Chapter 8, Problem 12PA

a

Summary Introduction

Interpretation: Tasks for each workstation, total time and idle time are to be calculated.

Concept Introduction:

Longest task time first rule:In this rule of line balancing, the task which has highest cycle time is assigned first to the workstation.

Explanation of Solution

Cycle Time:

It is the time between two consecutive outputs come out from a process.

  $\text{CT=}\frac{\text{A}}{\text{R}}$

Here CT = Cycle time; A = Available time to produce output; R = Demand Forecast

Assembly line:

It is a manufacturing process in which goods or services created in previous steps are joined together. Assembly line efficiency can be calculated by following formula:

  $\text{Assembly line efficiency=}{\displaystyle \sum \frac{\text{t}}{\text{N×CT}}}$

Here t = Task time; N= Actual no. of workstations; CT = cycle time

Idle time:

It is time for which machines or employees do not work due to the stoppage or work for any causes.

  ${\text{N}}_{\text{t}}\text{=}{\displaystyle \sum \frac{\text{t}}{\text{CT}}}$

Here N= Actual no. of work stations; CT= cycle time; t = task time

Assembly time per shift (A) = 480 min

No. of shifts each day = 1

Output forecasted per shift(R) = 60 units

  $\text{CT=}\frac{\text{A}}{\text{R}}$

  $=480/60$

  $=8\min$

The theoretical minimum no. of the workstation can be obtained as follows:

  ${\text{N}}_{\text{t}}\text{=}{\displaystyle \sum \frac{\text{t}}{\text{CT}}}$

Here CT= Cycle time; t= Sum of tasks time; $N_1$ = Minimum no. of work station required

  ${\displaystyle \sum \text{t=Task}\text{time}\text{1+Task}\text{time2+Task}\text{Time3+Task}\text{time4}}$

  $\text{+Task}\text{time5+Task}\text{time6+Task}\text{time7+Task}\text{time8}$

  $=4+5+1+2+6+3+3+8$

  $=32\min$

Theoretically, minimum no. of the workstation is:

  $N_t=\frac{32}{8}$

  $\text{=4}\text{workstation}$

Line balancing by longest task time first rule can be calculated as follows:

Lines balanced by the assignable task are determined by longest task time first rule. It has following steps:

Task 2 has highest processing time and no prerequisite is required therefore it is assigned on station A.

Although task 1 has highest processing time after task 2 in that case, total time exceeds the cycle time, therefore, task 4 and 3 is assigned on station A along with task 2 as no prerequisite is required for these tasks.

Task 1 is assigned to station B and there is a tie between task 6 and 7 as both task time is same.

Task5 is assigned to station C as its preceding task 1, 2, 3 has already assigned.

Either task 6 or 7 is assigned to station D.

Finally, task 8 is assigned to station E.

In actual situation no. of workstation are 5 in order to follow precedence and cycle time rule.

Station	Tasks	Total Time(min)	Idle Time(min)
A	2,4,3	8	0
B	1,6 (or 7 tie)	7	1
C	5	6	2
D	7 (or 6)	3	5
E	8	8	0
	Total	32	8

Total time at station A= Task Time of task 2+ Task Time of task 3+ Task Time of task 4

Total time at station A = 5+1+2

  = 8 min

Idle time at station A= Cycle time-Total tasks at station A

  = 8-8

  = 0 min

Total time at station B = Task Time of task 1+ Task Time of task (6 or 7)

Total time at station B = 4+3

  =7 min

Idle time at station B= Cycle time-Total tasks time at station B

  =8-7

  =1 min

Total time at station C = 6 min

Idle time at station C= Cycle time-Total tasks time at station c

  = 8-6

  = 2 min

Total time at station D= Task time of task 7 or 6

Total time at station D= 3 min

Idle time at station D= Cycle time-Total tasks time at station D

  = 8-3

  = 5 min

Total time at station E= Task Time of task D

Total time at station E= 8 min

Idle time at station E= Cycle time- Total tasks time at Station E

  = 8-8

  = 0 min

Total time $\left({\displaystyle \sum \text{t}}\right)\text{=Total}\text{time}\text{at}\text{station}\text{A+Total}\text{time}\text{at}\text{station B}\text{+}$

  $\text{Total}\text{time}\text{at}\text{station C+Total}\text{time}\text{at}\text{station D+}$

  $\text{Total}\text{time}\text{at}\text{station E}$

  $\text{Total}\text{time=8+7+6+3+8}$

  $=32\min$

Total idle time:

It can be calculated as follows:

  $\text{Total}\text{idle}\text{time=}\left(\text{N}\right)\text{×}\left(\text{CT}\right)\text{-}{\displaystyle \sum \text{t}}$

  $HereN=5;CT=8\min ;{\displaystyle \sum t=32\min }$

  $=5\times 8-32$

  $=40-32$

  $=8\min$

b

Summary Introduction

Interpretation:Assembly line efficiency is to be determined.

Concept Introduction:

Assembly line:It is a manufacturing process in which goods or services created in previous steps are joined together. Assembly line efficiency can be calculated by following formula:

  $\text{Assembly line efficiency=}{\displaystyle \sum \frac{\text{t}}{\text{N×CT}}}$

Here t = Task time; N= Actual no. of workstations; CT = cycle time

Explanation of Solution

Assembly line efficiency can be calculated as follows:

  $\text{Assembly line efficiency=}{\displaystyle \sum \frac{\text{t}}{\text{N×CT}}}$

  $Here{\displaystyle \sum t=32\min };N=5;CT=32\min$

  $\text{Assembly}\text{line}\text{efficiency=}\frac{\text{32}}{\text{(5×8)}}$

  $=32/40$

  $=.8$

  $=80.00\%$

c

Summary Introduction

Interpretation:Whether assembly line balance solution is good or bad and a criterion for assessment is to be explained.

Concept Introduction:

Idle time:It is time for which machines or employees do not work due to the stoppage or work for any causes.

Explanation of Solution

The assembly line balancing goodness is assessing its assembly line efficiency and idle time. In present case, although the assembly line efficiency is 80% the idle time is 8 min which is very high. It is highest for station D i.e. 5 min and at station B and C has 1 and 2 min idle time respectively. Therefore a fair chance that station B and C becomes bottleneck station for the assembly line. So it is not the very good balancing solution.