C. A teacher (it is always good to consult an adult regarding rumors or school wrong-doings)
Answer:
PLAN A:
(120 * 0.39) + (40 * 0.19) + 20 = $74.40
PLAN B:
(120 * 0.49) + (40 * 0.14) + 20 = $84.40
PLAN C:
$20 + $75 = $95 ;
PLAN A is optimal from 0 to 192 minutes
PLAN C is optimal from 192 minutes onward ;
Explanation:
PLAN A :
Service charge = $20
Daytime = $0.39 per minute
Evening = $0.19 per minute
PLAN B :
Service charge = $20
Daytime = $0.49 per minute
Evening = $0.14 per minute
PLAN C :
Service charge = $20
225 minutes = $75
Minutes beyond 225 = $0.36 per minute
A.)
Determine the total charge under each plan for this case: 120 minutes of day calls and 40 minutes of evening calls in a month.
PLAN A:
(120 * 0.39) + (40 * 0.19) + 20 = $74.40
PLAN B:
(120 * 0.49) + (40 * 0.14) + 20 = $84.40
PLAN C:
$20 + $75 = $95
b. If the agent will use the service for daytime calls, over what range of call minutes will each plan be optimal?
PLAN A:
20 + 0.39D = 95
0.39D = 95 - 20
D = 75 / 0.39
D = 192.31
Answer:
Kanban container size = 73
Number of kanbans needed = 5
Explanation:
Kanban container size (Q):
Q = SQRT [(2 x D x S) / H x (1 - d/p)]
where,
D = Annual demand
S = Setup cost
H = Holding cost
d = Daily usage
p = Daily production
Putting the given values in the above formula,
CONTAINER SIZE = SQRT ((2 * ANNUAL DEMAND * SETUP COST) / (HOLDING COST * (1 - (DAILY USAGE / DAILY PRODUCTION))))
Q = SQRT [(2 x 4,000 x $30) / $125 x (1 - 16/25)]
Kanbans container size = 73 units (Rounding off to the nearest whole number)
NUMBER OF KANBANS = DEMAND DURING LEAD TIME + SAFETY STOCK / SIZE OF CONTAINER
K = ((16 * 16) + (4 * 25) / 73 = 5
