ISyE 3104: Introduction to Supply Chain Modeling ...



ISyE 3104: Introduction to Supply Chain Modeling: Manufacturing and Warehousing

Instructor: Spyros Reveliotis

Spring 2002

Homework #2

Due Date: Wednesday, 2/6/02

Problem Set:

1. Problems from textbook:

1.5 What is the purpose of a process plan? List the types of information normally included in the plan.

▪ The purpose of process plan is to transform part or product definition data into detail production instructions; it can also be used to show step-by-step directions for process achievement.

▪ In general, the process plan includes the following types of information:

• Sequence of operations to transform raw material into finished goods. Operations can be fabrication, assembly, kitting, separation, testing, etc.

• Production time information: setup time, unit-processing time and/or batch-processing time.

• Material handling information: amount of unit load, process batch, transfer batch, etc.

2.6 A product is produced in batches of 100 units. The machine requires 1 hour of setup time. Unit processing time is 10 minutes but the machine can process up to 5 units at a time. After processing, each unit must spend 2 hours on a cooling rack before it can be used. There is no limit to the number of units that can be cooled at a time. Assuming there is no additional waiting time for the machine, find the production time for batches of this product.

▪ Production time = Setup time + Processing time + Cooling time

= 1 hr. + (100/5)*10min + 2 hrs.

= 6 hrs and 20 min.

2.7 Describe the various types of physical layouts used by production systems and for each give an example where it would be used.

▪ There are four types of physical layouts: product, process, cellular, and fixed position layouts, each defined as follows:

• Product layout – the product goes through a sequence of processes inside the plant designed for and dedicated to that product. This type of layout is suitable for repetitive manufacturing where demand is high and can support high utilization of the underlying equipment. Product layout is also typical for the continuous-flow production of commodity products processed in liquid or in bulk form, e.g., the production of paper from pulp, the production of gas and other derivatives from crude oil, the production of aluminum from bauxite.

• Process layout – this layout groups the same or similar machines together according to the operation they provide, establishing functional departments. Products go through the different department in different sequences indicated by the corresponding process plans. This type of layout provides more flexibility than product layout; however, it gives lower throughput rate, needs skilled work at production processes, and, in general, it is more difficult to schedule and control the underlying material flow. It might be suitable for low volume, high variety, and/or high production cost. It is used extensively for the fabrication of the various metallic or plastic parts of modern cars and appliances.

• Cellular layout – separate cells are created within the production system, with each cell producing a family of related parts or products. Each cell is essentially a mini-factory, with short material moves and easier coordination. This layout is a compromise between product and process layouts. An example can be garment manufacturing in which cell can be classified by product types, i.e. shirt cell, pants cell, etc.

• Fixed position layout – resources are allocated next to the ongoing product, instead of the product going through a set of resources. An example of this type of layout might be ship or space shuttle manufacturing.

2.23 A batch of material must visit five workstations in series to be converted into finished product. Each batch makes 100 units. Setup time at each workstation is 1 hour. Variable processing time is 1.2 minutes per unit. After completing an operation the batch has to wait an average of 1 hour for the material handler to move it to the next machine. Assuming there is no time lost waiting for machines; find the time it takes for a batch of material to go through the system.

We consider two different scenarios:

1. There will occur a setup time upon the arrival of a batch on each machine

▪ The total time for a batch = Setup time + Processing time + Material handling time

= 5 hr. + (1.2 min/unit/machine)(100 units)(5 machines) + 4 hrs.

= 300 + 600 + 240 = 1140 min. = 19 hrs.

2. All five workstations are setup together at the beginning of the process

▪ The total time for a batch = Setup time + Processing time + Material handling time

= 1 hr. + (1.2 min/unit/machine)(100 units)(5 machines) + 4 hrs.

= 60 + 600 + 240 = 900 min. = 15 hrs.

You can see that the second scenario results in a shorter total processing time for the considered batch, but on the other hand, it enforces substantial idleness on the downstream machines (these machines could have been used for some other jobs while the considered lot is processed in its first stages).

Repeat the previous problem assuming units are moved between operations in transfer batches of size 10 and that operations are close together so no time is lost waiting for material handling.

▪ The point here is that by reducing the transfer batch size from 100 to 10, multiple machines can work on different sub-batches at the same time. This element of concurrency or parallelization will eventually reduce the required total processing time for the entire batch. Again we consider two different scenaria regarding the machine set-ups:

1. Each machine is set up only upon the arrival of the first sub-batch:

▪ Let’s consider at the end of the process line, the time until the first sub-batch is completed is 5(12+60) = 360 min. However, the time until the second batch is completed is the previous amount of time plus the processing time of the last station = 360 + 12 = 372 min. Therefore, the amount of processing time (the time until the last batch is completed) is 360 + 12*9 = 468 min. Thus, the total processing time is 468 min = 7 hrs and 48 min.

|Units |Time until completed (min) |

|First 10 units |5(12+60) = 360 |

|Second 10 units |360 + 12 = 372 |

|Third 10 units |372 + 12 = 384 |

|Fourth 10 units |384 + 12 = 396 |

|Fifth 10 units |396 + 12 = 408 |

|Sixth 10 units |408 + 12 = 420 |

|Seventh 10 units |420 + 12 = 432 |

|Eight 10 units |432 + 12 = 444 |

|Ninth 10 units |444 + 12 = 456 |

|Tenth 10 units |456 + 12 = 468 |

2. All five workstations are setup together upon the release of the entire batch of 100 parts into the system:

▪ Now the required time for the completion of the first transfer batch is 60+5*12 = 120 min. Similarly to the previous case, the completion time of the second batch is equal to the completion time of the first transfer batch plus the processing time of the second batch on the last station = 120+12 = 132 min. The time for the completion of the last transfer batch is 120 + 12*9 = 228 min. or 3 hrs and 48 min. The different between this and previous scenario is 4 hrs, which is the additional setup time on the last four machines.

|Units |Time until completed (min) |

|First 10 units |60+5*12 = 120 |

|Second 10 units |120 + 12 = 132 |

|Third 10 units |132 + 12 = 144 |

|Fourth 10 units |144 + 12 = 156 |

|Fifth 10 units |156 + 12 = 168 |

|Sixth 10 units |168 + 12 = 180 |

|Seventh 10 units |180 + 12 = 192 |

|Eight 10 units |192 + 12 = 204 |

|Ninth 10 units |204 + 12 = 216 |

|Tenth 10 units |216 + 12 = 228 |

3.1 Why is demand forecasting important? How are forecasts used?

• Demand forecasting is important because it allows the company to see early on the various trends developing into the market and plan accordingly. More specifically, demand forecasts taken over a short to medium term, can predict how the product demand will behave in the future, and allow the company to develop a set production plans describing the timing and quantity for manufacturing products. Forecasts taken over a longer horizon can determine the new products to be designed and technologies to be developed, the needs for part and material supply contracts, financing plans, and capacity expansion.

▪ We notice that good forecasts must communicate also the level of uncertainty regarding the forecasted quantities (typically in the form of confidence intervals around the forecasted values).

3.4 For checking model adequacy, why is the sum of absolute forecast errors a better measure than simply the sum of forecast errors?

▪ The sum of absolute forecast errors will sum the total amount of errors; in contrast, in the sum of forecast errors, the sums of positive and negative amounts of errors will tend to cancel each other, often leading to a misleading low total/average error. Another way to avoid this canceling effect is by using the sum of squared errors.

2. The Bill of Materials (BOM) for two end items, X and Y, and their sub-assemblies are as follows.

Classify the various components and sub-assemblies appearing in the above BOM’s into levels according to the classification logic presented in class (hint: start by placing the end items X and Y at level 0)

Expanding the complete structure for each of the two end items we get:

[pic]

Therefore, Level 0: X, Y

Level 1: components that feed only into level 0 items => A

Level 2: components that feed into level 1 and possibly level 0 items => D

Level 3: components that feed into level 2 and possibly level 1 and 0 items => B, C

Extra credit (30%)

Revise the theory on linear regression from your statistics classes, and apply it to problem 3.22 in your textbook.

3.22 U.S. production of hydrochloric acid over a twelve-year period is shown in Table. Plot the data. Hypothesize a model form. Estimate parameter values. Check adequacy of the model. Repeat until you are satisfied with your model, then forecast 1993 demand.

[pic]

Simple linear regression model:

[pic]

Regression line: Y = 2251 + 94.35*X

[pic]

Forecasted demand for 1993: Y = 2251 + 94.35*13 = 3477.55

[pic]

[pic]

Hypothesis:

Ho: βo = 0 Since the p-value = 1.04 E-08 is so small, the null hypothesis is rejected, which means that intercept is not equal to zero.

Ho: β1 = 0 Since the p-value = 0.000374 is so small, the null hypothesis is rejected, which means that the slop is not equal to zero

Adequacy:

[pic][pic]

[pic]

From the Normal plot, residual plot, and the R-square = 0.73, we can say that this model is “fairly” accurate.[pic]

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