The following tasks are to be performed on an assembly line: The workday is 8 ho

Question

The following tasks are to be performed on an assembly line:

The workday is 8 hours long. Demand for completed product is 800 per day.

a. Find the cycle time required to meet the desired output rate. (Round your answer to 1 decimal place.)

Cycle time             seconds per unit

b. What is the theoretical number of workstations required to meet the desired output rate? (Round up your answer to the next whole number.)

Number of workstations            

d. Balance the line using the longest-operating-time rule. (Leave no blank - be certain to enter "0" wherever required. Round your answers to 1 decimal place.)

e. What is the efficiency of the line balanced as in part (d) assuming it is running at the cycle time determined in part (a)? (Round your answer to 1 decimal place.)

Efficiency             %

f. Suppose that demand rose from 800 to 850 units per day. What would you do? (Round your answer for cycle time down to the nearest whole number.)

(Reduce or Increase) cycle time to ___ seconds, which may require rebalancing the line.

g. Suppose that demand rose from 800 to 1,400 units per day. What would you do? (Round your answer for cycle time down to the nearest whole number.)

(Increase or Reduce) cycle time to ___ seconds, which may require rebalancing the line.

TASK SECONDS TASKS THAT
MUST PRECEDE A 28 — B 14 A C 20 B D 19 B E 15 C F 12 D G 25 E, F H 10 G

Explanation / Answer

a) Cycle time = Time available/Demand = 8*60*60/800 = 36 seconds

b) Number of workstation = Total task time/Cycle time = (28+14+20+19+15+12+25+10)/36 = 143/36 = 3.97. Hence 4

d)

WS 1: A. As time left is 36-28 = 8 sec, no other task can be assigned. Idle time is 8 sec

WS 2: Assigning B & C. Idle time = 36-20-14 = 2 sec

WS 3: Assigning D & E. Idle time = 36 - 19 - 15 = 2 sec

WS 4: Assigning F. Idle time = 36 - 12 = 24 sec ( G cannot be assigned due to higher time and H because it depends on G)

WS 5: Assigning G & H. Idle time = 36-25-10 = 1 sec

e) Efficiency = Total task time/(Number of workstation*Cycle time)*100 = 143/(36*5)*100 = 79.4%

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