Industrial Processes I Manufacturing Economics

Transcription

1 Industrial Processes I Manufacturing Economics Equipment Cost Rate (Example 1) A production machine is purchased for an initial cost plus installation of $500,000. Its anticipated life = 7 yrs. The machine is planned for a two-shift operation, 8 hours per shift, 5 days per week, 50 weeks per year. The applicable overhead rate on this type of equipment = 35%. Determine the equipment cost rate. H = (50 weeks per year)*(5 days per week)*(8 hours per shift)*(2 shifts per day) C eq = equipment cost rate, $/min IC = initial cost of the equipment, $ N = anticipated number of years of service H = annual number of hours of operation, hr/yr R OH = applicable overhead rate for the equipment, % IC = $500,000 N = 7 yr H = (50 weeks per year)*(5 days per week)*(8 hours per shift)*(2 shifts per day) H = hr/yr R OH = 35% = 0.35 Solve for C eq C eq = $/min C eq = $/hr 1

2 Industrial Processes I Manufacturing Economics Cycle Time and Cost per Piece (Example 2) There is a production machine used to produce a batch of parts that each has a starting material cost of $2.35. Batch quantity = 100. The actual processing time in the operations = 3.72 min. Time to load and unload each workpiece = 1.60 min. Tool cost = $4.40, and each tool can be used for 20 pieces before it is changed, which takes 2.0 minutes. Before production can begin, the machine must be setup, which takes 2.5 hours. Hourly wage rate of the operator = $16.50/hr, and the applicable labour overhead rate = 40%. Determine (a) the cycle time for the piece, (b) average production rate when setup time is figured in, and (c) cost per piece. T c = cycle time of the unit operation, min/pc T o = actual processing time in the operation, min/pc T h = work part handling time, min/pc T t = tool handling time if that applies in the operation, min/pc T o = 3.72 min T h = 1.60 min T t = 2.00 min / 20 pc = min/pc Solve for T c T c = min/pc 2

3 Industrial Processes I Manufacturing Economics T p = average production time per piece, min/pc T su = setup time, min/batch Q = batch quantity, number of pieces (pc) T c = cycle time of the unit operation, min/pc T su = (2.5 hours/batch)*(60 min/hour) = min/batch Q = 100 pc T c = min/pc Solve for T p T p = min/pc 3

4 Industrial Processes I Manufacturing Economics R p = average hourly production rate, pc/hr T p = average production time per piece, min/pc T p = min/pc Solve for R p R p = pc/hr 4

5 Industrial Processes I Manufacturing Economics C L = labor cost rate, $/min R H = worker s hourly wage rate, $/hr R LOH = labor overhead rate, % R H = $/hr R LOH = 40 % = 0.40 Solve for C L C L = $/min C L = $/hr 5

6 Industrial Processes I Manufacturing Economics C o = cost rate of operating the work cell, $/min C L = cost rate of labor, $/min C eq = cost rate of equipment in the work cell, $/min C L = $/min C eq = $/min Solve for C o C o = $/min C pc = cost per piece, $/pc C m = starting material cost, $/pc C o = cost rate of operating the work cell, $/min T p = average production time per piece, min/pc C t = cost of tooling used in the unit operation, $/pc If applicable in the particular manufacturing process, the cost of tooling C t must be determined by dividing the actual cost of the tooling by the number of pieces between tool changes. C m = 2.35 $/pc C o = $/min T p = min/pc C t = $4.40 per 20 pc = $/pc Solve for C pc C pc = $/pc 6

7 Industrial Processes I Manufacturing Economics Scrap Rate (Example 3) A customer has ordered a batch of 1000 parts to be produced by a machine shop. Historical data indicate that the scrap rate on this type of part = 4%. How many parts should the machine shop plan to make in order to account for this scrap rate? q = the scrap rate In batch production more than the specified batch quantity is often produced to compensate for the losses due to scrap. Q = the required quantity of parts to be delivered Q o = the starting quantity q = 4% = 0.04 Q = 1000 parts Solve for Q o Q o = starting parts (round answer up) 7

8 Industrial Processes I Manufacturing Economics Cycle Time and Cost per Piece (Example 4) A high-production operation manufactures a part for the automotive industry. Starting material cost = $1.75, and cycle time = 2.20 min. Equipment cost rate = $42.00/hr, and labour cost rate = $24.00/hr, including overhead costs in both cases. Availability of the production machine in this job = 97%, and the scrap rate of parts produced = 5%. Because this is a long-running job, setup time is ignored, and there is no tooling cost to be considered. (a) Determine the production rate and finished part cost in this operation. (b) If availability could be increased to 100% and scrap rate could be reduced to zero, what would be the production rate and finished part cost. a) R p = average hourly production rate, pc/hr T p = average production time per piece, min/pc T p = 2.20 min/pc Solve for R p and multiply the result by the availability of the production machine R p = pc/hr (based on the production machine being 97% available) 8

9 Industrial Processes I Manufacturing Economics C o = cost rate of operating the work cell, $/min C L = cost rate of labor, $/min C eq = cost rate of equipment in the work cell, $/min C L = ($24.00/hr)(1 hr per 60 minutes)/availability = $/min C eq = ($42.00/hr)(1 hr per 60 minutes)/availability = $/min Solve for C o C o = $/min C pc = cost per piece, $/pc C m = starting material cost, $/pc C o = cost rate of operating the work cell, $/min T p = average production time per piece, min/pc C t = cost of tooling used in the unit operation, $/pc C m = (1.75 $/pc)/(1-q) = $/pc C o = $/min T p = (2.20 min/pc)/(1-q) = min/pc C t = 0.00 $/pc Solve for C pc C pc = $/pc 9

10 Industrial Processes I Manufacturing Economics (b) Assume 100% availability and q=0 C o = cost rate of operating the work cell, $/min C L = cost rate of labor, $/min C eq = cost rate of equipment in the work cell, $/min C L = ($24.00/hr)(1 hr per 60 minutes) = $/min C eq = ($42.00/hr)(1 hr per 60 minutes) = $/min Solve for C o C o = $/min C pc = cost per piece, $/pc C m = starting material cost, $/pc C o = cost rate of operating the work cell, $/min T p = average production time per piece, min/pc C t = cost of tooling used in the unit operation, $/pc C m = 1.75 $/pc C o = $/min T p = 2.20 min/pc C t = 0.00 $/pc Solve for C pc C pc = $/pc 10

11 Production Scheduling with Machine-Machine Diagrams Problem MM1: The cell represented in the following figure is formed by three machines. The setup time between operations can be considered null. The company works 5 days a week, 8 hours per shift, with two shifts per day. 1. Calculate the cycle time. 2. Determine the maximum weekly production. 3. Show the machine-machine diagram, assuming a stable state. First, all the cycles must be calculated. The cycles can be obtained from the represented production process. Cycles: M1 = 5 min + 2 min + 1 min = min M2 = 10 min The larger cycle time fixes the cycle = min 11

12 Production Scheduling with Machine-Machine Diagrams Problem MM1 Continued: Maximum weekly production can be easily obtained by calculating the amount of minutes included in a week under problem working conditions: 5 day 2 shift 8 hr 60 min 1part P weekly = = parts/week week day shift hr cycle time min In order to draw the diagram, it is important to start with the machine that fixes the working cycle (in a stable situation). Next, the other machines operations must be included trying to minimize the product lead time. Finally, it is important to draw one product cycle. 12

13 13 MM1 Machine-Machine Diagram M1 OP M Cycle time min

14 Production Scheduling with Machine-Machine Diagrams Problem MM2: The cell represented in the following figure is formed by two machines. The setup time between operations can be considered null. The company works 7 days a week, 8 hours a day. 1. Determine the maximum weekly production. 2. Calculate the cycle time. 3. Show the machine-machine diagram, assuming a stable state. First, all the cycles must be calculated. The cycles can be obtained from the represented production process. Cycles: M1 = min + min = min M2 = min + min = min The larger cycle time fixes the cycle = min 14

15 Production Scheduling with Machine-Machine Diagrams Problem MM2 Continued: Maximum weekly production can be easily obtained by calculating the amount of minutes included in a week under problem working conditions: 7 day 8 hr 60 min 1part P weekly = = parts/week week day 1hr cycle time min In order to draw the diagram, it is important to start with the machine that fixes the working cycle (in a stable situation). Next, the other machines operations must be included trying to minimize the product lead time. Finally, it is important to draw one product cycle. 15

16 MM2 Machine-Machine Diagram OP1 0 M2 M Cycle time min 16

17 Production Scheduling with Machine-Machine Diagrams Problem MM4: The cell represented in the following figure is formed by two machines. The setup time between operations can be considered null. The company works 5 days a week, 8 hours a day. 1. Calculate the cycle time and the lead time. 2. Determine the maximum weekly production. 3. Show the machine-machine diagram, assuming a stable state. First, all the cycles must be calculated. The cycles can be obtained from the represented production process taking into account that one final product needs two components C2. Cycles: M1 = min + min = min M2 = min + min = min The larger cycle time fixes the cycle = min 17

18 Production Scheduling with Machine-Machine Diagrams Problem MM4 Continued: Maximum weekly production can be easily obtained by calculating the amount of minutes included in a week under problem working conditions: 5 day 8 hr 60 min 1part P weekly = = parts/week week day 1hr cycle time min In order to draw the diagram, it is important to start with the machine that fixes the working cycle (in a stable situation). Next, the other machines operations must be included trying to minimize the product lead time. Finally, it is important to draw one product cycle. The lead time has to be obtained looking at the diagram because it depends on the operation order. Lead time = min (lead time can be minimized, but this is not the goal of the problem) 18

19 MM4 Machine-Machine Diagram Cycle Time min OP2 OP2 0 M2 M LT Lead time min 19

20 Production Scheduling with Machine-Machine Diagrams Problem MM5: The cell represented in the following figure is formed by two machines. The company works 5 days a week, 8 hours a day. The company uses the one-piece flow concept to manufacture the product. As a result, the batch size is one part. Besides, the part must be in the machine to start the setup process. The setup time in M2 is 1 minute, and the setup time in M1 (which depends on the operation sequence) is shown in the following table: 1. Calculate the cycle time. 2. Determine the maximum weekly production. 3. Show the machine-machine diagram, assuming a stable state. 20

21 Production Scheduling with Machine-Machine Diagrams Problem MM5 Continued: First, all the cycles must be calculated. The cycles can be obtained from the represented production process. Cycles: M2 (including setup times) = 1 min + min + 1 min + min = min M1 cycles depend on the selected sequence. Processing times are fixed (2 min + 2 min + 1 min + 3 min = 8 min) Many sequences are possible. It important not to choose a sequence which makes M1 being the machine that fixes the working cycle. M1 OP1-OP3-OP4-OP6-OP1 = 8 min + 2 min + 1 min + 1 min + 2 min = 14 min The larger cycle time fixes the cycle = min Maximum weekly production can be easily obtained by calculating the amount of minutes included in a week under problem working conditions: 5 day 8 hr 60 min 1part P weekly = = parts/week week day hr cycle time min In order to draw the diagram, it is important to start with the machine that fixes the working cycle (in a stable situation). Next, the other machine operations must be included trying to minimize the product lead time. Finally, it is important to draw one product cycle. 21

22 22 MM5 Machine-Machine Diagram M1 S OP1 OP1 S M2 7 S S S S Cycle time min S S S S 27 S S S S S 40 S

23 Additive Manufacturing Build Cycle Time in Stereolithography (Example 1) The square cup-shaped part shown below is to be fabricated using stereolithography. The base of the cup is 40 mm on each side and 5 mm thick. The walls are 4 mm thick, and the total height of the cup = 52 mm. The SL machine used for the job uses a spot diameter = 0.25 mm, and the beam is moved at a speed = 950 mm/s. Layer thickness = 0.10 mm. Repositioning and recoating time for each layer = 21 s. Compute an estimate of the cycle time to build the part if the setup time = 20 min. The geometry of the part can be divided into two sections: (1) base and (2) walls The cross-sectional area of the base A 1 = mm 2 The thickness of the base = mm The cross-sectional area of the walls A 2 = mm 2 The height of the walls = mm Building the base with a layer thickness = mm / = layers to build the base Building the walls with a layer thickness = mm / = layers to fabricate the walls Total number of layers = + = layers 23

24 Additive Manufacturing T i = time to complete layer i, s (sec) where the subscript i is used to identify the layer A i = area of layer i, mm 2 (in 2 ) v = speed of the moving spot on the surface, mm/s (in/sec) D = diameter of the spot (assumed circular), mm (in) T r = repositioning and recoating time between layers, s (sec) Time per layer for the base Time per layer for the walls A i = mm 2 A i = mm 2 v = mm/s D = mm T r = s v = mm/s D = mm T r = s Solve for T i for the base T i = s for the base Solve for T i for the walls T i = s for the walls 24

25 Additive Manufacturing T c = build cycle time, s (sec) T su = setup time, s (sec) n l = number of layers used to approximate the part T i = time to complete layer i, s (sec) T su = s n l = number of layers to approximate the base n l = number of layers to approximate the walls T i = s for the base T i = s for the walls Solve for T c T c = s T c = min T c = hr 25

26 Additive Manufacturing Cost per Piece in Additive Manufacturing (Example 2) The cost of a stereolithography machine is approximately $100,000 installed. It operates 5 days per week, 8 hours per day, 50 weeks per year, and is expected to last 4 years. Material cost for the photopolymer = $120/liter. Assume that all of the photopolymer in the container that is not used for the part can be reused. Labor rate = $24.00/hr, but labor will be used for only 25% of the build cycle, mostly for setup. The post-processing time = 6.0 min/part. Assume that the cycle time is hr and assume that there is no overhead rate for the equipment. Determine the part cost. C eq = equipment cost rate, $/min IC = initial cost of the equipment, $ N = anticipated number of years of service H = annual number of hours of operation, hr/yr R OH = applicable overhead rate for the equipment, % IC = $ N = yr H = hr/yr R OH = % Solve for C eq = equipment C eq = $/min C eq = $/hr 26

27 Additive Manufacturing C pc = cost per piece that sums the material, labor, and machine operating costs, $/pc C m = material cost, $/pc C L = labor cost, $/hr C eq = equipment cost, $/hr T c = cycle time, hr/cycle U L = utilization rate should be applied to the labor rate during the build cycle, % T pp = post-processing time required for support removal, cleaning, and/or finishing, hr Solve for C m Assume that the volume of the part is 35,072 mm 3 C m = $/pc C L = $/hr C eq = $/hr T c = hr/cycle U L = % = converted to decimal format T pp = hr Solve for C pc = cost C pc = $/pc 27

28 Additive Manufacturing Additive Manufacturing (Problem 1) 3D Printing is used to fabricate a prototype part whose total volume = 1.17 in 3, height = 1.22 in, and base area = 1.72 in 2. The printing head is 5 in wide and sweeps across the 7-in worktable in 3 sec for each layer. Repositioning the worktable height, recoating powders, and returning the printing head for the next layer take another 13 sec. Layer thickness = in. Compute an estimate for the time required to build the part. Ignore setup time. T c = build cycle time, s (sec) T su = setup time, s (sec) n l = number of layers used to approximate the part T i = time to complete layer i, s (sec) T c = sec T c = min T c = hr 28

29 Additive Manufacturing Additive Manufacturing (Problem 2) A tube with a rectangular cross section is to be fabricated by stereolithography. Outside dimensions of the rectangle are 38 mm by 60 mm, and the corresponding inside dimensions are 30 mm by 52 mm (wall thickness = 4 mm except at corners). The height of the tube (z-direction) = 40 mm. Layer thickness = 0.10 mm, and laser spot diameter = 0.25 mm. The beam velocity across the surface of the photopolymer = 800 mm/s. Compute an estimate for the cycle time to build the part, if 20 s are lost each layer for repositioning and recoating. Ignore setup time. T i = time to complete layer i, s (sec) where the subscript i is used to identify the layer A i = area of layer i, mm 2 (in 2 ) v = speed of the moving spot on the surface, mm/s (in/sec) D = diameter of the spot (assumed circular), mm (in) T r = repositioning and recoating time between layers, s (sec) T i = s 29

30 Additive Manufacturing T c = build cycle time, s (sec) T su = setup time, s (sec) n l = number of layers used to approximate the part T i = time to complete layer i, s (sec) T c = sec T c = min T c = hr 30