Question

Question: Consider the following tasks, times, and predecessors for an assembly of set-top cable converter boxes:

Task Element	Time (Minutes)	Element Predecessor
A	1
B	1	A
C	2	B
D	1	B
F	3	C, D
F	1	A
G	1	F
H	2	G
I	1	E, H

Given a cycle time of four minutes, develop two alternative layouts. Use the longest task time rule and the largest number of following tasks as a secondary criterion. What is the efficiency of your layouts, assuming the 4-minute cycle time?

Short answer

Answer

The efficiency of the workstation will be 100% and 81.2% respectively.

Step-by-step solution

The precedence diagram

Using the below table we can form the precedence diagram:

The project paths and duration would be:

A-B-C-E-I= 1+1+2+3+1= 8 minutes

A-B-D-E-I= 1+1+1+3+1= 7 minutes

A-F-G-H-I= 1+1+1+2+1= 6 minutes

Thus, we can say that project will be completed in a maximum of eight minutes.

Longest task time rule

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{workstations}=\frac{\mathrm{Total}\mathrm{task}\mathrm{time}}{\mathrm{Cycle}\mathrm{time}} \\ =\frac{8}{4} \\ =2\mathrm{workstations}. \end{gathered}$$

To assign tasks according to the workstations to find the layout for the longest task time:

$$\begin{gathered} \mathrm{Efficiency}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{task}\mathrm{times}}{(\mathrm{Number}\mathrm{of}\mathrm{workstations}\times \mathrm{Workstation}\mathrm{cycletime}} \\ =\frac{1+1+2+3+1}{(2\times 4)} \\ =\frac{8}{8} \\ =1\mathrm{OR}1\% \end{gathered}$$

Theefficiency of the longest task timeis 100%.

For the largest number of tasks

To assign tasks according to the workstations to find the layout for the largest number of tasks:

$$\begin{gathered} \mathrm{Efficiency}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{task}\mathrm{times}}{(\mathrm{Number}\mathrm{of}\mathrm{workstations}\times \mathrm{Workstation}\mathrm{cycletime}} \\ =\frac{1+1+2+1+3+1+1+2+1}{\left(4\times 4\right)} \\ =\frac{13}{16} \\ =0.8125\mathrm{OR}81.2\% \end{gathered}$$