School of Industrial Engineering & Management (IEM) Inventory Management

International University, VNU-HCM Instructor: Dr. Nguyen Van Hop

Homework 2
Problem 1: Input data of AAC in excel for the case of discount
 Change the range of quantity to the levels of 500 units
 Change the discount price to the levels of 0.5USD
 Observe the results and explain
Problem 2: Input data of SCANLON in excel and change the lead time to be longer, observe
the results and explain
Problem3: problem 5.4 page 189 Silver et.al. (1998) book
We have:
D = 4000 unit/yr
A = $5
v = $4 per 100 unit = $0.04 per unit
r = 0.25$/$/yr
a. The economic order quantity of the item is:
2 × D ×Co 2 × 4000 ×5
EOQ =
√ Ch
=
√ 0.04 ×0.25
= 2000
b. The time between consecutive replenishment of the item when the EOQ is used is”
Q∗¿ 2000
T= ¿= = 0.5 yr ( = 6 months)
D 4000
c. The EOQ at A = $5 í 2000.
4000 2000
TRC(2000) = A × +0.04 × 0.25 × = 2A + 10
2000 2
4000
Under a three-month supply rule is: Q = = 1000
4
4000 1000
TRC(1000) = A × +0.04 × 0.025× = 4A + 5
1000 2
We can see: 2A + 10 ≤ 4A + 5 where 2A ≥ 5 or A ≥ $2.5

Problem 4: problem 5.6 page 189 Silver et.al. (1998) book

We have:
v = $0.03
D = 200 cookies per day
Order quantity is: Q = 200 ×7 = 1400 cookies
a. Two missing parameters are cookies fixed cost (A) and interest rate (r). If Ernie is
implicitly following an EOQ policy, then:
2×D× A A Q2 × r 14002 × 0.03
Q=
√ v×r
=> =
r 2×D
=
2 ×200
= 147
We can see the implicit values that the ratio of ordering cost and holding cost is 147.
b. The cost savings from a special order for 10000 at the discount price is:
10000
F(10000) = √ 2 A × 200× 0.03 r +10000× 0.03− A−10000 ×0.02
200
−100002 × 0.02r
2× 200
With A = 147r, we have:

1
 10000
F(10000) = √ 2× 147 ×200 ×0.03 r +10000 ×0.03−147 r−¿
200
100002 ×0.02 r
10000 ×0.02− Let us set this expression equal
2 ×200
to 0 to find the interest rate r for which Ernie would be indifferent to the special offer:
100
0 = 2100r + 300 – 147r – 200 -5000r or r = = 0.0328 per day
3,047
 R = 11.972 year
Surely, Ernie’s interest rate is less than per year. And if the interest is less than this,
the savings from taking the special offer is positive. For example, for an interest rate
of 15% per year (0.041% per day), the saving from the special order is:
2100×0.00041 + 300 - 147×0.00041 – 200 - 5000×0.00041 = $98.75
So from a financial viewpoint, the special offer is worth taking. However, a purchase
of 10000 cookies represents a 50-day supply of cookies. The shelf life if the cookies
could be a problem here, i.e., perhaps 50-day-old cookies would be too stale for the
Cookie Monster.

Problem 5: problem 5.9 page 189 Silver et.al. (1998) book

We have:
D = 40 units/week = 2080 units/year
a. Applying this formula to calculate Q*:
2 AD
Q=
√ vr
Applying this formula to calculate Total Cost:
AD vr Q ¿
TC = ¿ + + CD
Q 2
Order size Unit Price ($) Q* Modified Q* Total Cost ($)
200.0
0<Q<300 10 0 200 21320.00
203.0
300<=Q 9.7 7 300 20727.63

Because of Total Cost (20727.63 < 21.320), the replenishment size should be 300
units
a) Applying this formula to calculate Total Cost when Q*= 500 units:
AD vrQ
TC = + +CD
Q 2
TCQ=500 = $ 20910.5
The TC when Q=300 and Q=500 with new price should be equal
The differential between TCQ=500 and TCQ=300:

2
 $20910.5 - $20727.63 = $182.87
The unit price should be decreased:
182.87
= 0.0879
2080
Therefore, the max price when Q=500 should be:
$9.7 - $0.0879 = $9.612

Problem 6: problem 5.22 page 192 Silver et.al. (1998) book

 Item 1

D = 10,000 (units/year)
v 0 = $ 5,00
A = $ 25
r = 0.3 $/$/year
2 AD
Q* =
√ v0 r
AD Qvr
TC = + + vD
Q 2
Order Size Price Discount Q* Modify Q* TC
0-999 $ 5.00 0% 577.3503 577.35 $ 50,866.03
1000-1999 $ 4.90 2% 583.2118 1000.00 $ 49,985.00
2000+ $ 4.75 3% 592.3489 2000.00 $ 49,050.00
Therefore, the appropriate order quantity of Item 1 to use is 2000. The minimum total
cost is $ 49,050.00
 Item 2
D = 1,000 (units/year)
v 0 = $ 5,00
A = $ 25
r = 0.3 $/$/year
2 AD
Q* =
√ v0 r
AD Qvr
TC = + + vD
Q 2
Order Size Price Discount Q* Modify Q* TC
0-999 $ 5.00 0% 182.5742 182.57419 $ 5,273.86
1000-1999 $ 4.90 2% 184.4278 1000 $ 5,660.00
2000+ $ 4.75 3% 187.3172 2000 $ 6,187.50
Therefore, the appropriate order quantity of Item 2 to use is 182.5742
The minimum total cost is $ 5,273.86
1 Item 3
D = 4,000 (units/year)
v 0 = $ 5,00
A = $ 25

3
 r = 0.3 $/$/year
2 AD
Q* =
√ v0 r
AD Qvr
TC = + + vD
Q 2
Order Size Price Discount Q* Modify Q* TC
0-999 $ 5.00 0% 365.1484 365.148372 $ 20,547.72
1000-1999 $ 4.90 2% 368.8556 1000 $ 20,435.00
2000+ $ 4.75 3% 374.6343 2000 $ 20,475.00
Therefore, the appropriate order quantity of Item 3 to use is 1000. The minimum total
cost is $ 20,435.00
2 Item 4
D = 130,000 (units/year)
v 0 = $ 5,00
A = $ 25
r = 0.3 $/$/year
2 AD
Q* =
√ v0 r
AD Qvr
TC = + + vD
Q 2
Order Size Price Discount Q* Modify Q* TC
0-999 $ 5.00 0% 2081.666 *** ***
1000-1999 $ 4.90 2% 2102.8 *** ***
2000+ $ 4.75 3% 2135.744 2135.7443 $ 620,543.44
Therefore, the appropriate order quantity of Item 4 to use is 2135.744
The minimum total cost is $620,543.44