# MGT613 Assignmnet 1 Solution and Discussion

• PRODUCTION/OPERATIONS MANAGEMENT (MGT613) FALL 2019
Assignment No: 1
DUE DATE: 28th January, 2020 Learning Outcome:
Marks: 10
After attempting this activity student will be able to have an understanding of various types of numerical problems related to inventory management.
Assignment:
Javed manufacturers limited is a manufacturing firm that deals in the production of different sorts of toys in Sialkot city. The firm requires 24,000 wheels per year for the manufacturing of its newly launched Caravan truck series toys. The firm makes its own wheels for the manufacturing of the newly launched toy series at a rate of 600 per day. There is a uniform production of toy trucks over the entire year. The company is incurring a carrying cost of Rs. 3 per wheel a year. The setup cost for the production run of wheels is Rs. 2400. It is believed that the firm operates for 240 days per year.
Questions:
Considering the above information, you are required to compute the following for Javed Manufacturing Ltd.

1. What is Optimal Run Size? (Marks 3)
2. Compute the minimum total annual cost for carrying and setup (Marks 3)
3. What is cycle time for the optimal run size? (Marks 2)
4. Compute the Run Time? (Marks 2)

Important Note:
Provide proper FORMULAS and CALCULATIONS. Failure to this, will lead to deduction of Marks.
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• Solution idea
D=Demand= 24000
S=Ordering/Setup cost= 2400
H=Holding/Carrying Cost= 3
U= Usage rate=24000/240= 100
P= Production rate= 600

1. Optimal Run Size:
Q0= √2DS/H√p/(p-u)
Q0= √(2240002400)/3√600/(600-100)
Q0= √115200000/3√600/500
Q0= √38400000* √1.20
Q0= 6196.77*1.0954
Q0= 6787.94 ≅6788

2. Minimum total annual cost for carrying and setup cost.
= Carrying Cost + Set up Cost
=( I max/2)H+ ( D/Q0)S
Where,
I max= Q0/p ((p-u))
I max= 6788/600 (600-100)
I max= 11.31 (500)
I max= 5655
TC min= Carrying Cost+ Setup Cost
TC min= (I max/2) H+ (D/Q0) S
TC min= (5655/2) 3 + (24000/6788) 2400
TC min= (2827.5) (3) + (3.53) (2400)
TC min= 8482.5 + 8472
TC min= 16954.5

3. Cycle time for the Optimal Run Size.
Q0/U=6788/100
= 67.88 ≅ 68 days

4. Run time
Q0/p=6788/600
= 11.31 days

• MGT613 assignment solution required

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