56:166 Production Systems

Homework No. 3

Posted: T, Sept 13, 2005

Due: T, Sept20, 2005

The data in Table 1 through Table 4 has been provided to a facility engineer. Two alternative layout designs are to be considered, single and double-row layout. An assumption has been made that all six machines will be served by an AGV.

Table 1. Frequency matrix.

[fij] / 1 / 2 / 3 / 4 / 5 / 6
1 / 0 / 40 / 30 / 80 / 20 / 40
2 / 0 / 70 / 6 / 70 / 15
3 / 0 / 20 / 50 / 30
4 / 0 / 20 / 50
5 / 0 / 40
6 / 0

Table 2. Clearance matrix.

[cij] / 1 / 2 / 3 / 4 / 5 / 6
1 / 0 / 1 / 2 / 2 / 2 / 1
2 / 0 / 1 / 1 / 2 / 1
3 / 0 / 1 / 1 / 1
4 / 0 / 1 / 1
5 / 0 / 1
6 / 0

Table 3. Matrix of machine constraints.

[Sij] / 1 / 2 / 3 / 4 / 5 / 6
1 / - / O / O / X / O / O
2 / - / A / O / O / A
3 / - / O / O / O
4 / - / O / O
5 / - / O
6 / -

X: Not desirable

O: Ordinary closeness (no limit on closeness)

A: Absolutely necessary

Table 4. Clearance matrix.

Machine No. / Dimension
li / wi
1 / 4 / 1
2 / 6 / 2
3 / 4 / 2
4 / 5 / 4
5 / 6 / 2
6 / 5 / 2
  1. Determine the best single-row machine layout and draw the layout.
  2. Determine the best double-row machine layout and draw the layout knowing that the width of the isle is 2 units.
  3. What is the impact of constraints in Table 3onthe solution cost of layouts in (1) and (2) above, if any?
  4. Which of two layouts in (1) and (2) would you recommend and why?

Solutions:

1. Step 0, from the Matrix of machine constraints, we find that machine 2 and 3 have to be put together; machine 2 and 6 have to be put together. Thus we can put these 3 machines together at first.

Step 1, from the frequencymatrix, we will find the machine with highest frequency with machine 3 or machine 6, which is max (fk6, fl3), k, l=1, 4, 5. Thus f46=f53=50. Thus there are 4 possible solutions.

The corresponding cost for this solution is:

f53*c53+f52*(c53+c23)+f56*(c53+c23+c26)=50*1+70*(1+1)+40*(1+1+1)=310

The corresponding cost for this solution is:

f56*c56+f52*(c56+c26)+f53*(c56+c26+c23)=40*1+70*(1+1)+50*(1+1+1)=330

The corresponding cost for this solution is:

f46*c46+f42*(c46+c26)+f43*(c46+c26+c23)=50*1+6*(1+1)+20*(1+1+1)=122

The corresponding cost for this solution is:

f43*c43+f42*(c43+c23)+f46*(c43+c23+c26)=20*1+6*(1+1)+50*(1+1+1)=182

So the best solution is:

Step 2, from the frequencymatrix, we will find the machine with highest frequency with machine 3 or machine 4, which is max (fk4, fl3), k, l=1, 5. Thus f41 =80 or f53=50. Also note that machine 1 and machine 4 can not be put together. Thus there are only 3 solutions:

The corresponding cost for this solution is:

f13*c13+f12*(c13+c23)+f16*(c13+c23+c26)+f14*(c13+c23+c26+c64)=30*2+40*(2+1)+40*(2+1+1)+80*(2+1+1+1)=740

The corresponding cost for this solution is:

f53*c53+f52*(c53+c23)+f56*(c53+c23+c26)+f54*(c53+c23+c26+c64)=50*1+70*(1+1)+40*(1+1+1)+20*(1+1+1+1)=390

The corresponding cost for this solution is:

f54*c54+f56*(c54+c46)+f52*(c54+c46+c26)+f53*(c54+c46+c26+c23)=20*1+40*(1+1)+70*(1+1+1)+50*(1+1+1+1)=510

Thus the best solution is:

So the final layout is

The layout of the machines is below:

2. Suppose the aisle width for the AGV is 2 units

Step 0, from the Matrix of machine constraints, we find that machine 2 and 3 have to be put together; machine 2 and 6 have to be put together. Also note that f26=15 < f23=70, we should put machine 2 and 3 on the opposite sites of each other along the AGV path. We get the initial solution:

Step 1, from the frequency matrix, we find max(f3l, f6k), k, l = 1, 4, 5, which is f35=f64=50; thus machine 5 is selected, and consider 2 possible solutions and calculate the corresponding costs:

The corresponding cost is calculated based on the frequency and the distance among the machines: f62*d62 + f63*d63 + f53*d53 +f52*d52=15*(3+1+2.5)+30*(3+1+2.5)+50*(2+1+3)+70*(2+1+3)=1012.5

The corresponding cost is calculated based on the frequency and the distance among the machines: f52*d52 + f53*d53 + f63*d63 +f62*d62=70*(3+2+3)+50*(3+2+3)+30*(2+1+2.5)+15*(2+1+2.5)=1207.5

So choose the best solution:

Step 2, from the frequency matrix, find the max(f2i, f3j, f6k, f5l), where i, j, k, l = 1, 4. Thus we find f64=50 is the max value. So we choose machine 4 and the layout is:

Since machine 1 can not be placed near the machine 4, so there are only 2 possible solutions:

The corresponding cost is: f12*d12+f16*d16+f14*d14+f13*d13+f15*d15=40*(3+1+2)+40*(3+1+2+3+1+2.5)+

80*(3+1+2+3+1+2.5+2.5+1+2.5)+30*(3+1+2)+20*(3+1+2+2+1+3)=2640

The corresponding cost is: f12*d12+f16*d16+f14*d14+f13*d13+f15*d15=40*(2+2+2)+40*(2+2+2+3+1+2.5)+80*(2+2+2+3+1+2.5+2.5+1+2.5)+30*(2+2+2)+20*(2+2+2+2+1+3)=2640

So both of the above 2 layouts are OK.

3. The impact of the Table 3 will increase the solution cost. As we can see from the frequency table f14=80 is the max value across the table, thus we should put machine 1 and 4 together, and reduce the total flow. But since we can not put them together, the final solution will have higher cost. Also f26=15, which is a low frequency, but according to the constraint, machine 2 and 6 have to be put together, which eliminates the possible good solutions of inserting other machines between them and reduce the solution cost.

Similarly, for double-row layout, put machine 1 and 4 on the opposite sides of each other will reduce the solution cost, but due to the constraint, they can not be put that way.

4. Obviously, the double-row layout will be chosen because it has lower total solution cost. Double-row layout is more compact and can reduce the total energy cost of AGV running across the machines.