An assembly line with 17 tasks is to be balanced. The longest task is 2.4 minutes, and the total time for all tasks is 18 minutes. The line will operate for 450 minutes per day. Show all relevant calculations. a. What are the minimum and maximum cycle times? b. What range of output is theoretically possible for the line? c. What is the minimum number of workstations needed if the maximum output rate is to be sought? d. What cycle time will provide an output rate of 125 units per day? e. What output potential will result if the cycle time is (1) 9 minutes? (2) 15 minutes?

a. What are the minimum and maximum cycle times?

OT = 450 minutes

a. Minimum cycle time = length of longest task, which is 2.4 minutes.

Maximum cycle time = Σ task times = 18 minutes.

b. What range of output is theoretically possible for the line?

Range of output:

2.4 min: = 187.5 units

18 min: = = 25 units

c. What is the minimum number of workstations needed if the maximum output rate is to be sought?

N = DxΣt / OT = 187.5(18) / 450 = 7.5, rounded off to 8.

d. What cycle time will provide an output rate of 125 units per day?

Output = OT/ CT,

Solving for CT, CT = 450 / 125 = 3.6 minutes per cycle

e. What output potential will result if the cycle time is (1) 9 minutes? (2) 15 minutes?

Potential output:

CT = 9 min: OT / CT = 450 / 9 = 50 units.

CT = 15 min: OT / CT = 450 / 15 = 30 units.

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

what do u need to know

step-by-step explanation: