---
# System prepended metadata

title: PID-739

---

a. **Precedence** diagram:

By using Tikz package we have the following code

```
\begin{tikzpicture}
\begin{scope}[every node/.style={circle,thick,draw}]
    \node (A) at (0,2) {A};
    \node (B) at (2,2) {B};
    \node (C) at (4,2) {C};
    \node (D) at (6,2) {D};
    \node (E) at (6,0) {E};
    \node (F) at (6,-2) {F} ;
    \node (G) at (8,-1) {G} ;
    \node (H) at (8,2) {H} ;
    \node (I) at (10,0) {I} ;
    \node (J) at (12,0) {J} ;
\end{scope}

\begin{scope}[>={Stealth[black]},
              every node/.style={fill=white,circle},
              every edge/.style={draw=red,very thick}]
    \path [->] (A) edge  (B);
    \path [->] (B) edge  (C);
    \path [->] (C) edge  (D);
    \path [->] (D) edge  (H);
    \path [->] (H) edge (I);
    \path [->] (I) edge  (J);
    \path [->] (E) edge  (G);
    \path [->] (F) edge  (G);
    \path [->] (G) edge (I); 
\end{scope}
\end{tikzpicture}
```
where we have to put at the begining of the document

```
\usepackage{tikz}
\usetikzlibrary{arrows.meta}
```
And the required diagram is 

![](https://hackmd.io/_uploads/ry4v0u1A2.png)

b. To produce $200$ units over a ten-hour day, we need to produce $20$ units per hour. Assuming that each unit takes the same amount of time to produce, the cycle time would be $3$ units per minute ($60$ minutes / $20$ units). Thus, 
$$ \text{cycle time}=3$$

c. The theoretical number of workstations needed can be calculated by dividing the sum of task times by the cycle time. The sum of task times is:

$$0.5 + 0.7 + 0.4 + 0.3 + 0.3 + 0.2 + 0.35 + 0.4 + 0.6 + 0.3 = 4.05 \text{ minutes.}$$

$$\text { Number of workstations }\left(\mathrm{N}_t\right) =\frac{\text { Sum of all Task time }}{\text { Cycle Time }}$$

$$=\frac{4.05}{3}$$

$$=1.35$$

$$=2$$

it is rounded up.
Hence, the number of workstations is $2$.

d. **By using most following task rule, we have**

|  | $\begin{array}{c}\text { Total time } \\\text { available } \\\text { (minutes) }\end{array}$ | $\begin{array}{c}\text { Task } \\\text { eligible }\end{array}$ | $\begin{array}{l}\text { Assigned } \\\text { task }\end{array}$ | $\begin{array}{l}\text { Task time } \\\text { (minutes) }\end{array}$ | $\begin{array}{c}\text { Time } \\\text { remaining } \\\text { (minutes) }\end{array}$ | $\begin{array}{l}\text { Idle time } \\\text { (minutes) }\end{array}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\begin{array}{c}\text { Station } \\1\end{array}$ | 3 | $A$ | $A$ | 0.50 | 2.50 | 0.20 |
|  |  | $B$ | $B$ | 0.70 | 1.80 |  |
|  |  | $C$ | $C$ | 0.40 | 1.40 |  |
|  |  | $D, E, F$ | D | 0.30 | 1.10 |  |
|  |  | $E, F$ | $E$ | 0.30 | 0.80 |  |
|  |  | $F$ | F | 0.20 | 0.60 |  |
|  |  | $G, H$ | $H$ | 0.40 | $\mathbf{0.20}$ |  |
| $\begin{array}{c}\text { Station } \\2 \\\end{array}$ | 3 | $G$ | $G$ | 0.35 | 2.65 | 1.75 |
|  |  | $I$ | $I$ | 0.60 | 2.05 |  |
|  |  | $J$ | $J$ | 0.30 | $\mathbf{1.75}$ |  |
|  |  |  |  |  | $\begin{array}{c}\text { Total idle } \\\text { time }\end{array}$ | $\begin{array}{c}1.75+0.20 \\=1.95\end{array}$ |

Note that if there are more than one eligible task, the task taking maximum time will be allocated firs


For the **layout using longest task rule,** we get the same table.
For this case we have if longest task is preceded by the less lengthy task, then less lengthy task will be completed first because without that successor task cannot be completed. Also, in case of more than one eligible task, the task taking maximum time will be allocated first. In case of equal time, either activity can be assigned.

e. By using both criteris, we have a total of 1.95 minutes as the idle time. Hence, both are equally efficient. One can therefore, select either of the criteria.