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ASSIGNMENT-1

Chapter-2

19) Given data:

month	demand	month	demand
Jan	89	july	223
feb	57	aug	286
mar	144	sep	212
april	221	oct	275
may	177	nov	188
june	280	dec	312

Compute the MAD for the forecasts obtained in problems 17 and 18. Which method gave better results? Based on forecasting theory, which method should have given better results?

Sol: We know that Dt =m + et

Where m= constant and ET is random variable

For a one step ahead forecast, it is given as:

And MAD is devined as :

Now from the above formulae from march to july( One step a head fore cast)

Fmarch = F7=(144+221+177+280)/4 = 206

Similarly, Fapril = F8 = 225, F (may) = F9= 242 F(june) = f10= 250, F11= 249, f12= 240.

Error is defined as = Fn – Demand

Implies, for july= -17, aug= -61, sep=30, oct =-25, nov= 61, dec= -72

Sum of all these = E(t) = 266 negleting the negative sign.

In t he above MAD formulae, = 266/6 = 44.3…………………………………..1

Now for One step a head fore cast,

Calculating the Fn values from Feb to may

Gives,

F7 = 150, F8 = 206, F9 = 225, similarly 242, 250 and 249 are the remaining.

By using the E(t) it os calculated as= 326

MAD = 54.32 from the above formulae………………………………………..2

From 1 and 2 equations we observe that the one step a head forecast method is giving better results with low MAD when compared with the two step.

21) What would an MA (1) forecasting method mean? Compare the accuracy of MA (1) and MA (4) forecasts for July through December 2013.

Using the same formulae’s used in the question-1

I will need to calculate MA (1) and MA (2) using the above formulae.

Where MA(1) depicts that the demand of previous month is considered as demand of current month.

Lets calculate the four-month average of one step a head forecast from july to dec.

MA(1) of F7 =280

MA(1) of F8 =223 similarly

MA(1) of F9 = 286

MA(1) of F10 =212 similarly 275, 188.

Error = MA (1) – Demand

So, Error for july, aug, sep, oct, nov, dec, jan= 57, -63, 74, -63, 87, -124

Ignoring the negative value, Et = 468

Now from formulae of MAD, It is given as MAD = 468/6 = 78…………………………1

Lets Calculate for MAD (4):

The fore point average is for period 7 is obtained by averaging demand from march to june.

Lets calculate the four-month average of one step a head forecast from july to dec.

MA(4) of F7 =206

MA(4) of F8 =225 similarly

MA(4) of F9 = 242

MA(4) of F10 =250 similarly 249, 240

Error = MA (4) – Demand

So, Error for july, aug, sep, oct, nov, dec, jan= -17, -61, 30, -25, 61, -72

Ignoring the negative value, Et = 468

Now from formulae of MAD, It is given as MAD = 266/6 = 44 appx……………..2

When compared 1 and 2 , MA(4) is better than MA(1)

Chapter-3

24) Mr. Meadows Cookie Company makes a variety of chocolate chip cookies in the plant in Albion, Michigan. Based on orders received and forecasts of buying habits, it is estimated that the demand for the next four months is 850, 1260, 510 and 980, expressed in thousands of cookies. During a 46-day period when there were 120 workers, the company produced 1.7 million cookies. Assume that the number of workdays over the four months are respectively 26, 24, 20 and 16. There are currently 100 workers employed, and there is no starting inventory of cookies.

a) What is the minimum constant workforce required to meet demand over the next four months?

b) Assume that CI= 10 cents per cookie per month, CH =$100, and Cf = $200. Evaluate the cost of the plan derived in part (a).

Sol: Given values:

The demand for the next four months is 850, 1260, 510 and 980.

During a 46-day period when there were 120 workers, the company produced 1.7 million cookies.

Number of workdays over the four months are respectively 26, 24, 20 and 16

Number of workers employed currently = 100.

a) Minimum number of workers required for all the next months is calculated as:

(Cumulative forecast demand /cumulative units per worker at that month)

So,

For the first month = 850/8.0072 = 106.77

Where Cumulative unit per work is dderived from unit per work.

Workers cannot be in points so = 107

Similarly,

For the second month = 2110/7.39 = 137.65

Where Cumulative unit per work is derived from unit per work ( this value is added from the first month)

Workers cannot be in points so = 138

Now,

For the Third month = 2620/6.1592 =121.77

Where Cumulative unit per work is derived from unit per work ( this value is added from the first month +second month)

Workers cannot be in points so = 122

Finally,

For the fourth month = 3600/4.92753

Where Cumulative unit per work is derived from unit per work ( this value is added from the first month second month third)

Workers cannot be in points so = 136

24 (b) Now,

Assume that CI= 10 cents per cookie per month, CH =$100, and Cf = $200. Evaluate the cost of the plan derived in part (a).

The cost of the plan derived = CH *(Total number of hired workers) + CF *(Total number of fired workers) + CI *(Total number of Inventory)………………….(1)

Now,

Total number of hired workers = 38 (obtained from 100- max workers)

Similarly,

Total number of fired workers = 0

Total inventory = 680000 (ending inventory *1000)

Now ,

Substitute all the values in the equation 1,

We get,

The total cost of the plan = $71,800

32) A local firm manufactures children’s toys. The projected demand over the next four months for one particular model of toy robot is:

month	Work days	Forecasted demand(in agg units)
july	23	3,825
aug	16	7,245
sep	20	2770
oct	22	4440

Assume that a normal work day is eight hours. Hiring costs are $350 per worker and firing costs (including severe pay) are $850 per worker. Holding costs are $4.00 per aggregate unit held per month. Assume that it requires an average of 1 hour and 40 minutes for one worker to assemble one toy. Shortages are not permitted. Assume that the ending inventory for June was 600 of these toys and the manager wishes to have at least 800 units on hand at the end of October. Assume that the current workforce level is 35 workers. Find the optimal plan by formulating as a linear program.

Sol:

Given details:

Hiring costs are $350 per worker, firing costs (including severe pay) are $850 per worker

Holding costs are $4.00 per aggregate unit,

Assume that it requires an average of 1 hour and 40 minutes for one worker to assemble one toy

Assume that the ending inventory for June was 600 of these toys and the manager wishes to have at least 800 units on hand at the end of October

The current workforce level is 35 workers

As discussed in the above problem,

The cost of the plan derived = CH *(Total number of hired workers) + CF *(Total number of fired workers) + CI *(Total number of Inventory)………………….(1)

Given,

Hiring cost = 350, firing cost = 850

Total hired = Number of people hired from July to oct

= 21+0+0+0 =21

Total fired = Number of people hired from July to oct

= 0+0+16+0 =16

Total Number of Hired Inventory =july( initial inventory + production of that month-forecast demand) +August(july inventory+ production with previous month- forecast demand)+………..oct.

= 4921.

Now place all the values in the equatiom ,

We get = 21*350+16*850+4*4793.3

= $ 40,268.2