Rap star Xzibit has signed an endorsement deal with athletic footwear manufacturer Dada to produce shoes. The line named "Pimp Y
1 answer:
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Answer:
A)Minimum = 0.2 minutes
Maximum = 4.6 minutes
Cycle time = = 2 minutes per unit
B) Number of Stations = 3
D) the percentage of idle time for the assignment in decimal = 0.43
Explanation:
A) Operating output = 240 units per 8hours.
Converting to per hour, we get; 240/8 = 30 units per hour.
Minimum cycle time = longest task time
Maximum cycle time = sum of task times
Minimum = 0.2 minutes
Maximum = 0.2 + 0.4 + 0.2 + 0.4 + 1.2 + 1.2 + 1.0 = 4.6 minutes
Now, operational time(OT) is 8 hours a day. Converting that to minutes, we have =60x8 = 480 minutes
Cycle time(CT) = OT (per day) /desired output
Therefore, calculated cycle time = 480/240
= 2 minutes per unit
B) Number of stations = Maximum time / CT
= 4.6/2 = 2.3 which is approximately 3 stations
C) The task diagram is drawn in image 1 attached. Therefore, we can assign tasks based in the greatest number of following tasks as follows;
Task Number of following tasks
a 4
b 3
c 2
d 2
e 1
f 1
g 0
D) for the idle time from the assigned tasks in "c" above, from the "image 2" attached; the percentage of the idle time will be=
Total remaining time of tasks without remaining feasibility tasks/ {(number of these tasks) x(CT)}
Which gives percentage of idle time = (0.8 + 0.8 + 0.8 + 1.0)/{(4)x(2)}
= 3.4/8 = 42.5%
Since we are to convert to decimal,
= 0.425 approximately 0.43
Answer:
optimal order quantity is 11450
Explanation:
solution
we will use here news vendor model here
and profit means that if we print less we will incur a certain opportunity loss
so it is known as cost of understocking (Cu) i.e = 35
and
production cost = $5
when product is not sold
we incur cost of (Co) i.e = 5
so
fixed cost is incurred either way in a year
so we need to consider the critical fractile value
that is express as
CF =
CF =
CF =
CF = 0.875
so value of Z at 0.875 is 1.15
so
means the optimal order per production will be here = 8000 + 1.15 ×3000 = 11450
so
optimal order quantity is 11450