Metal Containers Inc. is reviewing the way it submits bids on U.S. Army contract
ID: 2986684 • Letter: M
Question
Metal Containers Inc. is reviewing the way it submits bids on U.S. Army contracts. The Army often requests open-top boxes, with square bases and of specified volumes. The Army also specifies the materials for the boxes, and the base is usually made of a different material than the sides. The box is assembled by riveting a bracket at each of the eight corners. For Metal Containers, the total cost of producing a box is the sum of the cost of the materials for the box and the labor costs associated with affixing each bracket.
Instead of estimating each job separately, the company wants to develop an overall approach that will allow it to cost out proposals more easily. To accomplish this, company managers need you to devise a formula for the total cost of producing each box and determine the dimensions that allow a box of specified volume to be produced at minimum cost. Use the following notation to help you solve this problem.
Cost of the material for the base = A per square unit
Cost of the material for the sides = B per square unit
Cost of each bracket = C
Cost to affix each bracket = D
Length of the sides of the base = x
Height of the box = h
Volume specified by the army = V
1. Write an expression for the company%u2019s total cost in terms of these quantities.
2. At the time an order is received for boxes of a specified volume, the costs of the materials and labor will be fixed, and only the dimensions will vary. Find a formula for each dimension of the box so that the total cost is a minimum.
3. The Army requests bids on boxes of 48 cubic feet with base material costing the container company $12 per square foot and side material costing $8 per square foot. Each bracket costs $5, and the associated labor cost is $1 per bracket. Use your formulas to find the dimensions of the box that meet the Army%u2019s requirements at a minimum cost. What is this cost?
Metal Containers asks you to determine how best to order the brackets it uses on its boxes. You are able to obtain the following information: The company uses approximately 100,000 brackets a year, and the purchase price of each is $5. It buys the same number of brackets (say, n) each time it places an order with the supplier, and it costs $60 to process each order. Metal Containers also has additional costs associated with storing, insuring, and financing its inventory of brackets. These carrying costs amount to 15% of the average value of inventory annually. The brackets are used steadily, and deliveries are made just as inventory reaches zero so that inventory fluctuates between zero and n brackets.
4. If the total annual cost associated with the bracket supply is the sum of the annual purchasing cost and the annual carrying costs, what order size n would minimize the total cost?
5. In the general case of the bracket-ordering problem, the order size n that minimizes the total cost of the bracket supply is called the economic order quantity, or EOQ. Use the following notations to determine a general formula for the EOQ.
Fixed cost per order = F
Unit cost = C
Quantity purchased per year = P
Carrying cost (as a decimal rate) = r
Explanation / Answer
1.
Area of base = x^2
Area of 4 sides = 4*x*h
Cost of base = A*(x^2)
Cost of 4 sides = B*(4*x*h)
Cost of 8 brackets = 8*C
Cost of fixing 8 brackets = 8*D
Total cost of each box = A*x^2 + B*(4hx) + 8*(C+D).................................1
2.
Volume V = x^2 *h = constant
So, x^2 = V/h
x = sqrt(V/h).....................................................2
Therefore, Total cost of each box = A*x^2 + B*(4hx) + 8*(C+D)
= A*(V/h) + B*(4*sqrt(Vh)) + 8*(C+D)
Cost will be minimum when d Cost / dh = 0
Thus, -A*V/(h^2) + B*4*(sqrt V)*(1/2)/(sqrt h) = 0
-A*V / (h^2) + B*2*sqrt (V/h) = 0
A*V / (h^2) = B*2*sqrt (V/h)
h^(3/2) = [A/(2B)]*(sqrt V)
h = [A*sqrt (V) / (2B)]^(2/3)..................................................3
x = sqrt (V/h) = (sqrt V) / [A*sqrt (V) / (2B)]^(1/3).......................................................4
3.
V = 48 ft^3
A = $12 / ft^2
B = $8 / ft^2
C = $5
D = $1
Putting values in eqn 3 we get,
h = [12*sqrt (48) / (2*8)]^(2/3)
h = 3 ft
Putting values in eqn 2 we get,
x = sqrt (48 / 3) = 4 ft
Putting values in eqn 1 we get
Cost of each box is = 12*4^2 + 4*8*4*3 + 8*(5+1)
Cost of each box = $ 624
4.
No. of annual orders = 100000 / n
Ordering cost per year = 60*(100000 / n)
Average inventory = (0+n)/2 = n/2
Value of average inventory = 5*n/2
Carrying Cost of average inventory = 0.15*5*(n/2) = 0.375*n
Total cost = ordering cost + carrying cost
Total cost = 60*(100000 / n) + 0.375*n
For min. cost, d Cost / dn = 0
-6,000,000 / n^2 + 0.375 = 0
n = sqrt (6,000,000 / 0.375)
n = 4000
5.
No. of orders = P / EOQ
Ordering cost = F*(P / EOQ)
Average inventory = (0 + EOQ)/2 = EOQ / 2
Cost of average inventory = C*(EOQ/2)
Carrying cost = r*C*(EOQ/2)
Total cost = F*(P / EOQ) + r*C*(EOQ / 2)
For min. cost, d Cost / d(EOQ) = 0
-F*P / EOQ^2 + r*C/2 = 0
EOQ = sqrt [2*F*P / (r*C)]