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Consider a company using the Power of Two policy to replenish the inventory of a

ID: 389187 • Letter: C

Question

Consider a company using the Power of Two policy to replenish the inventory of a subset of its SKUs. Which of the following statements is/are true?

A) The Power of Two policy assumes ordering costs are the same for all SKUs using the policy

B) All SKUs that are part of the policy will have replenishment periods of length 2k base periods, where k is some non-negative integer.

C) None of the SKUs that are part of the policy will have a replenishment period length, T, more than 6% from its optimal replenishment period length,T*.

D) None of the above.

Explanation / Answer

ANSWER IS B.

DESCRIPTION AS FOLLOWS :-

     T = nT           where T is the base planning period (similar like a shift, day, week) and n is an integer which can take only power of two values. For instances n can be 20, 21, 22, 23, 24, etc. (therefore as simplified notification let n=2^l)

Implementing a PO2 policy:

min Z (T)         = S/T   + kT                                         where T => 0

subject to

T = (2^l)T           l = {0, 1, 2, 3…}

Let an optimal re-order interval be equal to (2^l*)T. Then l* should satisfy the following inequality:

            Z (2l*T)                       = <       Z (2l*+1T)

This inequality can be re-written as follows,

S/(2^l*)T + k.2l*T = <      S/2^[(l*+1)]T + k. 2l*+1T

S/(2^l*)T (1- ½) = <      k.(2^l*) (2-1)

S/k                                           = <      2((2^l*)T) 2

S/2k                                       = < (2^l*)T

Accordingly l* is the smallest non-negative integer such

            2l* > = 1/ TS/2k = (1/2T) T

Consider the above example and assume T is 1 day. T is 1/365 of a year. From the solution T = 0.091333 years. Hence the optimal power – of – two multiple 2l* is the smallest non-negative integer that

            2^l* > = (1/2T) T

(1/2*(1/365))* 0.0913         = 33.3245/2 = 23.56

Therefore 2^l* = 32. The optimal PO2 re-order interval is 2^l*T= (32) (1/365) = 0.0876 years = 32 days.

By using T=Q/D, let’s find the Q

Q = T*D = (32/365)*1500      132 units

Total cost = (D * C) + (H * Q/2) + (S * D/Q)

= 1500* 2500 + (40*132/2) + (250*1500/132)

                        = 3,750,000 + 2640 + 2840.9

                        = 3,755,480.9 Rupees

Cost difference = 3,755,480.9 - 3,755,477.2   = 3.7 rupees.

On conclusion it is clear that using PO2 policy will not increase the total cost significantly whereas it provides a realistic answer for the resource planner to make the schedules.

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