The University of Chicago Booth School of Business Harold s Hot Dog Stand Part I: Deterministic Process Flows December 28, 2011 Harold runs a hot dog stand in downtown Chicago. After years of consulting and a stint on wall street, Harold decided to simplify his life. While the hot dog business is good, he wonders if he could improve his operation. He knows that customers will sometimes not stop if the line is too long could he improve his service? 1 System I Harold has an analytical mind. He divided up the service process into 3 main tasks. A) Retrieving a warm steamed bun and placing a hot dog inside. B) Putting on the toppings. C) Getting the beverage, and collecting payment. He had his brother Bernie measure how long it took to perform each task. The flow diagram in Figure 1 shows the time required for each task. The total time required for all three tasks is 90 sec. Harold performs all three tasks. He notes that the rate at which he can make money, his revenues, cannot exceed his maximum speed or capacity. He denotes his service capacity for System I by µ I, so that Capacity: µ I = 1 cust/90 sec. This case was prepared by Donald D. Eisenstein, Booth School of Business, University of Chicago, Chicago, IL 60637. E-mail: don.eisenstein@chicagobooth.edu Copyright c 2010 Donald D. Eisenstein 1
System I (A) Dog & Bun 15 secs (Harold) (B) Toppings 45 secs (Harold) (C) Drink&Pay 30 secs (Harold) Figure 1: The original 3 task process that Harold performs to serve hot dogs to customers. Harold opens his hot dog stand between 11:00am 2:00pm we consider a day for Harold to be 3 hrs. No day or time is the same, queues fluctuate sometimes forming long lines, and at other times the queue is empty. Harold collects receipts for three weeks, and calculates that the average customer spends $5.00. Thus Harold s capacity can be expressed in terms of the maximum rate he could generate revenues: µ I = 1 cust/90 sec = 120 cust/day (using 10800 Sec in a 3 hour day.) = $600/day. In total, Harold actually averages $500.00 per day of revenues. And thus he calculates that he averages $500/day $5/cust = 100 cust/day. He now lets λ represent his throughput the average rate at which customers move through his system: Throughput: λ = $500/day $5/cust = 100 cust/day = 1 cust/108 sec. Harold realizes that on most days there are times when he is idle, even though there are other times when he has a number waiting in line. He Eisenstein 2
measures his Capacity Utilization, denoting it by ρ, to be Capacity Utilization: ρ I = λ/µ = 100 cust/day 1 cust/108 sec = 120 cust/day 1 cust/90 sec = 5/6. Harold realizes that his Capacity Utilization must always be less than 1, since it is impossible for him to serve more customers in a day than he has capacity. He jots down: λ µ, and 0 ρ 1. He also notes that he might calculate his Slack Capacity to be the difference between his service rate capacity and the rate customers move through his system: Slack Capacity : µ I λ I = 120 cust/day 100 cust/day = 20 cust/day. 2 System II Harold wondered what would be the effect of adding some more workers to his hot dog stand. He thinks his niece Sarah would help out, and of course his brother Bernie. They each would handle one of the tasks, such as the following: System II (A) Dog & Bun 15 secs (Harold) (B) Toppings 45 secs (Bernie) (C) Drink&Pay 30 secs (Sarah) Figure 2: Harold now shares the tasks with Bernie and Sarah. Eisenstein 3
Now how fast can they together process customers that is, what would be the new system capacity? Harold realized that the Topping task was the Bottleneck, the task that took the longest time (45 secs). It is the task that would throttle the process. That is, even though Dogs&Buns is faster (15 secs), it did not do much good piling up naked dogs in front of the Topping process would only serve to get them cold. And similarly, Drink&Pay was faster (30 secs), but again, that extra speed will not help the process Drink&Pay would be idle 15 secs for every customer. The capacity of System II, µ II, would then be determined by the bottleneck process, Toppings, and would therefore be µ II = 1 45 cust/sec. Or as Harold explained, each step, A, B, and C for System II had a capacity as follows: µ A = 1 15 cust/sec, µ B = 1 45 cust/sec, and µ C = 1 30 cust/sec. Each capacity is expressed as a rate of flow in units of customers per second. Since each customer must process through all three tasks, the bottleneck is the task with the smallest capacity. So the system capacity is determined by the task with the smallest capacity: µ II = min{µ A, µ B, µ C } = min{1/15, 1/45, 1/30} cust/sec = 1/45 cust/sec. Harold tried to explain this to Bernie, but Bernie failed to grasp the concept. Harold then drew a diagram that maps out the maximum rate of Eisenstein 4
flow of customers through his system over time and showed it to Bernie. This finally convinced Bernie. This system, in which all three combined efforts for each customer did improve the capacity of the system; from µ I = 1/90, to µ II = 1/45 cust/sec. The capacity doubled but the labor tripled! Both Task A (Harold) and Task B (Sarah) have slack in System II whenever the system is operating at capacity (λ = 1/45 cust/sec). Slack capacities for System II when flow of customers is exactly λ = 1/45 cust/sec: Slack A = 1/15 1/45 = 2/45 cust/sec. Slack B = 1/45 1/45 = 0 cust/sec. Slack C = 1/30 1/45 = 1/90 cust/sec. The capacity utilization of each worker is: Capacity Utilization for System II when flow of customers is exactly λ = 1/45 cust/sec: ρ A = 1/45 1/15 = 1/3 ρ B = 1/45 1/45 = 1 ρ C = 1/45 1/30 = 2/3 The diagram Harold drew has been found and is more generally known as a Gantt Chart. Your professor may be good enough to share this diagram with you during class. Eisenstein 5
Harold was not pleased that so much of his labor would be idle when the system was so busy. If he could somehow keep all his workers busy then they could equally share the total 90 sec time required to produce each dog. The system capacity would increase to 3/90 sec. Another way to view this, is that his system has some Cost of Imbalance: For System II: Theoretical Capacity of Cost of Imbalance = a Balanced System = 3/90 1/45 cust/sec = 1/90 cust/sec. Actual Capacity Harold could regain this capacity if he could re-engineer the tasks somehow to evenly spread the load among the workers. Or, perhaps set up a line like the local Chipotle, where workers tend to stay busy by multi-tasking, moving to wherever the work is. But such a re-engineering task would have to wait for Harold. 3 System III Harold wondered if instead he could operate some processes in parallel. That is, couldn t a customer s hot dog be assembled at the same time their drink and payment is done? But that would require them to write out the entire order first, send the hot dog to assembly and simultaneously the drink and payment could be made. This is in contrast to the current system where the customer places his order as the hot dog is assembled A bit more mustard, but no onions!. Harold devised the following process shown in Figure 3 he estimated the processing times from some experiments he ran with Sarah and Bernie. Now, what is the system capacity? Each customer order must pass through all 3 Tasks. Again, the system bottleneck is the task with the Eisenstein 6
System III Place Order 30 secs (Harold) Dog, Bun, and Toppings 35 secs (Bernie) EXIT Drink&Pay 25 secs (Sarah) Figure 3: The processes are re-engineered with two of them performed in parallel. largest processing time; Dog, Bun, and Toppings at 35 secs. So, the capacity of System III is: µ III = 1/35 cust/sec He was able to increase his capacity by a bit in System III, but not too much. He realized that even though he parallelized some tasks, his capacity was still at the mercy of the bottleneck process. He was not able to smooth the load evenly over all three of his workers. But something else did seem significantly better about the new process indeed, there was a measurement he has yet to take! How long does it take a customer to go through his system? He called this the Cycle Time. Now certainly customers often suffered some time to wait in a queue, but he decided he would tackle that later. So for now, Harold will think about the cycle time just through the production process. And furthermore, Harold In fact, your professor has documented this exploration of queues in a Part II companion to this write-up. Eisenstein 7
assumes that once a customer enters service, he is not delayed by another customer. So for the original serial system, shown in Figure 1, the cycle time of a customer is 90 secs. For the new process, shown in Figure 3 the cycle time of a customer is reduced to 65 secs. It is the top path through the network, 30 secs to place an order plus 35 secs to assemble the hot dog. This top path is the longest path, it is known as the critical path through the network. So Harold realized that placing tasks in parallel may allow one to reengineer some of the tasks so as to balance the system better and thus improve its capacity. But parallelizing tasks is primarily a tool to reduce the critical path through the network and thereby decrease the cycle time of a customer through the process. 4 Little s Law Bernie and Harold then went to the library and picked up a book on Process Flows. They found the following Little s Law: INV = λ CT. They soon understood that Little s Law equates the Average Inventory (INV) in a system to the product of the Average Throughput (λ) and average Cycle Time (CT). The Inventory in the system is the number of jobs or customers in the system. INV is the time-average number of customers in the system. Harold was not sure he really understood the law, so he collected some data from four customers who entered and exited his system. When the first customer entered, the system was empty; and when the last customer exited, there were no other customers waiting. He produced the following table of each customers arrival and departure times (all units of time are in minutes). Eisenstein 8
Cust Arrival Departure Cycle Time 1 0.0 1.5 1.5 2 1.0 3.0 2.0 3 1.0 4.5 3.5 4 3.0 6.0 3.0 In the last column he computed the Cycle Time of each customer this is the time each customer spent in the queue plus service. And so, the average cycle time, CT is easily computed as: CT = 1.5 + 2.0 + 3.5 + 3.0 4 = 5/2 min. The average throughput can also be calculated. If one considers the 6.0 min as the time frame of interest, then the average throughput, λ, is simply: λ = 4 cust 6 min = 2 3 cust/min. So, by Little s Law, he would compute the average number of customers in the system, INV to be: INV = λ CT = (2/3 cust/min) (5/2 min) = 5/3 cust. Almost 2 customers in the system on average. Harold produced the following plot shown in Figure 4 of Customers (Inventory) over time to help him understand Little s Law. From the plot in Figure 4, and a recall of a lecture in Calculus by his high school teacher, Harold realized he could calculate INV directly from the plot as follows: Eisenstein 9
Customers (Inventory) 3 3 2 2 3 4 1 1 2 3 4 0 1 2 3 4 5 6 Minutes Figure 4: A plot of customers in the system over time. Each customer is a series of blocks, as they occupy a place in service over time. The number in each block represents the customer number. Area Under the Curve INV = Time (Area of Bottom Row) + (Area of Middle Row) + (Area of Top Row) = = Time ((1 6) + (1 3.5) + (1 0.5)) cust-min 6 min = 10 6 = 5 3 cust Harold was impressed with himself. He indeed has verified Little s Law on this small example. Eisenstein 10
far: Harold and Bernie decided to make a list of what they have learned so µ : The average service capacity of the system (customers/time). λ : The average throughput rate of customers in the system (customers/time). λ µ: The average rate of throughput of customers cannot exceed the average capacity of the system. CT: The average time a customer spends in the system, queueing plus service (time). Little s Law: INV = λ CT: The average number of customers in the system (INV) equals the average Throughput (λ) times the average Cycle Time (CT). Eisenstein 11
Harold and Bernie s Practice Exercises 1. After a busy week under System I, Harold determines that he averaged $550 per day in revenues. Estimate Harold s slack capacity for this week. 2. Harold is considering adding a new cash register which would change System II by decreasing Sarah s task time from 30 secs to 20 secs. What is the improvement to system capacity? 3. Back to the original System II. Harold is now considering adding a station in which customers will self-serve mustard and relish after purchase. As a result Harold estimates that Bernie can now complete the Toppings task in 20 seconds. What is the improvement to system capacity? Suppose customers arrive at a rate of 1 cust every 35 sec, what is the new capacity utilization for Task A, B, and C? What is the new Cost of Imbalance? 4. For a one hour period under System II, Harold asks his son Jim to walk by the stand every 5 minutes and write down how many people he sees in the system. At the end of the period, Jim reports data of 0, 1, 2, 2, 1, 3, 4, 5, 3, 2, 1, and 0 customers observed. In addition, over the one hour time period, Harold s receipts indicated that he served 20 customers. Estimate of the average cycle time for customers during this period. 5. Suppose in Harold s experiment (described in the Section 4) one more customer was observed, arriving at minute 2.5 and departing at minute 5. What is this customer s cycle time? What is the system s new average cycle time? What is the new average number of customers in the system? Verify that Little s Law holds. Eisenstein 12
SOLUTIONS 1. Harold s service capacity under System I is 120 customers per day. At $5 per dog, we estimate that Harold averaged a throughput of 110 customers per day. Thus his average slack capacity is estimated to be 120-110 = 10 cust/day. 2. There is no improvement. Bernie is still the bottleneck with a processing time of 45 secs. 3. Task (C) is now the bottleneck with 30 secs. So system capacity improves to 1/30 cust/sec. The capacity utilizations are: ρ A = 1/35 1/15 = 15/35 ρ B = 1/35 1/20 = 20/35 ρ C = 1/35 1/30 = 30/35 The updated Cost of Imbalance is 3/65-1/30 = 0.0128 cust/sec. 4. Within the fidelity of an observation every 5 min, we estimate: INV = 0 + 1 + 2 + 2 + 1 + 3 + 4 + 5 + 3 + 2 + 1 + 0 12 = 24/12 = 2 cust, and we have: λ = 20/60 = 1/3 cust/min. And thus by Little s Law we have: CT = INV/λ = 2 cust/(1/3 cust/min) = 6 min NOTE: In calculating INV above, the units were sloppily left out. Be sure you can verify the calculation. Eisenstein 13
5. The new customer s cycle time is 2.5 min. So we now have: CT = 1.5 + 2.0 + 3.5 + 3.0 + 2.5 5 = 25/10 min. and λ = 5 6 cust/min. And, after updating the graph of inventory over time we have: ((1 6) + (1 4.0) + (1 2.5)) cust-min INV = 6 min = 25 12 cust. So we can verify Little s Law with: INV = λ CT 25 12 cust = 5 25 cust/min 6 10 min Eisenstein 14