Table 9.23
The Terminal Company is attempting to balance its assembly line of high-voltage electrical connectors. The desired output for the line is 50 connectors per hour, and the information on the work elements for this assembly line is as follows.
$$\begin{array} { | c | c | c | }
\hline \begin{array} { c }
\text { Work } \
\text { Elements }
\end{array} & \begin{array} { c }
\text { Time } \
\text { (sec) }
\end{array} & \begin{array} { c }
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 40 & - \cdot \
\mathrm { B } & 36 & \mathrm {~A} \
\mathrm { C } & 20 & \mathrm {~A} \
\mathrm { D } & 2 & \mathrm {~A} \
\mathrm { E } & 30 & \mathrm {~B} , \mathrm { C } \
\mathrm { F } & 34 & \mathrm { D } \
\mathrm { G } & 35 & \mathrm { E } \
\mathrm { H } & 5 & \mathrm {~F} \
\mathrm { I } & 15 & \mathrm { E } , \mathrm { H } \
\mathrm { J } & 40 & \mathrm { H } \
\mathrm { K } & 38 & \mathrm { G } , \mathrm { I } , \mathrm { J } \
\hline
\end{array}$$
-Use the information from Table 9.23,and the longest work-element rule to balance this line.What is the efficiency of the solution?
A) more than 90%
B) 80-90%
C) 70-79%
D) less than 70%

Question

Table 9.23
The Terminal Company is attempting to balance its assembly line of high-voltage electrical connectors. The desired output for the line is 50 connectors per hour, and the information on the work elements for this assembly line is as follows.
 $$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Elements }
\end{array} & \begin{array} { c } 
\text { Time } \
\text { (sec) }
\end{array} & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 40 & - \cdot \
\mathrm { B } & 36 & \mathrm {~A} \
\mathrm { C } & 20 & \mathrm {~A} \
\mathrm { D } & 2 & \mathrm {~A} \
\mathrm { E } & 30 & \mathrm {~B} , \mathrm { C } \
\mathrm { F } & 34 & \mathrm { D } \
\mathrm { G } & 35 & \mathrm { E } \
\mathrm { H } & 5 & \mathrm {~F} \
\mathrm { I } & 15 & \mathrm { E } , \mathrm { H } \
\mathrm { J } & 40 & \mathrm { H } \
\mathrm { K } & 38 & \mathrm { G } , \mathrm { I } , \mathrm { J } \
\hline
\end{array}$$
-Use the information from Table 9.23,and the longest work-element rule to balance this line.What is the efficiency of the solution?
A) more than 90%
B) 80-90%
C) 70-79%
D) less than 70%

Quizplus · Accepted Answer

Using the longest work-element rule, we first find the cycle time. Starting with work element A, we calculate the time required for each subsequent work element, adding the times of any immediate predecessors. We continue this process until we have calculated the total time required for one cycle through the entire assembly line. The cycle time is then the longest of these total times.
\begin{align*}
&\text{Cycle time}= \max\{40,40+36,40+20,40+2,40+36+30,40+2+34+5,40+36+30+35,}\ 
&\qquad \qquad \max\{40+36+30+35+15,40+2+34+5+15\},40+36+30+35+15+40+38\} \
&\qquad \qquad \qquad = \max\{128,138,82,42,146,83,141,156,204,139\} = 204 \ \text{sec}
\end{align*}

Next, we calculate the number of stations required at each work element based on the desired output of 50 connectors per hour and the time required for each work element. We round up to the nearest integer to ensure that each station has sufficient work to meet the desired output.

\[\begin{array} { | c | c | c | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Elements }
\end{array} & \begin{array} { c } 
\text { Time } \
\text { (sec) }
\end{array} & \begin{array} { c } 
\text { Number of } \
\text { Immediate} \
\text { Predecessors }
\end{array} & \begin{array} { c } 
\text {Total time} \
\text {to this point } \
\end{array} & \begin{array} { c }
\text{Required Stations} \
\text{(rounded up)} \
\end{array} & \begin{array} { c } 
\text {Actual Output} \
\text {per hour} \
\end{array} & \begin{array} { c } 
\text {Efficiency} \
\text {(in decimal form)} \
\end{array} \
\hline \mathrm { A } & 40 & 0 & 40 & 1 & 1800 & 0.28 \
\mathrm { B } & 36 & 1 & 76 & 2 & 1388 & 0.28 \
\mathrm { C } & 20 & 1 & 96 & 2 & 1875 & 0.27 \
\mathrm { D } & 2 & 1 & 98 & 2 & 15000 & 0.33 \
\mathrm { E } & 30 & 2 & 128 & 3 & 1250 & 0.28 \
\mathrm { F } & 34 & 1 & 162 & 4 & 882 & 0.26 \
\mathrm { G } & 35 & 1 & 197 & 4 & 857 & 0.25 \
\mathrm { H } & 5 & 1 & 202 & 5 & 6000 & 0.29 \
\mathrm { I } & 15 & 2 & 217 & 5 & 1667 & 0.25 \
\mathrm { J } & 40 & 1 & 257 & 6 & 750 & 0.21 \
\mathrm { K } & 38 & 3 & 295 & 8 & 425 & 0.17 \
\hline \end{array}\]

Based on this calculation, the bottleneck is at work element K, which requires 8 stations. Since the total required number of stations (including those at the bottleneck) is 38, we can reduce the number of stations at some non-bottleneck work elements to improve efficiency. The best choice is to reduce the number of stations at work element I from 5 to 4, which reduces the total number of required stations to 37 and increases efficiency to 0.28. This is the highest efficiency that can be achieved with integer numbers of stations. Therefore, the efficiency of the solution is in the range of 70-79%, which corresponds to choice C.

Table 923 the Terminal Company Is Attempting to Balance Its Assembly

Table 923
the Terminal Company Is Attempting to Balance Its Assembly