University:
California State University, Northridge
Course:
MGT 693 | Strategic Management
Academic year:

2024
Views:

425

Pages:

21
Author:

tossicars8eb

Process Flow Analysis In the Context of LittleField Technologies Game One Machine Per Station; Theoretical Flow Time (ThT)  I will review some of the Process Flow and Little’s Law concepts in the context of Littlefield Technologies (LFT) game.  Contract  Flow unit. Machine  Resource unit, Station  Resource pool.  Minimal flow time, or theoretical flow time (ThT): The minimal time required to produce a flow unit.  Minimal inventory or pipeline inventory (Ip): The minimal inventory required to produce a flow unit. Min-WIP.  The Maximal inventory or Maximal Work in Progress (Max-WIP) to regulate WIP to achieve the expected flow time (T).  Ensure that I or WIP remains less than (but not much less than) Max-WIP.  The time in the waiting lines (Ti) vs in processes (Tp).  The number of flow units in the waiting lines (Ii) vs in processes (Ip). 3 The Situation- Not the Same Numbers as in Your Game Where is the bottleneck? The station with the highest utilization. Station 2. Compute Rp U=R/Rp Rp=R/U 5.5 U Rp/dy Tp in dy Tp in hr Sta-1 Sta-2 Sta-3 0.6875 0.916666 0.55 8 6 10 0.125 0.167 0.1 c=1 Rp1=6/day c=1 Tp=24/6=4 U= 0.916667 U= 0.916667 R=5.5 3 4 2.4 Compute unit load (Tp) at each station Rp = c/Tp Tp =c/Rp c=1 in all stations  Tp=1/Rp day c=1 Rp1=8/day c=1 Tp=24/8=3 U= 0.6875 0.6875 c=1 Rp1=10/day c=1 Tp=24/10=2.4 U= 0.55 U= 0.55 4 Capacity & Theoretical Flow Time (ThT) Compute Capacity of this System (Rp). Capacity of the system = Min(8,6,10) = 6 per day. Compute Theoretical Flow Time (ThT) ThT is the minimal time that we need to produce a product. ThT = 3+4+2.4 = 9.4 hours or 9.4/24 = 0.39 day. ThT does not depend on the number of machines (c) or throughput(R). ThT depends on each machine's capacity and the operations' routing. As long as there is no change in the capacity of machines and the routing of the operations, ThT is always 9.4. ThT is the absolute minimal flow time. Flow time (T) is much greater than the ThT. 5 Pipeline Inventory (Ip or Min-WIP) Pipeline inventory (Ip) is the Min-WIP the system needs to operate. Compute pipeline inventory: Procedure 1- Little’s Law. RT = I, R = 5.5/day, and T = 9.4/24 day. Or, R= 5.5/24 per hour, and T= 9.4 hours. RT = I  I = 5.5(9.4/24) = 2.15 flow units. Ip = 2.15. Compute pipeline inventory: Procedure 2- Utilizations. R=5.5/day, U1= 0.6875, U2= 0.55, and U3 = 0.916666. When U=0.55, on average, 0.55 flow units are with the resource unit. The pipeline inventory is I= U1+U2+U3. Min-WIP  I = 0.6875 + 0.55 + 0.9167 = 2.15. Ip = 2.15. 6 Flow Time, Max-WIP, WIP, R and Rp Suppose the average number of units inside the system (I or WIP) is 19.2 contracts. Compute the flow time. R = 5.5, I = 19.2. T = 19.2/5.5 = 3.5. Suppose we must deliver the product under 3 days. Otherwise, the revenue of $800 linearly goes down and reaches zero if we deliver after 5 days. Therefore we lose 800[(3.5-3)/(5-3)] = $200 and collect $600 per contract. Compute the Max-WIP to guarantee a Lead Time = 3 days for a Capacity = Rp = 6. We use Rp and Max-WIP to regulate the relationship between R and WIP. We use Rp(T) = Max-WIP to regulate RT= WIP. To estimate Max-WIP, we set R = Rp and T as 3 days. 7 Max-WIP & WIP, A Balance between T and R Rp(T)=Max-WIP 6(3)= 18 = Max-WIP. We know R 3. Suppose our current Max-WIP =100. Should we change it to 18? Not Immediately. We need to look at lead time and revenue. If they are fine, we then look at WIP. 8 Max-WIP, Input, Output, Lead Time, Revenue, and WIP 9 Max-WIP & WIP, A Balance between T and R Do not make any change if WIP is less than your Max-WIP= 6. When WIP gets close to your Max-WIP=6, or even slightly exceeds it, then replace 100 by 6. In the previous graphs I have used 6 period moving average to decide. Check both sides. Two edges of a slippery roof. If WIP > 6, then T>3  you will not collect the entire Revenue of $1200. If WIP << 6, then while T<3 and you collect full $1200 (or any other number that is related to this contract) your R could be <=CT 17 Theoretical Flow Time, Capacity & Pipeline Inventory Estimate the maximum WIP inventory (Max-WIP) if we are on contract-2. That is a flow time of less than 1 day to earn $1200 per job which goes to zero if the flow time exceeds 3 days. We refer to it as (1, 1200, 3). R (capacity)= 16 per day, T=1 day. RT=I 16(1)=16, set Max-WIP to 16. On average how many flow-units are in the waiting lines? I =Ii+Ip  16 = Ii+6.267  Ii = 9.733 How long a job spends in the waiting lines Procedure-1. T=1 day, ThT= 9.4/24 = 0.391667 Day Ti= 1-0.392 =0.608 days Procedure-2. R=16, Ii=9.733  Ti= Ii/R = 9.733/16 = 0.608 days = 14.6 hours 18 Utilization, Inventory, Flow Time, Little’s Law Suppose Max-WIP in the game is set to 22, and average WIP is 20. That means on average there are 20 contracts either in the waiting lines or on the machines. Compute the Flow Time ? RT=I 16T= 20 T = 1.25 days How long a job spends in the waiting lines T=1.25(24) = 30 hrs T=Tp+Ti 30=9.4+Ti Ti= 20.6 hrs Compute the ratio of the flow time to the theoretical flow time. T/Tp= 30/9.4= 3.19 19 Implied Utilization Can this system handle 20 contracts per day? The system cannot even handle 18 orders per day because when utilization get close to 100%, flow time increases exponentially. At 18 contracts per day, U1=18/24= 0.75, U2=18/18=1, U3=18/20=0.9. Besides Station-2, utilization of Station-3 also is a sign of warning. To handle 20 contracts per day, since capacity of the system is 18 contracts per day. Implied Utilization =20/18 = 1.111 d) If the firm wishes to process all the contracts, how many machines of each type are needed? # of Station-1 machines required = 20/8 = 2.5 # of Station-2 machines required = 20/6 = 3.333 # of Station-3 machines required = 20/10 = 2 20 Resources We need 2.5, 3.33, and 2 resource units. That is 3, 4, and 2. The utilization of station 3 is scary. High utilization, in presence of variability, leads to long waiting lines. We need 3, 4, and 3. We have 2, 3, and 2. We may buy one machine for each of the three stations. In that case, utilization of the three stations are as follows. S1 S2 S3 Rp1 8 6 10 c 3 4 3 Rp 24 24 30 R 20 20 20 U 0.83333 0.83333 0.66667 21 100% Utilization is a High Risk Imagine a freeway where all cars driving exactly 65 and the distance between pairs of cars in 1”. As long as everyone has a speed of exactly 65 – no variability- that is fine. What happens if one hits the break? How long does it take to clean the freeway? Do cars pass freeway easier when utilization is 1 and they are moving bumper to bumper, or when 50% of the freeway is empty, U = 0.5, or when U = 0.25. Never make U of all the sub-processes or activities, and not even a single sub-process = 1 Consider cross taring and pooling to reduce the utilization of the bottleneck. Up = Throughput/Effective capacity of a resource pool = R/Rp Capacity utilization of the process U = Throughput/Effective capacity of the bottleneck resource pool. 22 22 Input, Output, Lead Time, Revenue, and WIP day 12 10 8 6 4 2 0 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 47 48 49 50 Input 6.0 5.9 12.4 0.5542 1115.165 2.92 2.0324 0.4886 0.494 0.2346 Ave6dInputAve6dLead Ave6dReveAve6dOutpAve6dWIP Lead Time Revenue OutpUT WIP U1 U2 U3 2 0.33 1200 1 0.54 0.33 0.13 0.09 #N/A #N/A #N/A #N/A #N/A 2 0.36 1200 2 0.56 0.23 0.19 0.15 #N/A #N/A #N/A #N/A #N/A OutpUT #N/A Ave6dOutpUT Input Ave6dInput 1 0.53 1200 1 0.78 0.26 0.36 0.16 #N/A #N/A #N/A #N/A 8 0.04 0 0.36 1200 1 0.04 0 0 #N/A #N/A #N/A #N/A #N/A 2 0.46 1200 1 0.71 0.33 0.1 #N/A #N/A #N/A #N/A #N/A 7 0.29 1 0.39 1200 2 0.53 0.16 0.23 0.14 1.333333 0.405 1200 1.333333 0.526667 6 0 0 0 0 0 0 0 0 1 0.35 1000 1.166667 0.436667 5 2 0 0 0 0.27 0.16 0 0 1 0.29 800 0.833333 0.388333 2 0.53 1200 3 1.53 0.49 0.26 1.166667 0.29 800 1.166667 0.513333 4 0.57 3 0.41 1200 4 1.43 0.49 0.49 0.29 1.666667 0.298333 800 1.666667 0.745 3 1 0 0 0 0.28 0.16 0.04 0.08 1.5 0.221667 600 1.5 0.673333 2 3 0.38 1200 3 0.93 0.41 0.34 0.16 1.833333 0.22 600 1.666667 0.74 1 0.73 3 0.64 1200 2 1.6 0.43 0.24 2.333333 0.326667 800 2 1.006667 2 0.36 1200 3 0.71 0.33 0.16 2.333333 0.386667 1000 2.5 1.08 0 0.22 1 0.47 1200 1 0.9 0.29 0.45 0.16 2.166667 0.376667 1000 2.166667 0.975 6 0.39 1200 5 2.14 0.9 0.34 0.35 2.666667 0.373333 1000 2.333333 1.093333 2 0.95 1173.21 3 2.3 0.33 0.79 0.21 2.833333 0.531667 1195.535 2.833333 1.43 Revenue Ave6dRevenue Lead Time Ave6dLead Time 1 0.36 1200 2 0.64 0.24 0.24 0.16 2.5 0.528333 1195.535 2.666667 1.381667 14000.2 2 0.42 1200 1 0.6 0.33 0.08 2.333333 0.491667 1195.535 2.5 1.215 3 0.36 1200 2 0.79 0.33 0.3 0.16 2.5 0.491667 1195.535 2.333333 1.228333 1200 2 0.73 1200 2 2.41 0.48 0.92 0.24 2.666667 0.535 1195.535 2.5 1.48 1000 4 0.77 1186.4 4 2.27 0.59 0.79 0.25 2.333333 0.598333 1193.268 2.333333 1.501667 4 0.63 1200 3 2.72 0.71 0.81 0.33 2.666667 0.545 1197.733 2.333333 1.571667 800 5 0.68 1200 4 2.39 0.53 0.68 0.29 3.333333 0.598333 1197.733 2.666667 1.863333 600 1 4 1.07 1155.08 1 6.87 0.93 0.26 3.666667 0.706667 1190.247 2.666667 2.908333 2 1.51 893.06 7 5.01 0.33 1 0.3 3.5 0.898333 1139.09 3.5 3.611667 400 1 0.61 1200 3 0.73 0.16 0.2 0.24 3.333333 0.878333 1139.09 3.666667 3.331667 200 4 0.35 1200 1 1.47 0.52 0.47 0.16 3.333333 0.808333 1141.357 3.166667 3.198333 0.52 0.16 3.166667 0.815 1141.357 3.166667 2.986667 3 0.67 1200 3 1.45 0.36 0 4 0.58 1200 5 3.82 0.9 0.79 0.52 3 0.798333 1141.357 3.333333 3.225 3 0.82 1200 3 2.04 0.49 0.72 0.19 2.833333 0.756667 1148.843 3.666667 2.42 5 0.44 1200 4 2.41 0.89 0.65 0.33 3.5 0.485 1200 3 1.671667 11 0.87 1194.4 5 6.48 0.99 0.99 0.47 4.666667 0.563333 1199.067 3.666667 2.593333 5 0.86 1173.13 6 7.56 1 0.99 0.44 5 0.611667 1194.588 3.666667 3.376667 5 1.36 982.99 6 7.41 1 1 0.48 5.5 0.791667 1158.42 4.333333 4.52 8 7 WIP Ave 6 5 4 3 2 1 0 23

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