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The Wod Chemical Com The compound can be produced at a rate of 10000 pounds per

ID: 370240 • Letter: T

Question

The Wod Chemical Com The compound can be produced at a rate of 10000 pounds per day. Annual demand for the compound is 0.6 million pounds per year. The fixed cost of setting up for a production run of the chemical is $1500. The variable cost of production is $3.5 per pound if the size of the production run is less than 40000lbs, $3 per pound if the size of the production run is greater than or equal to 40000lbs but less than 55000 lbs, and $2.50 per pound if the size of the production run is greater than or equal to 55000 lbs. The company uses an interest rate of 22 percent to account for the cost of capital, and the costs of storage and handling of the chemical amount to 12 percent of the value. Assume that there are 250 working days in a year. a) What is the optimal size of the production run for this particular compound? b) What proportion of each production cycle consists of uptime and what proportion consists of downtime? e) What is the average annual cost of holding and setup at the optimal size of the production run?

Explanation / Answer

What is the optimal size of the production run for this particular compound?

Answer (a)

Compound production = 10,000 pound/day

= 2,500,000 pound/year (250 working days)

Demand of compound = 600,000 pound/year

Setup cost = $ 1,500

Variable cost = $3.5 per pound

Annual charge: Cost of Capital = 22%

Cost of Storage = 12%

Total Interest charge = 34 %

Holding cost: Total interest charge * Variable cost

= 0.34*3.5

=1.19 pound per year

Because production rate is assume infinite and the company did not need to order or internally produce the compound, we need to modified the holding cost,

Modified the holding cost(H)= H (1- /P)

= 1/.19* (1-600000/2500000)

=1.19*(2500000-600000)/2500000

=1.19*(1900000/2500000)

=1.19*0.76

=0.9044

So, the optimal size of the production run could be calculate using EOQ,

Q* =square root of (2k )/ Modified the holding cost (H)

= square root of (2*1500*600000/0.9044)

= square root of (1800000000/0.9044)

= square root of(1990269792.127377)

= 44612.43 pound

What is the proportion of uptime and downtime?

Answer (b)

Cycle time, T= Q/ , 44612.43/600000=0.0743 year.

Uptime T1 =Q/P=44612.43/2500000=0.0178 year

Downtime T2 = T- T1

=0.0743-0.0178

=0.0565year

The proportion of uptime and downtime,

Uptime T1 / Downtime T2

=0.0178/0.0565

=23.95/76.05

=1/4

Answer C

Annual holding cost and setup cost can be written mathematically as:

G (Q*) =K. / Q*+ Modified the holding cost (H)* Q* /2

= (1500*600000)/44612.43+(0.9044*44612.43)/2

=20173.75+20173.75

=40347.48

If the compound sold $ 3.9 per pound, then the revenue from this sales,

Revenue = 3.9 *600000

Revenue =2340000

Capital budget

Compound cost, C= 3.5*600000

=2100000

Total capital cost for a year: = Annual holding cost and setup cost+ Compound cost

Total capital cost for a year: = 2100000+40347.48

Total capital cost for a year: =2140347.48

From this result as we see that Wod chemical company still get a profit from compound sales,

Profit =2340000-2140347.48

= $199652.52

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