The previous sections have assumed that the data required by a model are known exactly.

But in actual business life, you never know all the values with perfect certainty.

For a given item, the following factors are involved in the determination
of C_{1} and C

- Unit selling price (S)
- Unit purchase cost (C)
- Carrying cost for the entire period (C
h ) - Salvage value (V)
- Shortage penalty cost (C
s )

The unit costs of over-ordering and under-stocking are then

C

C2

A trader stocks a particular seasonal product at the beginning of the season and cannot reorder. The item costs him Rs. 25 each and he sells at Rs. 50 each. For any item that cannot be met on demand, the trader has estimated a goodwill cost of Rs. 15. Any item unsold will have a salvage value of Rs. 10. Holding cost during the period is estimated to be 10 percent of the price. The probability distribution of demand is given below.

Units stocked | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|

Probability of demand, p(D=Q) | 0.35 | 0.25 | 0.20 | 0.15 | 0.05 |

Determine the optimal number of items to be stocked.

Solution.

Given

S = Rs. 50, C = Rs. 25 , C_{h} = 0.10 X 25 = 2.5, V = Rs. 10,
C_{s} = 15.

The probability distribution of demand is given in the following table.

Units stocked | Probability of Demand p (D=Q) |
Cumulative probability P (D _{}≤Q) |
---|---|---|

2 | 0.35 | 0.35 |

3 | 0.25 | 0.60 |

4 | 0.20 | 0.80 |

5 | 0.15 | 0.95 |

6 | 0.05 | 1.00 |

C_{1} = 25 + 2.5 - 10 = 17.5

C_{2} = 50 - 25 - (2.5/2) + 15 = 38.75

The ratio,

C_{2}--------- C _{1} + C_{2} |
= | 38.75 ------------- 17.5 + 38.75 |
= | 0.69 |

In the above table, the ratio (0.69) lies between cumulative probabilities
of 0.60 and 0.80, which in turn reflect the values of Q as 3 and 4.
That is,

P(D ≤ 3) = 0.60 < 0.69 < 0.80
= P(D ≤ 4).

Therefore, the optimal number of units to stock is 4 units.

A newspaper boy buys papers for Rs. 0.35 each and sells them for Rs. 0.60 each. He can't return unsold newspapers. Daily demand has the following distribution:

No. of customers | 230 | 240 | 250 | 260 | 270 | 280 | 290 | 300 | 310 | 320 |
---|---|---|---|---|---|---|---|---|---|---|

Probability | 0.01 | 0.03 | 0.06 | 0.10 | 0.20 | 0.25 | 0.15 | 0.10 | 0.05 | 0.05 |

If each day's demand is independent of the previous day's demand, how many papers should he order each day?

Solution.

The probability distribution of demand is given in the following table.

No. of customers | 230 | 240 | 250 | 260 | 270 | 280 | 290 | 300 | 310 | 320 | |
---|---|---|---|---|---|---|---|---|---|---|---|

Probability | 0.01 | 0.03 | 0.06 | 0.10 | 0.20 | 0.25 | 0.15 | 0.10 | 0.05 | 0.05 | |

Cumulative Probability | 0.01 | 0.04 | 0.10 | 0.20 | 0.40 | 0.65 | 0.80 | 0.90 | 0.95 | 1.00 |

C

C

C_{2}--------- C _{1} + C_{2} |
= | 0.25 ------------- 0.35 + 0.25 |
= | 0.416 |

From the above table, we notice that the computed value of 0.416 lies between 0.40 and 0.65 corresponding to 270 and 280 customers respectively. Hence 280 being the higher value is the optimal no. of papers to be stocked by the newspaper boy.