Operations Management - Process Strategy and Sustainability

Process Strategy and Sustainability

Introduction

• Manufacturing and service businesses aim to improve efficiency and productivity.

• Operations management (OM) can help achieve this goal.

• Designing an efficient process strategy is a key OM decision.

McDonald's vs. French Dining Restaurant

• Both offer food, but their missions differ.

• McDonald's: Quick service, affordable prices.

• French Dining: High-quality, relaxed experience.

• McDonald's: High volume demand, low-medium variety.

• French restaurant: Low volume, high variety.

Lecture Overview

• Discuss different manufacturing/service processes.

• Explain how to evaluate processes through breakeven analysis.

• Discuss the importance of layout strategy.

Process or Transformation Strategy

• Operations management focuses on making the transformation system from input to output more efficient and responsive.

• Key questions:

  • How to produce output from inputs?

  • Where to assemble?

  • How to save time and increase responsiveness?

  • How to reduce costs and increase efficiency?

• Example: HP vs. Dell or FedEx vs. Canada Post - different competitive advantages lead to different process strategies.

Beer vs. Laptop

• Both have high demand, but production processes differ.

• Beer: Continuous process, low variety, high volume to reduce costs.

• Laptop: Mass customization, high variety based on customer needs.

Harley Davidson vs. Dell

• Harley Davidson: Repetitive process, customization through modules, lower demand.

• Dell: Mass customization, high demand, customer-specific customization.

Process Strategies

Process Focus

• Low volume, high variety.

• Skilled workers and flexible equipment in each department/workstation.

• Raw materials go through different directions for customization.

• Examples: Factories with welding, grinding, painting departments; banks with different departments; restaurants with bar, grill, and bakery.

• Flexible labor and equipment.

• Low equipment utilization due to lack of routine and difficulty in production scheduling.

Product Focus

• High volume, low variety.

• Long and continuous production for efficiency and economies of scale.

• May involve multiple shifts for non-stop production.

• Example: Frito Lay products (corn, potatoes, water, seasoning) with simple variations in taste and size.

Repetitive Focus

• Combines benefits of product and process focus.

• Products divided into modules allowing for many combinations and customization.

• Each module produced in high volume for economy of scale.

• Example: Harley Davidson motorcycles with different modules for customization.

Mass Customization Focus

• Unique products for different or even every customer.

• Offers variety with high volume production to reduce costs.

• Relies on build-to-order approach: wait for actual order before production.

• Example: Dell laptops - suppliers produce components in high volume, but manufacturing and customization occur after the order is received.

Process Strategy Selection

• Depends on production volume and product variety.

• High Volume: Mass customization or product focus.

• Low Volume: Process focus with high variety.

• Medium Volume: Repetitive focus using modules.

Crossover Charts

• Compares costs of different processes at possible volumes.

• Considers variable costs (C) and fixed costs (F).

• Variable Costs: Depend on production volume (e.g., raw materials, direct labor).

• Fixed Costs: Remain constant regardless of production volume (e.g., rent, equipment).

• Total Cost: F + C oldsymbl V (V = volume of production)

Example: Clever Enterprises

• Evaluating three accounting software products (A, B, C).

• Fixed cost and variable cost per accounting report for each software.

• Goal: Find the volume ranges for which each software is most economic.

Solution

• Find total cost function for each software.

• Software A:

  • Fixed cost = 200,000

  • Variable cost = 60 per report

• Determine volume ranges by finding intersection points (V1, V2) where total costs are equal.

• V1: Intersection of A and B: 200,000 + 60VA = 100,000 + 95VB

• V2: Intersection of B and C: 100,000 + 95VB = 0 + 110VC

• Solve for V1 and V2.

  • V_1 = 2,857

  • V_2 = 6,666

• Ranges:

  • 0-2,857 reports: Software A

  • 2,857-6,666 reports: Software B

  • More than 6,666 reports: Software C

Process Analysis and Design

• Questions to ask:

  • Does the process achieve a competitive advantage?

  • Does it eliminate non-value-added steps?

  • Does it maximize customer value?

Techniques and Tools

Flowchart

• Diagram showing steps and operations in a process.

• Also known as process flow diagram or process map.

• Can be high-level or detailed for process improvement.

Time Function Mapping

• Similar to a flowchart but includes time as the x-axis.

• Illustrates the duration of each step in the process.

• Example: American National Can Company - before and after time function mapping to reduce waiting times.

Value Stream Mapping

• Includes the full supply chain in the picture.

• Identifies non-value-added time due to high inventory and buffers.

• Helps in understanding the process within the supply chain.

• Example: Identifying how raw material inventory can be reduced by having delivery twice per week rather than once per week or Increase the reliability of those components, for example.

Process Chart

• Uses symbols for operations, transportation, inspection, delay, and storage.

• Identifies and focuses on value-added activities.

• Value-added Time = Operating Time / Total Time

• Example: Fast food restaurant - focus on reducing storage, delay time, and distance.

Service Blueprinting

• Applicable for high-service processes.

• Identifies three levels:

  • Level 1: Activities under customer control.

  • Level 2: Interactions between service provider and customer.

  • Level 3: Activities not visible to the customer (support activities).

• Identifies potential failure points and implements Poka-yoke mechanisms to avoid mistakes.

• Example: Car oil change service - installing a bell to notify of customer arrival, offering coffee and reading materials in the waiting room.

Layout Strategy

• Floor plan of a plant or service provider unit.

• Aims to improve efficiency by arranging equipment, people, and stations according to function.

• Different Types of Layouts

Office Layout

• Groups workers with similar jobs together for ease of movement of information.

• Considers comfort, safety, and ease of movement of information.

• Uses relationship charts to obtain a proper layout based on movement of information or individual desires.

• Technology increases layout flexibility.

Retail Layout

• Maximizes profitability per square foot of floor space.

• Sales profitability varies directly with customer exposure.

• Manufacturers may pay slotting fees to retailers to display products.

Warehouse and Storage Layout

• Optimizes trade-offs between handling costs and costs associated with warehouse space.

• Considers vertical moves.

• Locates high-volume items closest to docks.

Fixed Position Layout

• Product remains in one place, workers and equipment come to the site.

• Used for large products such as ships and aircraft.

• Scheduling and moving workers, materials, and machines can be challenging.

• Example: Operating room - patient remains stationary, and medical personnel and equipment are brought to the site.

Process Oriented Layout

• Arranges work centers to minimize the cost of material handling.

• Work centers and workstations like cutting department, bending department, drilling department, painting department.

• The layout of choice for a typical job shop.

• Lower matching investment costs are typically lower than those for a high volume assembly line product focus.

• Requires skillful labor.

• Example; Raw materials components depend on which product customer customizing. Each may go through different directions.

• Basic cost elements: number of loads and distance loads.

• Optimize by minimizing the number of loads and the distance loads that material moves.
  *Example: Walter’s company management wants to arrange the six departments of its factory in a way that minimizes interdepartmental material handling costs.

IMPROVEMENT

*Can compare whether it is an improvement.

*It should result in a lower layout cost according to the previous step.

Work Cells Layout

• Recognizes people and machines into groups to focus on single products or product groups.

• Requires identification of family of products.

• Advantages: reduced working process inventory, less floor space required, increased use of equipment, higher efficiency, and reduced investment in machinery.
  Some improvement:

Improvement for the work cells

For example, think about the current layout. Workers in a small closed area. Both workers are doing the similar job, for example, yeah, because we are talking about work cells, but we can improve this layout. Yeah, using this U type of layout. So this is an improved layout Cross trained workers can assist each other. For example, if the first guy needs to take a break for a few minutes, then that guy also can help on the others, for example. Or we may be able to add a third worker if we want to increase the rate of the output, for example. But in the current layout, adding the third worker is a very difficult task.

*Another type of improvement. The current layout is in straight lines. It makes it hard to balance tasks because work may not be divided evenly. So you see there are different workers in straight lines, and it's going to be a mess if they want to access a longer distance, for example. So it is not that much clear who is doing how many of the control, for example, quality control, for example, in this case. But we can take advantage of U shape of the layout in this case. Improve layout in U shape workers have better access so they can go back and forward in order to help each other to improve the efficiency and productivity of the system. So we can say that U shape line may reduce employee movement, space requirements, as you can see, while enhancing communication, reducing the number of workers, and facilitating inspection as well.

Line Balancing

• Assigns work/jobs/activities to stations in a line to achieve the desired output rate with the smallest number of workstations.

• Minimizes capacity costs.

• Similar to identifying bottlenecks.

• Key question: How to assign tasks to workstations so that the bottleneck operation time would be less than desired cycle time.

• Activity on node diagram.

• Example
  A company is setting up and assembling a line to produce 192 units per eight hour shift. So at the end of the day, which, includes eight working hours, we need to produce 192 units. That's the demand. That's what determines the cycle time in our problem. The following table identifies the work elements, times, immediate predecessors or the sequence between activities. So we have different work elements A, B, C, D, E, F, G, H, I, J. The second column gives you the time, the seconds. For example, if you want to produce one unit of this item, yeah, A, B, C, D all the way to J should be done so that you have one item of it. The first element of the production is elemenA, takes forty seconds. Second is B, eighty seconds. Third one is C, thirty seconds. D is twenty five seconds, and it goes like this. And what is the relationship, technological relationship between those work elements is the last column. You can right away start working on the first element A. If you want to work on element b, a should be finished. That's what we mean by immediate predec for b. For item c, if you want to work on item c, d, e, and f should be finished. If you want to work on d, b should be finished. That's what the last column means. Okay. Now here's the thing. What is the total time you require to produce one minute? Seven twenty seconds. Yeah. The summation of all the times in the second column. But in reality, it's not like this. Why? Because we may have different workstations to do to to to work in parallel. So we can reduce this significantly. Now that's the question. How many workstations we need to have? And which activities or which work elements should be assigned to each workstation? That's the question we are going to answer.
  *Recall that the total number of items we need to produce is 190 units. That's the demand at the end of every day. So it was given, yeah? Demand was 92 units per eight hour shift. So what does that mean? If you want to produce 192 units at the end of each day, what should be the completion time of each unit? So every day you have eight hours multiplied by sixty minutes, multiplied by sixty to half seconds. Yeah, this is the total time in terms of the seconds you have available. And if you want to produce 192 units, then each unit should not take more than one hundred and forty seconds. Yeah. Why? Because we have cycle time. The length of time, the duration, yeah, that each unit should be completed. That's the cycle time. So that you can have 192 units at the end of the day. That's in other words, production rate. What is the production rate? So that you can produce 190 units at the end of the day. So you have eight hours, each hour is three thousand six hundred seconds, so you need 92. Therefore, each unit shouldn't take more than one hundred and fifty seconds if you want to produce 92 units. Otherwise, if it is more than one fifty, you cannot produce 192. That's the meaning of cycle time. You need to understand it, guys. So again, cycle time is the duration for completion of one unit so that you can meet the demand which is 192. Having one shift, eight hours, sixty minutes, sixty seconds, the total time you have is eight x 30, 600 divided by 192. Therefore, the duration completion time for each unit is one hundred and fifty seconds. But from the information given, the total time for all the work elements was seven twenty seconds, just compare it with 150. This can help you to find the minimum number of workstations, which is five. Four point eight, just round it up. So you need at least. Why at least? Because there is a technological relationship between the work elements, but just for the sake of the time, since each item takes seven twenty seconds, but the cycle time is 150, the minimum number of workstations you need to have is five workstations. Okay, Having five workstations doesn't mean that everything is solved, yeah? You need to also understand which work elements should be assigned to first workstation, which of them to the second, to third, to fourth, and to fifth one. So in order to do that, we need to run an algorithm. Here is the algorithm. Use trial and error to work out a solution and show your solution on a precedence diagram. Just bear with me. It's very straightforward how to find the assignment for the workstations.

Algorithm

*First, create one station at a time. So think about the first workstation.
Then you need to identify a list of all the work elements or tasks that can be assigned to this workstation. Yeah?
Then eliminate those tasks that have been assigned already. If you have already assigned work elements or tasks to other workstations, remove from the list. You also need to eliminate any task whose president's relationship has not been satisfied. For example, in our, example, you cannot assign activity or work element B. Before A has been assigned. You can assign B as long as A has been already assigned. Also, you need to eliminate those tasks for which inadequate time is available at the workstation. So remember that the duration in which in every workstation is a cycle time. Yeah, one hundred and fifty seconds in our example. We have five workstations. The maximum for each of them is one hundred and fifty seconds. Why? Because if it is more than that, we cannot meet the demand, 192. So the maximum is 150. Therefore, you need to eliminate any task that is a candidate to be assigned to workstation if the time of the task is more than the availability of that workstation. Finally, if there is more than one task in the list, we need just to use a heuristic to find a task to assign. So let's consider the longest task time to assign. Just bear with me, you will see how to solve this problem in the next slide. Okay, let's apply the algorithm we just discussed in our example. So the first column here is the station, the second is the candidate to be assigned to that workstation, which one we choose, what is the work element time, what is the cumulative time for that workstation, and what is the availability or idle time of that workstation. Remember, the total time available for each workstation in is only hundred fifty seconds, and it is a cycle time. Why it is one fifty? Because if it is more than that, we cannot meet the daemon. We cannot produce 192 units. That comes from a cycle time. So workstation one is s one. So the only work element or task can be assigned is a because all the others are dependent on the others. So you cannot go with B or G. Why? Because A has not been assigned yet, so candidate is only A, and we choose A. What is the work element type? It's forty seconds, and then since this is the first test, cumulative times 40. So what is the availability then? One fifty minus 40. You will still have one hundred and ten seconds in this workstation one. Now, what is the second work element to be assigned to Station 1? In terms of precedence dependency, you can go with b and g. Yeah? Because a has been already assigned to Workstation 1. Can we assign B? Yes, because A has been assigned its immediate predecessor, and the time of the B is eighty seconds, which is less than idle time of workstation one, which is hundred and ten seconds. So B is one candidate. How about G? Can we assign G? Although A has been assigned its immediate predecessor, but the time of G is one hundred and twenty seconds, which is more than the idle time of Workstation 1 which is one hundred and ten seconds. Therefore, the only candidate here to assign to workstation one is B. Go with B, it takes eighty seconds, the cumulative time for a session one is now 40 plus 80, is one hundred and twenty. And you still have thirty seconds for work session one. Now, do we have any other candidate to be assigned to work session one? G cannot be the case because the available time for workstation one is thirty seconds and G requires 120. So G is not the case. However, D, E and F, yeah, all three can be the candidate because the immediate predecessor for them, which is B, has been already assigned, and the timing of D and E and F, they are all less than thirty seconds, this idle time for workstation one. But which one we choose? Based on the algorithm, let's take the one with the longest duration, which is activity D. Why? Because through this, we can make sure that the more capacity, the efficiency of Workstation 1 is higher. We have less idle time remain at Workstation 1. So go with d, the longest one, twenty five seconds, plus 120, it's 145. Compared the availability of the duration for Workstation 1, you only have five seconds left, which is quite good. The performance, the utilization workstation one is five, is quite high, yeah, 145 divided by 150. Now, can we assign more work element to Workstation 1? No, because the idle time for one is five seconds and all the tasks that can be assigned, which is E and F, is more than five seconds. Therefore, we can no longer assign any activity to Workstation 1. So let's go to Workstation 2. This is the second one. What are the candidates to be assigned? E, yes, F, yes, and G. Only these three, why? Because the timing of all three is less than 150 for Workstation 2, and the immediate predecessor of all three E, F, and G has been already assigned. So which one to choose? The longest one which is G. 120, one hundred and 20, and the idle time is thirty seconds. So thirty seconds is good for E and F but not H. You cannot assign H here because it takes 145 seconds, it is more than thirty seconds, the idle time of workstation two. So E and F, which one we choose? E because it's longer. 20 plus 120, one hundred and 40, and the idle time would be ten seconds. Now we need to go to the third workstation. Why? Because ten seconds is not enough to assign H or F. S three, candidates are F and H, you go with H, the longer one, hundred forty five seconds, and only there is five seconds left. So S three is done. The only activity to assign to S three is activity H, because five seconds idle time remain is not enough for f or I. So go with s four. We have f and I. We go with I, the longer one, hundred thirty seconds, twenty seconds left, you can assign F fifteen seconds, and the idle time with five seconds. Finally, we have S5, Workstation 5, you can assign C and J, and the total time is one hundred and forty five. Five seconds available. So if we want to summarize our findings, we have five workstations in this example for our line, production line. For the first workstation activity, anyone works in Workstation 1 should do activity A, B and D. The guy that is assigned to Workstation 2 should do the job G and E. The third guy in worker session three just do h. The fourth guy in worker session four will do I and f. And finally, the last guy in worker session five will do c and j.