1. Re-Order Point Problems Set 3: Advanced

2. Problem 1: Problem 7.2 – Average Inventory With Isafety
Weekly demand for DVD-Rs at a retailer is normally distributed with a mean of 1,000 boxes and a standard deviation of 150. Currently, the store places orders via paper that is faxed to the supplier. Assume 50 working weeks in a year. Lead time for replenishment of an order is 4 weeks. Fixed cost (ordering and transportation) per order is $100. Each box of DVD-Rs costs $1. Annual holding cost is 25% of average inventory value. The retailer currently orders 20,000 DVD-Rs when stock on hand reaches 4,200.
R/week = N(1000,150)
L = 2 week
Q = 20,000
ROP = 4,200
50 weeks /yr
Ave. R /year = 50,000
S =100
C = 1
H=0.25C = $0.25
a.1) How long, on average, does a box of DVD-R spend in the store?
Icycle =Q/2 = 20000/2
Isafety = ROP – LTD
LTD = L×R = 4×1000 = 4000

3. Problem 1: Problem 7.2 – Average Inventory With Isafety
Average inventory = Icycle + Isafety
I = Average Inventory = 20000/ = 10200
RT = I =  1000T =  T = 10.2 weeks
R/week = N(1000,150)
L = 2 week
Q = 20,000
ROP = 4,200
50 weeks /yr
Ave. R /year = 50,000
S =100
C = 1
H=0.25C = $0.25
a.2) What is the annual ordering and holding cost.
Number of orders R/Q
R/Q = 50,000/20,000 = 2.5
Ordering cost = 100(2.5) = 250
Annual holding cost = 0.25% ×1 ×10200 =2550
Total inventory system cost excluding purchasing = = OC ≠ CC for two reason, (i) Not EOQ, (ii) Isafety

4. Problem 1: Problem 7.2 – Average Inventory With Isafety
b.1) Assuming that the retailer wants the probability of stocking out in a cycle to be no more than 5%, recommend an optimal inventory policy: Q and R policy.
=6325
Z(95%) = 1.65
Z(95%) = 1.65
Isafety = zσLTD =1.65(300) = 495
ROP = = 4495
Optimal Q and R Policy: Order 6325 whenever inventory on hand is 4495
b.2) Under your recommended policy, how long, on average, would a box of DVD-Rs spend in the store?

5. Problem 1: Problem 7.2 – Average Inventory With Isafety
Average inventory = Icycle+Isafety = Q/2 + Isafety
Average inventory = I = 6325/ = 3658
RT = I  1000T =  T = 3.66 weeks
c) Reduce lead time form 4 to 1. What is the impact on cost and flow time?
Safety stock reduces from 495 to 247.5 units reduction
That is 0.25(247.5) = $62 saving
Average inventory = Icycle + Isafey = Q/2 + Isafety
Average Inventory = I = (6,325/2) = 3,410.
Average time in store  I = RT
3410 = 1000T  T = 3.41 weeks

6. Problem 2: Problem 7.8 – Centralization
R/week in each warehouse follows Normal distribution with mean of 10,000, and Standard deviation of 2,000. Compute mean and standard deviation of weekly demand in all the warehouse together – considered as a single central warehouse.
Mean (central) = 4(10000) = 40,000 per week
Variance (central) = SUM(variance at each warehouse)
Variance (central) = 4(variance at each warehouse)
Variance at each warehouse = (2000)2 = 4,000,000
Variance (central) = 4(4,000,000) = 16,000,000
Standard deviation (central) =
=40000
R (central) = Normal(40,000, 4,000)

7. Problem 2: Problem 7.8 – Centralization
The replenishment lead time (L) = 1 week.
Standard deviation of demand during lead time in the centralized system
Safety stock at each store for 95% level of service
Isafety (central) = 1.65 x 4,000 = 6,600.
Average inventory in the centralized system
I = Icycle +Isafety
I = Q/2 +Isafety
40000/
=26600
Average Centralized Inventory
= 26600
Average Decentralized Inventory = 53200
Average time spend in inventory (Centralized): RT =I
4000T =  T = weeks

8. Problem 2: Problem 7.8 – Centralization
S(R/Q) = 1000(4)(500,000/40,000) = 50,000
H(Isafety + Icycle) = 2.5(26,000) = 66,500
Purchasing cost = CR = 10(500,000) = 5,000,000
We do not consider RC because it does not depend on the inventory policy. But you can always add it.
Inventory system cost for four warehouse in centralized system
116,500
Inventory system cost for four warehouses in decentralized system
233,000

9. Problem 3: Problem 7.3- Lead Time vs Purchase Price
Home and Garden (HG) chain of superstores imports decorative planters from Italy. Weekly demand for planters averages 1,500 with a standard deviation of 800. Each planter costs $10. HG incurs a holding cost of 25% per year to carry inventory. HG has an opportunity to set up a superstore in the Phoenix region. Each order shipped from Italy incurs a fixed transportation and delivery cost of $10,000. Consider 52 weeks in the year.
R/week = N(1500,800)
C = 10
h = 0.25
H =0.25(10) = 2.5
52 weeks /yr
R = 78000/yr
S =10000
L = 4 weeks
SL = 90%
a) Determine the optimal order quantity (EOQ).
=24980
b) If the delivery lead time is 4 weeks and HG wants to provide a cycle service level of 90%, how much safety stock should it carry?

10. Problem 3: Problem 7.3- Lead Time vs Purchase Price
c) Reduce L from 4 to 1, Increase C by 0.2 per unit. Yes or no?
c.1) Rough computations.
Purchasing cost increase  0.2(78000) = 15600
Safety stock decrease  from 2048 to 1024  1024 units reduction
Safety stock cost saving = 2.5(1024) = 2560
= increase in cost

11. Problem 3: Problem 7.3- Lead Time vs Purchase Price
c.1) Detailed computations.
R/week = N(1500,800)
C =
H = .25(10+.2) = 2.55
52 weeks /yr
R = 78000
S =10000
L = 4 weeks
SL = 90%
For the original case of C = 10 and H =2.5
+CR
+HIs
Change in C  Change in H. Not only purchasing cost changes
But also cost of EOQ and Is Changes
Reduce L from 4 to 1
Purchasing cost increase = 0.2(78000) = 15600
Safety stock reduces from to
Safety stock cost saving = 2.5(2048)-2.55(1024) = 2509

12. Problem 3: Problem 7.3- Lead Time vs Purchase Price
Total inventory cost (ordering + carrying) will also increase
R/week = N(1500,800)
C =
H = .25(10+.2) = 2.55
52 weeks /yr
R = 78000
S =10000
L = 4 weeks
SL = 90%
Total inventory cost (ordering + carrying) increase = *62450 = 621
Total impact = = 13712

13. Problem 3: Problem 7.3- Lead Time vs Purchase Price
# of warehouses = 4
R/week at each warehouse
N(10,000, 2,000)
50 weeks per year
C = 10
H=0.25(10) = 2.5 /year
S = 1000
L = 1 week
SL = 0.95
=20000
z(0.95) = 1.65, σLTD = 2,000
Is = 1.65 x 2,000 = 3,300.
ROP = LTD + Is =
1×10, ,300 = 13,300.
Average inventory at each warehouse :
Each time we order Q=20000  Icycle = Q/2 = 20000/2=10000
Average Inventory = Ic + Is =
=13300
Average Decentralized inventory in 4 warehouses =
I = 4(13300) = 53200
Demand in in 4 warehouses =
4(10000) =40000/w
RT = I  40000T= 53200
T = 1.33 weeks

14. Problem 3: Problem 7.3- Lead Time vs Purchase Price
OC = S(R/Q) = 1000[(50×10,000)]/Q = 25,000.
CC = H (Icycle + Isafety) = H(Q/2+Isafety) = 2.5 (13300) =32,250.
We do not consider RC because it does not depend on the inventory policy. But you can always add it.
RC = 10 [(50×10,000)]= 5,000,0000
Inventory system cost for one warehouse excluding purchasing
TC = 25, ,000 = 58,200
Inventory system cost for four warehouses = 4(58,250)
Inventory system cost for four warehouses = 233,000
Inventory system cost for four warehouses including purchasing
= 4(5,000,000) + 233,000 = 20,233,000