In: Operations Management
A = Annual Production Requirement = 6000 cakes
O = Production Setup cost / Production order = $ 600
C = Holding cost / cake/ annum= $ 10
Optimal Production run Quantity = Q = Economic Production Quantity:
Q=2AOC
= √(2*6000*600)/10
=848.528
Q ≈ 849 cakes
Optimal number of production runs per year (P) = A/Q
P = 6000/849 = 7.06 ≈ 7 production runs per year
Total annual Holding Cost = (Q/2)*C = (849/2)*10 = $4245
Total Annual Setup cost = P*O = 7*600 = $4200
Total Annual Inventory cost = Total annual Holding Cost + Total Annual Setup cost
=4245 + 4200
Total Annual Inventory cost = $8445
Optimal cycle time (time between run starts)
Per day production capacity = 100 cakes
Therefore,
Run length in working days to produce one optimal production run (Q) total days required
= Q / Per day production capacity
= 849/100 = 8.49 days
Time taken to complete all production runs = 8.49 days * 7 production cycles = 59.43 days
Total weeks Ambrosia bakery operates = 50 weeks
Days per week the bakery operates = 6 days
Total no. of working days = 50*6 =300 days
Idle days = 300 – 59.43 = 240.57 days
1 |
Idle |
2 |
Idle |
3 |
Idle |
4 |
Idle |
5 |
Idle |
6 |
Idle |
7 |
Optimal cycle time (time between run starts) = Time between 2 production runs = 240.57 / 6 = 40 days
Therefore the answer can be summarized as: