Your approach is correct. However I teach to my student that 1W ANOVA (it is very laborious) may indicate more effects than the nW ANOVA. In your problem: you have 6x5x4=120 treatments (each one with two replicates). You can use 1W ANOVA (of the 120 treatments) with Tukey paired comparison in each response. The paired comparisons are very much (the combinatory of 120 in 2) but MINITAB performs it and gives to you the report. The treatments difference significance will indicate to you the whole differences between factors and levels. First you can perform a 1W MANOVA with the four responses simultaneously in order to verify the significance obtained individually for each response.

 Prof.Miguel Angel,

Glad to recieve your valuable comments.

As per your direction, when I had applied 1Way MANOVA, it requires independent variable. In my case throughput, Workinprocess, Machine utilization, worker utilization, Average Waiting in Queue and Average Waiting, are out put responses from the simulation study. You have mentioned first MANOVA to be carried out with this four responses, but these are dependent variable. Which I have to select as a independent variable? Waiting for your valuable reply. Or I can perform direct that full facotiral design? If So how to get the inference from the results? 

Your responses are the dependent variables, and the independent variable represents your treatments. You have 6x5x4=120 treatments therefore your independent variable is MINITAB data column with 1,2,3,..,120 number repeated by each replicates. As example: treatment 1 may be F1 level 1 with F2 level 1 and F3 level 1 and your responses (of one of the responses) are y1,1 and y2,1 (by the replicates); treatment 2 is F1 level 1 with F2 level 1 and F3 level 2 and the responses are y1,2 and y2,2; and so on, the treatment 120 may be F1 level 6 with F2 level 5 and F3 level 4 and the responses are y1,120 y2,120. Then in MINITAB data you must write

Column

   A                       E   

y1,1                     1

y2,1                     1

y1,2                     2

y2,2                     2

   .                         .

   .                         .

   .                         .

y1,120             120

y2,120             120

The 1W ANOVA must be performed with A as dependent variables and E as independent variable. You can write the other 3 responses in columns B, C and D; and repeat the 1W MANOVA for each one of the four responses. The 1W MANOVA must be performed with A, B, C, and D as dependent variables and E as independent variable.

Prof.Miguel Angel,Sir,

As you mentioned I have applied the 1Way MANOVA, hereby attaching the results, please explain how to infer the result.Waiting for your valuable reply.

Sir, 

In Minitab full factorial, if the p value and f value comes zero what its mean.infer about the model.pls help.attaching the result

Dear Paramasivam

You have a problem. The attached ANOVA report indicates that you have only 1 response by treatment, that is, 119 degrees of freedom totals (120-1, 1 by the great average). Like your complete model requires 119 degrees of freedom there is not statistic, like you can observe in empty results for F and p values. You wrote in the original question “…Full factorial -with 2 replications, totally 240 experiments…” but you feed only 120 results (I guess the average of two replications). You must to feed the two replicates by treatment, MINITAB needs the 240 results. Like I wrote in my first answer: this 3W ANOVA is valid, but the multiple comparisons (when the analysis would be correctly done) will be very difficult. The 1W ANOVA that I recommended will be better but with the 240 results. I have another suggestion, you can use a complete second order model to perform the analysis:

Y= B0+B1x1+B2x2+B3x3+B11x1^2+B12x1x2+B13x1x3+B22x2^2+B23x2x3+B33x3^2

Dear Prof.Miguel Angel,Sir,

I have considered 2 replications and the minitab file I hereby attaching.

please suggest your valuable advice

Yours Sincerely

A.paramasivam

I was off-line during weekend. Your attached 3W-ANOVA report, show 119 as total degrees of freedom, which implies, at least in the classical ANOVA School, 120 results. Therefore, I do not know what´s happening, and I cannot read your attached file. However I can advise to you. Really, I do not like Minitab, because I like to program. Please, wait me a day or two and I will send to you a 1W ANOVA program ready to use for 120 treatments with 2 replicate by treatment and the Tukey comparison.

I attach the 1WANOVA program (1WANOVA.EXE), the data file (1WANOVA.DAT) a Tukey Table (Tukey.pdf), an auxiliary excel file (1WANOVA.XLSX) and the results files with an example (example1.res, example1.tuk) of 1W-ANOVA for a 6x5x4 factorial design with two replicates by treatment. The example results have the levels translated to English language, but when you run the program the results will be in Spanish language. I wrote this program 30 years ago, when my student only speak in Spanish, and the FORTRAN compiler used do not perform any more in the new versions of Windows. The *.EXE files performs in Windows or in a MS DOS environmental. The example was written in 1WANOVA.XLSX file. As you can see you must write your responses in the first column of DATA space; the second column indicate the treatment; and the third column indicate the factorial combination. 111 means factor 1 in level 1, factor 2 in level 1 and factor 3 in level 1 and so on. You only must to take the two first columns and copy paste to 1WANOVA.DAT file, to write the *.res file name and *.tuk file name, and performs the 1WANOVA.EXE. The Tukey value has taken from Tuket.pdf table at 0.95 and 120 degrees of freedom. Like the table has not values for 120 means, I must to extrapolate the Table values (the polynomial extrapolation is indicated in the 1WANOVA.XLSX file). The attached example ANOVA was reported in file example1.res, and the attached example Tukey comparison test is reported in file example1.tuk. Significance different between mean are identified when lower and upper limits of difference have the same sign. As example, in the attached example difference Mean1-Mean4 is significant (Tukey p