Question

of 1 component spare is $2000. If spare is not available during breakdowns, the cost of 0 3 procuring component is $ 3000. Company has estimated that the number of spares required during the past 6 months follows the probability distribution. Number of Spares required: Probability 0.2 0.3 0.4 0. Prepare and execute an extensive analysis to determine the number of spares to be stocked if the objective is to minimize expected cost

Show all formulas and calculations

Answer 1

Let Q be the number of spares that the company stocks and S be the number of spares that it needs

S=0,1,2,3 is required with the probabilities P(S=0)=0.2, P(S=1)=0.3, P(S=2)=0.4, and P(S=3) = 0.1

the cost of stocking 1 spare is $2000. The cost of stocking Q spares is $2000Q

The cost of spares when the quantity stocked Q is greater than or equal to S, the quantity of spares needed is given as )all the required quantity is already stocked at $2000 per piece)

C=2000Q

The cost of spares when the quantity stocked Q is less than S, the quantity of spares needed is given as (not all quantity of required spare are procured up front. Q have been procured up front at $2000 each, but the rest (S-P) have to be procured at a higher cost of $3000 per piece)

C= 2000Q + (S-Q)*3000

So the cost of spares is

$C(Q,S)=\begin{cases}2000Q,&Q\ge S\\ 2000Q&plus;3000(S-Q),&Q<S \end{cases}$

The expected cost for a given quantity of spares stocked is given by

$E(C(Q))=\sum_{s=0}^3C(Q,s)\times P(S=s)$

We will calculate this expected cost for different quantities Q=0,1,2,3 of spares stocked

The expected cost when Q=0, that is no spares are stocked

$\begin{align*} EX[C(Q=0)]&=\sum_{s=0}^3C(0,s)\times P(S=s)\\ &=C(0,0)\times P(S=0)&plus;C(0,1)\times P(S=1)&plus;C(0,2)\times P(S=2)&plus;C(0,3)\times P(S=3)\\ &=2000\times0\times0.2&plus;(2000\times0&plus;(1-0)\times3000)\times0.3&plus;(2000\times0&plus;(2-0)\times3000)\times0.4&plus;(2000\times0&plus;(3-0)\times3000)\times0.1\\ &=\$4200 \end{align*}$

The expected cost when Q=1, that is 1 spare is stocked

$\begin{align*} EX[C(Q=1)]&=\sum_{s=0}^3C(1,s)\times P(S=s)\\ &=C(1,0)\times P(S=0)&plus;C(1,1)\times P(S=1)&plus;C(1,2)\times P(S=2)&plus;C(1,3)\times P(S=3)\\ &=2000\times1\times0.2&plus;(2000\times1)\times0.3&plus;(2000\times1&plus;(2-1)\times3000)\times0.4&plus;(2000\times1&plus;(3-1)\times3000)\times0.1\\ &=\$3800 \end{align*}$

The expected cost when Q=2, that is 2 spares are stocked

$\begin{align*} EX[C(Q=2)]&=\sum_{s=0}^3C(2,s)\times P(S=s)\\ &=C(2,0)\times P(S=0)&plus;C(2,1)\times P(S=1)&plus;C(2,2)\times P(S=2)&plus;C(2,3)\times P(S=3)\\ &=2000\times2\times0.2&plus;(2000\times2)\times0.3&plus;(2000\times2)\times0.4&plus;(2000\times2&plus;(3-2)\times3000)\times0.1\\ &=\$4300 \end{align*}$

The expected cost when Q=3, that is 3 spares are stocked

$\begin{align*} EX[C(Q=3)]&=\sum_{s=0}^3C(3,s)\times P(S=s)\\ &=C(3,0)\times P(S=0)&plus;C(3,1)\times P(S=1)&plus;C(3,2)\times P(S=2)&plus;C(3,3)\times P(S=3)\\ &=2000\times3\times0.2&plus;(2000\times3)\times0.3&plus;(2000\times3)\times0.4&plus;(2000\times3)\times0.1\\ &=\$6000 \end{align*}$

We can see from these expected costs that the expected cost when stocking Q=1 is the lowest at $3800

Hence stocking 1 spare would have the minimum expected cost.