Question

Annual demand for number 2 pencils at the campus store is normally distributed with mean 1,000 and standard deviation 250. The store purchases the pencils for 6 cents each and sells them for 20 cents each. There is a two-month lead time from the initiation to the receipt of an order. The store accountant estimates that the cost in employee time for performing the necessary paperwork to initiate and receive an order is $20, and recommends a 22 percent annual interest rate for determining holding cost. The cost of a stock-out is the cost of lost profit plus an additional 20 cents per pencil, which represents the cost of loss of goodwill.

a. Find the optimal value of the reorder point R assuming that the lot size used is the EOQ.

b. Find the simultaneous optimal values of Q and R.

c. Compare the average annual holding, setup, and stock-out costs of the policies determined in parts (a) and (b).

d. What is the safety stock for this item at the optimal solution?

Answer 1

Lot size-reorder point system is one of the multi period models. This system is denoted by decision variables (Q, R). This multi period models are implemented when there is uncertain demand in inventory control. However, in the simple EOQ model, demand is known and fixed. But when the demand is random, these lot size-reorder point (Q, R) systems allow random demand. There are two decision variables in a (Q, R) system:

• Order quantity, Q and • Reorder point, R The following costs are assumed in lot-size reordering:

• Setup cost per order =$K

• Holding cost at per unit held per year =$h

• Proportional order cost of per item = $c

• Stock-out cost per unit of unsatisfied demand =$p

lead time in a year = 1 •mean lead - time demand = u = ht .standard deviation of annual demand = 0, •standard deviation of leadtime-timedemand = 0

Service levels in ( Q, R) systems

Type 1 service:

a. Determine R to satisfy the equation

F(z) = a; z =R-4

b. Set Q=EOQ Type 2 service:

Type 2 service:

n(R) 1-B n(R)= EOQ(1-B) L(Z)=(1-B)Qlo R=Oz+u B=F(R)

Consider the information given for the campus store:

Mean annual demand, 2 = 1000 Mean, u = 1000/6=166.67 Stock-outcostperunit, p=0.34 Setup cost perorder, K = 20 Annual interest rate, 1 = 0.22 Lead Time = 2 months ~ = year Cost of each pencil,c = 0.06 Holding cost, h=1xc=0.22 x 0.06 = 0.0132

a. Calculate the standard deviation of lead time, as shown below:

Standard deviation of lead time, o = 0,VT = 250, = 102.1

2.vn 2x20x1,000 0.0132 = 1,741

1- F(R)= 2h p2 1741 x 0.0132 1-F(R)= 0.34x1000 = 0.067

From this value of F(R), obtain z value from Appendix table z =1.5

R = σZ + μ = (102.1) (1.5)+166.67 = 319.8 - 320

Therefore, the optimal point of the reorder point R is 320.

b.
Calculate the iterative values of O and R.
We have obtained Iteration 1:

Q. = 1,741 R = 320 Obtain L(z) = 0.0293

Now Calculate:

n(R.)=OL(2) = (102.1)(0.0293) n(R) = 2.99

Iteration 2:

Calculate Q1 as shown below:

22[K + pn(R)] 2 AV | 2× 1,000 [ 20 + (0.34 × 2.99 ] 9 - 0.0132 Q = 1,785

1-F(R) - 20h pi 1,785 x 0.0132 1-F(R) = 0.34x1,000 = 0.069

From this value of F(R) obtain z value from Appendix table. The value of z =1.48.

R=oz+u R=(102.1)(1.48)+166.67 R=318 We obtain L(z) = 0.030 n(R)= L(2) = 102.1x0.030 = 3.063

Iteration 3:

22[K + pn(R)] 2 x 1,000/20+(0.343.06] Q2=V 0.0132 Q = 1,786

1-F(R)-gh р 1,786x0.0132 1-F(R) = 0.34x 1,000 = 0.069

As, both Q and R values are within one unit, terminate the iteration.
Therefore, (Q, R) = (1,786 , 318).
C- Consider the (Q, R) from policy (a) i.e. (Q, R) = (1 ,741 320). Calculate the annual holding and setup costs, as shown below:

Annual holding and set-up costs = h Anual holding and set up costs = n [+ (R– 4) * Kia M4 MB) = 0.0132 1,741 +(320–166.67)]20*1,000 1,741 0.34x1,000.00 1,741(2.99) = S(13.51+11.49+0.58) = $25.58

Therefore, the annual holding cost and set up cost are $25.58 for this policy.

Now, consider the (Q, R) = (11787, 318) policy. Calculate the annual holding and setup costs, as
shown below:

Annual holding and set-up costs = h Annual holding and set up costs = [ { + (8 + x)) + K JA MOR) = 0.0132[1,787+ (318–90))+ 2021,00 T 1,787 m 2 0.34x1,000 3063 1.787 (3.063) = $(14.80+11.19+0.58) = $26.57

Therefore, the annual holding cost and set up cost are $26.57 for this policy.

Therefore, the annual holding and set up costs for policy (Q, R) (1 ,741, 320) are lesser than that of policy (Q R) (1 ,787, 318).

d.
Safety stock for optimal solution is calculated as below:

Safety stock = R-u = 320-166.67 = 153.33

Therefore, the safety stock obtained is 153.33.