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Calculus of Warehouses. Part 4

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Edited by Valentin Fadeev, Thursday, 17 June 2010, 22:02

I am posting it in reply to a question I came across in one social network. There is no original material here. I am just trying to reproduce the argument from one university lecture, as far as I can remember it.

The task is to calculate optimal stock level at a warehouse. Stock is replenished to a certain value (which is to be determined) immediately at fixed equally spaced moments in time. Amount of stock units cnosumed during each period is random.

Let s be the stock level which is maintained. Assume that stock is replenished over a fixed period of time (take it to be unitary to simplyfy calculations). Assume that r units are used during each time period with probability cap p of r .

Let cap c sub one be unitary storage cost, cap c sub two is a “penalty” cost
assigned when stock is theoretically below 0, or below some
minimal admissible level (lost orders, emergency orders, etc).

Assume also that the stock is replenished immediately.

Consider two possible situations.

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1) r less than or equals s , i.e there is a positive remainder in the warehouse at
the end of the period. Storage cost is proportional to the area
under the stock graph.

equation sequence cap e sub one super r of s equals cap c sub one times s plus open s minus r close divided by two equals cap c sub one times open s minus r divided by two close

Expectancy of this value is:

cap m of cap e sub one super r of s equals cap c sub one times cap p of r times open s minus r divided by two close

Therefore the average cost is:

cap e sub one of s equals cap c sub one times n ary summation from r equals zero to s over cap p of r times open s minus r divided by two close

2) r greater than s . In this case the cost of both types is incurred. Again using the geometrical approach we find that cost elements are proportional to the areas of the triangles. The same argument as above gives the answer:

cap e sub two of s equals cap c sub one times n ary summation from r equals s plus one to normal infinity over cap p of r times s squared divided by two times r plus cap c sub two times n ary summation from r equals s plus one to normal infinity over cap p of r times open r minus s close squared divided by two times r

Thus the total cost is given by the following expression:

cap e of s equals sum with, 3 , summands cap c sub one times n ary summation from r equals zero to s over cap p of r times open s minus r divided by two close plus cap c sub one times n ary summation from r equals s plus one to normal infinity over cap p of r times s squared divided by two times r plus cap c sub two times n ary summation from r equals s plus one to normal infinity over cap p of r times open r minus s close squared divided by two times r

Marginal analysis is used to determine the optimal stock level. Optimal value of s minimizing cap e satisfies the relation:

cap e of s less than cap c sub one divided by cap c sub one plus cap c sub two less than cap e times open s plus one close

(here my memory is abit vague, so I am giving it without proof and without guarantee)

Practically this means calculating cap e of s for a series of the values of s until the above double inequality holds.

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