rpw method sdk

7
QUESTION :) Using Ranked Positional Weight (RPW) METHOD solves the problem given below: TASK PRECEDED BY TIME (in minutes) A - 0.8 B A 1.3 C A 3.4 D - 1.5 E B 1.8 F C, D 1.5 G E 1.0 H E, F 2.1 I G, H 1.1 CYCLE TIME – 5 MINUTES SOLUTION :) RPW: When you have a choice, select the feasible task with the highest POTENTIAL WEIGHT (PW). Where PW is equal to the sum of the time of the task and the times of the tasks following it. A- 0.8 B- 1.3 C- 3.4 E- 1.8 F-1.5 H- 2.1 G- 1.0 D-1.5 I- 1.1

Upload: md-shah-ahsan-leon

Post on 18-May-2017

215 views

Category:

Documents


3 download

TRANSCRIPT

Page 1: Rpw Method SDk

QUESTION :) Using Ranked Positional Weight (RPW) METHOD solves the problem given below:

TASK PRECEDED BY TIME (in minutes)

A - 0.8

B A 1.3

C A 3.4

D - 1.5

E B 1.8

F C, D 1.5

G E 1.0

H E, F 2.1

I G, H 1.1

CYCLE TIME – 5 MINUTES

SOLUTION :)

RPW: When you have a choice, select the feasible task with the highest POTENTIAL WEIGHT (PW). Where PW is equal to the sum of the time of the task and the times of the tasks following it.

Potential weight (PW):

PW for task A= Time of tasks (A+B+C+E+F+G+H+I)

= (0.8 +1.3+3.4+1.8+1.0+1.5+2.1+1.1) minutes = 13 minutes

A-0.8

B-1.3

1.3MIC-3.4

E-1.8

F-1.5 H-2.1

G-1.0

I-1.1

D-1.5

Page 2: Rpw Method SDk

PW for task B= Time of tasks (B+ E+G+H+I)

= (1.3+1.8+1.0+2.1+1.1) minutes = 7.3minutes

PW for task C= Time of tasks (C+F+ H+I)

= (3.4+1.5+2.1+1.1) minutes = 8.1minutes

PW for task D= Time of tasks (D+F+ H+I)

= (1.5+1.5+2.1+1.1) minutes = 6.2minutes

PW for task E= Time of tasks (E+G+ H+I)

= (1.8+1.0+2.1+1.1) minutes =6 minutes

PW for task F= Time of tasks (F+H+I)

= (1.5+2.1+1.1) minutes =4.7minutes

PW for task G= Time of tasks (G+I)

= (1.0+1.1) minutes =2.1minutes

PW for task H= Time of tasks (H+I)

= (2.1+1.1) minutes =3.2minutes

PW for task I= Time of task I

= 1.1 minutes

TASK PRECEDED BY TIME PW (POTENTIAL WIEGHT)

A - 0.8 13.0

B A 1.3 7.3

C A 3.4 8.1

D - 1.5 6.2

E B 1.8 6.0

F C,D 1.5 4.7

G E 1.0 2.1

H E,F 2.1 3.2

I G,H 1.1 1.1

∑ 14.5

Page 3: Rpw Method SDk

According to PW calculation the task should be ranked in that manner where first priority will be given to the task with the highest PW and then second priority to the task with the second most PW and the process will continue until all tasks are ranked. The following shows the RPW of the tasks:

Ranked Positional Weight (RPW)

A -1

C -2

B -3

D -4

E -5

F -6

H -7

G -8

I -9

Number of Work stations needed = Total task time / Desired Cycle time

= 14.5 minutes / 5 minutes = 2.9 = 3 work stations

So we need 3 Work stations (WS) instead of 9 and if we group tasks among these 3 work stations cleverly then we will be able to maximize efficiency by lowering idle time.

Before assigning the tasks in the work stations we need to answer three questions:

1. Which jobs are available?2. Who gets the priority?3. Where should we assign the job?

Task A is available as it has no predecessor and according to RPW A should get the priority as the rank of A is 1 and so we should assign A in work station 1 (WS 1). After A, tasks B and C are available but C should get the priority the rank of C is 2 according to RPW and so C should also be assigned in WS 1. Then task B should get the priority but it cannot be assigned to WS 1 as the total time of A and C is 4.2minutes and if we assign B there it would cross the cycle time and that cannot be allowed. So B will be assigned to WS 2. After B, task D should get the priority as the RPW is 4 and so D should also be assigned to WS 2. After D, E gets the priority as its RPW is 5 and so E will also be assigned to WS 2. The total time of WS 2 i.e. the time of B, D and E is 4.6 minutes, this means that no other tasks can be assigned in WS 2 as doing so would cross the cycle time. So the other tasks have to be assigned in WS 3. Task F gets the next priority and this task will be assigned to WS 3. After F, H gets priority and that would be assigned to WS 3 and after H, G will get the priority and will be assigned to WS 3. The total time of WS 3 is 4.6 minutes which means task I cannot be assigned to WS 3 as in doing so it would cross the cycle time. So in order to assign task I we need to open up a new Work station WS 4 and we need to assign task I there.

Page 4: Rpw Method SDk

WS-1 WS-2 WS-3 WS-4

A-0.8 B-1.3 F-1.5 I-1.1

C-3.4 D-1.5 G-1.0

E-1.8 H-2.1

4.2 minutes 4.6 minutes 4.6 minutes 1.1 minutes

Maximum Work station time = 4.6 minutes

Efficiency (E) = Total task time/ (actual no of stations*maximum work station time)

= 14.5 minutes / (4 * 4.6 minutes)

= 0.788

= 78.8% = 79%

This means that out of 100%, 79% of my workers will be efficient.

Page 5: Rpw Method SDk

PRODUCTION AND OPERATIONS MANAGEMENT (MGT 314)

ASSIGNMENT # 05

GROUP NAME: FORECASTERS

SECTION: 10

SUBMITTED TO:

Mr. Sayedul Karim

Lecturer

School of Business

North South University

SUBMITTED BY: ID

Monisha Qamrul Chowdhury 092 0648 530

Syeda Mushfequn Nahar 092 0253 030

Zafrin Jahan 092 0054 030

Mohammed Sayedul Bashar 092 0510 030

Md. Imran Faruque Rashed 093 0873 030

Zahin Tarannum Zamnan 092 0148 030

DATE OF SUBMISSION: 08 / 04 / 2012