Chapter 11, Problem 2PA

Anvils Works’ requires, on average, 2800 tons of aluminum each week, with a standard deviation of 1000 tons. The lead time to receive its orders is 10 weeks. The holding cost for one ton of aluminum for one week is $11. It operates with a 0.98 in-stock probability.

1. a. On average, how many tons does it have on order? [LO11-5]
2. b. On average, how many tons does it have on hand? [LO11-5]
3. c. If its average inventory was 5000 tons, what would be its average holding cost per week? [LO11-5]
4. d. If its average inventory was 10,000 tons, what would be its average holding cost per ton of aluminum? [LO11-5]
5. e. Suppose its on-hand inventory is 4975 tons, on average. What instock probability does it offer to its customers? [LO11-5]

Summary Introduction

To determine: The number of tons that are on order.

Explanation of Solution

Given information:

Weekly Demand (D)                         = 2,800 tons

Standard deviation (S)                     = 1,000 tons

Lead time (L)                             = 10 weeks

Holding cost per one ton of aluminum per week is (H)     = $11

In-stock probability (P)                     = 0.98

Number of tons on order:

$\begin{array}{c}\text{Number of tons on order}=L\times D\\ =10\times 2,800\\ =28,000\text{ tons}\end{array}$

The number of tons on order is 28,000 tons.

Summary Introduction

To determine: The number of tons that are on hand.

Explanation of Solution

Given information:

Weekly Demand (D)                         = 2,800 tons

Standard deviation (S)                     = 1,000 tons

Lead time (L)                             = 10 weeks

Holding cost per one ton of aluminum per week is (H)     = $11

In-stock probability (P)                     = 0.98

Number of tons on hand:

For an in-stock probability of 0.98, the Z-value is 2.06

$\begin{array}{c}\text{Number of tons on hand}=\left(\sqrt{\left(L+1\right)}\right)\times \left(S\times Z\right)\\ =\left(\sqrt{\left(10+1\right)}\right)\times \left(1,000\times 2.06\right)\\ =3.316\times 2,060\\ =6,830.96\\ =6,831\text{ tons}\end{array}$

The number of tons on hand is 6,831 tons.

Summary Introduction

To determine: The average holding cost per week.

Explanation of Solution

Given information:

Weekly Demand (D)                         = 2,800 tons

Standard deviation (S)                     = 1,000 tons

Lead time (L)                             = 10 weeks

Holding cost per one ton of aluminum per week is (H)     = $11

In-stock probability (P)                     = 0.98

Average inventory (I)                         = 5,000 tons

Average holding cost per week:

$\begin{array}{c}\text{Average holding cost per week}=H\times I\\ =11\times 5,000\\ =\$55,000\end{array}$

The average holding cost per week is $55,000.

Summary Introduction

To determine: The average holding cost per ton of aluminum.

Explanation of Solution

Given information:

Weekly Demand (D)                         = 2,800 tons

Standard deviation (S)                     = 1,000 tons

Lead time (L)                             = 10 weeks

Holding cost per one ton of aluminum per week is (H)     = $11

In-stock probability (P)                     = 0.98

Average inventory (I)                         = 10,000 tons

Average holding cost per ton of aluminum:

$\begin{array}{c}\text{Average holding cost per ton of aluminum}=\frac{H\times I}{D}\\ =\frac{11\times 10,000}{2,800}\\ =\frac{1,10,000}{2,800}\\ =39.285\\ =\$39.29\end{array}$

The average holding cost per ton of aluminum is $39.29.

Summary Introduction

To determine: The average holding cost per ton of aluminum.

Explanation of Solution

Given information:

Weekly Demand (D)                         = 2,800 tons

Standard deviation (S)                     = 1,000 tons

Lead time (L)                             = 10 weeks

Holding cost per one ton of aluminum per week is (H)     = $11

In-stock probability (P)                     = 0.98

Average on-hand inventory (I)                 = 4,975 tons

Calculation of in-stock probability:

$\begin{array}{c}\text{Number of tons on hand}=\left(\sqrt{\left(L+1\right)}\right)\times \left(S\times Z\right)\\ 4,975=\left(\sqrt{\left(10+1\right)}\right)\times \left(1,000\times Z\right)\\ Z=\frac{4,975}{\left(\sqrt{\left(10+1\right)}\right)\times 1,000}\\ =\frac{4,975}{3.316\times 1,000}\\ =\frac{4,975}{3,316}\\ Z=1.50\end{array}$

The Z-value is 1.50. From the table, the probability for a Z value of 1.50 is 0.93319

Therefore, the in-stock probability is 93.32%.