Set 3 Advanced Weekly demand for DVDRs at a retailer is normally distributed with a mean of 1000 boxes and a standard deviation of 150 Currently the store places orders via paper that is faxed to the supplier Assume ID: 158404
Download Presentation The PPT/PDF document "Re-Order Point Problems" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Re-Order Point Problems Set 3: AdvancedSlide2
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.
Problem 1: Problem 7.2 – Average Inventory With Isafety
R/week
= N(1000,150)L = 2 week Q = 20,000ROP = 4,200 50 weeks /yrAve. R /year = 50,000S =100C = 1H=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 =
4000Slide3
Problem 1: Problem 7.2
– Average Inventory With
Isafety
Isafety
= 4200 - 4000 = 200
Average inventory = Icycle + IsafetyI = Average Inventory = 20000/2 + 200 = 10200RT = I = 1000T = 10200 T = 10.2 weeks
R/week
= N(1000,150)
L = 2 week Q = 20,000ROP = 4,200 50 weeks /yrAve. R /year = 50,000S =100C = 1H=0.25C = $0.25a.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 = 250+2550
=
2800. OC ≠ CC for two reason,
(i) Not EOQ, (ii)
IsafetySlide4
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
Problem 1: Problem 7.2
– Average Inventory With Isafety
Z(95%) = 1.65
Z(95%) = 1.65
Isafety
= z
σ
LTD
=1.65(300
) =
495
ROP = 4000+195 = 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
?Slide5
Average
inventory =
Icycle+Isafety
= Q/2 + IsafetyAverage inventory = I = 6325/2 + 495 = 3658 RT = I 1000T = 3658 T = 3.66 weeksc) Reduce lead time form 4 to 1. What is the impact on cost and flow time?
Problem 1: Problem 7.2 – Average Inventory With Isafety
Safety stock reduces from
495
to
247.5
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)
+ 247.5 =
3,410.
Average time in store
I
= RT
3410 =
1000T
T
= 3.41 weeksSlide6
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 = 4,000,000Variance (central) = 4(4,000,000) = 16,000,000Problem 2: Problem 7.8 – Centralization
=
40000
Standard deviation (central) =
R
(central) = Normal(
40,000
,
4,000
)Slide7
The replenishment lead time (L) =
1 week
.
Standard deviation of demand during lead time in the centralized system
Problem 2: Problem 7.8
– CentralizationSafety stock at each store for 95% level of service Isafety (central) = 1.65 x 4,000 = 6,600.Average inventory in the centralized system
=26600
= 26600
I = Icycle +Isafety
40000/2 +6600
Average Centralized Inventory
I = Q/2 +
Isafety
Average Decentralized Inventory = 53200
Average time spend in
inventory
(
Centralized):
RT =I
4000T = 26600
T
=
0.67
weeks Slide8
S(R/Q) = 1000(4)(500,000/40,000) = 50,000
H(
Isafety
+
Icycle) = 2.5(26,000) = 66,500Purchasing cost = CR = 10(500,000) = 5,000,000We 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 system116,500Inventory system cost for four warehouses in decentralized system233,000Problem 2: Problem 7.8 –
CentralizationSlide9
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
.
Problem 3: Problem 7.3-
Lead Time vs Purchase PriceR/week = N(1500,800)C = 10
h = 0.25 H =0.25(10) = 2.552 weeks /yrR = 78000/
yrS =10000L = 4 weeksSL = 90%
a) Determine the optimal order quantity (EOQ).=24980b) 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?Slide10
c) Reduce L from 4 to 1, Increase C by 0.2 per unit. Yes or no?
Safety stock decrease
from 2048 to 1024
1024 units reduction
Safety stock cost saving = 2.5(1024) = 2560
15600-2560 = 13040 increase in cost
Purchasing cost increase
0.2(78000
) = 15600
c.1) Rough computations.
Problem 3: Problem
7.3- Lead Time vs Purchase PriceSlide11
R/week
= N(1500,800)
C = 10 +
0.2
H = .25(10+.2) = 2.5552 weeks /yrR = 78000S =10000L = 4 weeksSL = 90%
+
CR
+HIsFor the original case of C = 10 and H =2.5
c.1) Detailed computations.
Change in
C
Change in H. Not only purchasing cost
changes
But also cost of EOQ and I
s
Changes
Reduce L from 4 to 1
Purchasing cost increase = 0.2(78000) = 15600
Safety stock reduces from 2048@2.5 to 1024@2.55
Safety stock cost saving = 2.5(2048)-2.55(1024) = 2509
Problem 3: Problem
7.3- Lead Time vs Purchase PriceSlide12
R/week
= N(1500,800)
C = 10 +
0.2
H = .25(10+.2) = 2.5552 weeks /yrR = 78000S =10000L = 4 weeksSL = 90%
Total inventory cost (ordering + carrying) increase = 0.00995*62450 = 621
Total impact = +621+15600-2509 = 13712
Total inventory cost (ordering + carrying) will also increase
Problem 3: Problem
7.3- Lead Time vs Purchase PriceSlide13
=20000
#
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 /yearS = 1000 L = 1 weekSL = 0.95z(0.95) = 1.65, σLTD = 2,000
ROP = LTD +
Is
=
1×10,000
+ 3,300 = 13,300.
I = 4(13300
) =
53200
Average inventory at each warehouse :
+
Is
=10000+ 3300
=
13300
Average Decentralized inventory in 4 warehouses
=
Demand in in 4 warehouses
=
4(10000)
=40000/w
RT = I
40000T= 53200
Is =
1.65 x 2,000 = 3,300.
T = 1.33 weeks
Each time we order Q=20000
Icycle
= Q/2 = 20000/2=10000
Average Inventory
=
Ic
Problem 3: Problem
7.3- Lead Time vs Purchase PriceSlide14
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)]= 5,000,0000Inventory system cost for one warehouse excluding purchasingTC = 25,000 + 32,000 = 58,200Inventory 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
Problem 3: Problem 7.3- Lead Time vs Purchase Price