Table 9.21
Table 7.21 The following information is given about an assembly line.
The desired output rate is 90 units per hour.
$$\begin{array} { | c | c | c | }
\hline \begin{array} { c }
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c }
\text { Immediate } \
\text { Predecessor } ( \mathrm { s } )
\end{array} \
\hline \mathrm { A } & 10 & - - \
\mathrm { B } & 6 & \mathrm {~A} \
\mathrm { C } & 15 & \mathrm {~A} \
\mathrm { D } & 20 & \mathrm {~B} \
\mathrm { E } & 12 & \mathrm {~B} \
\mathrm {~F} & 14 & \mathrm { C } , \mathrm { E } \
\mathrm { G } & 8 & \mathrm { D } \
\mathrm { H } & 20 & \mathrm {~F} , \mathrm { G } \
\hline
\end{array}$$
-Use the information in Table 9.21.Suppose that the desired output rate is increased to 120 units per hour and a solution has been found that has 4 workstations.What is the efficiency of this new assembly line?
A) less than 80%
B) between 80% and 85%
C) between 85% and 90%
D) more than 90%

Question

Table 9.21
Table 7.21 The following information is given about an assembly line.
The desired output rate is 90 units per hour.
 $$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor } ( \mathrm { s } )
\end{array} \
\hline \mathrm { A } & 10 & - - \
\mathrm { B } & 6 & \mathrm {~A} \
\mathrm { C } & 15 & \mathrm {~A} \
\mathrm { D } & 20 & \mathrm {~B} \
\mathrm { E } & 12 & \mathrm {~B} \
\mathrm {~F} & 14 & \mathrm { C } , \mathrm { E } \
\mathrm { G } & 8 & \mathrm { D } \
\mathrm { H } & 20 & \mathrm {~F} , \mathrm { G } \
\hline
\end{array}$$
-Use the information in Table 9.21.Suppose that the desired output rate is increased to 120 units per hour and a solution has been found that has 4 workstations.What is the efficiency of this new assembly line?
A) less than 80%
B) between 80% and 85%
C) between 85% and 90%
D) more than 90%

Quizplus · Accepted Answer

To calculate the theoretical minimum number of workstations, use the following formula:

$$N = \dfrac{\sum t_i}{C\times T}$$

where:

N = theoretical minimum number of workstations
$\sum t_i$ = total processing time for all work elements
C = desired hourly output rate
T = length of the working time (in minutes)

Using the given data, we get:

$\sum t_i$ = 10 + 6 + 15 + 20 + 12 + 14 + 8 + 20 = 105

C = 90 units per hour, or 1.5 units per minute

T = 60 minutes (assuming continuous operation)

Therefore, N = 105 / (1.5 x 60) = 1.17

This means that we need at least 2 workstations to achieve an output rate of 90 units per hour.

To increase the output rate to 120 units per hour, we need to reduce the cycle time. One possible solution with 4 workstations is:

Station 1: A, B
Station 2: C, E
Station 3: D, F
Station 4: G, H

The cycle time for this solution is 0.5 minutes (30 seconds), calculated as follows:

$$0.5 = \dfrac{T}{C \times N}$$

Using this cycle time, the processing times for each station can be calculated as follows:

Station 1: 10 + 6 = 16 seconds
Station 2: 15 + 12 = 27 seconds
Station 3: 20 + 14 = 34 seconds
Station 4: 8 + 20 = 28 seconds

The efficiency of the assembly line is calculated as follows:

Efficiency = (total actual working time) / (total idle time + total actual working time)

To calculate the total actual working time, we add up the processing times for each station:

16 + 27 + 34 + 28 = 105 seconds

To calculate the total idle time, we subtract the total actual working time from the total cycle time:

0.5 x 60 - 105 = 15 seconds

Therefore, Efficiency = 105 / (15 + 105) = 0.875 or 87.5%

Since the efficiency is between 85% and 90%, the answer is C.

Table 921 Table 7 -Use the Information in Table 9

Table 921
Table 7 -Use the Information in Table 9