Line Boloncing Use this information for questions 15-18 An assembly operation ha

ID: 350796 • Letter: L

Question

Line Boloncing Use this information for questions 15-18 An assembly operation has a product layout with 7.5 hours per day to meet planned demand of 900 units per day. Here are the tasks needed to make the product: TASK PREDECESSOR. - TMES econds) CESSOR TIME (seconds) product 16 14 25 b,e 15 Considering the information given above, what is the shortest feasible cycle time, regardless of demand? 15 A 5 B 15 C 25 D 30 E 75 16 What is the minimum number of workstations needed to balance the line? A 1 station B. 2 stations C. 3 stations D. 4 stations E 5 stations Using the best eyeball method, balance the line. With the minimum number of workstations what is the efficiency of this line? A, 70% 7 B. 86.50% 88% D. 90% 93%

Explanation / Answer

The precedence diagram is shown above.

15. The shortest feasible cycle time regardless of demand is equal to the length of the longest task in the critical path. Here the critical path is A-C-D-E-F.
Hence the cycle time = Duration of D = 25

Answer: C (25)

16.

The required cycle time = (Production time available)/(Number of units to be produced)

= (7.5 * 60 * 60)/900

= 30 secs

Theoretical minimum number of workstations needed = (Sum of total task time)/(Required Cycle time)
= (16+14+5+25+6+15)/30
= 2.7 i.e. 3 (rounded)

Answer: C (3)

17. Using Best eyeball method, we can have the following 3 workstations

Workstation Name

Tasks Grouped

Total Time

Workstation 1

A & B

Workstation 2

C & D

Workstation 3

E & F

Efficiency of the line = (Sum of all tasks)/[Cycle Time * Actual number of workstations]

= 81/(30*3) = 0.9 = 90%

Answer: D (90%)

18. From the above table, the last task in the second workstation is task D.

Answer: D (d)