Linear Progamming Problem

[BrettD]

A new sporting company is purchasing a new assembly system for producing basketballs. However, in 3 years a new upgraded system will be installed, therefore the assembly system being purchased will not be needed beyond 3 years. Due to heavy usage, the assembly system may need to be replaced after 1 or 2 years. The following table gives the total cost associated with purchasing an assembly system at the end of year 'i' and trading it in at the end of year 'j' (where year 0 is now)

	j​
	1​	2​	3
0 ​	$8,000	$18,000	$31,000
1​		10,000	21,000
2​			12,000

Formulate an LP to Min. Costs.


I am struggling to determine the constraints for this problem. There's no capacity or demands listed, can anyone help?

 

[Ishuda]

A new sporting company is purchasing a new assembly system for producing basketballs. However, in 3 years a new upgraded system will be installed, therefore the assembly system being purchased will not be needed beyond 3 years. Due to heavy usage, the assembly system may need to be replaced after 1 or 2 years. The following table gives the total cost associated with purchasing an assembly system at the end of year 'i' and trading it in at the end of year 'j' (where year 0 is now)

	j​
	1​	2​	3​
i​	0​	$8,000	$18,000	$31,000
1​		$10,000	$21,000
2​			$12,000

Formulate an LP to Min. Costs.


I am struggling to determine the constraints for this problem. There's no capacity or demands listed, can anyone help?

This is harder to write the constraints that to figure out the answer. Basically you want to minimize the sum from i=1 to n of a_ji,ki where j₁=0; k_i>j_i; j_i=k_i-1, n>i>1 [note: this may be the empty set]; and k_n=3 over all possible combinations. However, I'm not sure right now how to put it otherwise.

BTW: I assume your matrix is
row 1: (8, 16, 31); a_0j, j= 1, 2, 3
row 2 = (10²³, 10, 21); a_1j, j= 1, 2, 3
row 3 = (10²³, 10²³, 12); a_2j, j= 1, 2, 3
where amounts are in thousands of dollars and the 10²³ indicates not possible.

 

[BrettD]

Sorry, I edited the original post as the table originally posted went weird after I posted it.

For the LP here is what I have written and I'm curious if that would clarify whether I'm close or way off.

Min Z 8000x0,1 + 1800x0,2 + 31,000x0,3 + 10,000x1,2 + 21,000x1,3 + 12,000x2,3

S.t.

x0,1 + x0,2 + x0,3 <=1
x1,1 + x1,2 + x1,3 <=1
x2,1 + x2,2 + x2,3 <=1
x0,1 + x1,1 + x2,1 =1
x0,2 + x1,2 + x2,2 =1
x0,3 + x1,3 + x2,3 =1

xi,jE{0,1}Vi,j


E=Greek E
V=Upside down A (Greek)

 

[Ishuda]

Sorry, I edited the original post as the table originally posted went weird after I posted it.

For the LP here is what I have written and I'm curious if that would clarify whether I'm close or way off.

Min Z 8000x0,1 + 1800x0,2 + 31,000x0,3 + 10,000x1,2 + 21,000x1,3 + 12,000x2,3

S.t.

x0,1 + x0,2 + x0,3 <=1
x1,1 + x1,2 + x1,3 <=1
x2,1 + x2,2 + x2,3 <=1
x0,1 + x1,1 + x2,1 =1
x0,2 + x1,2 + x2,2 =1
x0,3 + x1,3 + x2,3 =1

xi,jE{0,1}Vi,j


E=Greek E
V=Upside down A (Greek)

I'm sorry but I don't understand. If we let a(i,j); i=0,1,2; j=1,2,3 stand for the element in the matrix, then the possibilities for the costs are
1 year each: a(0,1)+a(1,2)+a(2,3)
1 year, then two years: a(0,1) + a(1,3)
2 years, then 1 year: a(0,2)+a(2,3)
3 years: a(0,3)
Notice that the first i is always 0 [we need a system immediately]; the i of a value is the j of the last value [we can't have a gap where we have no system, again there may not have been a previous value], the j is always greater than the i [we will keep a system for at least one year], and finally the last j is 3 [the new system will be installed in three years].

So it looks like if we keep the first system for two years and the second system for one year [2 years, then 1 year] it will be the cheapest.