MSCI 432/633: Production and Service Operations Management, Winter 2010

MSCI 432.001/633: Assignment # 3; MSCI 432.002: Assignment # 2

Q1. Discounting schemes.

a. Find the optimal ordering quantities for each of the items and the corresponding annual total cost. [3 marks X3]

b. What is the primary difference between the ordering decisions made in a? Illustrate your answer graphically. [2 marks]

Q2. Limit on shelf life.

Simply find TEOQ=EOQ/� and if TEOQ>SL then QSL=SL*λ

Q3. Uncertain demand.

a. Assume that the firm has centralized all inventories in a single warehouse and that the probability of stocking out in a cycle can still be no more than 5%. Ideally, how much average inventory can the company now expect to hold, and at what cost? In this case, how long will a unit spend in the warehouse before being sold? [5 marks]

Solution: (Since pipeline inventory is not influenced, we do not include it in the calculations below.) (a) To determine the optimal order quantity, we will use the EOQ formula. Observe that H = $10 * 25%/yr = $2.5/yr, d = 10,000 /wk, (i.e., annual demand, D = 500,000/yr), and S = $1000. Optimal order quantity at each warehouse

Q = 5.2

1000000,50022 ××=H

DS

= 20,000. Demand is normally distributed with: Mean weekly demand = 10,000 units Std dev. = 2000 units Since replenishment lead time is 1 week: LT = 1 week And, the standard deviation of demand during lead time at each warehouse: σLT = √LT*σd = 2,000. For a 95% desired service level, z = 1.65. Safety stock (SS) at each warehouse for 95% level of service = z*σLT =1.65*2,000 = 3,300. Average lead time demand (σLT) = 10,000. Then, the reorder point (ROP) = σLT*z+ SS = 13,300. (Not necessary to calculate ROP in this question.) Average inventory at each warehouse (I) = Q/2 + SS = (20,000/2) + 3,300 = 13,300 units. Average inventory holding cost per warehouse = H*I = $2.5*13,300 = $33,250. Number of orders per year at each warehouse = D/Q = 500,000/20,000 = 25 order/year. Annual Order cost per warehouse = S*R/Q = $1,000 * 25 = $25,000. Average time unit spends in warehouse (by Little’s Law) = I/R = 13,300/10,000 = 1.33 weeks. Since each warehouse is identical, the total average inventory across four warehouses = 4* Average inventory in each warehouse = 4*(13,300) = 53,200. Annual order cost for all four warehouses = 4* Annual order cost per warehouse = 4* $25,000 = $100,000. Annual holding cost for all four warehouses = 4 * Average annual holding cost per warehouse = 4* $33,250 = $133,000. [b] The total average demand that the centralized warehouses faces = 4* Average demand per warehouse = 4 * 10,000 = 40,000 units per week. We assume that the centralized warehouse has the same cost structure as the individual warehouses. The optimal order quantity for the centralized warehouse is determined using the EOQ formula with d = 40,000 per week (i.e., Annual Demand D = 50*40,000 = 2,000,000) and the remaining parameters as in [a]. This gives Q = 40,000 units. Standard deviation of weekly demand at the central warehouse = SQRT(# of warehouses) * std dev of weekly demand at one warehouse = SQRT(4)* 2000 = 4,000 / week. With a replenishment lead time of 1 week, the standard deviation of lead time demand at central warehouse (σLT) = SQRT(LT)*( σd) = 4,000. Safety stock (SS) at central warehouse for 95% level of service

= z*σLT = 1.65*4,000 = 6,600. Average lead time demand (σLT) = 40,000 units. Reorder point (ROP) = σLT + SS = 46,600 units. (Not necessary to calculate ROP in this question.) Average inventory in central warehouse = Q/2 + SS = (40,000/2) + 6,600 = 26,600. Number of orders per year the central places = D/Q = 50*40,000/40,000 = 50 orders/year. Annual order cost for central warehouse = 50*$1,000 = $50,000. Annual holding cost for central warehouse = H*I = $2.5 * 26,600 = $66,500. Average time unit spends in warehouse = I/R = 26,600/40,000 = 0.67 weeks.

Q4 is a bonus question!

Q4. Order quantity under inflation.

Q5 and Q6 are required only from MSCI432.001/633 students

Q5 [Nahmias 4.20] EPQ.

(a) Calculate the number of JJ39877 filters that Filter Systems should produce in each production run of this particular part, so as to minimize annual holding and set-up costs? [2 marks]

(b) Assuming that it produces the optimal number of filters in each run, compute the maximum level of on-hand inventory of these filters that the firm has at any point in time. [1.5 marks]

(c) Assuming that the policy in part (1a) is used, calculate the percentage of the working time that the company spends producing these particular filters. [1.5 marks]

Solution a) h' = I*c(1-λ/P)=(.22)(2.50)[1 - (200*12)/(50*6*20*12)] = .5317 Setup cost, K = (100+55)(1.5)

Q* = 5317.

)2400)(5.232)(2( = 1,449

b) H = Q (1 - λ/P) = (1449)(1 - 2400/72,000) = 1,401. c) T = Q/λ = 1449/2400 = .60375 years T1 = Q/P = 1,449/72,000 = .0201 years .0201/.60375 = .033 or 3.3% of each cycle is up-time.

Q6.[based on Nahmias 4.22].

(a) Which source should be used, and what is the size of the optimal order?

(b) If the replenishment lead time for wafers is three months, determine the reorder point based on the on-hand inventory of wafers.

(c) Assume that two years have passed, and the purchasing agent is reconsidering the optimal number of wafers to purchase and from which source to purchase them. Source A still has the same pricing policy. Source B now accepts orders of any size, but sells the wafers for $2.55 each for orders of up to 3000 wafers, and $2.25 each for the incremental amount ordered over 3000 wafers. Source C went out of business. Which source should now be used, and what is the new size of the optimal order?

Solution: a) λ = 20,000 All Units Discount K = 100 I = 0.20 c0 = $2.50 c1 = $2.40 c2 = $2.30

Q(0) = )50.2)(20(.

)000,20)(100)(2(2

0

=Ic

Kλ = 2828(realizable)

Q(1) = )40.2)(2(.

)000,20)(100)(2(2

1

=Ic

Kλ = 2887(not realizable)

Q (2) = )30.2)(2(.

)000,20)(100)(2(2

2

=Ic

Kλ= 2949(not realizable)

Only Q (0) is realizable. Cost at Q = 4,000

(20,000)(2.30) + )000,4(

)000,20)(100(

2

)000,4)(30.2)(2(. +

= $47,420 Cost at Q = 3,000

(20,000)(2.40) + 000,3

)000,20)(100(

2

)000,3)(40.2)(2(. +

= $49,386.67 Cost at Q = Q(0) = 2828

(20,000)(2.50) + 0

2 ICKλ > 50,000

It follows that the optimal order size is Q = 4,000. Hence, source C should be selected.

b) T = Q/λ = 4000/20000 = 0.02 years; τ = 3 months = 0.25 years τ/T = 0.25/0.2 = 1.25 cycles > 1. Therefore, we use 1.25 – 1 = 0.25 cycle for the purpose of calculating reorder point R. 0.25 cycle = (0.25)(0.2) = 0.05 year

R = λτ = (20,000)(0.05) = 1000 units

c) Incremental schedule for source B.

For Q < 3,000, c (price/unit) charged by source B ($2.55) is greater than that charged by source A, it is, therefore, never optimal to use source B for Q < 3,000. Therefore, we need only compare source A and source B for Q > 3000. C(Q) = 2.50Q for Q ≤ 3,000 (Source A) C(Q) = (2.55)(3,000) + 2.25(Q - 3,000) for Q ≥ 3,000 = 7650 + 2.25Q - 6750 = 900 + 2.25Q for Q ≥ 3,000 C(Q)/Q = 2.55 for Q ≤ 3,000 = 900/Q + 2.25 for Q ≥ 3,000 (Source B) For source B for Q > 3000,

G(Q) = 2

25.2900

20.0000,20*100

25.2900

000,20Q

QQQ

+++

+

= 090,45$2

)25.2)(20(.000,000,2000,000,18 +++ Q

QQ

= 090,45$225.0000,000,20 ++ Q

Q

The minimizing Q occurs where d(C(Q))/d(Q) = 0

�Q* =225.

000,000,20 = 9428

Cost = 20,000,000

9428 + (.225) (9428) + $45,090

= 2121.34 + 2121.3 + 45,090 = $49,332.64 Note: There is an easier way to solve this by noting the structure of cost function, C(Q). C(Q) = 900 + 2.25Q for Q ≥ 3,000 This suggests that 900 is an additional setup cost under the new incremental discount price scheme. Therefore,

Q* = )25.2)(20(.

)000,20)(900100)(2('2 +=Ic

K λ = 9428

Since the annual cost for source A exceeds $50,000 (as shown in part a), now source B should be used and the new size of optimal order is 9428 units.