Key Revision Chapter 4 on Lecture Note Batch Size.xlsx

# Key Revision Chapter 4 on Lecture Note Batch Size.xlsx -...

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Unformatted text preview: Lesson Learnt (1) Assembly-Type Batch sell any of them till all co (a) Formula of Process C B/ [S + (B x P)] (b) Always have 1st setup hence, there are two set 60min x 2 = 120min. (c) Time per unit (P). The 1 steer support and 2 rib In short, Setup time and Example: E.g. Process Capacity of per min E.g. Process Capacity of B unit per min (1) Only Milling Machine such: (a) Process Capacity of B min (b) Process Capacity of B per min (2) Assembly process has (1/3) unit per min which (3) Interpretation: At bat bottleneck is assembly p (4) To find a good batch s min (b) Process Capacity of B per min (2) Assembly process has (1/3) unit per min which (3) Interpretation: At bat bottleneck is assembly p (4) To find a good batch s bottleness: Process Capacity of M Lesson Learnt (1) Assembly-Type Batch Process (i.e. components are dependent where you cannot sell any of them till all components are produced). (a) Formula of Process Capacity in batch size only applies for process with setup only = B/ [S + (B x P)] (b) Always have 1st setup time. Since 2 types of components (i.e. steer and ribs), hence, there are two setup time (one for steer and another for ribs). Therefore, S = 60min x 2 = 120min. (c) Time per unit (P). The unit here refers to per set whereby a set of scooter comprises 1 steer support and 2 ribs, hence P = 1min + (2 x 30sec) = 2min In short, Setup time and Time per unit are based on per production cycle. Example: E.g. Process Capacity of Batch Size 12 = 12/ [120 + (12 x 2)] = 12/ 144 = 0.0833 unit per min E.g. Process Capacity of Batch Size 300 = 300/ [120 + (300 x 2)] = 300/ 720 = 0.4166 unit per min (1) Only Milling Machine has setup time, hence, its process capacity is calculated as such: (a) Process Capacity of Batch Size 12 = 12/ [120 + (12 x 2)] = 12/ 144 = 0.0833 unit per min (b) Process Capacity of Batch Size 300 = 300/ [120 + (300 x 2)] = 300/ 720 = 0.4166 unit per min (2) Assembly process has NO setup time, hence, its process capacity is calculated as : (1/3) unit per min which is 0.33 units per min. (3) Interpretation: At batch size 12, bottleneck is milling machine. At batch size 300, bottleneck is assembly process (4) To find a good batch size where both miiling machine and assembly are co- min (b) Process Capacity of Batch Size 300 = 300/ [120 + (300 x 2)] = 300/ 720 = 0.4166 unit per min (2) Assembly process has NO setup time, hence, its process capacity is calculated as : (1/3) unit per min which is 0.33 units per min. (3) Interpretation: At batch size 12, bottleneck is milling machine. At batch size 300, bottleneck is assembly process (4) To find a good batch size where both miiling machine and assembly are cobottleness: Process Capacity of Milling Machine = Process Capacity of Assembly Process (1/3) = B/ [120 + (B x 2)] (1/3)B = 40 B = 120 B = summation of all capacity = 80 + 100 + 120 S = 30min (downtime) P = time per unit = time to produce one barrel per barrel. P is based on the process with set step produce 100 barrels per hr, and hence in produce (60/100) barrel in a min B = summation of all capacity = 80 + 100 + 120 S = 30min (downtime) P = time per unit = time to produce one barrel per barrel. P is based on the process with set step produce 100 barrels per hr, and hence in produce (60/100) barrel in a min capacity = 80 + 100 + 120 = 400 barrels e) me to produce one barrel = (60/100) min d on the process with setup time where the rrels per hr, and hence in a min, it can rrel in a min (1). Formula to use for proce (2). Given Batch size is 1000 comprises Part A and Pa Due to 1st setup and alt Time per unit (P) is 0.01 Given 7 hr operating hou Note: Make sure all unit (3). Answer for Q1(a): The p (1000 x 0.02 hr)] } x 7hr (4). For Q1b, 110% of 250 se Answer for Q1b: [B/ 8 + (B x 0.02)] x 7 = 2 1.5B = B= (5). Q (c):Hint: refer to the c Formula to use for process capacity per day basis = {B/ [S + (B x P)]} x operating hrs per day Given Batch size is 1000 (because the answer is required in terms of sets where one set comprises Part A and Part B i.e. 1000 of part A and 1000 of part B equals 1000 sets) Due to 1st setup and alternate setup to switch from Part A to B, hence the setup is 4hrs x 2 = 8hrs Time per unit (P) is 0.01 hr x 2 = 0.02 hr Given 7 hr operating hour per day. Note: Make sure all unit of measurement are same i.e. hours before using formula Answer for Q1(a): The process capacity of the company in terms of sets per day = {1000/ [8hr + (1000 x 0.02 hr)] } x 7hr = 250 sets per day For Q1b, 110% of 250 sets = 275 sets Answer for Q1b: [B/ 8 + (B x 0.02)] x 7 = 275 1.5B = 2200 B = 1467 sets Q (c):Hint: refer to the components of the formula remarks; WIP = FR x FT = PC x (TPT/ Operating hr per day) remarks; When batch size is increased, WIP is increased (higher carrying cost), while capacity increases or improves Q2a To find bottleneck and Max TPT (in term of finished good) per day. Given 400 min as daily operating hr (1). Determine the flow unit is a batch size of 300 finished goods (where one finished good comprises 1 x part A and 2 x part B, hence 300 finished goods comprises 300 x part A and 600 part B) (2). Draw Process Flow Diagram (note: some process stage like drying is in paragraph), and compute processing time of each process stage bec need to compute TPT Stamping Stage: Setup comprises 1st setup and alternate setup (i.e. 2 x 100min = 200min) and Time per unit = (Part A x 1min) + (Two Part B x 30sec) = 2min per finished good. Hence, processing time of Stamping = 200min + (300 finished good x 2min per finished good) = 800min Processing rate or Resource Capacity of Stamping Stage per day = 300/ 800 x 400min = 150 sets per day OR using the formula: {B/ [S + (B x P)]} x Operating hr = {300/ [200 +(300 x 2)]} x 400 = 150 sets per day. Note: As there is only one stamping machine, the two operators do not make any difference Painting Stage: Only 1st Setup bec (hint: robot can easily switch & change color only after 300 of part A and 600 of part B which is 300 sets of finished good). Hence, Setup = 20min. Time per unit = (Part A x 30 sec) + (Two Part B x 15sec) = 1min per finished good. Hence, processing time of Painting = 20min + (300 finished good x 1min) = 320min. Resource Capacity of Painting Stage per day = {300/ [20 +(300 x 1)]} x 400 = 375 sets per day Drying Stage: As no resource is involved, processing time is 120min, and no resource capacity Assembly Stage: No setup time as not given. Time per unit = 26.67min. Hence, processing time = (300 finished good x 26.67min) / 12 workers = 667 min. Note: need to divide by the 12 assembly workers as they carry out the task concurrently Resource Capacity of Assembly Stage per day = (300/ 667) x 400min = 180 sets per day OR {300/ [0 +((300 x 26.67)/ 12)]} x 400 = 180 sets per day Remarks For c, use the answer fou and then find the batch s Among these stages, the minimise inventory with Remarks For d, Flow Rate = min {D Question provide DR. Then, choose the FR, and painting stages. Among t chosen as it will minimis rate. Remarks For c, it is saying that 50 parts must go thru step 1 and 2 (bec have to stay together), before 20 parts can just go through step 3. Remarks For d, it is saying an ongoing process. Thus, TPT + (n-1)CT TPT = 20min (setup) + 1min + 2min + 1.5min = 24.5min (i.e. FT) Hence, 24.5 + (20-1) x 2min = 62.5min Remarks For e, solve the equation by getting the two lowest capacity to be co-bottleness i.e. Step 1 Vs Step 2 or Step 3? Step 2 is (1/2) unit per min while Step 3 is (1/3) unit per min, hence Step 2 has lower capacity Remarks For b, solve the equation by getting the two lowest capacity to be co-bottleness For c, use the answer found in Qa which is the process capacity and then find the batch size for moulding AND painting stages. Among these stages, the higher batch size is chosen as it will minimise inventory without decreasing the process capacity. For d, Flow Rate = min {DR, PC}. Question provide DR. Then, choose the FR, and find the batch size for moulding AND painting stages. Among these stages, the higher batch size is chosen as it will minimise inventory without decreasing the flow rate. Author: Remember to set unit of measurement the same for DR and PC t the same for in one production cycle, there are three setup and three production for \$5, \$10 and \$50 note. Batch size = 10 notes per sec x 60 sec x 90 min = 54000 notes [remark: 90 min bec of three production peri Setup = 90min x 60 sec = 5400 sec [remark: is expressed in sec as sec is the basis] FR = 10 notes per sec therefore, P = (1/10) sec per note 24hrs per day = 24*60smin* 60 sec = 86400 seconds Q (a) Capacity = B/ S + (B*P) x 24hrs per day (i.e. 86400 sec) 432000 notes per day 144000 Q (b) \$5 + \$10 + \$50 = \$65 therefore, 432000/ 3 types of note = 144000 notes. Hence, value is 144000 x \$65 \$ 9,360,000 9360000 Q (c) Setup = 45min x 60 sec = 2700 sec [remark: from 90min setup to 45min as to reduce the changeover time Capacity = 576000 notes per day therefore, 576000/ 3 types of note = 192000 notes. Hence, value is 192000 x \$65 = 12480000 Therefore, answer \$ 3,120,000 [remark: \$12480000- \$9360000] or \$5, \$10 and \$50 note. k: 90 min bec of three production period, each 30 min. Is expressed in sec bec sec is the basis] is the basis] therefore, P = (1/10) sec per note 44000 x \$65 min as to reduce the changeover time by half in Question] 92000 x \$65 = 12480000 \$12480000- \$9360000] 2700 192000 12480000 ...
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• Fall '16
• gina
• International System of Units, Hour, Only Milling Machine

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