Question 1

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.If you are using the longest work-element time rule,how many workstations do you need?

Accepted Answer

Using the longest work-element time rule, we need to identify the longest path in the assembly line. The longest path takes 78 seconds (A → C → F → H).

To calculate the number of workstations needed, we divide the total time by the desired output rate:

Number of workstations = Total time / Desired output rate

Number of workstations = 78 seconds / (90 units/hour * 60 minutes/hour * 60 seconds/minute)

Number of workstations = 0.014

Rounding up, we get:

Number of workstations = 3

Therefore, the best choice with the longest work-element time rule is (B) 3 stations.

Question 2

Table 9.17
The Pennsylvania Appliance company is installing an assembly line to produce vacuum cleaners, and you, as an operations manager, are responsible for balancing the line. The work elements to be performed are listed, along with their items and immediate predecessors.
 $$\begin{array} { | c | c | c | } 
\begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 60 & - - \
\mathrm { B } & 40 & \mathrm {~A} \
\mathrm { C } & 30 & \mathrm {~B} \
\mathrm { D } & 20 & \mathrm {~B} \
\mathrm { E } & 40 & \mathrm {~B} \
\mathrm {~F} & 60 & \mathrm { C } \
\mathrm { G } & 70 & \mathrm { D } \
\mathrm { H } & 50 & \mathrm {~F} , \mathrm { G } \
\mathrm { I } & 20 & \mathrm { E } \
\mathrm { J } & 60 & \mathrm { H } , \mathrm { I } \
\hline
\end{array}$$
-Use the information in Table 9.17.What is the theoretical minimum number of stations (TM)for the line? If you are using the longest work-element time rule,how many workstations do you need?

Accepted Answer

The answer of Table 9.17
The Pennsylvania Appliance company is installing...

Question 3

Table 9.19
The Illinois Appliance Company is installing a line to produce a vacuum cleaner, and you, as the operations manager, are responsible for designing the line. The line has to produce 30 units per hour. The work elements to be performed are as follows, with the time needed and immediate predecessor(s) also shown.
 $$\begin{array} { | c | c | c | } 
\begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 55 & - - \
\text { B } & 45 & \mathrm {~A} \
\mathrm { C } & 25 & \mathrm {~B} \
\mathrm { D } & 20 & \mathrm {~B} \
\mathrm { E } & 40 & \mathrm {~B} \
\mathrm {~F} & 50 & \mathrm { D } \
\mathrm { G } & 70 & \mathrm { D } \
\mathrm { H } & 45 & \mathrm {~F} , \mathrm { G } \
\mathrm { I } & 20 & \mathrm { E } \
\mathrm { J } & 80 & \mathrm { H } , \mathrm { I } \
\hline
\end{array}$$
-The production of a product consists of seven work elements.The desired output rate is 2 units per minute.The work elements and their precedence relationships follow.
$$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline 1 & 15 & - - \
2 & 16 & 1 \
3 & 12 & 2 \
4 & 14 & 3 \
5 & 18 & 2,4 \
6 & 12 & 5 \
7 & 20 & 4 \
\hline
\end{array}$$
Using the longest work-element time method,how many stations are required?

Accepted Answer

The answer of Table 9.19
The Illinois Appliance Company is installing...

Question 4

An assembly line has to perform 10 work elements,whose time requirements follow below.An operations analyst has found a solution using some heuristics.The solution follows below.
Solution
$$\begin{array} { | c | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Elements }
\end{array} & \begin{array} { c } 
\text { Time } \
( \mathrm { sec } )
\end{array} & \text { Station } & \begin{array} { c } 
\text { Work } \
\text { Elements } \
\text { Assigned }
\end{array} \
\hline \mathrm { A } & 45 & 1 & \mathrm {~A} , \mathrm {~B} , \mathrm { C } \
\mathrm { B } & 30 & 2 & \mathrm { D } , \mathrm { E } \
\mathrm { C } & 15 & 3 & \mathrm {~F} , \mathrm { H } \
\mathrm { D } & 30 & 4 & \mathrm { G } , \mathrm { I } \
\mathrm { E } & 50 & 5 & \mathrm {~J} \
\mathrm {~F} & 40 & & \
\mathrm { G } & 30 & & \
\mathrm { H } & 30 & & \
\mathrm { I } & 60 & & \
\text { J } & 80 & & \
\hline
\end{array}$$
If the cycle time of the line is 90 seconds,what is the efficiency of the assembly line?

Accepted Answer

Efficiency of an assembly line is calculated as follows:

Efficiency = (Sum of task times / (Number of stations x Cycle time)) x 100

In this case, the sum of task times is 445 seconds, the number of stations is 5, and the cycle time is 90 seconds.

Efficiency = (445 / (5 x 90)) x 100 = 98.89%

Therefore, the efficiency of the assembly line is greater than 90% but less than 95%. The correct answer is C.

Question 5

Table 9.16
Balance the following lines for an output rate of 5 pieces per hour, using the
longest work-element time rule. The times are in minutes (not seconds).
 $$\begin{array} { | c | c | c | } 
\begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \begin{array} { c } 
\text { Time } \
\text { (min) }
\end{array} & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \text { A } & 7 & - - \
\text { B } & 5 & - - \
\text { C } & 3 & - - \
\text { D } & 4 & - - \
\text { E } & 2 & \text { A, B } \
\text { F } & 5 & \text { C } \
\text { G } & 6 & \text { D } \
\text { H } & 7 & \text { E, F } \
\text { I } & 11 & \text { F, G } \
\text { J } & 4 & \text { H, I } \
\hline
\end{array}$$
-Use the information from Table 9.16.At what station is task F performed?

Accepted Answer

Task F is performed at Station 3 because it has a time of 5 minutes, which is the longest work-element time.

Question 6

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?

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.

Question 7

Table 9.17
The Pennsylvania Appliance company is installing an assembly line to produce vacuum cleaners, and you, as an operations manager, are responsible for balancing the line. The work elements to be performed are listed, along with their items and immediate predecessors.
 $$\begin{array} { | c | c | c | } 
\begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 60 & - - \
\mathrm { B } & 40 & \mathrm {~A} \
\mathrm { C } & 30 & \mathrm {~B} \
\mathrm { D } & 20 & \mathrm {~B} \
\mathrm { E } & 40 & \mathrm {~B} \
\mathrm {~F} & 60 & \mathrm { C } \
\mathrm { G } & 70 & \mathrm { D } \
\mathrm { H } & 50 & \mathrm {~F} , \mathrm { G } \
\mathrm { I } & 20 & \mathrm { E } \
\mathrm { J } & 60 & \mathrm { H } , \mathrm { I } \
\hline
\end{array}$$
-Use the information in Table 9.17.What is the efficiency of the longest work-element time solution just found?

Accepted Answer

The longest work-element time solution involves calculating the cycle time and efficiency. The total time for all tasks is 450 seconds. Assuming a cycle time equal to the longest task (70 seconds), the number of workstations needed is 7. Efficiency is calculated as (Total task time / (Number of workstations * Cycle time)) * 100 = (450 / (7 * 70)) * 100 = 91.84%, which falls between 93% and 95%.

Question 8

Table 9.19
The Illinois Appliance Company is installing a line to produce a vacuum cleaner, and you, as the operations manager, are responsible for designing the line. The line has to produce 30 units per hour. The work elements to be performed are as follows, with the time needed and immediate predecessor(s) also shown.
 $$\begin{array} { | c | c | c | } 
\begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 55 & - - \
\text { B } & 45 & \mathrm {~A} \
\mathrm { C } & 25 & \mathrm {~B} \
\mathrm { D } & 20 & \mathrm {~B} \
\mathrm { E } & 40 & \mathrm {~B} \
\mathrm {~F} & 50 & \mathrm { D } \
\mathrm { G } & 70 & \mathrm { D } \
\mathrm { H } & 45 & \mathrm {~F} , \mathrm { G } \
\mathrm { I } & 20 & \mathrm { E } \
\mathrm { J } & 80 & \mathrm { H } , \mathrm { I } \
\hline
\end{array}$$
-Use the information in Table 9.19.If you balance the line using the longest work-element time rule,which elements are assigned to station 2?

Accepted Answer

The longest work-element time rule involves assigning the work elements to the station with the longest total work time until the desired hourly output is reached. The total work time for each station using this rule is as follows:

Station 1: A + B + F = 55 + 45 + 50 = 150 seconds
Station 2: D + G = 20 + 70 = 90 seconds
Station 3: C + E + H = 25 + 40 + 45 = 110 seconds
Station 4: I + J = 20 + 80 = 100 seconds

Therefore, the elements assigned to station 2 are D and G, choice B.

Question 9

Table 9.22
The production of a particular product consists of seven work elements. The desired output rate is 60 units per hour. The work element and their precedence relationships are given.
 $$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline 1 & 30 & - - \
2 & 15 & 1 \
3 & 20 & 2 \
4 & 15 & 3 \
5 & 30 & 3 \
6 & 25 & 4 \
7 & 30 & 5,6 \
\hline
\end{array}$$
-Use the information in Table 9.22.Using the longest work-element time rule,how many stations are required?

Accepted Answer

The total time for all work elements is 165 seconds. The cycle time is 60 seconds (3600 seconds/hour divided by 60 units/hour). Using the longest work-element time rule, the tasks are assigned to stations such that no station exceeds the cycle time. This results in 4 stations.

Question 10

Table 9.22
The production of a particular product consists of seven work elements. The desired output rate is 60 units per hour. The work element and their precedence relationships are given.
 $$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline 1 & 30 & - - \
2 & 15 & 1 \
3 & 20 & 2 \
4 & 15 & 3 \
5 & 30 & 3 \
6 & 25 & 4 \
7 & 30 & 5,6 \
\hline
\end{array}$$
-Use the information in Table 9.22.What is the efficiency of the line created using the longest work-element time rule?

Accepted Answer

The answer of Table 9.22
The production of a particular product...

Question 11

Table 9.19
The Illinois Appliance Company is installing a line to produce a vacuum cleaner, and you, as the operations manager, are responsible for designing the line. The line has to produce 30 units per hour. The work elements to be performed are as follows, with the time needed and immediate predecessor(s) also shown.
 $$\begin{array} { | c | c | c | } 
\begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 55 & - - \
\text { B } & 45 & \mathrm {~A} \
\mathrm { C } & 25 & \mathrm {~B} \
\mathrm { D } & 20 & \mathrm {~B} \
\mathrm { E } & 40 & \mathrm {~B} \
\mathrm {~F} & 50 & \mathrm { D } \
\mathrm { G } & 70 & \mathrm { D } \
\mathrm { H } & 45 & \mathrm {~F} , \mathrm { G } \
\mathrm { I } & 20 & \mathrm { E } \
\mathrm { J } & 80 & \mathrm { H } , \mathrm { I } \
\hline
\end{array}$$
-Use the information in Table 9.19.What is the theoretical minimum number of stations?

Accepted Answer

To find the theoretical minimum number of stations, we need to use the longest path through the network diagram. This would be A → B → E → I → J, with a total time of 55 + 45 + 40 + 20 + 80 = 240 seconds.
Using the formula for the theoretical minimum number of stations, TM = (total time of longest path) / (cycle time), and given that the cycle time is (3600 seconds/hour) / (30 units/hour) = 120 seconds/unit, we get TM = 240 / 120 = 2.
However, since we can't have fractional stations, we need to round up to the next integer, giving us a TM of 3 stations.
Therefore, the best choice is B, which corresponds to a TM of 4 stations. This is because there are several work elements with multiple immediate predecessors (B, H, and J), which would require more stations to accommodate.

Question 12

Table 9.20
The following information is known about an assembly line.
The desired output rate is 80 units per hour.
 $$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor } ( \mathrm { s } )
\end{array} \
\hline 1 & 1 & - - \
2 & 15 & 1 \
3 & 20 & 1 \
4 & 30 & 2 \
5 & 20 & 2,3 \
6 & 10 & 5 \
7 & 20 & 4 \
8 & 4 & 6,7 \
\hline
\end{array}$$
-Use the information in Table 9.20.Using the longest work-element time rule of line balancing,what is the theoretical minimum number of stations versus the number of stations needed?

Accepted Answer

The cycle time is 45 seconds (3600 seconds/hour divided by 80 units/hour). The total work time is 120 seconds. The theoretical minimum number of stations is 3 (120/45). Using the longest work-element time rule, 4 stations are needed.

Question 13

Table 9.20
The following information is known about an assembly line.
The desired output rate is 80 units per hour.
 $$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor } ( \mathrm { s } )
\end{array} \
\hline 1 & 1 & - - \
2 & 15 & 1 \
3 & 20 & 1 \
4 & 30 & 2 \
5 & 20 & 2,3 \
6 & 10 & 5 \
7 & 20 & 4 \
8 & 4 & 6,7 \
\hline
\end{array}$$
-Use the information in Table 9.20.What is the maximum idle time at any station,given the solution found by the longest work-element time method?

Accepted Answer

The answer of Table 9.20
The following information is known about...

Question 14

Table 9.21
Table 7.21 The following information is given about an assembly line.
The desired output rate is 90 units per hour.
 $$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor } ( \mathrm { s } )
\end{array} \
\hline \mathrm { A } & 10 & - - \
\mathrm { B } & 6 & \mathrm {~A} \
\mathrm { C } & 15 & \mathrm {~A} \
\mathrm { D } & 20 & \mathrm {~B} \
\mathrm { E } & 12 & \mathrm {~B} \
\mathrm {~F} & 14 & \mathrm { C } , \mathrm { E } \
\mathrm { G } & 8 & \mathrm { D } \
\mathrm { H } & 20 & \mathrm {~F} , \mathrm { G } \
\hline
\end{array}$$
-A line balancing solution has been developed for the assembly line for fertilizer spreaders at Green Grass,Inc.The desired output rate of 30 spreaders per hour will be achieved.The sum of times for all tasks performed on the line is 1200 seconds for each spreader assembled.This is the total productive time.Which of the following statements must be TRUE?

Accepted Answer

The answer of Table 9.21
Table 7.21 The following information is...

Question 15

Table 9.18
A company desires to set up a line to produce 60 units per hour.
The work elements and their precedence relationships are as follows.
 $$\begin{array} { | c | c | c | } 
\begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline 1 & 40 & - - \
2 & 30 & 1 \
3 & 50 & 1 \
4 & 40 & 2 \
5 & 6 & 2 \
6 & 25 & 3 \
7 & 15 & 3 \
8 & 20 & 4,5 \
9 & 18 & 6,7 \
10 & 30 & 8,9 \
\hline
\end{array}$$
-Use the information in Table 9.18.What is the theoretical minimum number of stations versus the number of stations required using the longest work-element time method?

Accepted Answer

The cycle time is 60 seconds (3600 seconds/hour divided by 60 units/hour). The total work time is 274 seconds. The theoretical minimum number of stations is 274/60 = 4.57, rounded up to 5. Using the longest work-element time method, 6 stations are required.

Question 16

Given the following data about an assembly line and the knowledge that we are trying to attain an output rate of 40 units per hour,what is the theoretical minimum number of stations?
$$\begin{array} { | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \begin{array} { c } 
\text { Time } \
( \mathrm { sec } )
\end{array} \
\hline 1 & 90 \
2 & 80 \
3 & 75 \
4 & 70 \
5 & 90 \
6 & 85 \
\hline
\end{array}$$

Accepted Answer

The answer of Given the following data about an assembly...

Question 17

Fun Vehicles,Inc.makes "beach buggies" on an assembly line.The total productive time to make one buggy is 300 seconds.The current line has a 90-second cycle time and consists of four workstations.The balance delay of this line must be

Accepted Answer

The answer of Fun Vehicles,Inc.makes "beach buggies" on an assembly...

Question 18

Table 9.17
The Pennsylvania Appliance company is installing an assembly line to produce vacuum cleaners, and you, as an operations manager, are responsible for balancing the line. The work elements to be performed are listed, along with their items and immediate predecessors.
 $$\begin{array} { | c | c | c | } 
\begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline \mathrm { A } & 60 & - - \
\mathrm { B } & 40 & \mathrm {~A} \
\mathrm { C } & 30 & \mathrm {~B} \
\mathrm { D } & 20 & \mathrm {~B} \
\mathrm { E } & 40 & \mathrm {~B} \
\mathrm {~F} & 60 & \mathrm { C } \
\mathrm { G } & 70 & \mathrm { D } \
\mathrm { H } & 50 & \mathrm {~F} , \mathrm { G } \
\mathrm { I } & 20 & \mathrm { E } \
\mathrm { J } & 60 & \mathrm { H } , \mathrm { I } \
\hline
\end{array}$$
-Use the information in Table 9.17.The company is planning to operate two shifts per day,eight hours per shift.If the desired output rate of the line is 480 units per day,what is the cycle time?

Accepted Answer

To find the cycle time, we need to first calculate the total available production time per day:

2 shifts x 8 hours per shift = 16 hours

16 hours x 60 minutes per hour x 60 seconds per minute = 57,600 seconds per day

To achieve a desired output rate of 480 units per day, we can use the following formula:

Cycle Time = Total Production Time / Desired Output Rate

Cycle Time = 57,600 seconds per day / 480 units per day

Cycle Time = 120 seconds

Therefore, the cycle time is 120 seconds. Option (B) is the correct answer.

Question 19

Table 9.18
A company desires to set up a line to produce 60 units per hour.
The work elements and their precedence relationships are as follows.
 $$\begin{array} { | c | c | c | } 
\begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \text { Time (sec) } & \begin{array} { c } 
\text { Immediate } \
\text { Predecessor(s) }
\end{array} \
\hline 1 & 40 & - - \
2 & 30 & 1 \
3 & 50 & 1 \
4 & 40 & 2 \
5 & 6 & 2 \
6 & 25 & 3 \
7 & 15 & 3 \
8 & 20 & 4,5 \
9 & 18 & 6,7 \
10 & 30 & 8,9 \
\hline
\end{array}$$
-Use the information in Table 9.18.Suppose that a solution has been found to this problem with five stations.What is its efficiency?

Accepted Answer

The answer of Table 9.18
A company desires to set up...

Question 20

The production of a particular product consists of the following work elements.
$$\begin{array} { | c | c | } 
\hline \begin{array} { c } 
\text { Work } \
\text { Element }
\end{array} & \begin{array} { c } 
\text { Time } \
\text { (min) }
\end{array} \
\hline 1 & 2.4 \
2 & 0.5 \
3 & 2.1 \
4 & 2.0 \
5 & 2.7 \
6 & 1.1 \
7 & 2.0 \
8 & 2.7 \
9 & 1.6 \
10 & 1.4 \
\hline
\end{array}$$
If the cycle time is 4 minutes and the work-element times are as follows,what is the theoretical minimum number of stations?

Accepted Answer

The total time for all work elements is 18.5 minutes. Dividing this by the cycle time of 4 minutes gives 4.625, which means at least 5 stations are needed.

Quiz 9: Location and Layout

Fun Vehicles,Inc.makes "beach buggies" on an assembly line.The total productive time to make one buggy is 300 seconds.The current line has a 90-second cycle time and consists of four workstations.The balance delay of this line must be