Optimizing Daily Flower Inventory with Poisson Demand: A Python Simulation Approach

Inventory Management Problem

Problem

A vendor purchases fresh flowers at a cost of a yuan per bunch and sells them at b yuan per bunch. Unsold flowers at the end of the day are lost. Daily customer demand follows a Poisson distribution with parameter λ.

The question is: how many bunches should the vendor order each day to maximize expected profit?

Modeling

This is a stochastic decision problem. Let the daily order quantity be u . If the realized demand D is less than or equal to u , profit is (b‑a)·D . If demand exceeds u , profit is (b‑a)·u – a·(D‑u) . The expected daily profit is therefore a function of u and the Poisson distribution of D .

The optimal order quantity u* satisfies the condition that the marginal expected profit of ordering one more bunch becomes non‑positive, which can be expressed using the Poisson cumulative distribution function.

Solution

By consulting a Poisson table or using Python, the optimal order quantity u* can be computed.

Code

<code>from scipy.stats import poisson
a=2; b=3; lamda=10; p=1-a/b
u=poisson.ppf(1-a/b, lamda)  # optimal order quantity
p1=poisson.cdf(u-1, lamda)    # verification values
p2=poisson.cdf(u, lamda)
print(u, p1, p, p2)
</code>

A Monte‑Carlo simulation can further verify the optimal decision. The algorithm iterates over candidate order quantities, generates many Poisson‑distributed demand samples, computes the average profit for each candidate, and selects the order quantity that yields the highest average profit.

<code>import numpy as np
a=2; b=3; lamda=10; M1=0;
u=1; n=10000;
for i in range(1, 2*lamda):
    d = np.random.poisson(lamda, n)  # generate n demand samples
    M2 = np.mean(((b-a)*u*(u<=d) + ((b-a)*d - a*(u-d))*(u> d)))  # average profit
    if M2 > M1:
        M1 = M2
        u = u + 1
    else:
        print('Optimal order quantity:', u-1)
        break
</code>

Reference

司守奎，孙玺菁. Python数学实验与建模 .