/pieces/month| 8328000| 138,8 x 60 000 pieces| This last value is the capacity per pieces per month but as we are asked to compute the capacity per pieces per day we must make the conversion.As each day has 15 hours of work time (calculations in the sub question above) and the time available for month is 347 hours, dividing this value by the 15 hours per day, we conclude that each month has 23, 1(3) days. Therefore, after the conversion of months into days we conclude that capacity per pieces per day equals 360000, meaning the maximum amount the resource, in this case the machines from the Die-Cut department, can produce per unit of time, in this question, day. c) All the others can be ganged In the case that all orders are ganged, the total process will include only one set up time.In order to know the number of orders of the process per month (Q), we have to match the total time available per month to the number of orders multiplying by the total time to produce one order. Time available per month = Setup time + Q x Run time The total time available per month remains the same, 41640 minutes. The setup time will be independent from the number of orders because there will be a single one for all of them considering that they are all ganged. Time available per month = 41640 min Setup Time per job = 150 minKnowing that one order has 10000 sheets, the run time per order will be 75minutes. Run time/order = Run time/she et x N? of sheets/order Run Time/order = 0, 0075 min x 10000 sheets Run Time/order = 75 min Regarding all of these values, it’s now possible to calculate Q – number of orders per month. Time available per month = Setup time + Q x Run time 41640 min = 150 min + Q x 75 min Q = 41640 min-150 min75 min = 553, 2 orders/month Keeping in mind that 1 order = 30 000 pieces =10 000 sheets we can convert Q – capacity per order per month – into capacity per sheets and also per pieces both per month.Value| Calculations| Capacity/order/ month| 553,2| Made above| Capacity/sheets/month| 5532000| 553,2 x 10 000 sheets| Capacity/pieces/month| 16596000| 553,2 x 30 000 pieces| As we are asked to compute the capacity per pieces per day we must make the conversion. Each day has 15 hours of work time (calculations in the sub question above) and the time available for month is 347 hours, dividing this value by the 15 hours per day, we conclude that each month has 23, 1(3) days. Therefore, after the conversion of months into days we conclude that hen orders can be ganged capacity per pieces per day equals 717406, 3, meaning the maximum amount the resource, in this case the machines from the Die-Cut department, can produce per unit of time, in this question, day. 4. Assume that 40 of the orders partialed in October each cause one broken production run in the Royal/Queen work center, resulting in two setups for these orders instead of one a) Capacity in October without these additional setups Assuming that 40 of the orders were partialed in October and that each induced one broken production run in Royal/Queen work center, resulting in two setups for these orders instead of one.First we calculated the capacity without these additional setups. Initially, we worked with the status of the process without these additional setups. We have already calculated the capacity production run, in pieces, in the Royal/Queen work center, an essential value for the computati on of the average time per order. So we calculated the ratio between the pieces scheduled and the orders scheduled of Royal/Queen machine, then we multiplied that value by the respective run time. Finally, we did the sum of the previous value with the respective setup time.Pieces Scheduled per Orders Scheduled = Sheets per Pieces scheduledOrders scheduled Pieces Scheduled per Orders Scheduled = 6. 209. 32977 Pieces Scheduled per Orders Scheduled =80640,(63) pieces Average time per order = Standard Setup Time + Pieces scheduled per Orders Scheduled x Standard Run Time Average time per order = 180+80640,(63) x 0. 0023 Average time per order = 365, 4734 min After that, we had to determine the capacity per order, where we calculate the ratio between total work time scheduled of the three machines in minutes and the average time per order.Total work time scheduled of the three machines in minutes= 347 x 60 x 3=62460min Capacity per order = 62460365,4734 = 170, 9016 orders/ min In the end , we calculated the capacity per piece multiplying the ratio between the pieces scheduled and orders scheduled by the capacity per order. Capacity per piece = 80640, (63) x 170, 9016 = 13781613, 7(68) pieces / min Therefore, we analyzed the capacity production run, in pieces, in the Royal/Queen work center but considering the additional setups. The company with the introduction of these setups the company loses time in the overall process. ) Capacity in fact In order to figure out what happened with the introduction of the additional setups, meaning that at this point the company had 40 orders partialed and as we have the information that there are 2 setups per order we consequently know that Bayonne had 80 setups in this work center. On the other hand, if there were no partialed orders, the work center would only have 40 setups. We conclude that when the setups increase the run time available will decrease. We know that there was a reduction of the capacity in this work center, aff ecting the overall process.Moreover we calculated the time spent in the production of those 80 orders partialed (we assume that they are equally distributed so 40 x 2), which is 21818,939 because we had to take into account the setup time and the run time of the 80 orders partialed. As we can see: Time to produce 80 partials = (80 x 180 + 80Ãâ€"0,0023 x 620932977) Time to produce 80 partials=21818,939 min We also calculated the time available to the company to produce the orders, considering the total time available in minutes, the time necessary to produce the 80 orders partialed and the additional orders produced in the available time of the total time per order:Available time = 62460 – 21818,939 Available time = 40641,061 Additional orders produced = 40641,061365,47355 =111,2011 Therefore, we calculated the additional number of pieces: Additional number of pieces =111,2011 x 80640,63 = Additional number of pieces = 8967329,736 So the total number of pieces produced in the end of the month was 12192955,191, since we had to consider the sum between the additional pieces produced and the pieces scheduled multiplied by the 40 orders partialed. Total number of pieces = 8967329,736 + (40Ãâ€"80640,63) Total number of pieces = 12192955,191 pieces/month 5.Size of orders route to the Royal/Queen work center and to the Staude work center Given the information on exhibit 2, we could calculate de size of orders to the Royal/Queen work center and to the Staude work center. In other words, we considered the setup standard times and the run standard time (the slope of line) of each work center while essential tools to create a graph where it is easier to take very useful conclusions about the size of the batch of these work centers. To have a clear comprehension of the graph, we considered two lines, one blue and one purple that represent, respectively, he Royal/Queen work center and the Staude work center. Royal/Queen Machine | Y=0,0023x +180| Staude Machine| Y=0 . 015x + 40| We know that Royal/Queen machine has a higher setup cost but in the other hand has a lower run cost. Comparing to the Staude machine, it has a higher run cost but a lower setup cost. With this data we can say that the Royal/Queen machine is indicated for big batches and the Staude machine for lower ones. As a result of these calculations, we obtained the intersection of the two lines (break-even point), with a value of 11. 23,62, that represents the point where is indifferent to use between the Royal/Queen machine and the Staude machine in the overall process. So, for batches with a size below than 11023,6 we choose the Staude machine, but if the batch has a size above the break-even point we will then choose the Royal/Queen machine. 6. Yield at each of the work centers Sheet, Print, Die-cut, and Royal/Queen and the cumulative yield for an other which the sheets starts with 40000 sheets Here we took into consideration the definition of yield which is the percentage of u nits lost of each work center.It was required to compute the yield of the following work centers: Sheet, Print, Die-cut, and Royal/Queen. In the tablet below, we organized the data provided and determined the values of the yield of each work center mentioned earlier. We can see that all of the work centers have a very low percentage of units lost because the values of the yield are very close to 100%. And as we know when the yield is 100% it means that there are no losses at all and the process reaches the maximum of efficiency possible. Work Center| Pieces in| Pieces out| Yield| Sheet| 9555097| 9488211| 99,300%|Print| 9488211| 9326912| 98,300%| Die-Cut| 9326912| 9233643| 99,000%| Royal/Queen| 6209329| 5588396| 90,000%| Besides this, we also had to compute the cumulative yield which is 86,972%. This value was calculated by multiplying the yields of each work center. In order to calculate the cumulative yield for another which the sheets start with 40000 sheets, we had to convert the number of pieces into sheets. If one sheet corresponds to 3 pieces, then: 40000 sheets x 3 pieces = 120. 000 pieces The input of the production process is 120000 sheets (100%). However we have found that the cumulative yield is 86,972%.So this tells us that 13,028% of the input is lost during the production process. Phases| Calculations| Values (in sheets)| Input| 120000| 120000| Losses| 120000 x 13,028%| -15633,6| Output| 120000 x 86,972%| 104366,646| 7. The data in exhibit 4 (value of actual shipments in October) After evaluating the graphic below that we have reached with the values of the variables Orders shipped, Late and Partialed, we can see that there is a direct effect of the number of orders Shipped to the Late ones, in other words, the more orders there are the more time is needed to deliver it, making them even more late.We can also extract that the number of orders partialed influences the number of late orders because when an order is divided, it turns into two and th is means that there would be another setup to be made instead of just one. These new setups interrupt the process flow, which consequently â€Å"steal† capacity. This intensifies depreciation of the machines and consequently increasing the costs. If Bayonne has several partialed orders that means that some quantity of those orders are still being produced, taking up resources that could be used to produce new orders.Because of this, these new orders will start the production process late, making almost impossible for them to be delivered at the scheduled time between Bayonne and the customers. 8. Recommendations of short-term and long-term After this report, one of the main goals of the BAYONNE PACKAGING, Inc is to reduce or minimize the problems that the company faces itself. Quality control, delivery time and cost problems are the major obstacles in the progress of the company. So in order to improve the management and the planning of the overall process, Dave Rand and the board of the company have to take some actions in short-term and medium term.So firstly, we suggest that they should have a closer careful oversight in Fold&Glue department, because there are in fact a significant portion of products that are defective and, consequently some of that portion is rejected by the customers. This issue have a huge impact in the image and reputation of the company, so this oversight is imperative and it can translated in more time spend in the inspection of this department. Also, we believe that the defective units need to be reworked or eliminated from the process.The company should to support the idea of reworking the defectives units, in the way that avoid the waste of the raw materials and the labor spent in the process of that units. In our opinion, Bayonne need to recycle the wastages and reutilizing them for further production. Also, we think that the company should be more organized in time schedules and deadlines, because it implicates delays in all the departments and therefore in the overall of the process. Finally, the company need to eliminate the setup time or at least try to reduce the time it takes to perform in the process, for the obvious reason that consequently it â€Å"steal† capacity.In other words, as nothing is produced at a resource during setup, the more frequently a resource is set up, the lower its capacity. So we believe that the company should be the increasing of the orders in a batch, with the objective that as more units there are in a batch, the more we can spread out the setup costs. And so to take advantage of the economies of scale in the entire process. In conclusion, if the company follows our suggestions with the expected results, maintaining everything else constant, they can conquer better results and diminish the difference between what is expectable and what actually occurs.Annex 255hrs ? 100% 347hrs=73,487031% Capacity utilization of composition 279hrs ? 100%347hrs = 80,40345821% C apacity utilization of Jagenburg 348hrs? 100%347hrs=100,2881844% Capacity utilization of Heidelberg press 272hrs ? 100%347hrs=78,386167% Capacity utilization of Bobst Die-Cut 156hrs ? 100%347hrs=44,9567723% Capacity utilization of Int. Royal/Queen 179hrs ? 100%347hrs=51,5850144% Capacity utilization of Staude Machines 145hrs ? 100%347hrs=41,7867435 % Capacity utilization of Int. 3A window