Operations

Solution to HW problems – Inventory Control

6.

a. = 1975.23 1975

From Standard normal distribution, z = 1.64

ROP = ( )*LT + z * ( ) * ( u)

= 515(1) + (1.64)25 = 556

b. Holding cost = = $3,258.75

Ordering cost = = $3,259.49

c. Holding cost = = $3,300.00

Ordering cost = = $3,218.75

Total annual cost with discount is $6,518.75 – 50(25750/2000) = $5,875.00, without discount it is $6,518.24. Therefore, the savings would be $643.24 for the year.

8. Just do part (a)

= 1224.74 1225 units

ROP = ( )*LT + z ( ) ( u)

= (10000/52)(4) + 55 = 824.23 824 units

15.

a. = 408.25 408 bottles

b. From Standard normal distribution, z = 1.64

ROP = ( )*LT + z ( ) ( u)

= 100(3) + (1.64 * (sqrt(3) * 30)) = 300.00 + 85.28 = 385.28 385 bottles

16.

a. = 77.46 77 sets

b. From Standard normal distribution, z = 2.05

ROP = ( )*LT + z ( ) ( u)

= (2400/365)(7) + (2.05)* sqrt(7)* 4 = 46.03 + 21.70 =67.73

21.

= 216.02 216 mufflers

From Standard normal distribution, z = 1.28

ROP = ( )*LT + z ( ) ( u)

= (3500/300)(2) + (1.28)* sqrt(2) * 6 = 23.33 + 10.87 = 34.20 34 sets

Order 216 sets when the on-hand inventory level reaches 34 sets.

24.

= 31.62 32 refrigerators

10 refrigerators

From Standard normal distribution, z = 1.88

ROP = ( )*LT + z ( ) ( u)

= (500/365)(7) + (1.88) 10 = 9.59 + 18.8 = 28.39 28 refrigerators

Order 32 refrigerators when the on-hand inventory level reaches 28 refrigerators.

NOTE: In this problem, you are given the standard deviation of demand during lead time which is

( ) ( u)

25.

= 75.59 76 tires

From Standard normal distribution, z = 2.05

ROP = ( )*LT + z ( ) ( u)

= (1000/365)(4) + (2.05) * sqrt(4)*3 = 10.96 + 12.3 = 23 tires

Order 76 tires when the on-hand inventory level reaches 23 tires.

31.

Quantity range Cost (C) EOQ Feasible

Less than 2500 pounds $0.82 per pound 4277 pounds No

2500 to 4999 pounds $0.81 per pound 4303 pounds Yes...