M451 Supply Chain Management- Tutorial 3 Solution
AssignmentTutorOnline
Dr Banafsheh Khosravi
Discrete and Continuous News Vendor models
1)
pp =1800 sp = 2500 s = 1700
Cu = 2500-1800 = 700 Cv = 1800-1700 = 100
P (D = 130) = 0.09
P (D ≤ 140) = 0.02 + 0.05 + 0.08 + 0.09 + 0.11 = 0.35
P (D > 140) = 1 – 0.35 = 0.65
P (D ≥ 140) = 1 – 0.35 + 0.11 = 0.76
2)
Profit = 5-3 = £2 Loss = 3-0.5 = £2.5
| Order quantity | Demand | Expected profit | |||||
| 10 | 20 | 30 | 40 | 50 | 60 | ||
| 10 | 2 | 2 | 6 | 6 | 2 | 2 | 20 |
| 20 | -0.5 | 4 | 12 | 12 | 4 | 4 | 35.5 |
| 30 | -3 | 1.5 | 18 | 18 | 6 | 6 | 46.5 |
| 40 | -5.5 | -1 | 10.5 | 24 | 8 | 8 | 44 |
| 50 | -8 | -3.5 | 3 | 16.5 | 10 | 10 | 28 |
| 60 | -10.5 | -6 | -4.5 | 9 | 7.5 | 12 | 7.5 |
| P(x) | 0.10 | 0.10 | 0.30 | 0.30 | 0.10 | 0.10 |
3)
Note: Find the closest probability values to 0.5883 in the standard normal distribution Table and do the interpolation which results in z’= 0.2231. Therefore, z = – 0.2231 as the complement probability was looked up in the table.
4)
Note: Find the closest probability values to 0.6 in the standard normal distribution Table and do the interpolation which results in z= 0.2533.
Note: Find the closest z values to 0.477 in the standard normal distribution Table and do the interpolation which results in P(Z ≤ 0.477) = 0.6833
