Question

Establishing a Product Line. National Metals Company (NMC) manufactures titanium shafts. Its equipment is capable of producing shafts in 10 lengths (in cm) as shown in the chart below, reflecting settings on its machinery. Setting up the machinery to produce one of these results costs $250. As a result, NMC has decided to make only a selected number of lengths. When a customer requests a given length, NMC may supply it from stock, if it happens to match one of the lengths in the production schedule. Otherwise, NMC trims a longer length to meet the order. The variable cost for producing the shafts is $20 per cm, and NMC receives revenue of $40 per cm. Trim waste can be sold to a recycler for $15 per cm. The demand requirements for the coming week are tabulated as follows; all demand must be satisfied.

LENGTH	32	34	36	38	40	42	44	46	48	50
DEMAND	12	4	7	8	16	7	12	5	8	3

A. WHAT IS THE OPTIMAL ASSORTMENT OF LENGTHS FOR NMC TO MANUFACTURE?

B. WHAT IS THE OPTIMAL PROFIT IN THE COMING WEEK?

THIS IS A LINEAR PROGRAMMING QUESTION. SO PLEASE EXPLAIN WITH OBJECTIVE FUNCTION, DECISION VARIABLES, CONSTRAINTS EQUATIONS AND ALSO WITH EXCEL SPREADSHEET

Answer 1

Let us consider x1,x2,x3,x4,x5,x6,x7,x8,x9 x10 be the quantities of shafts of lengths 32,34,36,38,40,42,44,46,48,50 respectively

As per the information in question set up cost as $250, Variable cost is $20 per cm and receives revenue of $40 per cm

In case of Trimming the trim waste can be sold for $15 per cm.

The objective is to find the maximum profit based on above data keeping in view the demands for each length.

Gross profit per shaft for different sizes shafts are worked out as follows:

Prod\Sold	l32	l34	l36	l38	l40	l42	l44	l46	l48	l50
l32	640
l34	630	680
l36	620	670	720
l38	610	660	710	760
l40	600	650	700	750	800
l42	590	640	690	740	790	840
l44	580	630	680	730	780	830	880
l46	570	620	670	720	770	820	870	920
l48	560	610	660	710	760	810	860	910	960
l50	550	600	650	700	750	800	850	900	950	1000

As per above workings length of 50 cm has only one option, therefore we need to have setup for this length to satisfy its demand.

All other lengths can be trim out of this. Next is 48 which has two options either to be trimmed from 50 or produce exactly of same size. .

In case of trimming there is savings of setup cost of $250 but loss of gross margin of $10 per unit giving total of $80 for 8 units so it is better to continue with 50.

To meet the demand for 46, we need to think of trimming from 50 and or 48 or going for exact size

Now the loss is $20 per unit in case of trimming from 50 size but saving setup cost of $250 justified as the number of units demanded are only 5.

For size of 44 loss is $30 per unit and units demanded are 12, therefore in this case better to have setup for the size 44.

Similarly let us compare costs of setting up for size 42 or get it by trimming 44. Per unit loss is 10 and demand is only seven so continue with size 44.

Next for size 40 the loss is 20 per unit and number of units demanded are 16, therefore go for new setup of size 40.

By above workings we found that there is need to have new setups for sizes of 34,36 and 38.

In case of 32 length we need new setup.

Now we have the following production plan

Length 32     Number of units 12

Length 40     Number of units 35   for 34, 36, 38 and 40

Length 44   Number of units 19 for 42 & 44

Length 50    Number of units 16 for 46, 48 and 50

Optimal Profit = 3*1000 +8*950 + 5*900 + 12*880 + 7*870 + 16*800 + 8*750 + 7*700 + 4*650+ 12*640 - 4*250 =65450-1000

=$64,450