Throughput CWF Flexibility and Quality Ardavan AsefVaziri CrossTraining

Throughput, CWF, Flexibility, and Quality Ardavan Asef-Vaziri

Cross-Training (Flexibility) Increases Capacity Suppose we have two resources and they work sequentially. They are both expert in their own tasks. Suppose the first resource completes its task in 5 mins (Tp=5 mins). Rp=60/5 =12 per hour. The second resource completes its task in 10 mins (Tp=10 mins). Rp=60/10 =6 per hour. Both resources are highly specialized in their task. The capacity of the system is 6 per hour. Now suppose the first resource is cross-trained (we increase flexibility of this resource) to do the second job, too. But under the new situation, due to moving back and forth, it takes 6 mins to do the first task and 12 mins to do the second task. Under the new situation, he can complete 7 units in the first task (7*6=42 min), and still have 18 minutes for the second task, where he can complete 18/12= 1. 5 task. By increasing the flexibility, the capacity increases from 6 to 7+. Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 2

Th. Cap, VTh. T, T, CWF Unit Load, Activity time (Tp) Capacity is computed based on the Unit Load Rp=c/Tp Theoretical Flow Time is computed based on Activity Time (Tp) Then What is Flow Time? 10 mins. 30 mins. Flow Time Ti + Tp Flow time includes time in buffers Capacity is not affected by time in buffers 3 days Theoretical Unit Load, Theoretical Activity Time (Th. Tp) Theoretical Capacity is computed based on the Theoretical Unit Load Rp=c/Th. Tp Theoretical Flow Time is NOT computed based on Theoretical Activity Time Very Theoretical Flow Time is computed based on Theoretical Activity Time Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 30 mins. 3

Th. Cap, VTh. T, T, CWF Unit Load (Tp) Capacity (Rp) Activity time (Tp) Theoretical Flow Time (Th. FT) 10 mins. 30 mins. Flow Time Capacity is not affected by the buffer times 3 days Theoretical Unit Load (Th. Tp) Theoretical Capacity (Th. Rp) Theoretical Activity Time (Th. Tp) Very Theoretical Flow Time (VTh. T) 30 mins. Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 4

Capacity Waste Factor and Theoretical Capacity An operating room (a resource unit) performs surgery every 30 min. Tp = 30 min. Tp includes all the distracts. Rp= Effective capacity =c/Tp= 1/30 per min or 60/30 =2/hr. On average, 1/3 of the time is wasted (cleaning, restocking, changeover of nurses and adjusting the equipment ). Capacity Waste Factor (CWF) = 1/3. Theoretical Unit load = Th. Tp= Tp(1 -CWF) =30(1 -1/3) = 20 min. Theoretical Capacity =Th. Rp = c/Th. Tp Effective Capacity = Rp = c/Tp Th. Rp=Rp/(1 -CWF) Theoretical Capacity = Th. Rp= 1/20 per minute or 3 per hour. Th. Tp= Tp(1 -CWF) Tp=Th. Tp/(1 -CWF) Th. Rp= Rp/(1 -CWF) Rp=Th. Rp(1 -CWF) Capacity Improvement Potential % = CWF/(1 -CWF) Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 5

Th. Tp to Tp Through Repair, Setup, and Rework A process works 24 hours a day. We consider Station-1. Average uptime (mf=190 hours), average downtime (mr=10 hour). A= Availability = A= mf/(mf+mr)= 190/(190+10)= 95%. T 0 = Th. Tp = 20 seconds. Ns = Setup batch =Ns=100 products Ts = Setup time =Ts=2 mins %rew = % of parts that need rework = %Rev =5% Trew = Rework time per product needing rework = Trew= 10 seconds per product needing rework. p Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 6

Capacity Improvement Poitrinal CIP = (Th. Rp-Rp)/Rp CIP(Rp)=Th. Rp-Rp Rp(1+CIP)=Th. Rp Rp=Th. Rp(1 -CWF) Rp/(1 -CWF)=Th. Rp Rp(1+CIP)=Rp/(1 -CWF) (1+CIP)(1 -CWF)=1 1+CIP-CWFCIP=1 CIP(1 -CWF)=CWD CIP=CWF/(1 -CWF) CIP=0. 124/(1 -0. 124) = 14. 2% Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 7

Quality Point Station-1 Tp=4 mins Station-2 Tp=6 mins Product Price at Factory is $20. Station-1 has 10% scrap. Input material to Station-1 costs $5 per unit. Station-2 has no scrap or rework. Input material to station-2 costs $2 per unit. Inspection cost: if it is after station-1 cost is $2 per unit, if it is after station-2 cost is $1 per unit. Inspection operations take 1 min. per unit. (a) Demand is 8. 1 units per hour. Where do you place the Inspection Station? After station-1 or after station-2? Evaluate in terms of profit and costs. What would be your profit per hour and per unit of good product? 90, 11. (b) Suppose demand is 10 units per hour. What is the impact? 102. 23, 10. 2 Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 8

Quality Point Demand = 8. 1 R= 8. 1/(1 -Scrap)= 8. 1/0. 9= 9 Place inspection after Sta-1 9 units pass Sta-1, at costs of 9($5 raw material + $2 inspection)= 9(7) = 63 8. 1 units pass Sta-2, at costs of 8. 1($2 raw material) = 16. 2 Total costs 63+16. 2 =79. 2 [increases inspection costs by 9(2 -1), but decreases purchasing cost by 0. 9(2)=1. 8 9 -1. 8= 7. 2. ] Total revenue 8. 1(20)= 162 Profit per hour = 162 -79. 2 = 82. 8, profit per unit = 82. 8/8. 1 = 10. 22 Place inspection after Sta-2 9 units pass Sta-1 and Sta-2, at costs of 9($5 RM 1 + $1 Insp+$2 RM 2)= 9(8) = 72 $72 is $79. 2 -$7. 2. Total revenue 8. 1(20)= 162 Profit per hour 162 -72 = 90, profit per unit = 90/8. 1 = 11. 11 Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 9

Quality Point Demand = 10, Place inspection after Sta-1 R= 8. 1/(1 -Scrap)= 10/0. 9= 11. 11 units pass Sta-1, at costs of ($5 raw material + $2 inspection) 11. 11(7) = 77. 10% is rejected and 10 units pass Sta-2, at costs of 10($2 raw material) = 20 Total costs 77. 77+20 =97. 77 Total revenue 10(20)= 200 Profit per hour = 200 -97. 77 = 102. 23, profit per unit = 102. 23/10 = 10. 22 Place inspection after Sta-2 10 units pass Sta-1 and Sta-2, at costs of ($5 RM 1 + $1 Insp+$2 RM 2) That is because capacity of Sta-2 is limited to 10 per hour 10(8) = 80 $80 is $17. 77 less than $97. 77 BUT Total revenue is not 10(20) but 9(20) which is $20 less Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. Profit per hour 180 -80 = 100, and profit per unit = 100/9 = 11. 11 10

Fixed Costs Per Product is Not Fix Station-1 Station-2 Station-3 Tp= 3 hrs Tp= 4 hrs Tp=2. 4 hrs The process is working 24 hours a day Each station has one machine. Annual fixed cost of each machine is $90, 000. Assume 300 working days per year. (a) Suppose demand is 3 products per day. Compute the total fixed cost per product. Rp 1=24/3=8, Rp 2=24/4=6, and Rp 3=24/2. 4= 10 Process Capacity is 6 per day. Throughput is 3 per day. Fixed Capital cost is $300 per day. The total fixed cost of each product = $300/3 =$100 per product. (b) Suppose demand is 6 products per day. Also suppose variability is extremely low. Therefore, R=Rp is not far from reality. Compute the total fixed cost of each product. Process Capacity is 6 per day. Throughput is 6 per day. The total fixed cost of each product = $300/6 =$50 per product. Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 11

Fixed Costs Per Unit of Product (c) Suppose demand is 8 product per day. Furthermore, suppose CWF in the three stations are 30%, 60%, and 30%, respectively. Suppose by (i) improving the methods and (ii)training the workers you have removed 50% of the waste in each station. Compute the total fixed cost of each product. Tp 1 = 8(1 -0. 15) =2. 55, Tp 2= 4(1 -0. 30)= 2. 8, Tp 3= 2. 4(1 -0. 15)= 2. 8. Rp 1 = 9. 41, Rp 2 = 8. 57, and Rp 3= 11. 76 The total fixed cost of each product = $300/8 =$37 per product. (d) Suppose you can move 10% of Station with the maximum utilization to the station with the minimal utilization. Demand is 9 product per day. Compute the total fixed cost of each product. The total fixed cost of each product = $300/9 =$33. 33 per product. Throughput Analysis-Basics. A. Asef-Vaziri, Department of Operations, Information, and Analytics. 12