In this exercise (inspired by Nahmias, 2013) the issue of reducing the number of different components in assembly production is considered.

This issue is very relevant in methods such as Product Modularity, Design for Assembly, Design for Assembly and Manufacture (see Ulrich, 1991, Gershenson and Prasad 1997, *Boothroyd, 1980 and 2002)*

Basic Idea is to increase the possibility to use the same component in different Assembled Models. In this manner it is possible to keep an high number of assembled models (high product diversification) while reducing the inventories levels.

In the following a very simple case is considered with a PC that is assembled from only 3 components: the processor, the hard disk and the memory.

References* *

*Nahmias, Production and Operation Analysis, 2013.*

*Boothroyd, “Design for Assembly – A Designer’s Handbook”, 1980.*

*Boothroyd, Dewhurst and Knight.“Product Design for Manufacture and Assembly”, 2002*

Text of the Excercise

A company assembles 3 separate components of PCs: the processor, the memory and the hard drive.

The number of models of PCs is 27. At the present (Present Case) each model uses a specific component.

This means that the number of different components is 81 (27 processors X 27 memories X 27 hard disks).

There are plans to use the same component in 9 different models (Future Case), reducing different components to 27 (3 X 3 X 3, keeping PC models=27).

The monthly demand of each PC is the same for each model and is independent each other and normally distributed,

with an average of D=5000 units/month and a standard deviation of σ=3000 units/month.

The cost of each order is 300 €/order.

The lead time of delivery is 1 month.

The component costs are: 10 €/piece for the processor, 5 €/piece for the memory and 8 €/piece for the hard disk.

The carrying costs are 10% €/€-y.

The company establishes a Cycle Service Level of CSL=95%.

a) Determine the safety stock for each component in both Present and Future cases;

b) Determine the total cost of the stocks in both cases;

c) Evaluate the ETSOPY (Expected Total Stock Out Per Year) in both cases;

d) Only in the Future Case, minimize the ETSOPY , redistributing the total value of safety stocks between stocked components.

Solution to point a)

Present Case

Let us consider the Present Case. CSL=95% means that for each component the safety stocks factor k=1.6411.

Standard Deviation demand during LT for each component is σ=3000 units/month.

This means that for each component the Safety Stocks are= 4923.49 units

Future Case

The difference with the Present Case is that in this case the number of models where each component is used is 9 and not 1.

Consequently the demand for each component is D_{9models}= D X 9 = 45000 units /month with standard deviation σ_{9models}=σ X 9^{0.5}= σ X 3 = 9000 units/month.

For each component then Safety Stocks are = k X σ_{9models} =14770.46 units/month.

Solution to point b)

In both cases the cost of stock (carrying costs) for a generic component is:

(1)

where vi depends on the type of component (processor, memory of hard drive), while σ_{LT} depends on the Demand of component (see point a) above) and Q_{i} on both Demand and type.

Since in both cases the number of different component per each type, n_{dct}, is the same (27 in Present Case, 9 in Future Case), we can calculate the total Carrying Cost using this formula:

(2)

Present Case

In the following index i is used for indicating the type of component:

i=1 means processor;

i=2 means memory;

i=3 means hard disk.

To evaluate Qi for each type of component, the EOQ formula is used. Results are:

Order Quantity | Value | Unit |

Q |
6000 | items/order |

Q_{2} |
8485 | items/order |

Q_{3} |
6708 | items/order |

By using 1 it is then possible to evaluate the carrying cost of each component type:

Carrying cost for each component type | €/year |

-Carrying costs_{1} |
7923.49 |

-Carrying costs_{2} |
4853 |

-Carrying costs_{3} |
6622 |

By using 2 Total costs of Stocks is:

Number of Different component Type | 27 |

Total cc | 516472.784 |

Future Case

Qi Results are (3 X Q_{i} Present Case):

Order Quantity | Value | Unit |

Q |
18000 | items/order |

Q_{2} |
25456 | items/order |

Q_{3} |
20125 | items/order |

carrying cost of each component type (3 X carrying costs_{i} Present Case):

Carrying cost for each component type | €/year |

-Carrying costs_{1} |
23770 |

-Carrying costs_{2} |
13749 |

-Carrying costs_{3} |
19866 |

By using 2 Total costs of Stocks is:

Number of Different component Type | 3 |

Total cc (Total cc Present Case/3) | 172157.6 |

Solution to point c)

Present Case

Considering that P1=95% (the probability to go in stock out in one cycle) and that Qi and D are known it is straighforward:

Component Type | number of Stock Out/year |

1 | 6 |

2 | 4.24 |

3 | 5.37 |

Considering that for each type there are 27 different component the ETSOPY=421.45

Future Case

Similarly ETSOPY= 140.48 (1/3 respect Present Case).

Solution to point d)

To find the mimimum of Expected Total Stock Out per her we guess a starting penalty cost for stock out incident B1.

Afterward safety stock for any component is determined and total cost verified. It the cost are too high B1 decreases, while if cost are too low B1 is increased.

After some iteration, we stop when we reach a good approximation.

Found value of safety factors are

K1=1.565

K2=1.772

K3=1.649

As it may be expcted safety factors increases for low value items.

in this excel file you find calculation for this exercise.