OPERATIONS MANAGEMENT LL W/CONNECT CODE
OPERATIONS MANAGEMENT LL W/CONNECT CODE
2nd Edition
ISBN: 9781266520037
Author: CACHON
Publisher: MCG CUSTOM
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Textbook Question
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Chapter 11, Problem 6PA

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]

a)

Expert Solution
Check Mark
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:

Number of tons on order=L×D=10×2,800=28,000 tons

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

b)

Expert Solution
Check Mark
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

Number of tons on hand=((L+1))×(S×Z)=((10+1))×(1,000×2.06)=3.316×2,060=6,830.96=6,831 tons

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

c)

Expert Solution
Check Mark
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:

Average holding cost per week=H×I=11×5,000=$55,000

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

d)

Expert Solution
Check Mark
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:

Average holding cost per ton of aluminum=H×ID=11×10,0002,800=1,10,0002,800=39.285=$39.29

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

e)

Expert Solution
Check Mark
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:

Number of tons on hand=((L+1))×(S×Z)4,975=((10+1))×(1,000×Z)Z=4,975((10+1))×1,000=4,9753.316×1,000=4,9753,316Z=1.50

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%.

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