Table 9.15
Balance the following line for an output rate of 3 units per minute,
using the longest work-element time rule.
$$\begin{array} { | c | c | c | }
\hline \begin{array} { c }
\text { Work } \
\text { Element }
\end{array} & \text { Time } ( \mathrm { sec } ) & \begin{array} { c }
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 12 & - - \
\mathrm { B } & 6 & \mathrm {~A} \
\mathrm { C } & 12 & \mathrm {~A} \
\mathrm { D } & 4 & \mathrm {~B} \
\mathrm { E } & 14 & \mathrm {~B} , \mathrm { C } \
\mathrm { F } & 10 & \mathrm { E } \
\mathrm { G } & 6 & \mathrm { D } , \mathrm { F } \
\hline
\end{array}$$
-Use the information in Table 9.15.What is the balance delay for your solution?
A) less than or equal to 15%
B) more than 15% but less than or equal to 25%
C) more than 25% but less than or equal to 35%
D) more than 35%

Question

Table 9.15
Balance the following line for an output rate of 3 units per minute,
using the longest work-element time rule.
 $$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time } ( \mathrm { sec } ) & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 12 & - - \
\mathrm { B } & 6 & \mathrm {~A} \
\mathrm { C } & 12 & \mathrm {~A} \
\mathrm { D } & 4 & \mathrm {~B} \
\mathrm { E } & 14 & \mathrm {~B} , \mathrm { C } \
\mathrm { F } & 10 & \mathrm { E } \
\mathrm { G } & 6 & \mathrm { D } , \mathrm { F } \
\hline
\end{array}$$
-Use the information in Table 9.15.What is the balance delay for your solution?
A) less than or equal to 15%
B) more than 15% but less than or equal to 25%
C) more than 25% but less than or equal to 35%
D) more than 35%

Quizplus · Accepted Answer

Using the longest work-element time rule, we start by assigning the work element with the longest time (E) to the first station. Then, we look for the work element with the longest time among the remaining unassigned ones (C) and assign it to the next station that has the fewest predecessors, which is the first station. Continuing this process, we get the following assignment:

Station 1: E (14 sec)
Station 2: A (12 sec), C (12 sec)
Station 3: B (6 sec), F (10 sec)
Station 4: D (4 sec), G (6 sec)

The cycle time is the sum of the work element times at each station, which is (14+12+12)+(6+10)+(4+6) = 64 seconds. The theoretical minimum number of stations required to achieve an output rate of 3 units per minute is 64/3 = 21.33, which rounds up to 22 stations. The actual number of stations in the balanced line is 4, so the balance delay is (22-4)/22 = 81.82%, which is more than 15% but less than or equal to 25%. Therefore, the answer is B.

The best choice is to add one or more stations to reduce the balance delay. In this case, adding stations may not be practical since the line is already very short with only 4 stations. Instead, we can consider reducing the workload of the work element with the longest time (E) by breaking it down into smaller elements or finding ways to perform it more efficiently. This would shorten the cycle time and reduce the balance delay.

Table 915 Balance the Following Line for an Output Rate of of 3

Table 915
Balance the Following Line for an Output Rate of of 3