Order sequences and order release, Mechanical Engineering

Assignment Help:

Order Sequences and Order Release

For this system, sequences of orders for particular parts (batch size 1) are produced in the following way: for each machine, here is, for each shift, a prescribed maximal load level ζ is calculated as ζ (machine type, shift number) = ζ0 (machine type) + Δ ζ (shift number) Δ ζ is arbitrary parameter, the load variation.

Because of the processing sequences structure, the average load ζ0 can only be selected freely on three of the four categories of machines. The selected values are as:

For case 1: 98 percent on HV, DOER and 98 percent on ECOC

For case 2: 98 percent on HV, DOER and 70 percent on ECOC

The load amplitude Δ ζ is understood to be either 0, - 30 percent, or + 30 percent. The actual part mix for simulation is produced from a basic random part type sequence along with average element type frequencies of 10 percent for part 1 and 9 percent for other parts. For all shift, there is a load account for all machines. An order for an exact part type is accepted as extended as the load accounts for all needed machines and for this shift it must not exceed the maximum load level ζ for this shift and these machines. Also, this is skipped. If no more element types can be accepted for this shift, the order sequence for this shift is finish. The subsequent order in the basic sequence becomes the initial new order for the subsequent shift. The actual average part frequencies are hence various from the values in the basic sequence.

As load level can be above 100 percent machine capability, there are backlogs. They are carried above to the subsequent shift. Over the initial 5, 10, 15 and so on, consecutive shifts, the average load level variations are put to zero.

The real order sequences are produced from five various basic element type sequences and five fundamental load variation sequences for all of the conditions illustrated below.

In case 1, orders are released along with equivalent spacing over the duration of a shift within every 19 minutes. In case 2, all orders are released at the starting of a shift. This offers rise to a considerably longer queues for the initial half shift, in exacting at the loading station.

Hence, the characteristics of case 1 can be summarized as: evenly distributed high loads on the machines along with temporary overload; at the stations, short to moderate queues. Within case 2, the characteristics can be specified as: part types 1-5 only need machines that are not bottlenecks; only element types 6-11 utilize machines that may be overloaded; the queues are long; the loading station is a dangerous resource.

Additionally to above conversation, the due date distribution is a critical parameter also that affects the presentation of the  scheduling policies. The due dates are produced as multiples of four hour (1/2 shift) intervals from order free. The percentage of orders for every value of this demanded throughput time was found such as under scheduling as per to FIFO a prescribed value of percentage of late jobs of as 30, 50, 70, and 85% was attained. The resulting distributions are shown in Table no.2 (a) and (b).  The  resulting  ratios  of  demanded  to  real  throughput  times  for  the various situations under the FIFO rule have also been specified in aforementioned tables.

The conditions A-D correspond to case 1, the percentage of late jobs raising from 30-85, the conditions E-H correspond to case 2 along with similar levels of late jobs under the FIFO rule.

Table no.2 (a): Due Dates Distribution for Situation A-D

Situation A : Case 1, 30% late jobs under the FIFO scheduling rule

Demanded Throuhput Time       %      of

Parts

240                            10

480                            10

720                            10

960                            35

1200                          35

Average Throughput Time: 900 min

Demanded/achieved flow factor: 1.29

 

Situation  A  :  Case  1,  50%  late  jobs under the FIFO scheduling rule

Demanded Throughput Time     %          of

Parts

240                            10

480                            25

720                            30

960                            25

1200                          10

Average Throughput Time: 720 min

Demanded/achieved flow factor: 1.03

 

Situation A : Case 1, 70% late jobs under the FIFO scheduling rule

Demanded Throuhput Time       %      of

Parts

240                            24

480                            34

720                            32

960                            5

1200                          5

Average Throughput Time: 559 min

Demanded/achieved flow factor: 0.80

 

Situation  A  :  Case  1,  85%  late  jobs under the FIFO scheduling rule

Demanded Throuhput Time       %          of

Parts

240                            60

480                            25

720                            5

960                            5

1200                          5

Average Throughput Time: 408 min

Demanded/achieved flow factor: 0.57

 


Related Discussions:- Order sequences and order release

Battery charger-electrical instruments , Battery Charger: As suggested by ...

Battery Charger: As suggested by the name, it is used to charge a battery. The detailed procedure for charging a battery is given in later units. List the uses of following too

Show the design of the control rooms, Q. Show the design of the control roo...

Q. Show the design of the control rooms? The design of the control rooms in any plant is very important. The control room operator(s) must be able to control and manage safety

Frictional force acting on the ladder, Frictional force acting on the ladde...

Frictional force acting on the ladder: A uniform ladder having length 13m and weighing 250N is placed against smooth vertical wall with its lower end 5m from wall. The coeffi

Resolving all of the forces normal to the inclined plane, Resolving all of ...

Resolving all of the forces normal to the inclined plane: A body weighing 500 N is resting on an inclined plane making an angle of 30 o with the horizontal. The coefficient o

Define the disadvantages of abrasive jet cutting machines, Define the Disad...

Define the Disadvantages of Abrasive Jet Cutting Machines Slow material removal rate. Stray cutting and hence accuracy is not good. Abrasive powder cannot be reused

Determine the reactions at the supports, Determine the reactions at the sup...

Determine the reactions at the supports: A smooth sphere weighing 200 N is resting as shown in Figure. Determine the reactions at the supports. Solution Let us fir

Calculate the kic from the data - thick cylinder, Calculate the KIC from th...

Calculate the KIC from the data - thick cylinder: In a fracture test a 3-point bend specimen of thickness 20 mm and depth 25 mm is supported over a span of 100 mm. The specime

What is the mean of off time, Q.What is the mean of Off time? It is the...

Q.What is the mean of Off time? It is the interval from the end of the hold time to the beginning of the squeeze time for the next (resistance) welding cycle. In automatic m

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd